# 1Disclaimer

The following are my notes that I am taking at our seminar meetings. I do not claim that they accurately represent what happens at the seminar. In fact, they probably misrepresent quite a few things. In particular, they have not been edited in any way and contain many errors. Moreover, no effort has been done to attribute results correctly. None of the errors should be attributed to the speakers and I bear all responsibility for them. If you spot some error do let me know so that I can correct it.

The notes are generated from a sourcefile written in a variant of markdown. At the top you should find a pdf version for print.

# 26. 11. 2013

2.1

Let $\mathbb B$ be a Boolean algebra and $P\in\mathbb B^\omega$ a sequence. For a finite $A\subseteq\omega$ we define $p(A,P)=\frac{\mbox{max}\ \{|K|:K\in[A]^{<\omega}\ \&\ P[K]\ \mbox{is centered}\}}{|A|}$ and the intersection number of $P$ to be $i(P) = \mbox{inf}\ \{ p(A,P): A\in[\omega]^{<\omega}\}$ The weak intersection number is defined similarly if we consider sequences instead of sets.

2.2

A filter $\mathcal F$ is a measure centering filter if for every Boolean algebra $\mathbb B$ and for every $P\in\mathbb B^\omega$ with positive intersection number there is a set $A\in\mathcal F$ such that $\{p(n):n\in A\}$ is centered.

2.3Foreman

Every rapid filter is a measure centering filter.

2.4

If a sequence has a positive intersection number then there are infinitely many centered subsequences.

2.5

What about the weak intersection number?

# 320. 11. 2013

## 3.1E. Thuemmel: Pure decision property

3.1

$\mathbb M_{\mathcal A}$ has the pure decision property (or Prikry property ), if for each formula $\varphi$ and a condition $[s,A]$ there is a stronger condition of the form $[s,B]$ deciding $\varphi$.

3.2

This is equivalent to saying that for each $a\in RO(\mathbb M_{\mathcal A})$ and $[s,A]$ there is a stronger condition of the form $[s,B]$ such that $[s,B]\leq a$ or $[s,B]\leq -a$.

3.3Farah

A coideal $\mathcal A$ is semiselective iff $\mathbb M_{\mathcal A}$ has the pure decision property.

3.4

The coideal corresponding to FIN, an ideal generated by a MAD family and a selective ultrafilter are selective. However note that the other two are not definable.

3.5

An ideal $\mathcal I$ on $\omega$ is Ramsey (we also write $\mathcal I^+\rightarrow(\mathcal I^+)^2_2$) if for each $X\in\mathcal I^+$ and each coloring $\chi: [X]^2\to 2$ there is a $Y\in\mathcal I^+, Y\subseteq X$ which is monochromatic.

3.6

An ideal is selective if it is Ramsey and $\mathcal P(\omega)/\mathcal I$ is $\sigma$-closed. It is semi-selective if it is Ramsey and $\mathcal P(\omega)/\mathcal I$ is $\omega$-distributive.

3.7

We will sometimes say that an ideal or a co-ideal is $\sigma$-closed, $\omega$-distributive etc. and will mean, in fact, that the forcing $\mathcal P(\omega)/\mathcal I$ is $\sigma$-closed etc.

3.8

If we add a Sacks real then groundmodel infinite reals form a semi-selective coideal which is not selective.

3.9

selective $\rightarrow$ semiselective $\rightarrow$ Ramsey

3.10

If $\mathcal A$ is left-shift invariant (i.e. $A\in\mathcal A\rightarrow A\setminus\{\mbox{min} A\}\in\mathcal A$) and $\mathbb M_{\mathcal A}$ has the pure decision property then it adds a dominating real.

3.11

A semiselective coideal cannot be definable.

3.12

Can we have a definable Ramsey coideal?

3.13

The ideal conv is the ideal generated by Cauchy sequences of rationals in the interval $[0,1]$.

3.142.15, Hrušák, Thuemmel & spol.

If an ideal is not $\omega$-distributive and is proper then there is $X\in\mathcal I^+$ such that $\mathcal I\upharpoonright X$ is Katětov above conv.

3.15

Can we ommit properness in the above lemma?

3.16Thuemmel

No.

3.17

An ideal has $P^I$ iff for all descending sequences $\langle A_i:i<\omega\rangle$ of $\mathcal I$ postive sets such that $A_i\setminus A_{i+1}\in I$ there is a positive pseudointersection.

The following is a combinatorial reformulation of 3.14.

3.18

An ideal $I^+$ is somewhere Katětov above conv iff it is either not $(\omega,2)$-distributive or not $P^I$.

It will be first instructive to notice the following reformulation of $(\omega,2)$-distributivity+$P^I$:

3.19

An ideal $\mathcal I$ is $(\omega,2)$-distributivity+$P^I$ iff for each matrix $\{A^j_i:i<\omega,j there is a disjoint \(\subseteq^*$-refinement.

To see this, distributivity gives us a $\mathcal I$-disjoint $\subseteq^{\mathcal I}$ refinement and $P^{I}$ allowes to improve this to an almost disjoint $\subseteq^*$-refinement).

of Proposition

"$\Rightarrow$". Fix an $X\in\mathcal I^+$ and $\varphi:X\to\mathbb Q\cap[0,1]$ witnessing that $I^+$ is somewhere above conv (i.e. for all $F\in$ conv, $\varphi^{-1}[F]\in\mathcal I$). Let $A^j_i = \varphi^{-1}\left(\left[\frac{j}{2^i},\frac{j+1}{2^i}\right)\right),\quad i<\omega,j<2^i$ The $A^j_i$s form a matrix. If $\mathcal I$ is not $(\omega,<\omega)$ distributive we are done (since then it is also not $(\omega,2)$-distributive). So assume it is. Then there is a function $f:\omega\to\omega$ and a positive set $X$ such that $X\setminus A_i^{f(i)}=I_i$ is in $\mathcal I$. Then $X_n = X\setminus \bigcup_{i is a descending sequence of positive sets with small differences. If \(\mathcal I$ had $P^I$ then we would have a positive $Y\subseteq X$ such that $Y\subseteq^* A^{f(i)}_i$ for each $i<\omega$. Then $Y\subseteq \varphi^{-1}[\varphi[Y]]$ however $\varphi[Y]$ is Cauchy (since $Y\subseteq A^{f(i)}_i$ for each $i<\omega$) so it is in conv --- a contradiction with $Y$ being positive.

"$\Leftarrow$": Assume $\mathcal I$ is either not $(\omega,2)$-distributive or not $P^I$. By the above observation (3.19) we may assume that there is a matrix $\{A_i^j:i<\omega,j<2^i\}$ which has no almost disjoint refinement. We will define $\varphi:\omega\to [0,1]$ as follows: $\varphi(n) = \sum_{i<\omega}\frac{i(n)}{2^{2i+1}},$ where $i(n)$ is the unique number such that $n\in A_i^{i(n)}$. Then we compose $\varphi$ with a suitable function such that its image is a subset of $\mathbb Q$. This will witness that $\mathcal I$ is Katětov above conv.

We will now show the proof of Lemma 3.14.

Assume $\mathcal I$ is not $\omega$-distributive and that it is proper. First notice that if $\mathcal I$ is Proper and $(\omega,2)$-distributive then it is $\omega$-distributive. Now apply proposition 3.18.

Next we turn to give an example showing 3.16.

3.20

Farah has an ideal $\mathcal I$ which is Ramsey and not semiselective and not $\omega$-distributive. We will improve it so that it is $(\omega,2)$-distributive and $P^I$. For $u\in {}^{<\omega}\mathbb R$ we define $A_u\subseteq\omega$ such that $A_\emptyset=\omega$ and $\{A_{u^{\smallfrown}\xi}:\xi\in\mathbb R\}$ is a MAD on $A_u$. Let $\mathcal H$ consist of all $A\subseteq\omega$ which have some $A_u\subseteq^*A$. If we are careful enough when constructing $A_u$ this will be a coideal. (E.g. enumerate $\mathcal P(A_u)$ as $\{B_\xi:\xi\in\mathbb R\}$ and choose $A_{u^{\smallfrown}\xi}$ so it is contained in $B_\xi$ or its complement). Now $\mathcal H$ will not be $\omega$-distributive by construction. If we are even more careful, it will be Ramsey: Assume we have constructed $A_u$. Enumerate all colorings of $[A_u]^2$ as $\{\chi_\xi:\xi\in\mathbb R\}$ and choose $A_{u^{\smallfrown}\xi}$ so that it is homogeneous with respect to $\chi_\xi$. In a similar way we will ensure that $\mathcal H$ has $P^I$ and is $(\omega,2)$-distributive.

3.21Balcar

In the above we could have everywhere replaced properness with $(\omega,\cdot,\omega_1)$-distributivity.

# 427. 11. 2013

## 4.1Wolfgang Wohofsky

Most of the following is part of Wolfgang's thesis and joint work with Michael Hrušák and Ondřej Zindulka.

4.1

A set $A\subseteq\mathbb R$ has strong measure zero if for each sequence of positive real numbers $\langle \varepsilon_n:n<\omega\rangle$ there is a sequence of intervals $\langle I_n:n<\omega\rangle$ which covers $A$ and such that $\lambda(I_n)<\varepsilon_n$.

4.2

It is important that we take intervals (or open balls in general Polish spaces) instead of general open sets. We would just get measure zero with this.

4.3

A strong measure zero cannot contain a perfect set (as it is preserved under uniform continuous images).

4.4

Strong measure zero sets form a $\sigma$-ideal.

4.5

A set $X\subseteq 2^\omega$ is meager shiftable if it can be translated away from each meager set, i.e. for each meager set $M\in\mathcal M$ there is $t\in\mathbb R$ such that $X\cap M+t=\emptyset$.

4.6

A set $X$ is shiftable away from $M$ iff $X+M\neq 2^\omega$

4.7

If $\mathcal I$ is any collection, let $\mathcal I^*$ be the collection of sets which are shiftable away from each set in $\mathcal I$.

4.8Příkrý

Every meager shiftable has strong measure zero.

4.9Galvin-Mycielski-Solovay

In $\mathbb R$ the strong measure zero sets are exactly $\mathcal M^*$.

This motivates the following question.

4.10

Given a polish topological group, does the Galvin-Mycielski-Theorem 4.9 hold in it?

The following is a partial answer.

4.11

If $(G,+)$ is a locally compact polish group then $\mathcal M^*(G) = SMZ(G)$.

The $\subseteq$ inclusion is easy and follows from separability. The hard inclusion is the other one. From this it follows that under the Borel Conjecture (i.e. there is no uncountable strong measure zero set), the GMS holds for all polish groups.

(idea) Say that a group $G$ has the strong GMS property if for each meager set $M$ and there is a sequence $\langle\varepsilon_n:n<\omega\rangle$ of positive reals such that for each sequence $\langle U_n:n<\omega\rangle$ such that $d(U_n)<\varepsilon_n$ there is a $t\in G$ such that $\left( \bigcap_{n<\omega}\bigcup_{m\geq n} U_m\right) + t\cap M=\emptyset$ One now shows that a compact group has the strong GMS property. Suppose now that $X$ is strong measure zero and $M$ is meager. Apply the strong GMS property to $M$ to find the sequence $\overline{\varepsilon}=\langle\varepsilon_n:n<\omega\rangle$ . Apply strong-measure zero to $X$ and this sequence to get a sequence $\langle U_n:n<\omega\rangle$ of open balls with diameters bounded by $\overline{\varepsilon}$. WLOG we may assume that $X\subseteq \left( \bigcap_{n<\omega}\bigcup_{m\geq n} U_m\right)$ Now...

4.12

Under CH there is an uncountable meager-shiftable set in $\mathbb Z^\omega$ and a SMZ set which is not meager-shiftable, i.e. $[\mathbb Z^\omega]^\omega\subsetneqq\mathcal M^*(\mathbb Z^\omega)\subsetneqq SMZ(\mathbb Z^\omega)$ The second part does not need CH.

The first $\subsetneqq$ is shown by simple induction. Works for a general ideal $\mathcal J$ if $cov(\mathcal J)=cof(\mathcal J)=\kappa$. Then there is a set $X$ of size $\kappa$ which is in $\mathcal J^*$. For the second, let $T=\mathbb Z^{<\omega}\setminus\bigcup_{s\in\mathbb Z^{<\omega}} s^{\smallfrown}\underbrace{0\cdots0}_{m(s)},$ where $m(s)$ is the maximum of the range of $s$ or 2. Then $[T]$ is a counterexample to the strong GMS property for $\mathbb Z^\omega$.

4.13

An ideal $\mathcal J$ of subsets of some group $G$ is $\kappa$-translatable if for each $M\in\mathcal J$ there is a (possibly larger) $M^\prime\in\mathcal J$ such that for each $T\in[G]^{<\leq\kappa}$ there is $t\in G$ such that $M+T\subseteq M^\prime+t$.

Notice that if an ideal is $\kappa$-translatable then it is a $\kappa^+$-ideal.

4.14Příkrý

We say that a set is strongly meager if it is in $\mathcal N^*$, i.e. shiftable away from measure null sets.

The Dual Borel Conjecture says that there are no uncountable strongly meager sets. Bartoszynski showed that, in ZFC, $\mathcal N$ (measure zero sets) are not $2$-translatable, in particular, not $\omega$-translatable.

4.15

The meager-shiftable ideal on a locally compact Polish group is $\omega$-translatable.

4.16

It might be that $[T]$ from the proof of 4.12 is a counterexample to the $\omega$-translatability of $\mathcal M^*(\mathbb Z^\omega)$.

4.17

If $(G,+,0)$ is a compact group and $\Theta$ an open cover of $G$ there is a (Lebesgue neighbourhood) $U$ of $0$ such that for each $x\in G$ there is $O\in\Theta$ such that $U+x\subseteq O$.

4.18

The diameter of the Lebesgue neighbourhood is the Lebesgue number of the open cover.

## 4.2Honza Starý / BB

4.19

A submeasure $\mu$ on a Boolean algebra $\mathbb B$ is exhaustive if for each countable antichain $\{a_n:n<\omega\}\subseteq\mathbb B$ and each $\varepsilon > 0$ the set $\{n:\mu(a_n)>\varepsilon\}$ is finite.

4.20Kalton,Roberts

Let $\mu$ be a uniformly exhaustive submeasure on a Boolean algebra. Then there is a finitely additive measure $m\leq\mu$ having the same null sets, i.e. $\mathcal N(m)=\mathcal N(\mu)$.

4.21

Let $\mathbb B\leq \mathbb C$ be Boolean algebras and $mu$ an exhaustive submeasure on $\mathbb B$. Then $\mu$ can be extended to an exhaustive submeasure on $\mathbb C$. If $\mu$ was a measure, this extension can also be chosen to be a measure.

(Basically, an application of the Sikorski theorem) Consider $\mathcal N(\mu)$ and the quotient algebra $\mathbb B/\mathcal N(\mu)$. Then $\mu$ is strictly positive exhaustive submeasure on the quotient and induces a metric $d$ on the quotient algebra: $d([a],[b])=\mu(a\triangle b)$ Let $(\mathbb M,\rho)$ be the metric completion of $(\mathbb B/\mathcal N(\mu),d)$. This completion is a complete Boolean algebra. By the Sikorski extension theorem (i.e. complete B.A.s are injective), we can extend the mapping from $\mathbb B$ into $\mathbb M$ to a mapping $h:\mathbb C\to\mathbb M$. Now define $\overline{\mu}(c) = \rho(h(c),\mathbf 0)$ Then $\mu$ is the required extension. We will show that it is exhaustive. Let $\{a_n:n<\omega\}$ be an antichain in $\mathbb C$ and fix an $\varepsilon >0$. By induction find $\langle b^\prime_n:n<\omega\rangle\subseteq\mathbb B$ such that $\rho(b_n,a_n)<\varepsilon/2^n$. Let $b_n=b^\prime_n- b^\prime_0\vee\cdots b^\prime_{n-1}$.

# 54. 12. 2013

## 5.1Honza Grebík: Asymptotic Density in Generic Extensions

5.1

Let $\overline{s}=\langle a_n:n<\omega\rangle$ be a sequence of numbers such that for each $i\neq j$ the greatest common divisor $gcd(a_i,a_j)$ is equal to one. Then the set $A_{\overline{s}}=\{k\in\mathbb N: (\forall n<\omega)(a_n\not|\ k)\}$ has density $d(A)=\prod_{i<\omega}\left(1-\frac{1}{a_n}\right).$

5.2

Given a finite sequence $\sigma$ of digits $\{0,\ldots,d\}$ the set $A_\sigma$ of numbers which start with $\sigma$ when written in base $d$ does not have density.

5.3

For $k<\omega$ define the set of exponent-k-free numbers $P_k = A_{\langle p^k:p\ \mbox{ is prime}\rangle}=\{n:(\forall p)(p^k\not|\ n)\}$

5.4

$d(P_k)=1/\zeta(k)$, where $xi$ is the Riemann zeta function, i.e. $\zeta(x) = \sum_{n=1}^\infty \frac{1}{n^x} = \prod_{p\in P}\left( 1- \frac{1}{p^x}\right)^{-1}$

5.5

$P_2^i = \left\{ k: k=\prod_{i=1}^m p_i\ \&\ m=i\ \mbox{mod}\ 2\right\}$

5.6
1. $d(P_2^1)=3/\pi^2$
2. $\pi(x)\sim x/\log x$
3. $\zeta(x)=0\rightarrow Re(x)\neq 1$

## 5.2Balcar: Torturing Honza Starý

Recall ...

5.7

Let $(X,\rho)$ be a metric space and $\mathcal O$ an open cover of this space. Then the lebesgue number of $\mathcal O$ is defined as $l(\mathcal O) = \sup\{\varepsilon>0:(\forall x\in X)(\exists O\in\mathcal O)(B(x,\varepsilon)\subseteq O)\}$

And now for something entirely different...

5.8

A $T\subseteq\mathbb B$ is a tree in the Boolean algebra $\mathbb B$ if it is a tree in the (inverted) Boolean ordering such that disjointness in $T$ implies disjointness in $\mathbb B$.

5.9

Let $\mathbb B$ be a Boolean algebra, $f$ a strictly positive exhaustive functional on $\mathbb B$ (i.e. $f(\mathbf 0)=0$) which is nondecreasing. Then each tree $T\subseteq\mathbb B$ is countable.

5.10

In case $f$ is a measure, this is due to/can be found in S. Koppelberg (CMUC).

First, since $\mathbb B$ is ccc (by exhausitivty), each level and branch is countable. Assume, aiming towards a contradiction, $T$ is uncountable. Then $ht(T)\geq\omega_1$, i.e. $T_\alpha\neq0$ for uncountably many $\alpha$s. Let $r_\alpha = \max\{f(x):x\in T_\alpha\}$. This maximum exists by exhaustivity. The following claim will finish the proof, since there is no uncountable strictly decreasing sequence of real numbers.

5.11

The sequence $\langle r_\alpha:\alpha contains a strictly decreasing cofinal sequence. Note that the sequence is decreasing. It is sufficient to show that for each \(\alpha$ the set $\{\beta:r_\alpha=r_\beta\}$ is countable. This follows from the fact that $f$ is exhaustive, that the witnessing elements from the $T_\beta$s cannot form a chain and from Erdös-Dushnik-Miller.

5.12E. Thuemmel

This is equivalent to the fact that there is no strictly positive, nondecreasing exhaustive functional on the Suslin tree. The fact that Suslin does not carry a measure is folklore.

5.13Prikry

Does every ccc forcing add a random real or a Cohen real?

5.14

If the measure algebra cannot be embedded into the Talagrand algebra, it might be a candidate for a ZFC answer to the above problem.

5.15

Let $\mathbb (B_n,\mu_n)$ be a sequence of Boolean algebras with finitely additive measures on them and $\mathcal U$ a free ultrafilter on $\omega$. . For $x\in\prod B_n$ define $\mu_{\mathcal U}(x)=\mathcal U-\lim\mu_n(x(n))$ Then $\mu_{\mathcal U}$ is a finitely additive measure on $\prod B_n$. Moreover the $[\mathbf 0]_{\mathcal U}\subseteq \mathcal Null(\mu_{\mathcal U})=\mathcal Null$, hence $\prod B_n/\mathcal Null(\mu_{\mathcal U})$ is a quotient aglebra of the ultraproduct. Since the ultraproduct is complete then $\prod\mathbb B_n/\mathcal Null$, as a quotient algebra, has the CSP. By the Smith-Tarski theorem it must be complete (since it also carries as strictly positive finitely additive measure). We shall show that $\mu_{\mathcal U}$ is in fact $\sigma$-additive.

We will show that it is continuous (and hence $\sigma$-additive. So let $b_k$ be a sequence of elements of $\mathbb B_{\mathcal U}=\prod \mathbb B/\mathcal Null$. such that $\bigwedge b^k = \mathbf 0$. We may assume that $b^k\in \prod\mathbb B$ and that they are strictly decreasing. Assume, aiming for a contradiction, that $\lim \mu_{\mathcal U}(b^k) = \varepsilon > 0$ Then $U_k = \left\{ n<\omega\setminus k:\mu_n(b^k(n)) > \varepsilon/2\right\}\in\mathcal U$ Then $\bigcap U_k=\emptyset$. We define $b\in\prod\mathbb B_n$ as follows. For $n\in U_k\setminus U_{k+1}$ let $b(n)=b^k(n)$. Then $b\leq b^k$ and, since $\mu_{\mathcal U}(b)>\varepsilon/2$, $\mathbf 0. A contradiction with \(\bigwedge b^k = \mathbf 0$.

5.16
5.17

If $\mathbb B_n$ is the Cohen algebra with a measure extending the standard measure on clopen sets, then $\mathbb B_{\mathcal U}$ is the measure algebra of length $\mathfrak c$. {} Proof: This follows, by Fremlin, from the fact that $(\mathbb B_{\mathcal U},\rho)$ has open-hereditary density $\mathfrak c$. To show this, consider in $\mathbb B_n$ a measure$1/2$-independent system (i.e. $\mu(x_k)=\mu(x_k\triangle x_n)=1/2 = 1/2\mu(x_k\wedge x_n)$, $\langle x_k:k<\omega\rangle$. This gives us $\mathfrak c$-many pairwise different $\langle b_n^\alpha:n<\omega\rangle$ such that $\rho(b^\alpha,b^\beta)=\mathcal U-\lim\rho(b^\alpha_n,b^\beta_n) = \mathcal U-\lim 1/2 = 1/2.$ Hence $\mbox{hd}((\mathbb B_{\mathcal U},\rho))\geq\mathfrak c$.

and the story continues...

## 5.3Balcar: Torturing Honza Grebík

5.18

$\mathcal P(\mathbb N)/\mathcal Z_0\simeq\mathcal P(\mathbb N)/fin*\mathbb B(\mathfrak c)$.

Let $I_n=[2^n,2^{n+1})$. Given $X\subseteq\omega$ let $f([X]_{fin}) = \left[\bigcup_{n\in X}I_n\right]_{\mathcal Z_0}$ Then, since $X\in\mathcal Z_0 \iff \lim\frac{X\cap[2^n,2^{n+1})}{2^n}=0,$ $f$ is a regular embedding. Now consider $\mathcal P(\mathbb N)/fin\simeq\prod \mathcal P(I_n)$ and use the ultrapower technique (5.15).

5.19

Let $\mathcal F$ be a rapid filter. Then $\mathcal G=\left\{X:\left(\exists F\in\mathcal F\right)\left(\lim_{n\in F} \frac{X\cap[n^2,(n+1)^2)}{2n+1} = 1\right)\right\}$ is a filter which cannot be extended to a rapid but, after adding a random real, can.

# 611. 12. 2013

## 6.1E. Thuemmel: Exponent

### 6.1.1Pure decision

Recall the following question which motivates the subsequent.

6.1

Does pure decision imply that the respective forcing adds a dominating number?

However, this question is not well defined since there is no general definition of pure decision. As an example of pure decision consider the following classical theorem:

6.2

Mathias forcing $\mathbb M$ has the pure decision property (Prikry property), i.e. for each sentence $\varphi$ and condition $[s,A]$ there is a stronger condition $[s,B]$ which decides $\varphi$.

Our aim is to show that, in a suitable sense, pure decision in fact does impliy that the forcing adds a dominating number.

We now aim to introduce the exponent. The following is based on E. Thuemmel's paper Open mappings on extremally disconnected compact spaces published in AUC Math. et Ph., Vol 47 (2006), No. 2, 73-105.

6.3

Let us identify each condition $[s,A]$ in the Mathias forcing with the set $[s,A] = \left\{B:s\sqsubseteq B\ \& B\in[A]^\omega\right\},$ thus we can consider each $[s,A]$ to be a (Laver) tree. For each $s\in[\omega]^{<\omega}\setminus\emptyset$ define $sh(s)=s\setminus\min s$. Consider $\mathbb M^\prime=\{[s,A]\in\mathbb M: s\neq\emptyset\}$ and extend $sh$ to a map $sh:\mathbb M^\prime\to\mathbb M$ as follows: $sh([s,A]) = [sh(s),A].$

This map has the following property

6.4

For each $[s,A]\in\mathbb M^\prime$ and $[t,B]\in\mathbb M$ such that $[t,B]\leq sh([s,A])$ there is $[r,C]\leq [s,A]$ such that $sh([r,C])\leq [t,B]$.

Let $r= s\cup t$, $C=B$.

We shall generalize this observation into a definition:

6.5

A monotone map $\rho:P\to Q$ has property (*) if $(\forall p\in P)(\forall q\leq \rho(p))(\exists r\leq p)(\rho(r)\leq q)$

6.6

If a monotone map $\rho: P\to Q$ has property (*) we can extend it to a map $\rho^*:RO(Q)\to RO(P)$: $\rho^*(\mathcal I) = \rho^{-1}(\mathcal I)$ which is a complete homomorphism.

6.7

If $\rho[P]$ is dense in $Q$ then $\rho^*$ is a complete embedding. However, in our setting, the image of $P$ will not be dense.

We will now consider the extension of the $sh$ map $sh^*: RO(\mathbb M)\to RO(\mathbb M)$. This is a complete homomorphism. Consider the dual space $X = St(RO(\mathbb M))$, which is an extremally disconnected compact space, and the continuous function $f=st(sh^*):X\to X$, which is an open mapping. Given a condition $[s,A]\in\mathbb B$ we will write $U_{s,A}$ for the basic clopen set given by this condition.

6.8

1. $f[U_{s,A}] = U_{sh(s),A}$
2. $f[U_{\emptyset, A}]\subseteq U_{\emptyset, A}$
6.9

The set of fixed points of $f$ is nowhere dense.

$\{U_{s,A}:[s,A]\in\mathbb M^\prime\}$ forms a $\pi$-base for the topology. Since $f[U_{s,A}]\cap U_{s,A}=\emptyset$ for each $[s,A]\in\mathbb M^\prime$, $U_{s,A}$ contains no fixed point for each element of the basis.

6.10

The map $f$ has a fixed point.

Aiming for a contradiction, assume that no $x\in X$ is a fixed point. So we can separate $x$ from $f(x)$ by an open neighbourhood. By compactness we can find finitely many $U_i$ covering $X$ such that $f[U_i]\cap U_i = \emptyset$. Since it is a cover, we can take an arbitrary $A$ and find $i$ such that $U_{\emptyset, A}\subseteq U_i$. However this is impossible by 2. of observation 6.8.

To summarize, starting from Mathias forcing, we have arrived to an extremally disconnected compact space $X$ with an open map $f:X\to X$ having a nowhere dense set of fixed points. It is surprising that we can, in some sense, reverse this process. This is the content of the following definition and theorem.

6.11

All of this works in the more general setting with Mathias forcing replaced by $\mathbb M_{\mathcal A}$ where $\mathcal A$ is a selective coideal, or by Prikry forcing.

6.12

Given a regular cardinal $\kappa$, a complete Boolean algebra $B$ and a mapping $r: B^\kappa\to B$ which is

1. a $\kappa$-complete homomorphism
2. uniform (i..e. if $|\{\alpha<\kappa:b_\alpha\neq\mathbf 0\}|<\kappa$ then $r(b)=\mathbf 0$).
3. a retract, i.e. $r(c_b) = b$ for each $b\in B$, where $c_b$ is the constant function having value $b$

Then we define $Exp_\kappa(B,r) = \left\{ T\in {}^{[\kappa]^{<\omega}} B:(\forall s\in[\kappa]^{<\omega}) r(\langle T(s^{\smallfrown}\{\alpha\}):\alpha<\kappa\rangle) = T(s) \right\}$ Then $Exp_\kappa(B,r)$ is a complete, $\kappa$-complete subalgebra of $B^{[\kappa]^{<\omega}}$ which completely embeds $B$ by assingning to $b\in B$ the tree $T_b$ such that $T_b(s)=b$.

6.13

$\kappa$ must either be $\omega$ or, at least, strongly inaccessible. It is conjectured that, if uncountable, it must be weakly compact and this is also sufficient.

6.14

Consider the characteristic function $\chi:\mathcal P(\kappa)\to B^\kappa$ and $r:B^\kappa\to B$. The composition gives the map $\varphi$. If $B$ is $kappa$-distributive, then each such $\varphi$ determines $r$: $r(\langle b_\alpha:\alpha<\kappa\rangle)=\bigvee\Big\{b:\varphi(\{\alpha:b\leq b_\alpha\})\geq b\Big\}$ If $\mathbb M$ is the Mathias forcing, then let $B=Compl(\mathcal(P)(\omega)/fin)$ and $\varphi_r(A)=[A]_{fin}$. The previous note gives us $r_\varphi$. Then $Exp_{\omega}(B,r_\varphi)=RO(\mathbb M)$. The regular embedding showing this is defined as follows $[s,A] \mapsto T\in Exp_\omega(B,r_\varphi),$ where $T(t) = \left\{ \begin{array}{ll} [A] & s\sqsubseteq t\ \&\ t\setminus s\subseteq A\\ \mathbf 0 & \mbox{otherwise } \end{array} \right.$

6.15

Exponent has fusion in the following sense. Given $T_1,T_2\in Exp_\kappa(B,r)$ and $n<\omega$ we define $T_1\leq_n T_2 \iff (\forall s, |s|\leq n)(T_1(s)=T_2(s))$ If $\langle T_n:n<\omega\rangle$ with $T_0(\emptyset)\neq\mathbf 0$ is a fusion sequence, then $\bigwedge_{n<\omega} T_n\neq\mathbf 0$.

6.16

Exponent has pure decision in the following sense. Given a sentence $\varphi$ a nonzero $T\in Exp_\kappa(B,r)$ and $s$ with $T(s)\neq\mathbf 0$ then there is a stronger $T^\prime$ with $T^\prime(s)\neq\mathbf 0$ which decides $\varphi$.

6.17

Given a general exponent $Exp_\kappa(B,r)$ we define $sh^*:Exp_\kappa(B,r)\to Exp_\kappa(B,r)$: $sh^*(T)(s) = \left\{ \begin{array}{ll} T(sh(s)) & s\neq \emptyset\\ T(\emptyset) & s=\emptyset \end{array} \right.$ Similarly as in the case of Mathias forcing, we continue to get an extremally disconnected compact space and an open mapping with a nowhere dense set of fixed points. That the set of fixed points is nonempty follows from pure decision, in general, $Fix(f)=St(B)$.

6.18

If $X$ is extremally disconnected compact space, $f:X\to X$ an open mapping with a nowhere dense set of fixed points, then there are $\kappa\in Ord$, a complete Boolean algebra $B$, a retract $r$ and an $f$-invariant clopen subset $X^\prime\subseteq X$ such that $Exp_\kappa(B,r)$ can be regularly embedded into $RO(X^\prime)$.

From now we will consider only $\kappa=\omega$.

6.19

$Exp_\omega(B,r)$ adds a dominating real.

Given $s\in[\omega]^{<\omega}$ define $T_s(t) = \left\{ \begin{array}{ll} \mathbf 1 & s\sqsubseteq t\\ \mathbf 0 & \mbox{otherwise} \end{array} \right.$ Then $\{T_s:|s|=n\}$ is a partition of $\mathbf 1$. If $G$ is a generic filter on $Exp_\omega(B,r)$ then for each $n<\omega$ there is exactly one $s\in[\omega]^n$ such that $T_s\in G$. Moreover if $T_s,T_t\in G$ then $s,t$ must be compatible. So we can let $f_G=\bigcup\{s:T_s\in G\}$. Then $f_G$ is a dominating function from $\omega$ to $\omega$. To show this, we consider an arbitrary $g\in{}^\omega\omega\cap V$, a condition $T$ and $s$ with $T(s)\neq\mathbf 0$. Define $T_{g,s}(t) = \left\{ \begin{array}{ll} \mathbf 1 & s\sqsubseteq t\ \&\ (\forall |s|\leq i < |t|)(g(i)\leq t(i))\\ \mathbf 0 & \mbox{otherwise} \end{array} \right.$ Now let $T^\prime = T\wedge T_{g,s}$. Then $T^\prime\leq T$ and $T^\prime\Vdash (\forall i>|s|)(g(i)\leq f_{\dot{G}}(i))$.

Now if a forcing has pure decision we will, as in the case of Mathias, define a shift and then get to the situation where we can apply theorem 6.18 and, afterwards, theorem 6.19.

6.20Matet Forcing

Matet forcing consists of pairs $[s,A]$ where $s\in[\omega]^{<\omega}$ and $A$ is a block sequence and the ordering is given by $[s,A]\leq[t,B]$ such that $t\sqsubseteq s$, $A$ is a "condensation" of $B$ and $t\setminus s$ is a union of blocks from $B$. Then Matet forcing has pure decision. If we define $sh([s,A])=[sh(s),A]$ and proceed as before we would like to show that $Fix(f)\neq\emptyset$. However, this will fail. And, indeed, Matet forcing does not add a dominating real.

The following can be found in the dissertation of Luz M. García Ávila.

6.21García Ávila

Let $\mathbb P_{FIN}$ consist of pairs $[\overline{s},\overline{A}]$ where $\overline{s}$ is a finite block sequence and $\overline{A}$ is an infinite block sequence.

6.22García Ávila

$\mathbb P_{FIN}$ has pure decision.

In this case, the shift function $sh$ which drops the first block of $\overline{s}$ , works nicely and makes it possible to apply our machinery. However, this is not surprising, as the following theorem shows:

6.23García Ávila

The set of minimal points of the generic block sequence is a Mathias real.

# 717. 12. 2013 M. Hrušák: Questions & stuff I've been working on

## 7.1Groups

7.1Malychin

Is there a separable Fréchet topological group which is not metrizable.

7.2Hrušák, Ariet

Consistently no.

(see Hrušák, Ariet: Malykhin's Problem, preprint)

7.3

Is it consistent that there is a countable Fréchet group of weight $\omega_2$ while there is none of weight $\omega_1$.

7.4

A $\gamma$-set is a set of reals where every $\omega$-cover has a $\gamma$-subcover. An infinite open cover $\mathcal U$ of a space $X$ is an $\omega$-cover if for every finite subset $F\subseteq X$ there is $U\in\mathcal U$ such that $F\subseteq U$. It is a $\gamma$-cover if each $x\in X$ is in all but finitely many $U\in\mathcal U$.

7.5

Is it consistent that there is a $\gamma$-set of size $\omega_2$ and no $\gamma$-set of size $\omega_1$

7.6

If $G$ is an extremally disconnected group and $h:G\to2^\omega$ is continuous then there is nonempty open $U\subseteq G$ such that $h[U]$ is nowhere dense.

7.7
1. A yes answer to the conjecture implies that there are no countable e.d. topological groups.
2. A yes answer is equivalent to saying that $RO(G)$ does not add Cohen reals for any e.d. group.
3. If $h$ is a group homomorphism, then the answer is yes.
7.8Group version of YefimoffvanDouwen?

Is there a countably compact topological group with no convergent sequences.

I have a reformulation of the above. Basically, if one tries to answer the above in a naive way by induction, one arrives at the following invariant. If it is equal to $\mathfrak c$, then the induction works.

7.9

$\mu = \min\big\{\kappa:(\exists\mathcal A\subseteq[2^\kappa]^\omega)(\forall A\in\mathcal A, 0\in\overline{A})(\exists C\in[2^\omega]^\omega,C\to0)(\forall \varphi:2^\kappa\to 2,\mbox{group homomorphism})(\varphi\ \mbox{"splits"} C\Rightarrow (\exists A\in\mathcal A)(0\not\in\overline{\{a\in A:\varphi(a)=0\}}))\big\}$

7.10

Is $\mu<\mathfrak c$ consistent?

## 7.2Countable Dense Homogeneity

7.11

Two countable sets $A,B$ are of the same type if there is a homeomorphism $h$ such that $h[A] = B$. A space is CDH if there is only one type of dense sets.

A typical example of a Polish space which has two types is the half-open interval. It turns out that, basically, this is the only possibility for finitely many types, i.e. all examples are of the form $X\cup F$ where $X$ is CDH and $F$ finite. The uncountable case is different and motivates the following question

7.12

Is there a Polish space with $\aleph_1$-many types of countable dense sets?

This is related to the Topological Vaughts conjecture. The following theorem can be found in the paper R. Hernandez-Gutierrez, M. Hrušák, J. van Mill: Countable dense homogeneity and lambda sets, preprint.

7.13vMill, Hrušák, Hernandez

If there is a $\lambda$-set of size $\kappa$ then there is $X\subseteq\mathbb R$ of size $\kappa$ which is CDH.

7.14

Is it consistent that there is a Baire CDH set of reals of size $<\mathfrak c$.

The following is a recent result published in the paper Hrušák, Guzmán: n-Luzin gaps, Top. Appl 160 (2013).

7.15

A system $\langle A_\alpha, B_\alpha\rangle$ is a Luzin gap if

1. $A_\alpha\cap B_\alpha=\emptyset$
2. $A_\alpha\cap B_\beta =^*\emptyset$
3. $(\forall \alpha\neq\beta)(A_\alpha\cap B_\beta\cup A_\beta\cap B_\alpha\neq\emptyset)$
7.16Hrušák, Osvaldo

Assume PFA. If $\mathcal A$ is AD then $\psi(\mathcal A)$ is normal iff $|\mathcal A|\leq\omega_1$ and $\mathcal A$ does not contain a Luzin gap.

7.17

Is MA enough in the above theorem?

## 7.4Cardinal invariants connected to AD families

7.18

$\mathfrak a_s$ is the minimal size of a maximal family of AD subsets of $\omega^2$ containing all vertical lines and otherwise only partial selectors.

7.19

$\mathfrak a, non(\mathcal M)\leq\mathfrak a_s$

7.20Brendle

It is consistent that $\mathfrak a=\omega_1,non(\mathcal M)=\omega_2,\mathfrak a_s = \omega_3$

7.21Hrušák & students

If $\mathfrak c\leq\omega_2$ then $\mathfrak a_s = \max\{a,non(\mathcal M)\}$.

7.22

If $\mathcal I$ is an ideal on $\omega$, let $a(\mathcal I) = \min\{|\mathcal A|:A\subseteq\mathcal I\ \mbox{is AD}$

7.23

If $\mathcal I$ is tall then $a(\mathcal I) = \min\{|\mathcal A|:\mathcal A\subseteq\mathcal I\ \mbox{and}\ \mathcal A\ \mbox{is MAD}\}$

7.24

Given a tall ideal $\mathcal I$ then $cov^*(\mathcal I) =\min\{|\mathcal J|:\mathcal J\subseteq I\ \&\ (\forall X\in[\omega]^\omega)(\exists I\in\mathcal I)(|I\cap X|=\omega)\}$

7.25

$\mathfrak a,cov^*(\mathcal I)\leq a(\mathcal I)$

The above theorem 7.23 follows from the following by choosing $\mathcal I=\mathcal{ED}$.

7.26

If $\mathfrak c\leq\omega_2$ and $\mathcal I$ is tall then $a(\mathcal I)=\max\{\mathfrak a,cov^*(\mathcal I)\}$.

Let $\{J_\alpha:\alpha<\omega_1\}$ be a witness of $cov^*(\mathcal I)$. We will construct a MAD $\mathcal A\subseteq \mathcal I$, where $\mathcal A = \bigcup_{\alpha<\omega_1}\mathcal A_\alpha$. Let $\mathcal A_0$ be an arbitrary MAD family on $J_0$ of size $\omega_1$. At step $\alpha<\omega_1$ consider $K=\{J_\beta\cap J_\alpha\}$ on $J_\alpha$. Then $K^\perp$ (on $J_\alpha$) is countably generated so we can disjointify it and consider it as columns in $J_\alpha\simeq\omega^2$. Then extend these columns to a MAD family $\mathcal A_\alpha$ of size $\omega_1$. Then these $\mathcal A_\alpha$s will have the following properties:

1. $\mathcal A_\alpha$ is an AD on $J_\alpha$
2. $\bigcup_{\beta\leq\alpha}\mathcal A_\beta$ is AD
3. $\bigcup_{\beta\leq\alpha}\mathcal A_\beta\upharpoonright J_\alpha$ is MAD on $J_\alpha$.
4. $\bigcup_{\beta\leq\alpha}\mathcal A_\beta$ has size $\omega_1$.

## 7.5Generic existence of MAD families

7.27

A MAD family having property $P$ exists generically if any AD family of size $<\mathfrak c$ can be extended to a MAD family with property $P$.

7.28
1. $\mathfrak a=\mathfrak c$ is equivalent to the fact that completely separable MAD families exist generically.
2. $\mathfrak b=\mathfrak c$ iff Cohen indestructible families exist generically.
3. In known models Sacks indestructible families exist (either because $\mathfrak a<\mathfrak c$ or because they exist generically due to a cardinal invariant).
7.29

Do Sacks indestructible MAD families exist (in ZFC).

## 7.6Fsigma-ideals

7.30

If $\mathcal I$ is $F_\sigma$ is it true that $\mathcal I$ is Tukey equivalent to $[\mathfrak c]^{<\omega}$ (i.e. Top) or $\mathcal I$ is Tukey below the summable ideal.

7.31

Does every tall Borel ideal contain a tall $F_\sigma$ subideal?

7.32

An ideal $\mathcal I$ is K-uniform if for every $X\in\mathcal I^+$ the ideal $\mathcal I\upharpoonright X\leq_K \mathcal I$ (note that it is always Katětov above)

7.33

Is $\mathcal{ED}_{fin}$ (i.e. the ideal generated by graphs of functions under the diagonal) the only tall $F_\sigma$ ideal which is K-uniform.

7.34

Is there a tall Borel (Analytic) ideal $\mathcal I$ such that $\mathcal I^+\to(\mathcal I^+)^2_2$.

7.35

The above fails for $F_\sigma$ ideals. However, there is an $F_\sigma$ ideal $\mathcal J$ such that $\omega\to(\mathcal J^+)^2_2$ and there is a co-analytic ideal for which it holds.

## 7.7Strong Measure Zero

Recall definition 4.1.

7.36Prikry

If $X\subseteq\mathbb R$ is meager-shiftable (see definition 4.5) then it is strong measure zero.

The same proof works for any separable polish group with left-invariant metric. He then asked about the reverse implication. A yes answer (on the reals) is due to Galvin-Mycielski-Solovay.

7.37

For which groups does the (Prikry-)Galvin-Mycielski-Solovay theorem hold?

From now on $G$ will always be a Polish group with a left-invariant metric and $X$ will be a separable metric space.

7.38

A set $M\subseteq X$ is uniformly nowhere dense if for every $\varepsilon > 0$ there is a $\delta > 0$ such that for each point $x\in X$ there is a point $y\in X$ such that the ball $B(x,\varepsilon)\setminus M$ centered at $x$ with diameter $\varepsilon$ contains $B(y,\delta)$. It is uniformly meager if its covered by countably many uniformly nowhere dense sets. We write $\mathcal{UM}$ for the collection of all uniformly meager sets.

7.39

A group is CLI if it carries a complete left-invariant metric.

7.40

Each Polish group carries a complete metric as well as a left-invariant metric. However there are groups, e.g. $S_\omega$, which are not CLI.

7.41

If $G$ is CLI and $X\subseteq G$ has strong measure zero, then it is uniformly meager-shiftable (i.e. $X+M\neq G$ for each uniformly meager $M$).

7.42

If $G$ is CLI and locally compact then meager sets are uniformly meager.

7.43

For which spaces/groups does meager imply uniformly meager. It should be exactly the locally compact but there is no proof yet.

7.44

$cpt\Rightarrow \mathcal M = \mathcal{UM} \Rightarrow$ GMS

7.45

A group has the strong GMS property if for every nowhere dense $N$ there is a sequence $\langle \varepsilon_n:n<\omega\rangle$ such that for every $\langle U_n:n<\omega\rangle$ with $\mbox{diam}\ U_n\leq\varepsilon_n$ there is $g\in G$ such that $\left(g+\bigcup_{n<\omega} U_n\right)\cap N = \emptyset$ It has the weak GMS property if the above is not dense in $N$.

7.46

strong GMS $\Rightarrow$ GMS $\Rightarrow$ weak GMS.

7.47

The strong and weak GMS properties are absolute properties (i.e. with respect to $\sigma$-closed extensions.

7.48

$\mathbb Z^\omega$ is not weakly GMS.

7.49

A group $G$ has the Bergman property if every compatible left-invariant metric is bounded.

7.50

Compact $\Rightarrow$ strong GMS $\Rightarrow$ Bergman.

of the second implication

Suppose G is not Bergman and fix $d$ a compatible unbounded left-invariant metric and a sequence $\langle W_n:n<\omega\rangle$ of disjoint balls whose union is dense and $\mbox{diam}\ W_n\to 0$ and let $N=\{x_n:n<\omega\}$ be the centers of the balls. And now work a little bit.

7.51Hrušák

A group $G$ is elastic if for each nonempty open $U$ there is a compatible left-invariant metric $d$ on $G$ such that $\mbox{diam}_d\ U=\infty$.

7.52

Bergman $\Rightarrow$ non-elastic.

7.53

GMS implies non-elastic.

So we have the following diagram: $\begin{array}{ccccccc} \mbox{compact} & &\Rightarrow&&\mbox{strong GMS} &\Rightarrow &\mbox{Bergman}\\ \Downarrow & &&&\Downarrow &&\Downarrow\\ \mbox{loc. compact} &\Rightarrow&\mathcal M=\mathcal{UM}\Rightarrow&&\mbox{GMS} &?&\mbox{non-elastic}\\ & &&&\Downarrow &&\\ &&&&\mbox{weak GMS}&&\\ \end{array}$

# 88. 1. 2014

8.1

Let $D\subseteq\omega^\omega$ be a countable dense set which is partitioned into countably many finite $D_k$s so that any infinite union of the $D_k$s is dense. Then there is $n_0$ such that for each choice $\{L_k\in[D_k]^{\leq n_0}:k\in\omega\}$ there is $x\in\omega^\omega$ such that $x\in\overline{\bigcup_{k}L_k}\setminus\bigcup_{k}L_k$

8.2Gryzlov

There is a dense $D\subseteq\omega^{\mathfrak c}$ of size $\omega$ which can be partitioned into countably many finite sets $D_k$ such that

1. $(\forall M\in[\omega]^\omega)(\bigcup_{k\in M} D_k\ \mbox{is dense})$
2. For each $n_0\in\omega$ and each choice $L_k\in[D_k]^{\leq n_0}, k\in\omega$ the set $\bigcup_{k\in\omega} L_k$ is closed discrete.

## 8.1E. Thuemmel: Exponent (cont.)

8.3

Does each tall Borel ideal contain an $F_\sigma$ (tall) ideal?

8.4

Let $\mathcal I$ be a tall analytic ideal on $\omega$ which is $\omega$-distributive (i.e. such that $\mathcal P(\omega)/\mathcal I$ is $\omega$-distributive). Then there is $A\in\mathcal I^+$ and a tall $F_\sigma$ ideal $\mathcal J$ on $A$ such that $\mathcal J\subseteq\mathcal I$.

We shall first show a proof by Hrušák. We will need some notation.

8.5

Given a set $A$ and its partition $\overline{A}=\langle A_n:n<\omega\rangle$ we let $\mathcal{ED}(\overline{A}) = \{ B\subseteq A:(\exists n_0)(\forall i>n_0 )(|A_i\setminus A_{i+1}|\leq n_0)\}$

There are two cases. Either there is an $\mathcal I$-positive set $A$ and its partition $\overline{A}=\langle A_n:n<\omega\rangle$ such that $\mathcal{ED}(\overline{A})\subseteq \mathcal I$. Then we are done.

Otherwise force with $\mathcal I^+$ and consider the generic ultrafilter $G$. Since the first case failed, this ultrafilter is selective. However, by Theorem 3.3 of M. Hrušák, J. Verner: Adding ultrafilters by definable quotients, Rend. Circ. Math. Palermo 60(3) 2011, $\mathcal I$ must be Fréchet, so is not tall --- a contradiction.

Let us now show a proof via exponents.

8.6

An ideal $\mathcal I$ on $\omega$ has property SEL if for each $\langle R_n:n<\omega\rangle$ partition of $\omega$ into sets in $\mathcal I$ there is an $\mathcal I$-positive selector. It is Ramsey if $\mathcal I^+\rightarrow(\mathcal I^+)^2_2$.

For the definition of selectivity and semiselectivity see the picture or take the following lemma as a definition.

8.7

Recall that $\mathcal I$ is Ramsey iff it has property SEL and is $(\omega,2)$-distributive. It is semiselective iff it has property SEL and is $\omega$-distributive and it is selective iff it has SEL and is $\sigma$-closed.

In the following we shall consider co-ideals instead of ideals.

8.8

Given a co-ideal $\mathcal H$ and $A\in\mathcal H$ a sequence $\langle A_i:i\in A\rangle$ is an $\mathcal H$-tower if $A_i \triangle A \not\in \mathcal H$ for each $i\in A$ and $A_i\subseteq A_j$ for $i>j$. Given such an $\mathcal I$-tower $\overline{A}$ we say that a $B\in[A]^{\leq\omega}$ is a diagonal such that $(\forall i\in B)(B\setminus(i+1)\subseteq A_i)$. Let $diag_\omega(\overline{A}$ be all infinite diagonals of $A$ and $diag_{<\omega}(\overline{A})$ the set of all finite diagonals.

8.9

Given a co-ideal $\mathcal H$ we define $Exp(\mathcal H) = \{ [s,\overline{A}]:s\in[\omega]^{<\omega}\ \&\ \overline{A}\ \mbox{is an}\ \mathcal H\mbox{-tower}\}$ with the extension relation $[s,\overline{A}]\leq[t,\overline{B}]$ defined as follows:

1. $t\sqsubseteq s$, $A\subseteq B$ and $A_i\subseteq B_i$ for all $i\in A$
2. $s\setminus t$ is a (finite) diagonal of $\overline{B}$

or, if we let $[s,\overline{A}] = \{B\in[\omega]^\omega:s\sqsubseteq B\ \&\ B\in diag_\omega(\overline{A})\}$ then the extension is just inclusion.

If $\mathcal H$ is distributive then this is a special case of the general exponent (see 6.12), i.e. $Exp(\mathcal H)$ embeds as a dense subset of $Exp_\omega(Compl(\mathcal P(\omega)/\mathcal I),r)$, where $\mathcal H=\mathcal I^+$ and the retract $r$ is defined: $r(\overline{b}) = \bigvee\bigg\{ [A]_{\mathcal H}:(\forall i\in A)([A]_{\mathcal H}\leq b_i)\bigg\}$.

The embedding takes $[s,\overline{A}]$ to the tree $T$ where $T(t) = [A]_{\mathcal H}$ for $s\sqsubseteq t$ such that $t\setminus s$ is a diagonal of $\overline{A}$ and $\emptyset$ otherwise.

Note that classical Ramsey theorems hold also in this setting.

8.10Nash-Williams

For all $\mathcal F\subseteq[\omega]^{<\omega}$ there is an infinite $A\in[\omega]^\omega$ such that either

1. $(\forall B\in[A]^\omega)(\exists s\in\mathcal F)(s\sqsubseteq B)$ (i.e. $\mathcal F$ is a barrier) or
2. $[A]^{<\omega}\cap\mathcal F = \emptyset$

Mathias generalized this to being able to select $A$ from a selective co-ideal and Farah generalized this further to semi-selective co-ideals. In our setting the following holds (basically with the same proof as Farah, only skipping the last step):

8.11generalized Nash-Williams

If $\mathcal H$ is an $\omega$-distributive co-ideal then For all $\mathcal F\subseteq[\omega]^{<\omega}$ there is an $\mathcal H$-tower $\overline{A}$ such that either

1. $(\forall B\in diag_\omega(\overline{A}))(\exists s\in\mathcal F)(s\sqsubseteq B)$ or
2. $diag_{<\omega}(\overline{A})\cap\mathcal F = \emptyset$

Consider next the Ellentuck theorem.

8.12

The Elentuck topology on $[\omega]^\omega$ is generated by sets of the form $[s,A]$, where $[s,A]=\{B\in[\omega]^\omega:s\sqsubseteq B\subseteq A\}$.

Note that a set $X$ has the Baire property in this topology if $\forall [s,A])(\exists [t,B]\leq [s,A])([t,B]\subseteq X\vee [t,B]\cap X=\emptyset)$. A set is completely Ramsey if $\forall [s,A])(\exists B\in [A]^\omega)([s,B]\subseteq X\vee [s,B]\cap X=\emptyset)$

8.13Ellentuck

The Baire property is equivalent to being completely Ramsey (in $[\omega]^\omega$ with the Ellentuck topology).

Mathias and Farah generalized this taking some selective and semiselective, respectively, $\omega$-distributive co-ideal $\mathcal H$ instead of $[\omega]^\omega$ and it also holds in our setting (taking conditions $[s,\overline{A}]\in Exp(\mathcal H)$). It is also true that $\mathcal H$-Ramsey sets ($\mathcal{R(H)}$) form a $\sigma$-algebra and $\mathcal H$-Ramsey null sets ($\mathcal R_0(\mathcal H)$) form a $\sigma$-ideal.

8.14

A Marczewski pair is a pair $(B,\mathcal I)$ where $\mathcal B$ is a $\sigma$-algebra of subsets of some Polish space $Y$ and $\mathcal I$ is a $\sigma$-ideal on $Y$ satsifying $(\forall X\subseteq Y)(\exists\Phi(X)\supseteq X,\Phi(X)\in B)(\forall Z)(Z\subseteq\Phi(X)\setminus X\ \&\ Z\in B\rightarrow Z\in\mathcal I)$

8.15Marczewski

If $(B,\mathcal I)$ is a Marczewski pair then B is closed under the Suslin operation $\mathcal A$.

8.16Pawlikowski

Assuming CH the $\mathcal H$-Ramsey sets and $\mathcal H$-Ramsey null sets form a Marczewski pair.

Using Marczewski's theorem it follows that, assuming CH, all analytic sets are $\mathcal H$-Ramsey. However, this statement is simple enough, so by some absolutness arguments (due to Platek) the CH assumption may be dropped.

Let us now give an alternative proof of theorem 8.4:

of theorem 8.4

Since $\mathcal I$ is analytic and $\omega$-distributive we can find an $\mathcal I^+$-tower $\overline{A}$ such that either

1. $[\emptyset,\overline{A}]\subseteq \mathcal I$ or
2. $[\emptyset,\overline{A}]\cap\mathcal I=\emptyset$

Since $[\emptyset,\overline{A}]$ is a tall, downwards-closed set it generates a tall $F_\sigma$-ideal. so in case 1. we are finished. Case 2, however, easily leads to a contradiction with the tallness of $\mathcal I$.

## 8.2J. Verner: Generalized Grigorieff forcing

J. Verner gave a (very) short overview of the paper R. Honzík, J. Verner: A Lifting Argument for the Generalized Grigorieff Forcing, to appear in Notre Dame J. of Formal Logic.

The article generalizes standard Grigorieff forcing to uncountable cardinals (see also B. M. Andersen, M. Groszek: Grigorieff Forcing on Uncountable Cardinals Does Not Add a Generic of Minimal Degree, Notre Dame J. of Formal Logic 50(2) 2009) and shows that it can be used to add unbounded subsets of $\kappa$ while, e.g., preserving measurability. In the following we introduce the basic notion and show that, under GCH, it does not collapse cardinals.

8.17

Given a an uncountable regular cardinal $\kappa$ and a $\kappa$-complete ideal $\mathcal I$ on $\kappa$ extending the nonstationary ideal $\mathcal{NS}$ we define the generalized Grigorieff forcing $P_{\mathcal I} = \{f\in {}^A2 : A\in\mathcal I\}$ which consists of partial characteristic functions of subsets of $\kappa$ ordered by extension.

8.18
1. We want the forcing to be $\kappa$ complete so that it does not add bounded subsets of $\kappa$.
2. Since we want the ideal to be $\kappa$ complete, it does not make sense to consider singular cardinals.
3. We want the ideal to extend $\mathcal{NS}$ because that allows us to use Fodors lemma in the arguments. The situation for $\kappa$-complete ideals not extending $\mathcal{NS}$ clear.

In the situation above (i.e. $\kappa$ uncountable, regular, $\mathcal I$ $\kappa$-complete, extending $\mathcal{NS}$) we have the following theorem.

8.19

Assume GCH and $2^{<\kappa}<\kappa$. Then $P_{\mathcal I}$ does not collapse cardinals iff $\mathcal I$ is closed under diagonal unions.

The diagonal union is the dual notion to the diagonal intersection and is defined as follows:

8.20

Given a sequence $\overline{A}=\langle A_\alpha:\alpha<\kappa\rangle$, its diagonal union is $\bigtriangledown \overline{A} = \{\beta<\kappa: (\exists\alpha<\beta)(\beta\in A_\alpha)\}$

If an ideal is not closed under diagonal unions we first show that it is not a "generalized P-ideal". Then we just copy the standard proof that Grigorieff forcing with a non-P-point collapses the continuum. This direction does not need the GCH. For the other direction, since $\mathcal I$ is $\kappa$-closed and, under GCH, $\kappa^{++}$-cc we need only show that it does not collapse $\kappa^+$. For this we use fusion.

8.21

Given two conditions $p,q\in P_{\mathcal I}$ we define $p\leq_{\alpha} q \Leftrightarrow p\leq q\ \&\ \mbox{dom}(p)\cap(\alpha+1) = \mbox{dom}(q)\cap(\alpha+1)$

A sequence $\langle p_\alpha:\alpha<\kappa\rangle$ is a fusion sequence if it satisfies:

1. $p_{\alpha+1} \leq_{\alpha} p_\alpha$ for all $\alpha<\kappa$
2. $p_\beta = \bigcup_{\alpha<\beta} p_\alpha$ for limit $\beta$

It is routine to check that if the ideal $\mathcal I$ is closed under diagonal unions then it is closed under limits of fusion sequences. Fusion is now used to show that each function $\dot{f}:\kappa\to\kappa^+$ in the extension is bounded. This is done by constructing a fusion sequence of length $\kappa$ such that its limit forces that the function is bounded. The crucial step is the following claim, which uses $\kappa$-completeness of $\mathcal I$:

8.22

If $p\in P_{\mathcal I}$ is a condition and $\alpha<\kappa$ then there is a set $A$ of size $2^\alpha$ (so $<\kappa$ by assumption) and a condition $q\in P_{\mathcal I}$ such that $q\Vdash \dot{f}(\alpha)\in A$ and $q\leq_{\alpha}p$.

8.23

The assumption that $2^{<\kappa}<\kappa$ is in fact superfluous. If $\kappa$ is a regular limit then it follows from GCH. If not, we can replace the assumption $2^{<\kappa}<\kappa$ with $\Diamond_{\kappa}$ at the cost of a more elaborate argument. If $\kappa$ is a successor $\Diamond_{\kappa}$ then follows from GCH. So the only case where the proof breaks down is $\kappa=\omega_1$ where we need to additionally assume $\Diamond$.

# 915. 1. 2014

## 9.1A. A. Gryzlov: On Dense subsets of Tichonoff products (preprint)

9.1

Let $D$ be a Tichonoff product of $\mathfrak c$-many infinite discrete countable spaces. Then there is a countable subset $Q\subseteq D$ which can be partitioned into finite sets $\{Q_k:k<\omega\}$ such that

1. For each infinite $C\subseteq\omega$ the set $\bigcup_{k\in C}Q_k$ is dense in D
2. If $F\subseteq Q$ si bounded in $Q_k$'s, i.e. if there is $n<\omega$ such that for each $k<\omega$ the set $Q_k\cap F$ has size at most $n$, then $Q\setminus F$ is dense in $D$
3. If $F\subseteq Q$ is bounded in $Q_k$'s (as in 2), then $F$ is closed discrete in $Q$.
9.2
1. condition 3. has no chance to hold in $\omega^\omega$.
2. (E. Thuemmel) condition 3. says that the ideal of sets having discrete closure (in Q) contains $\mathcal{ED}_{fin}$.

Recall the (special case of the) classical theorems

9.3Hewitt-Marczewski-Pondyczéry

The product of $\mathfrak c$-many separable spaces is separable.

9.4Rasiowa-Sikorski

There is an independent family $\{A_{\alpha,n}:\alpha<\mathfrak c, n<\omega\}$ of countable partitions of $\omega$ into infinite sets, i.e. such that whenever $A\in[\mathfrak c]^{<\omega}$ and $g:A\to\omega$ $|\bigcap_{\alpha\in A} A_{\alpha,g(\alpha)}| = \omega.$

Let $H=\{\sigma:\mathcal P(k)\to k : k<\omega\}$. It is sufficient to partition $H$. However, this independent partitioning can be easily defined as follows For $X\subseteq\omega$ let $A_{X,n} = \{ \sigma: \sigma(X\cap dom(\sigma)) = n\}.$

of HMP

To get a dense set of functions $f:\mathfrak c\to\omega$, define $f_\sigma(X)=\sigma(X\cap dom(\sigma))$ for each $\sigma\in H$.

of the main theorem

For $k\in\omega$ let $H_k = \{\sigma:\sigma\in {}^{\mathcal P(k)}(\mathcal P(k)\setminus \{\emptyset\}), (\forall i. Note that the \(H_k$s form a partition of $H$. For $X,Y\subseteq\omega$ and $k<\omega$ let $A_k(X,Y) = \{\sigma\in H_k:\sigma(X\cap k)= Y\cap k\}$ and $A(X,Y) = \bigcup_{k<\omega} A_k(X,Y).$ Define now a family $M=\{ A(X,Y):X,Y\subseteq\omega, Y\neq\emptyset\}\}$. Notice that for each $X\subseteq\omega$ the family $M_X=\{A(X,Y):Y\subseteq\omega,Y\neq\emptyset\}$ is an almost disjoint partition of $H$. Moreover the family $M$ is independent.

9.53.1

Let $k_0\in\omega$, $\sigma\in H_{k_0}$ and $F\subseteq H$ such that

1. $(\forall k<\omega)(|F\cap H_k|\leq 1)$
2. $(\forall k\leq k_0)(F\cap H_k = \emptyset)$

Then for each $X\subseteq\omega$ there is $Y\neq\emptyset$ such that $A(X,Y)\cap F=\emptyset$ and $\sigma_0\in A(X,Y)$.

Let $C=\{k\in\omega:F\cap H_k\neq\emptyset\}$ and $C_0=C\cup\{k_0\}$. For $k\in C$ define $\tau_k$ such that $H_k\cap F=\{\tau_k\}$. Construct Y by induction as an increasing union of $Y_k$'s for $k\in C$. Let $\sigma_0(X\cap k_0)=Y_{k_0}$. Suppose that $k\in C$ and, for $k^\prime , \(k^\prime\in C$ we have already constructed $Y_{k^\prime}$. An easy argument shows how to construct $Y_{k}$.

9.63.2

Let $\sigma,\tau\in H$ be distinct. Whenever $\Gamma\subseteq\mathcal P(\omega)$ has size $<\mathfrak c$ then there is $X\in\mathcal P(\omega)\setminus\Gamma$ and a nonempty $Y\neq\emptyset$ such that $\sigma\in A(X,Y)$, $\tau\not\in A(X,Y)$.

We have $\sigma\in H_{k(\sigma)}$ and $\tau\in H_{k(\tau)}$.

Case 1: If $k(\sigma) >k(\tau)$ then, by definition of $H_k$, there is $X^\prime\subseteq k(\sigma)$ such that $\sigma(X^\prime)=\{k(\tau)\}$. Choose $Y$ such that $Y\cap k(\sigma)=\{k(\tau)\}$. ...

Case 2: If $k(\sigma)=k(\tau)$ then, since $\sigma,\tau$ are distinct, there is $X_0\subseteq k$ such that $\sigma(X_0)\neq\tau(X_0)$. Since the set $\{ X:X\cap k= X_0\}$ has size $\mathfrak c$, there is an $X$ in this set, such that $X\not\in \Gamma$ ...

Case 3: If $k(\sigma) apply lemma 3.1 (9.5). 9.73.3 There is a set \(\Delta\subseteq\mathcal P(\omega)$ of size $\mathfrak c$ and a function $\pi:\Delta\to\mathcal P(\omega)\setminus\{\emptyset\}$ such that for $\sigma\in H$ and $F\subseteq H$ satisfying $|F\cap H_k|\leq 1$ for each $k$ there is a finite $\Delta(F,\sigma)\subseteq\Delta$ for which $\sigma\in\bigcap \{ A(X,\pi(X)):X\in\Delta(F,\sigma)\}$ while at the same time $F\setminus\{\sigma\}\cap \bigcap \{ A(X,\pi(X)):X\in\Delta(F,\sigma)\}=\emptyset.$

The presenter went too fast and used lemma 3.1 (9.5) and 3.2 (9.6).

Using the lemmas we obtain an independent matrix $M_2 = \{ A(X,Y): X\in\Delta, Y\subseteq\mathcal P(\omega)\setminus \{\emptyset\}\}.$ For each $X\in\Delta$ choose an infinite $\nabla_X\subseteq \mathcal P(\omega)\setminus \{\emptyset\}$ countable set such that $\pi(X)\in\nabla_X$ and $\bigcup \{ A(X,Y):Y\in\nabla_X\} = H.$ The $\nabla_X$ gives us a matrix $M_3 = \{ A(X,Y):X\in\Delta,Y\in \nabla_X\}$ We now disjointify this matrix to a matrix $M_3^\prime$ while maintaining $A^\prime(X,\pi(X)) = A(X,\pi(X))$ and $A^\prime(X,Y)=^*A(X,Y)$. This matrix is as required (i.e. finish the proof as when prooving HMP from Rasiowa-Sikorski).

## 9.2D. Chodounský: PID and Tukey Maximality

The following is based on the preprint D. Chodounský, P. Borodulin-Nadzeja: Hausdorff gaps and towers in $\mathcal P(\omega)$/Fin, arxiv:1302.4550.

9.8

Recall that a poset $P$ is Tukey-reducible to $Q$ if there is a cofinal function $f:Q\to P$. A poset $P$ of size (or cofinality) at most $\omega_1$ is said to have Tukey-type $[\omega_1]^{<\omega}$ if there is $P^\prime\in[P]^{\omega_1}$ such that any infinite part of $P^\prime$ is unbounded (in P).

We have the following Tukey-types of posets of size at most $\omega_1$. $1 \leq \omega\perp\omega_1\leq \omega\times\omega_1\leq [\omega_1]^{<\omega}$

Also recall that

9.9

PID is the following statement. Whenever $\mathcal I\subseteq[\omega_1]^{\leq\omega}$ is an ideal then there are exactly two possibilities:

1. $(\exists K\in[\omega_1]^{\omega_1})([K]^{\leq\omega}\subseteq\mathcal I)$ or
2. $\omega_1$ can be partitioned into countably many sets $\{A_n:n<\omega\}$ such that for each $I\in\mathcal I$ and $n<\omega$ $I\cap A_n=\emptyset$.
9.10

The ideal $\mathcal I$ consisting of subsets of a Suslin tree which hit every branch below some level only finitely often (or something like that) does not satisfy either 1. or 2. In particular, PID implies there are no Suslin trees.

Under PFA these five types are the only possible types.

9.11Todorčević, Raghavan

Assume PID. Then the existence of precisely 5 Tukey types is equivalent to $\omega_1 < \mathfrak b$ + something.

David and Piotr Borodulin-Nadzeja were interested in Tukey-types of ($\omega_1$-generated) ideals on $\omega$. They proved the following theorem.

Assume $\omega_1< \mathfrak b$ and PID. Then each (nontrivial) $\omega_1$-generated P-ideal has Tukey type $[\omega_1]^{<\omega}$, i.e. is maximal.

They were, in fact, interested mainly in ideals generated by towers. The following definition is relevant in this context

9.13

An ($\subseteq^*$-increasing) tower $\overline{T}=\{T_\alpha:\alpha<\omega_1\}$ has property (H) if for all $\alpha<\omega_1$ and $n<\omega$ the set $\{\xi<\alpha:T_\xi\setminus T_\alpha\subseteq n\}$ is finite.

9.14

An ideal generated by a tower $\overline{T}$ is Tukey maximal iff $\overline{T}$ has property (H).

Let $\overline{T}$ be a tower. We define $\mathcal I\subseteq[\omega_1]^{\leq\omega}$. A set $I\in\mathcal I$ iff the sets $C_\alpha^n(I)=\{\xi\in I\cap\alpha: T_\xi\setminus T_\alpha\subseteq n\}$ are finite for each $\alpha<\omega_1$ and each $n<\omega$. The idea is that we want to construct a subtower satisfying (H) and each $I\in\mathcal I$ is a potential initial segment of such a sequence, i.e. it has to satisfy (H) and, moreover, we want any such $I$ to always be extendible to a larger $I$ which is still a potential initial segment. We will show that $\mathcal I$ is a P-ideal and then apply PID. In fact, we will show something slightly more general. Consider the (bad) sets $B_\alpha^n = \{\xi<\alpha: T_\alpha\setminus T_\alpha\subseteq n\}$ and $\mathcal B = \{B_\alpha^n:\alpha<\omega_1,n<\omega\}$. Then $\mathcal B^\perp\cap[\omega]^{\leq\omega} = \mathcal I$. I.e. the ideal $\mathcal I$ is the orthogonal of $\mathcal B$ which has size $\omega_1$. It immediately follows that it is an ideal. As is typical in PID proofs, we will use the following lemma:

9.15

If $\omega_1<\mathfrak b$ then any orthogonal to a set $\mathcal B$ of size at most $\omega_1$ is a P-ideal.

Consider any countable sequence $\langle I_n:n<\omega\rangle\subseteq\mathcal I$ as a sequence of vertical lines. Since $\mathcal I$ is orthogonal to $\mathcal B$ each $B\in\mathcal B$ gives rise to a function $f_B:\omega\to \omega$. SInce $\omega_1<\mathfrak b$, they are dominated by some $f:\omega\to\omega$. Then $I=\bigcup_{n<\omega} I_n\setminus f(n)$ is also orthogonal to $\mathcal B$.

Now we are ready to apply PID to $\mathcal I$. There are two cases to consider. The first condition of PID immediately gives us the required subtower. It will, therefore, be sufficient to show that the second condition of PID is impossible, i.e. for each $A\in[\omega_1]^{\omega_1}$ we need to find an $I\in\mathcal I\cap[A]^\omega$.

9.16

There is $X\in2^\omega$ such that $X\in\overline{\{T_\alpha:\alpha\in A\}}$ (closure taken in the Cantor space) and $X$ is not in the ideal $\mathcal I(\overline{T})$ generated by the tower $\overline{T}$.

Otherwise $\overline{\{T_\alpha:\alpha\in A\}}$ is a closed subset of $\mathcal I(\overline{T})$. Moreover $\mathcal I(\overline{T})$ is generated from $\overline{\{T_\alpha:\alpha\in A\}}$ by Borel-operations, i.e. $\mathcal I(\overline{T})$ is Borel (Analytic will be sufficient for our purposes). However, using Solecki's characterization of analytic P-ideals, $\mathcal I(\overline{T})$ must be generated by at least $\mathfrak d$ many sets --- a contradiction.

Consider now $\{T_\alpha:\alpha\in I\}$ for some $I\in[A]^\omega$. Use the claim to get an $x$. Suppose $I\not\in\mathcal I$. Then there are $\beta<\omega_1, n<\omega$ such that $T_\alpha\subseteq T_\beta\cup n$ --- a contradiction.

9.17

The proof of Todorčević's and Raghavan's result is very similar however, they need to guarantee that not only analytic P-ideals are generated by $\mathfrak d$-many sets but all $F_\sigma$-ideals.

# 1022. 1. 2014

## 10.1Ch. Brech: Application of PID to Banach spaces

(see also the Winter School 2014 talk Ch. Brech: On PID and biorthogonal systems)

10.1

For a Banach space $X$ a family of pairs $\langle x_\alpha,f_\alpha:\alpha\in\Gamma\rangle\subseteq X\times X^*$ is a bi-orthogonal system if $f_\alpha(x_\beta)=\delta_{\alpha\beta}$.

10.2

The span of the first (or second) coordinates need not be the full space (dual space).

Considering

10.3

Separable infinite dimensional Banach spaces have infinite bi-orthogonal systems. (in fact even satisfying the stronger condition mentioned in the above note).

10.4

Suppose $X$ is a Banach space of density $\kappa$. Is it true that it has a bi-orthogonal system of size $\kappa$?

10.5Kunen, '80

Assuming CH, there is a scattered, compact, Hausdorff, non-metrizable space $K$ of weight $\omega_1$ such that all finite powers are hereditarily separable.

10.6Todorčević, '89

Assuming $\mathfrak b=\omega_1$ there is a space as in 10.5

Both of the above examples yield a nonseparable Banach space with no uncountable bi-orthogonal system, viz $X=C(K)$. $X$ will be Lindelof in the weak topology and, if $\langle x_\alpha,f_\alpha:\alpha<\omega_1\rangle$ is a bi-orthogonal system, then $\{x_\alpha:\alpha<\omega_1\}$ will be discrete in the weak topology --- a contradiction. (To see this note that the basis of the weak topology consists of sets of the form $V(x,f_0,\ldots,f_n,\varepsilon) = \{y\in X:(\forall i\leq n)(|f_i(y)-f_i(x)|<\varepsilon)\}$

10.7Todorčević, 2006

Assuming PID and $\mathfrak p>\omega_1$ then every nonseparable Banach space has an uncountable bi-orthogonal system.

See 9.9 for the definition of the P-ideal dichotomy (PID).

10.8Todorčević

Assume PID. Is $\mathfrak b=\omega_1$ equivalent to the existence of a nonseparable Banach space with no uncountable bi-orthogonal system.

10.9

An Asplund Banach space is a space whose separable spaces have separable duals.

10.10

If $Y$ is scattered then $C(Y)$ is Asplund.

10.11

For a Banach space $X$ a family of pairs $\langle x_\alpha,f_\alpha:\alpha\in\Gamma\rangle\subseteq X\times X^*$ is a $\varepsilon$-bi-orthogonal system if $|f_\alpha(x_\beta)|<\varepsilon$ for each $\alpha\neq\beta$.

10.12Brech

Assume PID and $\mathfrak b > \omega_1$ then every nonseparable Asplund space has an uncountable $\varepsilon$-bi-orthogonal system for each $0<\varepsilon<1$.

Hahn-Banach extension theorem (14.2) means we can extend bi-orthogonal from subspaces. In particular we may assume $d(X)=\omega_1$. Write $X=\bigcup_{\alpha<\omega_1} X_\alpha$ as an increasing union of separable closed subspaces. By induction construct $x_\alpha\in X_{\alpha+1}\setminus X_\alpha$ and $h_\alpha\in X^*$ such that $h_\alpha(x_\alpha)=\mathbf 1$, $||h_\alpha||=1$ and $h_\alpha\upharpoonright X_\alpha =0$. Let $S=\{h_\alpha-h_\beta:\beta\neq\alpha\}$. The proof now splits into three successive claims. First, fix a $D\subseteq X$ dense, $\mathbb Q$-linear subspace of $X$ of size $\omega_1$.

10.13

There is an uncountable subset $\{f_\alpha:\alpha<\omega_1\}\subseteq S$ such that for each $x\in D$ the sequence $\langle f_\alpha(x):\alpha<\omega_1\rangle\in c_0(\omega_1) (=(\{ t:(\forall\varepsilon>0)(|\{\alpha<\omega_1:t(\alpha)\geq\varepsilon\}|<\omega)\}$.

of claim

Apply PID to the ideal $\mathcal I = \{ A\in[S]^{\omega_1}:(\forall x\in D)(\langle f_\alpha(x):\alpha\in A\rangle\in c_0(A))\}$ (which is a P-ideal under $\mathfrak b>\omega_1$) to find the uncountable subset of $S$.

10.14

There is an uncountable $\Gamma\in[\omega_1]^{\omega_1}$ such that $(\forall x\in D)(\langle f_\alpha(x):\alpha\in\Gamma\rangle\in l_1(\Gamma)),$ where $l_1(\Gamma) = \Big\{ t\in{}^\Gamma X: \sum_{\alpha\in\Gamma} |h(\alpha)|<\infty\Big\}$

of claim

Let $\mathcal I=\Big\{A\in[\omega_1]^\omega:(\forall x\in D)\left(\sum_{\alpha\in A}|f_\alpha(x)<\infty\right)\Big\}$ First show, using $\mathfrak b>\omega_1$, that $\mathcal I$ is a P-ideal. Now apply PID to $\mathcal I$. The first possibility gives $\Gamma=K$ so it is sufficient to show that the second possibility is impossible :-) But that is easy, since any sequence in $c_0$ always contains an infinite summable subsequence.

The following claim now finishes the proof.

10.15

There is a sequence $\langle \alpha_\xi:\xi\in\omega_1\rangle\subseteq\Gamma$ and $\langle x_\xi:\xi\in\omega_1\rangle\subseteq X$ such that $\langle x_\xi,f_\xi:\xi\in\omega_1\rangle$ is an $\varepsilon$-bi-orthogonal system.

## 10.2J. Lopez-Abad: Families of finite sets

First a remark concerning Christina's talk.

10.16

An Auerbach base for a finite dimensional normed space $X$ is a sequence $\langle x_i,f_i:i which generatex \(X$ and such that $f_i(x_j)=\delta_{ij}$ and, crucially, $||f_i||=||x_i|| = 1$ for $i. 10.17 Every finite dimensional normed space has an Auerbach basis. 10.18Pelczynski If \(X$ is an infinite dimensional separable Banach space then it has, for each $\varepsilon>0$ a Markushevich basis (i.e. a sequence $\langle x_i,f_i:i<\omega\rangle\subseteq X\times X^*$, $f_i(x_j)=\delta_{ij}$, such that the first coordinates span a dense subspace of $X$) such that $1-\varepsilon < ||f_i||,||x_i||< 1+\varepsilon$).

Now we will consider families of finite subsets of $S$ (typically, $S=\mathbb N$) and the two relations $\subseteq, \sqsubseteq$.

10.19

Given $M\in[S]^\omega$ and a family $\mathcal F\subseteq [S]^{<\omega}$ we let $\mathcal F\upharpoonright M = \mathcal F\cap \mathcal P(M)$ (the restriction) and $\mathcal F[M] = \{s\cap M:s\in\mathcal F\}$ (the trace)

Let us consider when $\mathcal F\upharpoonright M$ or $\mathcal F[M]$ are $\subseteq$ or a $\sqsubseteq$ antichains. When are they $\subseteq$ or $\sqsubseteq$ hereditary.

10.20

The family $\mathcal F$ is compact iff every sequence in $\mathcal F$ has a $\Delta$-system subsequence with root in $\mathcal F$.

10.21

The family $\mathcal F$ is precompact iff every sequence in $\mathcal F$ has a $\Delta$-system subsequence.

10.22

$\mathcal F$ is precompact iff its closure (in $2^\omega$) is compact.

10.23

Given $l<\omega$ the family $[\omega]^l$ is a precompact antichail while its closure, $[\omega]^{\leq l}$, is hereditary and compact.

10.24Ramsey

For any coloring $\chi:[\omega]^l\to n$ there is an infinite $M\subseteq\omega$ such that $\chi\upharpoonright[M]^l$ is constant.

10.25

A family $\mathcal F$ is Ramsey if for any coloring $\chi:\mathcal F\to n$ there is an infinite $M$ such that $|\{i 10.26Nash-Williams For a family \(\mathcal F$ the following are equivalent: 1. $\mathcal F$ is Ramsey 2. there is an infinite $M$ such that $\mathcal F\upharpoonright M$ is a $\sqsubseteq$-antichain (it is thin) 3. there is an infinite $M$ such that $\mathcal F\upharpoonright M$ is a $\subseteq$-antichain (it is a Sperner system)

10.27Nash-Williams

Let $\mathcal F\subseteq [M]^{<\omega}$. It is called a barrier on M if it is Sperner and is unavoidable, i.e. every infinite subset of $M$ has an initial part in $\mathcal F$. It is called block on $M$ if it is a thin unavoidable family.

10.28

For every thin (Sperner) family $\mathcal F$ there is an infinite $M$ such that $\mathcal F\upharpoonright M=\emptyset$ or $\mathcal F\upharpoonright M$ is a barrier.

10.29

If $\mathcal F$ is a barrier on $M$ then for every coloring $\chi:\mathcal F\to n$ there is an infinite $N\subseteq M$ such that $\chi\upharpoonright(\mathcal F\upharpoonright N)$ is constant.

10.30

If $\mathcal F$ is precompact then there is an infinite set such that its trace $\mathcal F[M]$ is the closure of a barrier.

10.31

If $\mathcal B$ is a barrier, then its topological closure is equal to its $\subseteq$-closure and its $\sqsubseteq$-closure.

In particular, every precompact family has a $\subseteq$-hereditary trace. For example, if $\mathcal F=[\omega]^n$ then $\mathcal F[2\mathbb N]$ is hereditary.

10.32

Suppose $\mathcal F$ is arbitrary then there exists an infinite $M$ such that 1.$[M]^{<\omega} = \overline{\mathcal F[M]}^{\subseteq}$ or 2. $\mathcal F[M]$ is the closure of a barrier on M.

For each $n,l<\omega$ and a coloring $\chi:[\omega]^l\to n$ there is $I\subseteq l$ and an infinite $M$ such that for each $s=\{n_1<\cdots \(c(s)=c(t)$ iff $(\forall i\in I)(n_i=m_i)$.

How do coloring of barriers behave?

10.34Pudlák-Rödl

For each barrier $\mathcal B$ on $M$ and a coloring $\chi:\mathcal B\to X$ there is an infinite $M$ and a barrier $\mathcal C$ on M, a $\varphi:\mathcal B\upharpoonright M\to \mathcal C$ and a $\hat{\chi}:\mathcal C\to X$ such that $\hat{\chi}\circ \varphi=\chi$, $\hat{chi}$ is 1-1 and $\varphi(s)\subseteq s$.

10.35

Let $\{m_n:n<\omega\}=M$ be an increasing enumeration. A 0-uniform family on $M$ is $\{\emptyset\}$. An $\alpha+1$-uniform family on $M$ if $\mathcal F_{\{m_n\}} = \{s:m_n is $\alpha$-uniform in $\{m_l:l>n\}$. If $\alpha$ is limit, then a family $\mathcal F$ is $\alpha$-uniform if there is an increasing sequence $\alpha_n$ converging to $\alpha$ such that $\mathcal F_{\{m_n\}}$ is $\alpha_n$-uniform. 10.36Pudlák-Rödl TFAE 1. There is an infinite $M$ such that $\mathcal B\upharpoonright M$ is a barrier on M 2. There is an infinite $M$ such that $\mathcal B\upharpoonright M$ is uniform on M 10.37 If $\varphi:\mathcal B\to Fin$ where $\mathcal B$ is a barrier and $\varphi[\mathcal B]$ is precompact then there is an infinite $M$ such that for each $s\in\mathcal B\upharpoonright M$ we have $\varphi(s)\cap M\subseteq s$. 10.38 If $\varphi:\mathcal B\to c_0$ is a mapping of a barrier into $c_0$ whose image is relatively compact in the weak topology then for each $\varepsilon > 0$ there is an infinite $M$ such that $\forall s\in\mathcal B\upharpoonright M$ we have \[ \sum_{n\in M\setminus s} |\varphi(s)|_n <\varepsilon$

10.39

The set $\mathcal S=\{s:|s|=\min s\}$ is $\omega$-uniform, precompact and large in the following sense.

10.40

A family $\mathcal F$ is large in $M$ iff for each $N\subseteq M$ and for all $n<\omega$ there is $s\in\mathcal F$ such that $|s\cap N|\geq n$. It is $\lambda$-filling (for $\lambda\in[0,1]$) if for each $s\subseteq M$ there is $t\in\mathcal F\upharpoonright s$ such that $|t|\geq \lambda\cdot|s|$.

10.41

For a cardinal $\kappa$ the following are equivalent: 1. There is a compact, hereditary, large family on $\kappa$. 2. $\kappa$ is not $\omega$-Erdös. 3. some statement about Banach spaces

10.42

A cardinal $\kappa$ is $\omega$-Erdös iff for any coloring $\chi:[\kappa]^{<\omega}\to 2$ there is an infinite set $A\subseteq\kappa$ such that on $[A]^{<\omega}$ the coloring $\chi$ only depends on the cardinality.

The following is Fremlin's DU-problem.

10.43

Does there exist a $1/2$-filling compact (or precompact) family on $\omega_1$?

10.44

Taking the Cantor set $2^\omega$ instead of $\omega_1$ and wanting the family to be Borel (or Analytic?) the answer is no (see Dodos, P. and Kanellopoulos, K: On filling families of finite subsets of the Cantor set, Math. Proc. Cam. Phil. Soc. 145 (2008), pp 165-175, doi:10.1017/S0305004108001096)

## 10.3A. Avilés: Tukey classification of orthogonals

The following theorem can be found in Avilés, A., Plebanek, G., Rodriguez, J.: Measurability in $C(2^\kappa)$ and Kunen cardinals, Israel Journal of Math. 195 (2012), pp 1-30, doi:10.1007/s11856-012-0122-0.

10.45Avilés, Plebanek, Rodriguez

Assume Analytic Determinacy. If $\mathcal I$ is an analytic family of subsets of $\omega$ then the orthogonal $\mathcal I^\perp =\{A\subseteq\omega: (\forall B\in\mathcal I)(|A\cap B|<\omega)\}$ (which is co-analytic) is Tukey-equivalent (see 9.8) to either

1. $\{0\}$;
2. $\omega$;
3. $\omega^\omega$;
4. $K(\mathbb Q)$ (i.e.compact subsets of the rationals ordered by inclusion); or
5. finite subsets of $\mathbb R$.
10.46Todorčević

Let $\mathcal I$ and $\mathcal J$ be two orthogonal (i.e. $\mathcal J\subseteq\mathcal I^\perp$ analytic ideals on $\omega$. Then either $\mathcal I$ and $\mathcal J$ are countably separated (i.e. there is a countable $\mathcal C\subseteq P(\omega)$, such that for any disjoint $a\in\mathcal I,b\in\mathcal J$ there is a $c\in\mathcal C$ such that $a\subseteq c$ and $b\subseteq\omega\setminus c$). or there is a 1-1 function $u:2^{<\omega}\to\omega$ such that 0-chains go to elements of $\mathcal I$ and 1-chains go to elements of $\mathcal J$ (where an i-chain is a sequence $\{x_0,x_1,\ldots\}\subseteq 2^{<\omega}$ such that for all $n$ we have $x_{n+1}=x_n^{\smallfrown}(i,k_1,\ldots,k_{p_n})$ for some $k_1,\ldots,k_{p_n}$ ).

10.47

In the above theorem, analytic determinacy gives the same result for $\Sigma^1_2$ ideals.

Applying the theorem of Todorčević to the ideal $\mathcal I$ and $\mathcal I^\perp$, the second option gives $[\mathbb R]^{<\omega}$.

10.48Avilés, Todorčević

$\mathcal I,\mathcal J$ are countably separated iff there is a metrizable compactification $K$ of $\omega$ and a decomposition $K\setminus\omega=U\cup V$ such that $(\forall a\in\mathcal I)(\overline{a}^K\subseteq U)$ and $(\forall b\in\mathcal J)(\overline{b}^K\subseteq V)$

This theorem directly reduces 10.45 to Fremlin's characterization:

10.49Fremlin

If $\mathcal E$ is a co-analytic metric space then $K(E)$ is Tukey-equivalent to 1,2,3 or 4.

# 114. 2. 2014

## 11.1Arturo Antonio Martinez-Celis Rodriguez: Porous Sets

The origin of the following definition goes back to the beginning of the 20th century.

11.1

Let $(X,\delta)$ be a metric space. For a set $A\subseteq X$ define $P(x,R,A) = \sup\{r\geq 0: (\exists z)(B(z,r)\subseteq B(x,R)\setminus A)\}$ and, for $x\in X$ $\overline{p}_{x,A} = \overline{\lim_{R\to0^+}} \frac{P(x,R,A)}{R},\quad \underline{p}_{x,A} = \underline{\lim_{R\to0^+}} \frac{P(x,R,A)}{R}.$ We say that a set $A$ is upper (lower) porous if for every $x\in X$ we have $\overline{p}_{x,A}>0$ ($\underline{p}_{x,A}>0$). The ideals $UP(X)$ and $SP(X)$ are the $\sigma$-ideals generated by, respectively, upper porous and lower porous sets.

Recently, the cardinal invariants of these ideals were investigated by several people. The following is a short summary of the results:

The following is from J. Brendle: The additivity of porosity ideals , Proc. Amer. Math. Soc., vol 124 (1) 1996

11.2Brendle, 96

$add(UP)=\omega_1$, $cof(UP)=\mathfrak c$.

The following is from M. Repický: Cardinal invariants related to porous sets , Set theory of the reals (Ramat Gan, 1991), 433-438, Israel Mathematical Conference Proceedings 6.

11.3Repický, 91

$cov(UP)\leq cov(\mathcal N)$, $non(UP)\geq \mathfrak t$, $non(UP)\geq add(\mathcal N)$.

The following result is from the paper M. Hrušák, O. Zindulka: Cardinal Invariants of monotone and porous sets , Journal of Symbolic Logic, vol 77 (1), 2012

11.4Hrušák, Zindulka, 12

$Con(non(SP) < \mathfrak m_{\sigma-\mbox{centered}})$, $Con(cov(SP)>cov(\mathcal N))$.

We will work with a different definition of the ideal $SP$ which is easier to work with.

11.5

A subset $A\subseteq 2^\omega$ is strongly porous if there is an $n<\omega$ such that for each $s\in 2^{<\omega}$ there is a $t\in 2^n$ such that $[s^{\smallfrown} t]\cap A=\emptyset$.

11.6Hrušák, Zindulka

$SP(2^\omega)$ is the $\sigma$-ideal generated by strongly porous sets.

11.7 Hrušák, Zindulka

The additivity, covering, non and cofinality invariants for $SP(\mathbb R)$ and $SP(2^\omega)$ coincide.

11.8

For $\mathbb Q$ the cardinal invariants are different!

11.9

For $\sigma:2^{<\omega}\to2^n$ we let $X_{\sigma} = \{ x\in 2^\omega:(\forall k\in\omega)(x\not\in[x\upharpoonright k^{\smallfrown}\sigma(x\upharpoonright k)\}$

11.10

Each $X_\sigma$ is a closed strongly porous set, and each strongly porous set is contained in a set of the form $X_\sigma$ for some $\sigma$.

11.11

A forcing $\mathbb P$ strongly preserves $non(SP)$ if for each $\mathbb P$-name $\dot{X}$ for an old strongly porous set, i.e. such that $\mathbb P\Vdash \dot{X}\subseteq 2^{\omega}\cap V$, then $\mathbb P\Vdash (\exists Y\in SP\cap V)(\dot{X}\subseteq Y)$.

The next lemma gives some examples for strong preservation of $non(SP)$.

11.12

If $\mathbb P$ is $\sigma$-$n$-linked for every $n<\omega$ then $\mathbb P$ strongly preserves $non(SP)$.

Let $\dot{\varphi}$ be a name for a function $2^{<\omega}\to 2^n$ such that $\mathbb P \Vdash \dot{X}\subseteq X_{\dot{\varphi}}\cap V.$ Write $\mathbb P=\bigcup_{n<\omega} P_n$ be a decomposition into $2^n$-linked subsets. Embed $\mathbb P$ into its completion $\mathbb B$.

11.13

For every $n<\omega$ and for every $t\in 2^{<\omega}$ there is $s\in 2^n$ such that for each $p\in P_n$ $p\wedge||\dot{\varphi}(t)=s||\neq\emptyset$

of Claim

By the above claim we can fix functions $\varphi_n: 2^{<\omega}\to 2^n$ witnessing the claim. The following claim now finishes the proof:

11.14

$\mathbb P\Vdash X_{\dot{\varphi}}\cap V\subseteq\bigcup_{n<\omega} X_{\varphi_n}$

of Claim
11.15

Let $\mathbb A=\{B\in Borel(2^\omega):\mu(B)>1/2\}$ ordered by inclusion. Then $\mathbb A$ is $\sigma$-$n$-linked.

Let $n<\omega$ and for every clopen $C$ let $A_C=\{A\in\mathbb A:\mu(C\setminus A)< 1/n(\mu(C)-1/2)\},$ and then the blackboard was erased.

We will need the following result:

11.16Hrušák, Zindulka

Finite support iteration of forcings strongly preserving $non(SP)$ strongly preserves $non(SP)$

11.17

It is consistent that $non(SP). Start with a model of CH and let \(\mathbb P$ be a f.sp. iteration of length $\omega_2$ of the amoeba forcing $\mathbb A$. Then $V[G]\models add(\mathcal N)=\omega_2$. By the previous theorem (11.16), $\mathbb P$ strongly preserves $non(SP)$. So $V[G]\models 2^\omega\cap V\not\in SP$ so $V[G]\models non(SP)=\omega_1$.

11.18

Let $Mon$ be the $\sigma$-ideal generated by GO-subspaces of $\mathbb R^2$.

11.19Hrušák, Zindulka

$non(Mon)=non(SP)$ and $cov(Mon)=cov(SP)$.

The following is also true $add(Mon)=\omega_1$, $cof(Mon)=\mathfrak c$.

11.20

The above might not strictly be true, the definition of $Mon$ in Hrušák's and Zindulka's is slightly different and it is not clear, whether they coincide.

## 11.2U. Ariet Ramos-Garcia: Extremal disconnectedness and ultrafilters

Our work with Michael is motivated by an old question of Arhangelskii:

11.21Arhangelskii, 1967

Is there a non-discrete extremally disconnected topological group?

11.22Sirota, 1969

Yes, under CH.

The following is probably (the notetakers guess) from A. Louveau: Sur un article de S. Sirota, Bull. Sci. Math, 1972.

11.23Louveau, 1972

The group $([\omega]^{<\omega},\Delta)$ with topology given by the neighbourhood basis of $0$ specified by some Ramsey ultrafilter $\mathcal U$ (i.e. the basis consists of $\{[F]^{<\omega}:F\in\mathcal U\}$) is extremally disconnected.

11.24

Given a point $x\in X$ we define $spect(X,x)=\Big\{ p\in\omega^*: (\exists \langle U_n:n<\omega\rangle\ \mbox{open}) \left(p=\Big\{A\subseteq\omega:x\in\overline{\bigcup_{n\in A}U_n}\Big\}\right)\Big \}$ and $Spect(X)=\bigcup_{x\in X}spect(X,x)$. Moreover we define $0-Spect(X,x)$ to consists of ultrafilters whose witnessing sequence consists of Clopen sets and $c-Spect(X)$ if it consits of disjoint sets.

11.25

1. If $X$ is extremally disconnected, then $Spect(X)=0-Spect(X)=c-0-Spect(X)$. 2. $spect(X,x)$ is closed downward in the Rudin-Keisler order. 3. If $X$ is extremally disconnected then $spect(X,x)$ is upwards directed in the Rudin-Keisler order.

11.26

If $X$ is extremally disconnected and $\chi(X,x)<\mathfrak d$ then $spect(X,x)\subseteq \{\mbox{P-points}\}$.

Assume, aiming towards a contradiction, that some $p\in spect(X,x)$ is not a P-point. since $X$ is ED we have $spect(X,x)=0-c-spect(X,x)$, i.e. there is a sequence $\langle U_n:n<\omega\rangle$ of disjoint clopen sets witnessing this. Since $p\in\omega^*$ we have $x\in\overline{\bigcup_{n<\omega} U_n}\setminus \bigcup_{n<\omega} U_n$ Also, since $p$ is not a P-point there is a sequence $\{A_n:n<\omega\}\subseteq p$ with no pseudointersection in $p$. Let $\mathcal B$ be a local base at $x$ of size $<\mathfrak d$. For each $B\in\mathcal B$ define $f_B(n) = \max\Big\{k: B\cap \bigcup_{i\in A_n\cap k}U_i=\emptyset\Big\}\cup\{0\}$ Since $\{A_n:n<\omega\}$ do not have a pseudointersection in p, we know that for each $f:\omega\to\omega$ $x\not\in H_f=\overline{\bigcup_{n<\omega} \bigcup_{i\in A_n\cap f(n)} U_i},$ so, in particular, there is $B\in\mathcal B$ which is disjoint from $H_f$. But then $f\leq f_B$. It follows that $\{f_B:B\in\mathcal B\}$ form a dominating family which contradicts the fact that $|\mathcal B|<\mathfrak d$.

Recall the following proposition, which follows from Stone duality.

11.27

IF $X$ is extremally disconnected then $RO(X)$ does not add Cohen reals iff for each continuous function $f:X\to 2^\omega$ there is a nonempty open $U$ such that $f[U]\in nwd(2^\omega)$.

The previous proposition motivates the following conjecture.

11.28Hrušák

if $G$ is an ED topological group and $f:G\to 2^\omega$ is continuous then there is a nonempty open $U\subseteq G$ such that $f[U]\in nwd(2^\omega)$.

11.29

The conjecture holds for countable Boolean ED topological groups and continuous homomorphisms.

11.30

Suppose that $G$ is a Boolean topological group. Then for any open subset $U$ and for every neighbourhood $V$ of $0$ there is an open subset $W\subseteq U$ such that $W+W\subseteq V$

of Lemma

Clear.

11.31

There is a partition $U\cup V=G\setminus\{0\}$ into open sets such that for each $n<\omega$ there are nonempty open $U^0_n,U^1_n,V^0_n,V^1_n\subseteq[s_n]$, where $\{s_n:n<\omega\}$ is an enumeration of $2^{<\omega}$, such that $f^{-1}[U_n^0+U^1_n]\subseteq U$ and $f^{-1}[V_n^0+V^1_n]\subseteq V$

of Lemma

Recursively apply the previous lemma.

Again, using lemma 1, we can construct the $U,V$s so that if $F_n=\overline{U_n^1+U_n^1}$ and $E_n=\overline{V_n^1+V_n^1}$ then $\langle F_n,E_n:n<\omega\rangle$ form a cellular family on $2^\omega/\{0\}$.

And then I got lost...

# 1219. 2. 2014

## 12.1B. Balcar & J. Verner: h,b,g (recap.)

A typical problem in mathematics is the problem of classifying structures. Since the structures in question might be very complicated one can approach the problem by identifying some simpler relevant features of the structures and classify them according to these features, which are sometimes called invariants. E.g. in algebraic topology, each (pointed) space has an associated fundamental group. Spaces can then be classified according to their fundamental groups. The fundamental group may be a very complicated object however and we can expect that it carries a lot of information about the space. However, sometimes even very simple objects give a lot of information. E.g. the dimension of a vector space over a given field, already carries all the information about the structure even though it is just a number. Another example where a simple number already gives interesting information (although not all) is the chromatic number of a graph.

This motivates the definition of cardinal invariants, which can be used to classify models of set theory according to the properties of their real numbers. The following will only be a cursory glance at the topic, the interested reader is advised to consult the very readable chapter on Cardinal invariants in the Handbook of Set Theory (A. Blass: Combinatorial Cardinal Characteristics of the Continuum, chapter 6 of the Handbook of Set Theory, vol 1., Springer 2011).

Before introducing the two most well known cardinal invariants, $\mathfrak{b,d}$, we need some definitions.

12.1

For two sets $A,B\subseteq\omega$ we say that $A\subseteq^*B$ if $|A\setminus B|<\omega$, $A=*B$ if $A\subseteq^* B\ \&\ B\subseteq^* A$. The ${}^*$ superscript on the relations indicates that the relations hold up to, possibly, a finite number of exceptions. Similarly, for two functions $f,g:\omega\to\omega$ one defines $f\leq^*g$ if $|\{n:f(n)>g(n)\}|<\omega$

12.2

Given a partial order $(P,\leq)$ we say that a set $X\subseteq P$ is unbounded if for each $p\in P$ there is some $b\in X$ such that $b\not\leq p$. We say that it is dominating (or cofinal) if for each $p\in P$ there is some $d\in X$ such that $p\leq d$.

Now we are ready to define the first two invariants:

12.3

The bounding number $\mathfrak b = \min\{\mathcal B\subseteq{}^\omega\omega:\mathcal B\ \mbox{is unbounded in}\ ({}^\omega,\omega),\leq^*)$ is the smallest cardinality of an $\leq^*$-unbounded family of functions from $\omega$ to $\omega$. Similarly the dominating number $\mathfrak d = \min\{\mathcal B\subseteq{}^\omega\omega:\mathcal B\ \mbox{is dominating in}\ ({}^\omega,\omega),\leq^*)$

12.4

$\omega_1\leq\mathfrak b =\mbox{cf}\ \mathfrak b\leq\mbox{cf}\ \mathfrak d\leq\mathfrak d\leq\mathfrak c=|\mathbb R|=2^{\aleph_0}$

The last equalities are definitions and the last inequality is immediate. To see the first inequality notice that any countable collection of functions may be dominated by a single function, which can be built inductively (this is very similar to Cantor's diagonal argument). Although slightly more involved, it is not very hard to prove the other inequalities.

The following cardinal invariant was isolated in B. Balcar, J. Pelant and P. Simon: The space of ultrafilters on N covered by nowhere dense sets, Fund. Math. 110 (1) 1980, pp. 11-24. ( where it was called simply $\kappa$)

12.5

$\mathfrak h$ is the smallest cardinal $\kappa$ such that the Boolean algebra $\mathcal P(\omega)/fin$ is not $(\kappa,2)$-distributive.

In forcing language, the above definition of $\mathfrak h$ may be rephrased as follows: $\mathfrak h$ is the smallest cardinal $\kappa$ such that forcing with $\mathcal P(\omega)/fin$ adds a new subset of $\kappa$.

We will show another, equivalent, definition of $\mathfrak h$. First we need to define the notion of a MAD family. To motivate this, note that any family of disjoint subsets of $\omega$ must necessarily be (at most) countable. What happens if we weaken the requirement of disjointness:

12.6

A family $\mathcal A$ of infinite subsets of $\omega$ is called almost disjoint (AD for short) if each distinct $A,B\in\mathcal A$ are almost disjoint, i.e. $|A\cap B|<\omega$. An AD family $\mathcal A$ is called maximal almost disjoint (MAD for short) if it is infinite and cannot be extended to a larger AD family, i.e. for any infinite set $X\subseteq\omega$ there is some $A\in\mathcal A$ such that $|X\cap A|=\omega$.

The existence of MAD families easily follows from the Axiom of choice and it turns out that there always are uncountable AD families (and hence also MAD families). (For example, instead of $\omega$ consider the countable set $S=2^{<\omega}$ of finite sequences of zeroes and ones. For each $f\in2^\omega$ let $A_f=\{f\upharpoonright n:n<\omega\}\subseteq S$. Then $\{A_f:f\in \omega^\omega\}$ is an almost disjoint family of subsets of $S$ of size $\mathfrak c$.)

Let us now start with some MAD family $\mathcal A_0$. For each $A\in\mathcal A_0$ we can restrict to $A$ and find another MAD family $\mathcal A_A$ on $A$. Then it is not hard to see that $\mathcal A_1=\bigcup\{\mathcal A_A:A\in A_0\}$ will again be a mad family on $\omega$. We can repeat this process to eventually get a sequence $\{\mathcal A_n:n<\omega\}$ of MAD families with the nice property that for each $X\in\mathcal A_{n+1}$ there is some $Y\in\mathcal A_n$ such that $X\subseteq^* Y$ (we say that $\mathcal A_{n+1}$ refines $\mathcal A_n$ or, in symbols, $\mathcal A_{n+1}\preceq \mathcal A_n$.

Suppose we would like to continue and define a MAD family $\mathcal A_\omega$ which would refine each of the previous families. It is not immediately obvious, that this is possible. A small trick is needed. Consider some function $f:\omega\to[\omega]^\omega$ which picks, on each level a set from $\mathcal A_n$ such that this set is contained in the previous choices, i.e. $f(n+1)\subseteq^*f(n)$ --- we shall call such functions threads. Each such thread gives us a $\subseteq^*$ descending sequence of subsets of $\omega$. This sequence may have an empty intersection. However (using induction) it is easily shown that there are many infinite sets $A$ having the property that $A\subseteq^* f(n)$ for all $n<\omega$ (these $A$'s are never unique and are sometimes called pseudointersections of the sequence. We can consider an A.D. family $\mathcal A_f$ consisting of these pseudointersections. Using the Axiom of choice, we can assume that this A.D. family is maximal (not necessarily MAD, however, no pseudointersection can be added for it to remain A.D.). The situation can be seen in the following picture

Finally, we can let $\mathcal A_\omega$ be the collection $\{\mathcal A_f:f\ \mbox{is a thread}\}$. It is easy to see that $\mathcal A_\omega$ is an A.D. system and short argument shows that it must be MAD. Having defined $\mathcal A_\omega$ we can define $\mathcal A_{\omega+1}$ and carry on our construction, at limit steps of countable cofinality using the above trick. What happens, however, at limit steps of uncountable cofinality (actually at steps of cofinality $\geq\mathfrak t$? This is not clear. Of course, there might be some suitable MAD family, which will allow us to continue, but there might not --- the above construction does not work. It turns out that, the cardinal $\mathfrak h$ can be characterized as the first cardinal where this process can stop:

12.7

$\mathfrak h$ is the least cardinal $\kappa$ such that there is a refining sequence of MAD families $\mathcal \langle A_\alpha:\alpha<\kappa\rangle$ (i.e. for $\alpha<\beta<\kappa$ we have $\mathcal A_\alpha\preceq\mathcal A_\beta$) which have no common refinement.

Since MAD families naturally correspond to tall ideals, we can also rephrase the above theorem as follows:

12.8

$\mathfrak h$ is the least size of a family of tall ideals whose intersection is not tall (or, equivalently, whose intersection is just the ideal of finite sets).

(Recall, that an ideal $\mathcal I$ is tall, if for each infinite $A\in[\omega]^\omega$ there is an infinite $I\subseteq A$ such that $I\in\mathcal I$. The correspondance between MAD families and ideals comes from the following observations

12.9

An AD family $\mathcal A$ is MAD iff the ideal generated by $\mathcal A$ and all finite sets is tall.

12.10

An ideal $\mathcal I$ is tall iff there is a MAD family $\mathcal A\subseteq\mathcal I$.

)

Tall ideals are ideals which are, in some sense, not too small. What if we consider a different notion of smallness. Since ideals are subsets of $\mathcal P(\omega)\simeq 2^\omega$ and since $2^\omega$ carries a natural topology given by the metric: $\rho(f,g)= \left\{ \begin{array}{ll} 2^{-\min\{n:f(n)\neq g(n)\}}& f\neq g\\ 0 & f=g \end{array} \right.$ we can ask whether an ideal is meager (i.e. can be covered by countably many closed co-dense sets). A simple argument shows that maximal ideals are nonmeager: If $\mathcal I$ were maximal and meager, then $\mathcal P(\omega)=\mathcal I\cup \mathcal I^*$ by maximality. Since $A\mapsto \omega\setminus A$ is a homeomorphism of $\mathcal P(\omega)$ we have that $\mathcal I^*$, the dual filter to $\mathcal I$ is also meager. However this is a contradiction with Baire's theorem, which -- in this context --- says that $\mathcal P(\omega)$ is non-meager. A slightly more involved argument shows that each ideal which is not tall must be meager. So the notion of nonmeagerness sits between tallness and maximality. It is natural to ask, whether a witnessing family consisting of nonmeager ideals for $\mathfrak h$ can be constructed. It turns out that this is not always the case, however it leads to the following definition

12.11

The idealized groupwise density number $\mathfrak g_{id} (=\mathfrak g_{f})$ is the smallest size of a family of non-meager ideals whose intersection is meager (or, equivalently, the ideal of finite sets).

The reason why its called idealized groupwise density is because of the related notion of a groupwise dense family (and groupwise density number) introduced in A. Blass, C. Laflamme: Consistency Results About Filters and the Number of Inequivalent Growth Types, J. Symb. Logic 54(1), 1989, pp. 50-56

12.12Blass, Laflamme

A family $\mathcal G\subseteq[\omega]^\omega$ is called groupwise dense if 1. it is downwards closed, i.e. if $A\subseteq G\in\mathcal G$ , then $A\in\mathcal G$ 2. dense in $(\mathcal [\omega]^\omega,\subseteq^*)$, i.e. for each $X\in[\omega]^\omega$ there is an infinite $G\in[X]^\omega$ such that $G\in\mathcal G$. 3. for each interval partition $\langle I_n:n<\omega\rangle$ of $\omega$ there is an infinite $X\in[\omega]^\omega$ such that $\bigcup_{n\in X} I_n\in\mathcal G$

12.13Blass, Laflamme

The groupwise density number $\mathfrak g$ is the smallest size of a family of groupwise dense sets whose intersection is not groupwise dense (or, equivalently, empty).

The connection between $\mathfrak g$ and $\mathfrak g_{id}$ comes from the fact that, for ideals, being groupwise dense is the same as being non-meager. This follows from an old combinatorial characterization of non-meagerness for ideals discovered, independently, by S. A. Jalaili-Naini ('76) and M. Talagrand ('80):

12.14Jalaili-Naini, Talagrand

An ideal $\mathcal I$ is non-meager iff for each interval partition $\langle I_n:n<\omega\rangle$ of $\omega$ there is an infinite $X\in[\omega]^\omega$ such that $\bigcup_{n\in X} I_n\in\mathcal I$.

It is immediately clear that $\mathfrak g\leq\mathfrak g_{id}$, since $\mathfrak g$ considers more families. One might ask, whether there is really any difference. It turns out that there is

12.15Brendle

It is consistent that $\mathfrak g\leq \mathfrak g_{id}$.

(see J. Brendle: Distinguishing groupwise density numbers, Monatshefte für Mathematik 152(3), 2007, pp. 207-215)

It might not be clear, at first, that the cardinal invariant $\mathfrak g_{id}$ (or $\mathfrak g$) is in any way significant. However, it turns out that it is intricately connected with the structure of filters (or ideals) on $\omega$. To show this connection we must investigate more closely the structure of filters. Suppose we start with $\omega$ and a filter $\mathcal F$ on $\omega$. We now blow up each $n\in \omega$ into a larger set $L_n$. The filter $\mathcal F$ on $\omega$ naturally gives rise to a filter $\mathcal F^\prime$ on $L=\bigcup_{n<\omega} L_n$ --- a subset of $L$ is in $\mathcal F^\prime$ if it contains a union of filter many blocks $L_n$. Formally, $L\in\mathcal F^\prime$ iff $(\exists F\in\mathcal F)(\bigcup_{n\in F}L_n\subseteq L)$. The process can also be reversed. If we have a filter $\mathcal F$ on the larger set $L$, we can define a "condensed" filter $\mathcal F_*$ on $\omega$ by declaring $A\in\mathcal F^\prime$ iff $\bigcup_{n\in A} L_n\in \mathcal F$.

Suppose now we have a function $f:\omega\to\omega$. This function naturally gives rise to the above situation --- each number gets "blown-up" to its preimage, i.e. $L_n=f^{-1}(n)$. The following definition says, essentially, that $\mathcal F\leq_{K}\mathcal G$ if $\mathcal F$ is a "condensation" of $\mathcal G$

12.16Katětov, Rudin-Keisler

A filter $\mathcal F\leq_K\mathcal G$ if there is a function $f:\omega\to\omega$ such that $\mathcal F=f_*(\mathcal G) = \{ A\subseteq\omega: f^{-1}[A]\in\mathcal G\}.$ We call $f$ the witnessing function. If this function is finite-to-one (i.e. preimages of finite sets are finite) then we say $\mathcal F\leq_{KB}\mathcal G$.

12.17

The above ordering, when restricted to ultrafilters, is called the Rudin-Keisler (and Rudin-Blass, respectively) ordering and denoted $\leq_{RK}$ (and $\leq_{RB}$, respectively).

12.18

It is instructive to convince onself that the definitions can be repeated for ideals and (using the Jalaili-Naini-Talagrand characterization) that an ideal is meager iff it is Katětov-Blass above the ideal of finite sets.

We have shown above that ultrafilters cannot be meager. However, if a filter is non-meager, it does not necessarily mean that it is an ultrafilter. How far are non-meager filters from ultrafilters? The following principle says: not very much.

12.19Filter Dichotomy

The Filter dichotomy is the statement "Each filter is either meager (i.e. KB above fin) or KB above some ultrafilter".

Is this principle true (provable in ZFC)? Is it consistent? No and Yes. Under CH, e.g., it is easy to construct a nonmeager filter which is not KB above any ultrafilter. Showing that it is consistent is a little harder and we will not go into the details. However, it is interesting that the FD principle can be formulated in terms of cardinal invariants. First a definition:

12.20

The ultrafilter number $\mathfrak u$ is the smallest size of an ultrafilter base (equivalently, the smallest character of a point in $\omega^*=\beta\omega\setminus \omega$).

Now we can state the theorem, due to H. Mildenberger, which can be found in H. Mildenberger: Groupwise dense families, Arch. Math. Logic 40, 2001, pp. 93-112:

12.21

The FD is equivalent to $\mathfrak{u ## 12.2Honza Starý, inspired by the Winter School Let \(\mathbb B$ be a complete Boolean algebra and $X$ its Stone space. Then $X$ is a compact extremally disconnected $T_2$ space.

Recall that

12.22

A space $X$ is extremally disconnected (ED for short) if the closure of each open set is clopen.

Also recall definition 11.24 of the spectrum of a point and observation 11.25 showing an equivalent definition for ED compac spaces.

12.23

$Spect(p,\beta\omega) = \{q\in\omega^*:q\leq_{RK}p\}$

12.24

For points $x,y\in X$ define $x\preceq y$ if $Spect(x,X)\subseteq Spect(y,X)$.

Next, recall

12.25Frolík

Each infinite ED compact space is not homogeneous.

The above theorem does not give 'honest witnesses' (vanDouwen) to nonhomogeneity. The quest for honest witnesses motivates the following question

12.26

Given the stone space $S(\mathbb B)$ of some complete Boolean algebra $\mathbb B$, can we find $x,y\in S(\mathbb B)$) with $Spect(x)\neq Spect(y)$?

12.27

A point $x\in X$ is discretely untouchable if it is not a limit point of each countable discrete set.

12.28

What is $Spect(x,X)$ for a discretely untouchable $x$?

12.29

For an ED space $Spect(x,X)=\emptyset$ iff $x$ is a P-point.

# 1326. 2. 2014

## 13.1P.Simon: Shelah's proof that groupwise density is at most b+

Recall the definitions 12.13, 12.12 and 12.3.

13.1Shelah

$\mathfrak g\leq\mathfrak b^+$

see S. Shelah: Groupwise density cannot be much bigger than the unbounded number, Math. Logic Quarterly 54(4), pp. 340-344 2008

## 13.2J. Starý

For the following recall definition 12.27.

13.2

A topological space $X$ is homogeneous if for each $x,y\in X$ there is an automorphism $f:X\to X$ such that $f(x)=y$.

13.3Frolík

Let $X$ be an infinite extremally disconnected compact space and $g:X\to X$ continuous and injective. Then the set of fixed points $Fix(g) = \{ x:g(x)=x\}$ is closed and open.

13.4Frolík

If $X$ is an infinite extremally disconnected compact space then there is an injective continuous $f:X\to X$ such that $f[X]$ is nowhere dense.

13.5

Infinite extremally disconnected compact spaces are not homogeneous.

Take the $f$ from the above theorem, let $p\in X$ and let $q=f(p)$. We shall show that if $h$ is a homeomorphism then $h(p)\neq q$. Otherwise consider $g=f\circ h$ and apply theorem 13.3. Then the set of fixed points of $g$ is clopen and this contradicts the fact that the image of $f$ is nowhere dense.

13.6

For a complete Boolean algebra we define $\mathfrak g_c(\mathbb B)$ to be the minimal size of a set of complete generators of $\mathbb B$ and $\pi(\mathbb B)$ (also denoted $\pi w(\mathbb B)$ ) the minimal size of a dense subset of $\mathbb B$. Moreover $\pi\chi(\mathbb B)=\min\{\pi\chi(p):p\in S(\mathbb B)\}$, where $\pi\chi(p)$ is the minimal size of a local $\pi$-base at $p$.

13.7Simon,90

Let $\mathbb B$ be a complete, ccc Boolean algebra with density $\pi(\mathbb B)\leq\mathfrak c$. If $\mbox{cf}\,\mathfrak g_c(\mathbb B) >\omega$, then $S(\mathbb B)$ contains a discretely untouchable point.

(see P. Simon: Points in extremally disconnected compact spaces, Rend. Circ. Math. di Palermo (2) 1990)

13.8Balcar, Simon

If $\pi w(\mathbb B) = \pi\chi(\mathbb B)$ then $S(\mathbb B)$ contains a discretely untouchable point.

(see B. Balcar and P. Simon: On minimal $\ \pi$-character of points in extremally disconnected compact spaces, Top. Appl. 41 ( 1-2) pp. 133-145, 1991)

13.9

Is there some $\mathbb B$ such that $\pi\chi(\mathbb B)<\pi w(\mathbb B)$?

13.10E. Thuemmel

If $\pi w(\mathbb B) = \pi\chi(\mathbb B)$ then the same holds for $Exp_\omega(\mathbb B,v)$.

# 1412. 3. 2014

## 14.1J. Grebík: Extending asymptotic density to a measure

14.1

Given $A\subseteq\omega$ define the upper asymptotic density $d^*(A) = \lim_{n\to\infty}\sup \frac{|A\cap[1,n]|}{n}$ and lower asymptotic density $d^*(A) = \lim_{n\to\infty}\sup \frac{|A\cap[1,n]|}{n}$

14.2Hahn-Banach

Given a vector space $V$ with a seminorm $||\cdot||$, $W$ a linear subspace of $V$ and $f:W\to\mathbb R$ a linear function such that $|f(w)|\leq ||w||$ for each $w\in W$ then $f$ can be extended to a linear function $g:V\to\mathbb R$ such that $f(v)\leq||v||$ for each $v\in V$

(Recall that $||\cdot||$ is a seminorm if $||v||\geq 0$,$||v+w||\leq||v||+||w||$ and $||cv||=|c|||v||$ for each $v,w\in V$ and each scalar $c$. It is a norm if, moreover, $||v||=0\equiv v=0$.)

We will apply this theorem to the above situation. We let $V=\{f\in \mathbb Q^{\mathbb N}:|f[\mathbb N]|<\omega\}$. Then this is a vector space and for each $A\subseteq\omega$ its characteristic function $\chi_A$ is an element of $V$. Further, we define a seminorm on $V$ as follows: $||f|| = \lim_{n\to\infty}\sup\frac{\sum_{i=1}^n|f(i)|}{n}.$ Let $W = \bigg\{f\in V: \lim_{n\to\infty}\frac{\sum_{i=1}^nf(i)}{n}\in\mathbb R\bigg\}$ and define a linear functional $h:W\to\mathbb R$ by letting $h(f) = \lim_{n\to\infty}\frac{\sum_{i=1}^nf(i)}{n}.$ It is clear that $h(\chi_A)=d^*(A)$ for each $A\subseteq\omega$ with $\chi_A\in W$. Now we can use the Hahn-Banach theorem 14.2 to extend $h$ to $\overline{h}: V\to\mathbb R$. Then for each $A\subseteq\omega$ we have $\overline{h}(\chi_A)\geq0$ (since $h(\chi_{\mathbb N})=1$ so $\overline{h}(\chi_A)+\overline{h}(\chi_{A^C})=1$.

This approach to extending asymptotic density to a measure (instead of using an ultrafilter) has the advantage that we have more control over the values of $\overline{h}(A)$ which may be chosen arbitrarily (for independent sets) between $d_*(A)$ and $d^*(A)$.

The above is motivated by the following question:

14.3

All measures are elements of $[0,1]^{\mathcal P(\omega)}$. For each ultrafilter $\mathcal U$ consider $m_{\mathcal U}$ the $\mathcal U$-extension of asymptotic density. Is the set of all finite combinations of measures of the form $m_{\mathcal U}$ for some $\mathcal U$ dense in $[0,1]^{\mathcal P(\omega)}$.

## 14.2J. Verner: Zindulka/Hrušák's question

14.4

Given a real number $c\in\mathbb R$ we say that a metric space $(X,\rho)$ is $c$-monotonne if there is a linear order $\leq$ on $X$ such that

1. $d(x,y)\leq c d(x,z)$ for each $x\leq y\leq z$
2. open intervals in the ordering are open in the metric

A metric space is $monotonne$ if it is $c$-monotonne for some $c\in\mathbb R$. It is $\sigma$-monotonne if it can be covered by countably many monotonne metric spaces.

For the following, see Zindulka, O: Is Every Metric on the Cantor Set $\sigma$-Monotonne?, Real Analysis Exchange 33 (2007), no. 2, 485--486.

14.5

If a metric space $X$ has a dense monotonne subspace then it is monotonne.

14.6

If a metric space $X$ is $\sigma$-monotonne then $\mbox{dim}\,X\leq 1$.

In particular, $[0,1]^2$ is not $\sigma$-monotonne. So, under CH, there is an example of a non $\sigma$-monotonne metric space of size $\aleph_1$. This motivates the following question:

14.7

Is there, in ZFC, a non $\sigma$-monotonne metric space of size $\aleph_1$?

14.8

Hrušák remarked that such an example cannot be separable.

## 14.3M. Rubin

The following is probably known, but I would be interested to know the answer:

14.9

Suppose that $X$ is a space, $d:X\times X\to\mathbb R^+_0$ a symmetric map and $c>1$ such that $d(x,z)\leq c (d(x,y)+d(y,z))$ and $d(x,y)=0\equiv x=y$. Suppose, moreover, that the base of the topology of $X$ is given by $d$-balls. Is $X$ metrizable?

## 14.4J. Starý

Recall the definition 12.22 of an extremally disconnected space.

14.10

A space is $0$-dimensional if it has a base consisting of clopen sets.

14.11

A $T_2$ extremally disconnected space is $0$-dimensional.

14.12

A $0$-dimensional compact space is extremally disconnected iff the algebra of clopen sets is complete.

14.13

A Boolean algebra $B$ is homogeneous if $B\upharpoonright b\simeq B$ for each $b\in B^+$. A topological space $X$ is homogeneous if for each $x,y\in X$ there is a homeomorphism $f:X\to X$ such that $f(x)=y$.

In this setting, recall Frolík's theorem 12.25.

14.14

The completion of a homogeneous Boolean algebra is always homogeneous.

14.15

$Borel(\mathbb R)/Meager$ is a homogeneous complete Boolean algebra.

## 14.5P. Simon, the history of Frolík's theorem

14.16Frolík

If $X$ is extremally disconnected and compact and $f:X\to X$ is an embedding. Then we can partition $X$ into four clopen sets $X=C_0\cup C_1\cup C_2\cup C_3$ such that $f\upharpoonright C_0=id$ and $F[C_i]\subseteq C_j\cup C_k$ for $1\leq i\neq j\neq k\neq i\leq 3$.

The consequence of the above theorem is that the fixed points of embeddings of an extremally disconnected space form a clopen set in the space.

14.17

Consider $\beta\omega_1$. This is an extremally disconnected space. Consider the equivalence relation $\mathcal U_1\sim \mathcal U_2$ if both are uniform (i.e. do not contain countable sets). Then a bijection $f:\omega_1\to\omega_1$ of limit and successor ordinals generates a homeomorphism of $\beta\omega_1$ whose fixed-points are only uniform ultrafilters. So $\beta\omega_1/\sim$ has an automorphism which has only a single fixed point, in particular, by the above theorem, cannot be extremally disconnected.

Frolík's theorem 12.25 is a consequence of the following (due to Balcar and Franěk ?) and 14.16.

14.18Frolík, 1967

If $X$ is an infinite extremally disconnected space contains a nowhere dense copy of itself.

of 12.25

Let $X$ be infinite extremally disconnected and $Y\subseteq X$ a nowhere dense copy of $X$ with $h:X\simeq Y$. Choose $x\in X\setminus Y$ such that $h(x)\in Y$. If $X$ were homogeneous, there would be a homeomorphism $f:X\to\ X$ such that $f(y)=x$. Then $y$ is a fixed point of the mapping $h\circ f$. However, $h\circ f[X]\subseteq Y$ which is nowhere dense, so the set of its fixed points cannot be clopen --- a contradiction with 14.16.

14.19

Frolík's proof, apparently (see W. W. Comfort: Some recent applications of ultrafilters to topology, General Topology and Its Relations to Modern Analysis and Algebra IV, Lecture notes in mathematics vol. 609, Springer 1977, pp 34-42), was different and used a result of K. Kunen that there are incomparable ultrafilters on $\omega$.

All of the above was motivated by the following considerations. The algebra $\mathcal P(\omega)/fin$ is homogeneous. How about its Stone space, $\omega^*$?

14.20W. Rudin, 1956

Under CH, $\omega^*$ contains a P-point so, in particular, is not homogeneous.

## 14.6J. Starý: Ultrapowers of Boolean algebras

Let $B$ be a complete Boolean algebra (cBA for short) and $A$ an algebra. Consider the set of "$B$-names" $F=\{f:P\to A:P\subseteq B\,\mbox{is a partition of}\, 1\}$ for elements of $A$. Given an ultrafilter $\mathcal U$ on $B$ we say $f=_{\mathcal U} g$ if $\bigvee\{ r\in dom(f)\cap dom(g):f(r)=g(r)\}\in\mathcal U$. $=_{\mathcal U}$ is an equivalence relation and the set $F$ naturally carries a structure of a Boolean algebra, it will be denoted $A^B/\mathcal U$ and called the Boolean ultrapower of $A$.

14.21

Consider, moreover, the set $Part(B)$ of partitions of $1$ in $B$. Then the partial ordering $\preceq$ of refinement makes $Part(B)$ a directed set. Moreover $A^B/\mathcal U$ is a direct limit of $\{ A^P/\mathcal U_P:P\in Part(B)\}$ where $\mathcal U_P$ is the ultrafilter on $P$ generated by $\mathcal U$

Now let $B$ be a ccc, cBA. For an infinite $P\in Part(B)$ denote $B_P$ to be the algebra generated by $P$. Then, if $P$ is infinite, $B_P$ is isomorphic to $\mathcal P(\omega)$.

14.22

$B$ is a direct limit of the system $\{ B_P:P\in Part(B)\}$ (again using the refinement ordering $\preceq$ on partitions).

Aiming to define an ultrafilter sum, for $b\in B$ we say $b\in\sum_{\mathcal U_P} \mathcal F_n$ if $\{n\in\omega:b\cap p_n\in\mathcal F_n\}\in\mathcal U_P$, where $P=\{p_n:n<\omega\}$.

14.23

An ultrafilter $\mathcal U$ on a ccc atomless BA $B$ is discretely touchable if there is a partition $P\in Part(B)$, $P=\{p_n:n<\omega\}$, with $P\cap\mathcal U\neq\emptyset$ and ultrafilters $\mathcal V_n$ with $p_n\in V_n$ such that $\mathcal U = \sum_{\mathcal U_P}\mathcal V_n$.

14.24

If $B$ is a ccc, atomless cBA of weight $\leq \mathfrak c$ then $St(B)$ it contains a discretely touchable ultrafilter.

# 1519. 3. 2014

## 15.1D. Chodounský: Mathias forcing (Part I.)

The following is joint work with L. Zdomskyy, D. Repovš, based on D. Chodounský, D. Repovš, L. Zdomskyy: Mathias Forcing and Combinatorial Covering Properties of Filters, preprint 2014

15.1

Given a filter $\mathcal F$ on $\omega$ we define Mathias forcing with respect to $\mathcal F$: $M_{\mathcal F} = \{ (s,F):s\in[\omega]^{<\omega}, F\in\mathcal F\},$ where $(s,F)\leq (t,H)$ if $t\sqsubseteq s$, $F\subseteq H$ and $s\setminus t\in H$.

15.2

Mathias forcing is ccc (no uncountable antichains), even $\sigma$-centered since conditions having the same first coordinate are compatible.

We will be interested in the following properties with respect to forcing extensions by Mathias forcing.

15.3

Given a model $M$ of ZFC we say its extension $N\supseteq M$ is ${}^\omega\omega$-bounding if each function $f\in N\cap{}^\omega\omega$ is coordinate-wise bounded by some function $d\in M\cap{}^\omega\omega$. It is almost-${}^\omega\omega$-bounding if each unbounded subfamily of $(\omega,\leq^*)$ in $M$ remains unbounded in $N$. We say that $N$ contains a dominating real if there is a $d\in N\cap{}^\omega\omega$ such that for each $f\in{}^\omega\omega\cap M$ we have $f\leq^* d$, i.e. $|\{n:f(n)>d(n)\}|<\omega$.

15.4

The above definition of almost ${}^\omega\omega$-bounding is not standard; It is proved in the cited preprint that it is, in fact, equivalent to the standard definition, given below.

15.5

A forcing $P$ is almost ${}^\omega\omega$-bounding if for each condition $p\in P$ and each name $\dot{f}$ for a function from $\omega$ to $\omega$ there is a function $g:\omega\to\omega$ such that for each $A\in[\omega]^\omega$ there is a stronger $q\leq p$ such that $q\Vdash \dot{f}\upharpoonright A\not\geq g\upharpoonright A$.

One part of the following equivalence is Lemma 3 in the referenced paper.

15.6

A forcing $P$ is almost ${}^\omega\omega$-bounding iff forcing extensions by $P$ are almost ${}^\omega\omega$-bounding.

Assume $P$ is almost ${}^\omega\omega$-bounding, $U\subseteq{}^\omega\omega$ an unbounded family and, aiming towards a contradiction, $\dot{f}$ a is a $P$-name for a function from $\omega$ to $\omega$ and $p\in P$ a condition forcing that $dot{f}$ dominates $U$. Since $P$ is almost ${}^\omega\omega$-bounding there is a $g:\omega\to\omega$ satisfying the conditions from the definition. Since $U$ it is unbounded, there is $A\in[\omega]^\omega$ and $h\in U$ such that $g\upharpoonright A\leq h\upharpoonright A$. By assumption there is a condition $q\leq p$ and a name $\dot{B}$ for an infinite subset of $A$ such that $q\Vdash \dot{f}\upharpoonright\dot{B}\leq g\upharpoonright\dot{B}\leq h\upharpoonright\dot{B}$ contradicting the fact that $p$ forces that $\dot{f}$ dominates $U$.

Assume, on the other hand, that $P$ is not almost ${}^\omega\omega$-bounding. So there is a $p\in P$ and a $P$-name $\dot{f}$ for a function from $\omega$ to $\omega$ such that for each $g:\omega\to\omega$ there is an infinite $A_g\in[\omega]^\omega$ such that $p\Vdash g\upharpoonright A_g\leq \dot{f}\upharpoonright A_g$. Then the family $U=\{g\upharpoonright A_g\cup 0\upharpoonright(\omega\setminus A_g):g\in{}^\omega\to\omega\}$ is an unbounded family of functions in the ground model which is bounded by $\dot{f}$ in the extension.

15.7

Mathias forcing cannot be bounding.

15.8

The generic real for $M_{\mathcal F}$ is a pseudointersection of $\mathcal F$.

In particular if the filter $\mathcal F$ is rapid (i.e. the enumeration functions of elements of $\mathcal F$ form a dominating family) the generic real is a dominating real. If $\mathcal F$ is nonmeager (which, in particular, implies that the family of enumeration functions of elements of $\mathcal F$ is unbounded) then $M_{\mathcal F}$ is not almost ${}^\omega\omega$-bounding.

The motivation for the following topological definitions which characterize $M_{\mathcal F}$ being almost ${}^\omega\omega$-bounding and not adding dominating reals is as follows. A necessary condition for both properties is that $\mathcal F$ itself (or its enumeration functions) is bounded. However it might happen that $\mathcal F$ is bounded however a continuous image by $f$ of $\mathcal F\subseteq 2^\omega$ might be unbounded. Then it is reasonable to expect that the $f$ image of the generic real will contradict ${}^\omega\omega$ bounding.

15.9

A topological space $X$ is Hurewicz if every continuous image of $X$ in $\omega^\omega$ is bounded (i.e. a bounded family of functions in $({}^\omega\omega,\leq^*)$). It is Menger if no continuous image of $X$ in $\omega^\omega$ is dominating.

15.10

$\sigma$-compact spaces are Hurewicz.

15.11probably

Subsets of $2^\omega$ are Hurewicz iff they are strong measure zero.

15.12Arhangelskii, ?

Analytic subsets of $2^\omega$ are Hurewicz iff they are $F_\sigma$, i.e. $\sigma$-compact.

15.13

Preservation of these properties under products (or powers) is an area of active research.

The following theorem gives the standard definition of Menger property.

15.14

A space $X$ is Menger iff for each sequence $\langle \mathcal U_n:n<\omega\rangle$ of open covers of $X$ there is a sequence $\langle F_n:n<\omega\rangle$ such that $F_n\in[\mathcal U_n]^{<\omega}$ for each $n<\omega$ and $\bigcup_{n<\omega}F_n$ is a cover of $X$.

To rephrase the Hurewicz property in a similar language we need some additional notions concerning covers.

15.15

An open cover $\mathcal U$ of a space $X$ is an $\omega$-cover if for each $F\in[X]^{<\omega}$ there is a $U\in\mathcal U$ such that $F\subseteq U$, i.e. not only points are covered but even finite sets.

15.16

If $\mathcal U$ is an open cover then all finite unions of elements of $\mathcal U$ form an $\omega$-cover.

15.17

An open cover $\mathcal U$ is a $\gamma$-cover if for each $x\in X$ the set $\{U\in\mathcal U:x\not\in U\}$ is finite.

A typical example of a $\gamma$-cover comes from writing the space as an increasing union of open sets.

We are now ready to rephrase the Hurewicz property.

15.18

A space $X$ is Hurewicz iff for each sequence $\langle \mathcal U_n:n<\omega\rangle$ of $\omega$-covers of $X$ there is a sequence $\langle F_n:n<\omega\rangle$ such that $F_n\in[\mathcal U_n]^{<\omega}$ for each $n<\omega$ and $\{\bigcup F_n:n<\omega\}$ is a $\gamma$-cover of $X$.

We shall only show that a $X\subseteq{}^\omega\omega$ satisfying the alternative condition is bounded. Let $\mathcal U_k=\{U^k_n:n<\omega\}$ where $U^k_n=\{f\in X:f(k). Apply the condition to find a \(\gamma$-cover. This $\gamma$-cover gives a bound on $X$, i.e. via $d(k)=\max\{n:U^k_n\in F_k\}$.

15.19

Given $X\subseteq\omega$ let $\uparrow X=\{Y\subseteq\omega:X\subseteq Y\}$.

15.20

$\uparrow X$ is a closed set which is open iff $X$ is finite.

The following is core lemma for dealing with Hurewicz and Menger properties of filters.

15.21

Given a filter $\mathcal F$ and an open cover $\mathcal O$ of $\mathcal F\subseteq 2^\omega$ there is a $Q\subseteq[\omega]^{<\omega}$ such that $\mathcal F\subseteq\bigcup\bigg\{ \uparrow q:q\in Q\bigg\}\subseteq\bigcup\mathcal O.$

We are given $\mathcal O$ and we may assume that $\mathcal O$ consists only of basic open sets. Fix an $F\in\mathcal F$. Then $\uparrow F$ is a closed set, hence it is compact so there is a finite $\mathcal O^\prime\subseteq\mathcal O$ covering $\uparrow F$. We may assume that $\mathcal O^\prime=\{[s_n]:n with \(|s_n|=|s_m|=N$ for each $n,m. Now pick \(n such that \(F\cap N=s_{n} = s_F$. Then $\uparrow F\subseteq\uparrow s_F$.

Finally let $Q=\{s_F:F\in\mathcal F\}$.

15.22

When checking the menger property for filters we need only check open covers consisting of elements of the form $\uparrow q$ for a finite $q\in[\omega]^{<\omega}$.

Due to lack of time, we will only recall the following definitions which Hrušák et al. use to deal with Mathias forcing.

15.23

Given a filter $\mathcal F$ on $\omega$ we define $\mathcal F^{<\omega}$ to be a filter on $[\omega]^{<\omega}\setminus\{\emptyset\}$ generated by sets of the form $[F]^{<\omega}$ for $F\in\mathcal F$.

15.24

A set $X\subseteq[\omega]^{<\omega}$ is positive with respect to $(\mathcal F^{<\omega})^+$ iff $\{\uparrow x: x\in X\}$ is an open cover of $\mathcal F$.

15.25

A filter $\mathcal F$ on a countable set $S$ is P${}^+$ iff for each $\subseteq$-descending sequence $\langle X_n:n<\omega\rangle$ of $\mathcal F$-positive sets there is an $\mathcal F$-positive $X$ which is almost contained in each $X_n$.

15.26

Given a filter $\mathcal F$ on $\omega$ we let $\mathcal F^{<\omega}$ be the filter on $[\omega]^{<\omega}\setminus\{\emptyset\}$ generated by sets of the form $[F]^{<\omega}$, where $F\in\mathcal F$.

15.27Hrušák,Minami

$M_{\mathcal F}$ does not add dominating reals iff $\mathcal F^{<\omega}$ is a P${}^+$-filter.

Given that the above definition may be reformulated as follows

15.28

A filter $\mathcal F$ is P${}^+$ iff for any sequence $\langle X_n:n<\omega\rangle$ of $\mathcal F$-positive sets ther is a sequence $\langle Y_n:n<\omega\rangle$ with $Y_n\in[X_n]^{<\omega}$ such that $\bigcup_{n<\omega}Y_n$ is $\mathcal F$-positive.

this, together with Hrušák's and Minami's result, almost immediately gives a topological reformulation of the property that $M_\mathcal F$ does not add dominating reals.

# 1626. 3. 2014

16.1

Tomorrow at 10:40 in Karlín, seminar room of the MÚ, Yasunao Hattori (Shimane University, Matsue, Japan) will talk about The Separation Dimension and Infinite-Dimensional Spaces.

## 16.1W. Kubis: MAD families on singulars

The following is based on Kojman, M., Kubis, W. and Shelah, S.: On two problems of Erdös and Hechler: New Methods in Singular Madness, Proc. Amer. Math. Soc. 132 (2004), no 11, 3357-3365

16.2

Let $S$ be a set of size $\mu$. A family $\mathcal A\subseteq[S]^\mu$ is called $\mu$-almost disjoint if for $|A\cap B|<\mu$ for distinct $A,B\in\mathcal A$. It is maximal $\mu$-almost disjoint ($\mu$-MAD for short) if it is $\mu$-almost disjoint and it is maximal w.r.t. inclusion with this property.

16.3

A partition of $S$ into $<\mbox{cf}\,\mu$ many sets is always a maximal $\mu$-almost disjoint family.

For the remainder of this talk fix $\aleph_0\leq\kappa=\mbox{cf}\,\mu<\mu$ and say that a $\mu$-MAD family $\mathcal A$ is nontrivial if $|\mathcal A|\geq\kappa$

16.4

Use standard diagonalization to show that there is no $\mu$-MAD family of size $\mbox{cf}\,\mu$.

16.5

The (MAD) spectrum of $\mu$ is defined to be the set $MAD(\mu)=\{|\mathcal A|:\mathcal A\ \mbox{is nontrivial}\ \mu-MAD\}$

16.6

It is clear that $MAD(\mu)\subseteq[\kappa^+,2^\mu]$

16.7

$MAD(\mbox{cf}\,\mu)\subseteq MAD(\mu)$. {} Proof: Let $\mathcal A$ be a $\kappa=\mbox{cf}\,\mu$-mad family on $\kappa$. Fix an increasing sequence $\langle \mu_\alpha:\alpha<\kappa\rangle$ of regular cardinals with supremum $\mu$ and partition $\mu$ into sets $M_\alpha$ such that $|M_\alpha|=\mu_\alpha$. Finally for $A\in\mathcal A$ let $B_A=\bigcup_{\alpha\in A}M_\alpha$. Then $\{B_A:A\in\mathcal A\}$ is a $\mu$-mad family on $\mu$.

16.8

The above also follows from the fact that the Boolean algebra $\mathcal P(\kappa)/[\kappa]^{<\kappa}$ regularly embeds into $\mathcal P(\mu)/[\mu]^{<\mu}$.

16.9

$2^\kappa < \mu\Rightarrow \mathfrak a_\mu\leq 2^\kappa$.

16.10

$2^{<\aleph_\omega} < \aleph_\omega\Rightarrow 2^{\aleph_\omega}\in MAD(\aleph_\omega)$ {} Proof: Take the full binary tree of height $\aleph_\omega$ and extend it to a $\aleph_\omega$-MAD family (on the set of nodes).

16.11Comfort

When does $\mu\in MAD(\mu)$?

16.12Erdös, Hechler

Is it consistent that $\mu\not\in MAD(\mu)$?

16.13Erdös, Hechler

Does $2^\kappa<\mu$ imply $\mu\in MAD(\mu)$?

16.14

$[\kappa^+,\mu]\cap MAD(\mu)$ is either empty or is of the form $[\mathfrak a_\mu,\mu]$ for $\mathfrak a_\mu=\min MAD(\mu)$.

of theorem
16.15

Assume $\langle\lambda_\alpha:\alpha<\theta\rangle\subseteq MAD(\mu)$ with $\mbox{cf}\,\theta=\theta <\lambda_0$ and $\lambda=\sup\{\lambda_\alpha:\alpha<\theta\}$ singular. Then $\lambda\in MAD(\mu)$.

of lemma

Let $\mathcal A_0$ be a $\mu$-mad family of size $\lambda_0$. Take the first $\theta$-many sets $\{A_\alpha:\alpha<\theta\}$ of $A_0$ and on each $A_\alpha$ pick a $\mu$-mad family $\mathcal B_\alpha$ of size $\lambda_\alpha$. Then $\mathcal B=\bigcup_{\alpha<\theta}\mathcal B_\alpha$ is a $\mu$-mad family of size $\lambda$.

16.16

Every regular $\lambda\in[\mathfrak a_\mu,\mu)$ is in $MAD(\mu)$.

of lemma

First notice that $\lambda>\kappa$.
Let $\mathfrak a_\mu<\lambda=\mbox{cf}\,\lambda<\mu$. For each $\delta<\lambda$ of cofinality $\kappa$ (i.e. $\delta\in S^\lambda_\kappa$ ) fix a continuous increasing $\langle l^\delta_\alpha:\alpha<\kappa\rangle$ sequence with supremum $\delta$ and $l^\delta_0=0$. Notice, that whenever we partition $\mu$ into $\kappa$ many pieces of size $\kappa$ we can find a $\mu$-mad family of size $\mathfrak a_\mu$ which contains the pieces. Using this we fix $\mathcal A_\delta$ a $\mu$-ad family on $\mu\times\lambda$ of size $\mathfrak a_\mu$ such that $\mathcal A_\delta\cup \{\mu\times[l^\delta_\alpha,l^\delta_{\alpha+1}):\alpha<\kappa\}$ is $\mu$-mad on $\mu\times[0,\delta)$. Then $\mathcal A_{\delta_0}\cup\mathcal A_{\delta_1}$ is $\mu$-ad for each $\delta_0<\delta_1\in S^\lambda_\kappa$. Now let $\mathcal A=\bigcup_{\delta\in S^\lambda_\kappa}\mathcal A_\delta$. This is $\mu$-a.d.

16.17

$\mathcal A\cup\{\mu\times\{\alpha\}:\alpha<\lambda\}$ is $\mu$-mad.

of claim

Fix $X\subseteq\mu\times\lambda$ of size $\mu$. Then there is a minimal $\delta<\lambda$ such that $|X\cap\mu\times\delta|=\mu$. It is not hard to see that $\delta\in S^\lambda_\kappa$.

Recall the definition of the bounding number for regular cardinals:

16.18

$\mathfrak b_\kappa = \min\{|\mathcal B|:\mathcal B\subseteq \kappa^\kappa\ \mbox{is unbounded in}\ <^* \}$, where $f<^*g\equiv|\{\alpha:f(\alpha)\geq g(\alpha)\}|<\kappa$.

For singular cardinals it is convenient to modify the definition slightly:

16.19

$\mathfrak b_\mu = \sup\{\mathfrak b(\prod_{i<\kappa} \mu_i,<^*):\mu_i=\mbox{cf}\,\mu_i\ \&\ \mu_i\nearrow\mu\}$

16.20

$\min\{\mathfrak b_\kappa,\mathfrak b_\mu\}\leq \mathfrak a_\mu$

Assume $\mathcal A$ is $\mu$-a.d. on $\kappa\times\mu$ of size $<\min\{\mathfrak b_\kappa,\mathfrak b_\mu\}$. Fix an increasing sequence $\langle \mu_\alpha:\alpha<\kappa\rangle$ of regular cardinals with supremum $\mu$ such that $|\mathcal A|<\mathfrak b(\prod_{\alpha<\kappa}\mu_\alpha,<^*)$. Let $\mathcal B=\mathcal A\setminus\{\{\alpha\}\times\mu:\alpha<\kappa\}$. For each $B\in\mathcal B$ define $f_B:\kappa\to\kappa$ by letting $f_B(\gamma)=\min\{\alpha: |B\cap \{\gamma\}\times\mu|<\mu_\alpha\}$. Now there is a function $f:\kappa\to\kappa$ dominating the $f_B$s such that $|B\cap(\{\alpha\}\times \mu$ for each $B\in\mathcal B$. Then given $B\in\mathcal B$ the intersection $B\cap (\{\gamma\}\times f(\gamma))$ is bounded for all but boundedly many $\gamma$s, i.e. there is $g_B\in\prod_{\alpha<\kappa}\mu_{f(\alpha)}$ such that for all but boundedly many $\gamma$s we have $B\cap (\{\gamma\}\times f(\gamma))\subseteq g_B(\gamma)$. Since $\mathfrak b(\prod_{\alpha<\kappa}\mu_{f(\alpha)},<^*) \leq\mathfrak b(\prod_{\alpha<\kappa}\mu_\alpha,<^*)$ there is $g\in \prod_{\alpha<\kappa}\mu_{f(\alpha)}$ dominating the $g_B$s. Let $X=\bigcup_{\alpha<\kappa} \{\alpha\}\times [g(\alpha),f(\alpha))$. Then $X$ is almost disjoint from each $A\in\mathcal A$ so $\mathcal A$ is not $\mu$-mad.

16.21

Martin's axiom with $2^{\aleph_0}>\aleph_\omega$ implies that there is no $\aleph_\omega$-MAD family of size $\aleph_\omega$.

16.22

If $\mathfrak a_\mu < \mathfrak b_\mu$ then $[\mathfrak a_\mu,\mathfrak b_\mu)\subseteq MAD(\mu)$. Moreover, if $\mathfrak b_\mu$ is a successor of a singular then $\mathfrak b_\mu\in MAD(\mu)$.

16.23

Can $\mathfrak a_\mu$ be singular?

16.24

Brendle proved that $\mathfrak a$ can be singular, even of countable cofinality. (see Brendle, J.: The Almost-Disjointness Number May Have Countable Cofinality, Trans. Amer. Math. Soc. 355(7), 2003)

16.25

A $\lambda$-scale in $(\prod_{\alpha<\mu}\mu_\alpha,<^*)$, where $\langle \mu_\alpha:\alpha<\kappa\rangle$, is an increasing sequence of regular cardinals with $\kappa<\mu_0$, is a $<^*$-increasing sequence $\langle f_\gamma:\gamma<\lambda\rangle$ which is cofinal in $(\prod_{\alpha<\mu}\mu_\alpha,<^*)$.

16.26Shelah

There is a set $M\in[\omega]^\omega$ such that $(\prod_{n\in M}\aleph_n,<^*)$ has an $\aleph_{\omega+1}$-scale.

16.27Shelah

For each $\lambda \in [\mu^+,\mathfrak b_\mu)$ there is a (continuous) $\lambda$-scale in $(\prod_{\alpha<\mu}\mu_\alpha,<^*)$ for a suitable sequence $\mu_\alpha$.

16.28

If $\mathfrak b_\mu$ is a successor of a regular then there is also a $\mathfrak b_\mu$-scale. Also, there is always a $\mu^+$-scale.

The proof of theorem 16.22 heavily uses Shelah's PCF theory, in particular theorem 16.27. Then it roughly follows the proof of 16.14 where the role of the "slices" $\mu\times[l^\delta_\alpha,l^\delta_{\alpha+1})$ is taken by the space between the functions in the scale, i.e. $\bigcup_{\gamma<\kappa} [f_\alpha(\gamma),f_{\alpha+1}(\gamma))$.

This has a connection to:

16.29Balcar, Simon

$\mathcal P(\kappa)/[\kappa]^{<\kappa}$ collapses $\mathfrak b_\kappa$ to $\omega$ for regular $\kappa>\omega$.

16.30Balcar, Simon

If $\mu$ is singular of countable cofinality, then $\mathcal P(\mu)/[\mu]^{<\mu}$ is $\sigma$-closed and collapses $\mu^{\aleph_0}$ to $\omega_1$.

Shelah proved:

16.31Shelah

If $\lambda\in MAD(\kappa)\setminus\kappa^+$ then $\mathcal P(\kappa)/[\kappa]^{<\kappa}$ collapses $\lambda$ to $\omega$ (again, $\kappa>\omega$ is regular).

# 172. 4. 2014

## 17.1M. Fabián: Separable reductions and rich families in variational analysis

17.1

Let $X$ be a metric space (or a normed linear space). We will consider the family $\mathcal S$ of all closed separable subspaces of $X$.

17.2

Given a function $f:X\to\mathbb R$, let $\mathcal C_f$ denote the family of $Y\in\mathcal S$ such that if $f\upharpoonright Y$ is continuous at some point $x$ of $Y$ then $f$ is continuous at $x$.

17.3

$\mathcal C_f$ is cofinal in $(\mathcal S,\subseteq)$.

Fix $Z\in\mathcal S$. We shall construct a sequence of countable sets $\langle C_n:n<\omega\rangle$. Fix $C_0$ arbitrarily such that $\overline{C_0}=Z$. Assume we have constructed $C_n$. For each $c\in C_n$ and $q\in\mathbb{Q}^+$ find a countable $D_{c,q}\subseteq B(c,q)$ with $c\in D_{c,q}$, $\mbox{diam}(D_{c,q})\leq q$ and $\mbox{diam}(f[D_{c,q}])=\mbox{diam}(f[B(c,q)])$. Next let $C_{n+1}=\bigcup_{c\in C_n,q\in\mathbb{Q}^+} D_{c,q}$ (in case of linear spaces, we finish by taking the "rational span" of the union, i.e. all finite linear combinations .with rational coefficients). Finally let $Y=\overline{\bigcup C_n}$. It is not hard to see that this $Y$ works.

The following notion was defined in J. Borwein, W. Moors: Separable Determination of Integrability and Minimality of the Clarke Subdifferential Mapping, Proc. AMS, 128(1) 1999, pages 215-221.

17.4Borwein, Moors

A family $\mathcal R\subseteq\mathcal S$ is rich if

1. it is cofinal
2. closed under closures of countable increasing unions ($\sigma$-complete)
17.5

This corresponds to the notion of a ($\sigma$-)club in a partially ordered directed set.

17.6

The intersection of countably many rich families is rich.

17.7

Let $\mathcal R=\{Y\in\mathcal S: (\forall q\in\mathbb Q^+,y\in Y) (\mbox{diam}f[B(y,q)]=\mbox{diam}f[B(y,q)\cap Y])\}$

17.8

$\mathcal R$ is rich in $\mathcal S$.

The above observation can also be vaguely formulated saying: "continuity can be separably reduced".

17.9Lindenstrauss, 1965

Let $X$ be a reflexive Banach space (i.e. $X^{**}\simeq X$ ). Then there is a PRI sequence $\langle P_\alpha:\alpha of retractions (where \(d(X)$ is the density of $X$) satisfying some conditions.

(see J. Lindenstrauss: On reflexive spaces having the metric approximation property, Israel J. Math, 3(4), 1965, pages 199-204)

The following is a generalization due to W. Kubis in Banach spaces with projectional skeletons, J. Math. Anal. Appl. 350(2), 2009, pages 758-776.

17.10

Given two linear functionals$p,q\in L(X)$ on a Banach space $X$ we define $p\preceq q$ if $p[X]\subseteq q[X]$.

17.11Kubiś

There is $\mathcal P\subseteq L(X)$ system of projections such that

1. $p\circ p^\prime = p$ for each $p\preceq p^\prime\in\mathcal P$,
2. $p[X]$ is separable for each $p\in\mathcal P$
3. $\mathcal P$ is rich in $(L(X),\preceq)$.

## 17.2J. Grebík: Ultrafilter Question

Is there an easy way to show that there is an ultrafilter $\mathcal U\in\omega^*$ such that for each $U\in\mathcal U$ there is a $k\geq 3$ with $\bigcup_{n\in U} [n(k-1),nk]\in\mathcal U$. The only proof I know uses idempotent ultrafilters.

## 17.3D. Chodounský: Filter Mathias forcing (Part II.)

For Part I. see section sec15.1. Recall definitions 15.9 (Menger and Hurewicz properties), 15.17 ($\gamma$-cover) and 15.15 ($\omega$-cover). Also recall the corollary 15.22 useful for checking the Hurewicz property for filters.

Also recall definition 15.26 and theorem 15.27.

17.12Chodounský, Repovš, Zdomskyy

Assume $\mathcal F$ is Hurewicz. Then $\mathbb M_{\mathcal F}$ is almost ${}^\omega\omega$-bounding.

Assume not. So let $X\subseteq{}^\omega\omega$ be an unbounded family of functions and $\dot{g}:\omega\to\omega$ is a name for a function dominating $X$. Hence for each $f\in X$ there is some condition $(s^f,F^f)\in\mathbb M_{\mathcal F}$ and $n^f<\omega$ such that $(s^f,F^f)\Vdash (\forall n\geq n^f)(f(n)\leq\dot{g}(n)).$ If we partition $X$ into countably many pieces one of the pieces will be unbounded. In particular, we may WLOG assume that there is $s^*, n^*$ such that $s^f=s^*$ and $n^f=n^*$ for each $f\in X$. For $m\in\omega$ define $S_m=\{s\in[\omega]^{<\omega}:(\exists F_s\in\mathcal F, g_s\in\omega) ((s^*\cup s,F_s)\Vdash \dot{g}(m)=g_s)\}$ It is easy to see that $S_m\in\mathcal F^{<\omega}$., i.e. $\{\uparrow s:s\in S_m\}$ is a cover of $\mathcal F$ (i.e. an $\omega$-cover of $\mathcal F$, since $\mathcal F$ is a filter). We now use that $\mathcal F$ is Hurewicz to choose $U_m\in[S_m]^{<\omega}$ such that $\{\bigcup U_m:m<\omega\}$ is a cover of $\mathcal F$, i.e. $\bigcup_{m<\omega} U_m$ is $\mathcal F^{<\omega}$-positive. Define $d(m)=\max\{g_s:s\in U_m\}$. We will show that $d$ dominates $X$ which will be the required contradiction. Pick some $f\in X$. Then there is $m_0<\omega$ such that for each $m\geq m_0$ there is $s_m\in U_m$ such that $s_m\subseteq F^f$. Now pick $n>n^*,m_0$. Then $(s^*\cup s_n,F_{s_n}\cap F^f)\Vdash f(n)\leq \dot{g}(n)\leq d(m),$ i.e. $f(n)\leq d(n)$ for all $n>m_0,n^*$, i.e. $d$ dominates $f$.

A similar proof gives:

17.13Chodounský, Repovš, Zdomskyy

Assume $\mathcal F$ is Menger. Then $\mathbb M_{\mathcal F}$ does not add dominating reals.

17.14

The above theorems can be reversed, the suitable unbounded (or dominating) family which will be destroyed by $\mathbb M_{\mathcal F}$ can be defined from a sequence of covers of $\mathcal F$ violating the Hurewicz (or Menger) property. The dominating/unbounded function in the extension can be easily defined from the generic real (since each cover from the witnessing sequence covers the generic real).

# 189.4.2014

18.1

Next week, Tue 15. 4. , there will be a workshop in Banach spaces organized by Wieslaw Kubiś and Jerzy Kakol at the Mathematical institute.

18.2

Wed, 23. Apr, Honza Starý will have his thesis defense at 11 am. Afterwards, at 12:30, we will go to the Green Garden restaurant.

## 18.1J. Grebík: Ultrafilters and dynamical systems

18.3

A dynamical system with discrete time is a space $X$ together with a continuous map $f:X\to X$ (or, possibly, more maps).

18.4

The intention is that the map transforms the space $X$ in units of time, i.e. a point $x\in X$ is moved to $f^n(x)=\underbrace{f(f(\cdots f}_{n\times}(x)\cdots ) )$ in time $n<\omega$.

18.5

A point $x\in X$ is a fixed point if $f(x)=x$, it is periodic if there is $n<\omega$ such that $f^n(x)=x$, it is almost periodic (or recurrent) if for each open neighbourhood $U$ of $x$ there is $n<\omega$ such that $f^n(x)\in U$. It is strongly almost periodic if for each open $U$ there is $d_U<\omega$ such that for each $n<\omega$ the set $\{f^n(x),f^{n+1}(x),\ldots,f^{n+d_U}(x)\}\cap U$ is nonempty.

18.6

If $X$ is compact then it contains a strongly almost periodic point.

18.7

Two points $x,y\in X$, where $X$ is a metric space, are proximal if for each $\varepsilon > 0$ there is an $n<\omega$ such that $\varrho(f^n(x),f^n(y))<\varepsilon$

We will now consider a dynamical system where the space is composed of ultrafilters on natural numbers and the maps are derived from addition and multiplication on natural numbers.

18.8

Given $A\subseteq\mathbb N$ let $A-1 = \{n\in\mathbb N:n+1\in A\}$. Given $\mathcal U$ an ultrafilter on $\mathbb N$ let $sh(\mathcal U) = \{A:A-1\in\mathcal U\}$.

18.9

Given a sequence $\langle p_n:n<\omega\rangle\subseteq\beta\mathbb N$ and an ultrafilter $\mathcal U$ there is $p\in\beta\mathbb N$ such that $p=\mathcal U-\lim p_n$

18.10

Given $\mathcal{U,V}\in\beta\mathbb N$ it is not hard to see that $\mathcal V-\lim sh^n(\mathcal U) = \Big\{ A\subseteq\mathbb N:\{n<\omega:A-n\in\mathcal U\}\in\mathcal V\Big\}$

18.11

For $\mathcal{U,V}$ we define $\mathcal U+\mathcal V = \mathcal V-\lim sh^n(\mathcal U),$ and $\mathcal U\cdot\mathcal V = \mathcal V-\lim n\cdot\mathcal U=\Big\{A:\{n:A/n\in\mathcal U\}\in\mathcal V\Big\},$

18.12

The $+$ operation on $\beta\mathbb N$, the set of ultrafilters on $\mathbb N$, is

1. associative
2. its restriction to $\omega$ is the standard addition
3. given $n<\omega$ then $\mathcal U+n = n+\mathcal U$.
4. is continuous in the right coordinate, i.e. the map $\mathcal U\mapsto\mathcal U+x$ is continuous for each $\mathcal U\in\mathbb N$.
18.13Ellis-Numakura

Each $(X,\cdot)$ compact, $T_1$, right-semi-topological semigroup contains an idempotent, i.e. an element $x\in X$ such that $x\cdot x=x$.

Let $\mathcal K\subseteq X$ be the system of compact sub-semigroups such that $K\cdot K=K$ for each $K\in\mathcal K$ ordered by inclusion. Then $\mathcal K$ is nonempty since $X\in\mathcal K$. Let $K$ be a minimal element of $\mathcal K$. Take $e\in K$. Then $e\cdot K$ is a compact subsemigroup of $K$ so, by minimality of $K$, $e\cdot K=K$ and there is $s\in K$ such that $e\cdot s=e$. Consider $S=\{s:e\cdot s=e\}$. This is a compact subsemigroup of $K$ so, again by minimality, $S=K$. It follows that $e\cdot e=e$.

18.14Kalášek

There is no additive idempotent which would be, at the same time, a multiplicative idempotent.

(see his thesis)

18.15

$(X,f)$ is a subsystem of $(Y,f)$ if $X$ is a closed invariant subset of $Y$.

18.16

Every compact dynamical system contains a minimal dynamical subsystem.

18.17

Every two different minimal subsystems are disjoint.

18.18Hindman

Suppose $[\omega]^{<\omega}=\bigcup_{i. Then there is \(j and \(D\in[A_j]^\omega$ consisting of disjoint sets such that $\forall F\in[D]^{<\omega}$ $\bigcup F\in A_j$.

18.19Chodounský

Is there an interval partition $\langle I_n:n<\omega\rangle$ such that for each partition $[\omega]^{<\omega}=A_0\cup A_1$, there is $i<2$ and $D\in[A_j]^\omega$ consisting of disjoint sets such that $\forall F\in[D]^{<\omega}$ $\bigcup F\in A_j$ and $\bigcup D$ intersects all but finitely many interval $I_n$.

18.20
1. If the answer is no then each union ultrafilter has a non-meager core and is almost ordered.
2. Even a consistent answer (either way) would be interesting.
3. Todorčević heard the problem and his opinion is that the answer is no.

# 1916.4.2014

## 19.1J. Grebík: Extending density to all sets of integers.

Recall that, for $A\subseteq\mathbb N$, we define the density of $A$ $d(A)=\lim_{n\to\infty}\frac{|A\cap[1,n]|}{n}$ if it exists. We let $\mathcal D$ be the family of all sets which have a defined density. We let $\mathcal M$ denote the family of all measures on $\mathcal P(\mathbb N)$ extending $d$.

Let $V=\{f\,|\,f:\mathbb N\to\mathbb R, |rng(f)|<\omega\}$ and let $U=\Big\{f:\exists \lim_{n\to\infty}\frac{\sum_{i=1}^n f(i)}{n}\Big\}.$ If we apply the Hahn-Banach theorem to a function from $U$, we will necessarily get a shift-invariant measure. Moreover, this measure will be very regular, in particular, if $m$ is such a measure, $A\subseteq \mathbb N$ and $\varphi:A\to\mathbb N$ is one-to-one, then for each $B\in\mathcal D$ we have $m(\varphi^{-1}[B])=m(A)\cdot d(B)$, i.e. $m$ "respects" relative densities.

If we take a different approach we can define, for $\mathcal U\in\omega^*$, $d_{\mathcal U}(A) = \mathcal U-\lim\frac{|A\cap[1,n]|}{n}.$ Then each $d_{\mathcal U}$ will be an element of $\mathcal M_1$. We let $\mathcal M_U=\{m_{\mathcal U}:\mathcal U\in\omega^*\}\subseteq\mathcal M_1$.

We let $\mathcal M_2$ be the closure (in $[0,1]^{\mathcal P(\mathbb N)}$) of the convex hull of $\mathcal M_U$.

19.1

$\mathcal M_1$ and $\mathcal M_2$ are both convex and closed (in $[0,1]^{\mathcal P(\mathbb N)}$).

19.2

$\mathcal M_2\subseteq\mathcal M_1\subseteq\mathcal M$.

19.3

Are the above inclusions strict?

Let $\mathcal M_\sigma=\{m\in\mathcal M: m\ \mbox{is}\ \sigma-\mbox{additive}\}$.

19.4

Is $\mathcal M_\sigma$ a closed subset of $[0,1]^{\mathcal P(\mathbb N)}$?

19.5

Given $\langle a_n:n<\omega\rangle$, the Caesar-limit of this sequence is, if it exists, $\lim_{n\to\infty} \frac{\sum_{i=0}^n a_n}{n}.$

19.6

When is $d_{\mathcal U}$ $\sigma$-additive?

19.7

An ultrafilter is $*$-invariant if for each $U\in\mathcal U$ there is $k\geq 2$ such that $\bigcup_{n\in U} [k\cdot n,(k+1)\cdot n]\in\mathcal U$

19.8

The measure $d_{\mathcal U}$ is $\sigma$-additive iff $U$ is not $*$-invariant.

19.9

A set $A\subseteq\omega$ is thin if $\lim_{n\to\infty} \frac{e_A(n)}{e_A(n+1)} = 0,$ where $e_A$ is the (increasing) enumeration of $A$.

19.10

If an ultrafilter $\mathcal U$ contains a thin set, then it is not $*$-invariant so, in particular, $d_{\mathcal U}$ is $\sigma$-additive.

19.11

Let $T$ be the set of ultrafilters contining a thin set. Consider the set of $*$-invariant ultrafilters as a subspace of $\omega^*\setminus T$. What are its properties. Note that $\omega^*\setminus T$ is a closed nowhere dense subset of $\omega^*$.

19.12

An $\mathcal U$ ultrafilter is thin if there is a $U\in\mathcal U$ such that $\limsup_{n\to\infty} \frac{e_U(n)}{e_U(n+1)} < 1,$ where $e_U$ is the (increasing) enumeration of $U$.

19.13

There is a function $G:\omega^*\to\{\mathcal U:\mathcal U\ \mbox{is thin}\}$ such that for each $\mathcal U$ we have $d_{\mathcal U}=d_{G(\mathcal U)}$ and, moreover, $\mathcal U$ is $*$-invariant iff $G(\mathcal U)$ is $*$-invariant.

of theorem 19.8

Assume $\mathcal U$ is not $*$-invariant and let $U$ be a witness. For each $k\geq 2$ we know that $\bigcup_{n\in U} [k\cdot n,(k+1)\cdot n]\not\in\mathcal U,$ so also $\bigcup_{n\in U} [k\cdot 2n,(k+1)\cdot 2n]\not\in\mathcal U,$ By the previous proposition we may assume that $\mathcal U$ is thin. Then there is some $n<\omega$ such that $\frac{e_U(n)}{e_U(n+1)}<1/2.$ Define $\Lambda:\mathcal P(\mathbb N)\to\prod_{i=1}^\infty \mathcal P([1,i])$ by letting $\Lambda(A)(n) = A\cap[1,n]$. Similarly let $\Lambda_{\mathcal U}:\mathcal P(\mathbb N)/d_{\mathcal U}\to\prod_{i=1}^\infty \mathcal P([1,i])/m_{\mathcal U},$ where $m_{\mathcal U}(f) = \mathcal U-\lim m_n(f(n)),$ and $m_n$ is the normalized counting measure on $\mathcal P([1,n])$.

19.14

Consider $\mathcal P(\mathbb N)/d_{\mathcal U}$ as a metric space (with $\varrho(A,B)=d_{\mathcal U}(A\triangle B)$) and the target of $\Lambda_{\mathcal U}$ as a metric space. Then $\Lambda_{\mathcal U}$ gives rise to an isometry between these two spaces.

We shall show that $\Lambda_{\mathcal U}$ is onto. This will suffice to show that $d_{\mathcal U}$ is $\sigma$-additive since the target space of $\Lambda_{\mathcal U}$ together with its measure is $\sigma$-additive and $\Lambda_{\mathcal U}$ is measure preserving.

So let $f\in\prod_{i=1}^\infty \mathcal P([1,i])/d_{\mathcal U}$. We will find $X\subseteq\mathbb N$ such that $m_{\mathcal U}(\Lambda_{\mathcal U}(X)\triangle f)=0$. Let $X = \bigcup_{n<\omega} [e_U(n),e_U(n+1)]\cap f(e_U(n).$ We will show that this $X$ works. To do this, we will find a sequence $\langle U_k:k<\omega\rangle\subseteq\mathcal U$ satisfying $m_{\mathcal U}(f(n)\triangle \Lambda(X)(n))<1/k$ for each $k<\omega, n\in U_k$. Let $U_k = U\setminus\bigcup_{n\in U}[k\cdot2n,(k+1)\cdot 2n]$.

The other implication follows along the lines of Fremlin's proof.

# 2030. 4. 2014

## 20.1J. Starý: Pi-characters

20.1

Let $X$ be a topological space. We say that a family of nonempty open sets $\mathcal B$ is a local $\pi$-base at $x\in X$ if for each open $U$ containing $x$ there is $B\in\mathcal B$ such that $B\subseteq U$. We denote $\pi_\chi(x)$ the minimal cardinality of a local $\pi$-base at $x$ and call it the $\pi$-character of $x$. We let $\underline{\pi_\chi}(X)=\min\{\pi_\chi(x):x\in X\}$. For a Boolean algebra $\mathbb B$ we let $\underline{\pi_\chi}(\mathbb B) = \underline{\pi_\chi}(st(\mathbb B))$. When we talk about the $\pi$-weight of Boolean algebra $\pi w(\mathbb B)$, we mean the $\pi$-weight of its stone space.

The following is proved in B. Balcar, P. Simon: On minimal pi-character of points in extremally disconnected compact spaces, Top. Appl. 41(1,2) 1991, pp. 133-145.

20.2Balcar,Simon

Let $\mathbb B$ be a complete, ccc Boolean algebra which is homogeneous in $\pi w$. If $\underline{\pi_\chi}(\mathbb B)=\pi w(\mathbb B)$ then $St(\mathbb B)$ contains a discretely untouchable point.

20.3

Is there a ccc, complete Boolean algebra for which $\underline{\pi_\chi}(\mathbb B)<\pi(\mathbb B)$?

20.4

Mary Bell proved that the answer is yes if we drop the completeness requirement.

20.5

If $\mathbb B$ is a complete, ccc, atomless Boolean algebra such that $d(St(\mathbb B))\leq \underline{\pi_\chi}(\mathbb B)$. Then either it contains two points with different $\pi$-characters or the minimal $\pi$-character is actually equal to $\pi w(\mathbb B)$

20.6

Let $X$ be a topological space. Its tightness, $t(X)$, is the smallest $\kappa$ such that whenever some $x\in X$ is in the closure of a set $A\subseteq X$ then it is already in the closure of a subset $B\subseteq A$ of size $\kappa$.

20.7

If $t(X)\leq d(X)$ then any dense set contains a dense subset of size $d(X)$.

20.8

$[0,1]^{\omega_1}$ is a separable compact space whose tightness is uncountable, since the space is universal for Tychonoff spaces of weight at most $\omega_1$, in particular it contains a copy of $\omega_1+1$.

20.9

Is there an EDC space $X$ with $t(X)>d(X)$.

20.10

Yes, $\beta\omega$, since it contains weak $P$-points guaranteeing uncountable tightness. If we want a space without isolated points, start with the space from Dow, Gubbi, Szymański taking a weak P-point as the parameter. The Čech-Stone compactification of this space is a separable, EDC space without isolated points and uncountable tightness (since the space has an $\aleph_0$-bounded remainder).

(For the space see A. Dow, Gubbi, A. V. Gubbi, A. Szymański: Rigid Stone spaces within ZFC, Proc. Am. Math. Soc. 102(3), 1988.)

20.11

Let $\mathbb B$ be a complete, atomless, ccc Boolean algebra with $t(St(\mathbb B))\leq d(St(\mathbb B))$. If the set $\{p\in St(\mathbb B):\pi_\chi(x)=\underline{\pi_\chi}(\mathbb B)\}$ is dense then $\underline{\pi_\chi}(\mathbb B)=\pi w(\mathbb B)$.

## 20.2Torturing J. Grebík (bachelor thesis)

The outline of the thesis is as follows:

1. Motivations (Riemann $\zeta$ function, square-free numbers, ...)
2. Measure
3. Results

2.1 Density

2.2 $\mathcal P(\omega)/\mathcal Z$

Recall definition 19.12. Note that the ideal $I$ generated by sets $A$ such that $\limsup_{n\to\infty} \frac{e_A(n)}{e_A(n+1)} < 1,$ is contained in the density zero ideal so, in particular, any ultrafilter extending the dual to $\mathcal Z$ is not thin.

Take a countable sequence $\langle \mathcal U_n:n<\omega\rangle$ of ultrafilters on $\omega$ and take a measure $m:\mathcal P(\omega)\to[0,1]$ extending the asymptotic density.

# 217. 5. 2014

## 21.1B. Balcar & P. Simon: Nonhomogeneity of Stone spaces

Consider the additive group of integers $(\mathbb Z,+)$ and the left-shift $f:n\mapsto n-1$. We can extend $f$ to the ultrafilters on $\mathbb Z$ as follows: $f(\mathcal U) = \{A-1:A\in\mathcal U\}.$ Then the set of nontrivial ultrafilters on the natural numbers $\mathbb N^*(\subseteq\mathbb Z^*)$ is an $f$-invariant set. We can define $\mathcal U-\mathcal V = \mathcal V-\lim f^n(\mathcal U).$ then $A\in\mathcal{U-V}\iff \{n:A+n\in\mathcal U\}\in\mathcal V$.

21.1

Can we find an $\mathcal U\in\mathbb N^*$ such that $\mathcal U-\mathcal U=\mathcal U$?

## 21.2E. Thuemmel: Yorioka + Zapletal

21.2Todorčević

Let $X$ be a sequential topological space we define the Todorčević ordering $T(X)$ as follows: $T(X)=\{K:K\ \mbox{is a union of finitely many converging sequences with limits}\}$ where $K\leq H$ if $H\subseteq K$ and $K^\prime\cap H=H^\prime$ ($K^\prime$ is the set of limit points of $K$).

21.3

$x\in 2^\omega$ is a random real if it is not contained in any ground-model coded measure zero set.

21.4Yorioka

If $T(X)$ is ccc then $T(X)$ does not add a random real.

Assuming towards a contradiction assume that some $p\in T(X)$ forces that some $\dot{x}\in 2^\omega$ is random. Let $M$ be a countable elementary submodel of some (sufficiently large) $H(\kappa)$ containing everything we need (e.g. $p,\dot{x}, X, T(X)$, ...). Now $M\cap 2^\omega$ is countable, hence measure zero, so there is a sequence of open sets $\langle O_n:n<\omega\rangle$ with diameters going to zero (i.e. $\mu(O_n)\to 0$) such that $M\cap 2^\omega\subseteq\bigcap O_n$. It follows that there is a stronger $q\leq p$ and $n<\omega$ such that $q\Vdash\dot{x}\not\in O_n$. We shall construct a tree. First, for $s\in 2^{<\omega}$, let $u(s,\dot{x}) = \{p\in T(X):p\Vdash\dot{x}\not\in[s]\}.$ Now, for $b\in[X]^\omega$, define the tree $S(\dot{x},b)$: $S(\dot{x},b)=\{s\in 2^{<\omega}:u(x)\ \mbox{is}\ b\mbox{-small}\},$ where a set $A\subseteq T(X)$ is $b$-large if for each $b\subseteq c\in[X]^\omega$ there is $q\in A$ such that $q^\prime\cap c= b$ and it is $b$-small if it is not $b$-large. Let $b=q^\prime\cap M$. Since $q$ is a union of finitely many converging sequences, the set of its limit points $q^\prime$ is finite, so $b\in M$ so $S(\dot{x},b)\in M$.

21.5

### 22.1.2Trick II.

22.8

A $G_\delta$-cover of a Polish space $X$ is a system $C$ of $G_\delta$-subsets of $X$ such that for each countable $b\subseteq X$ there is $c\in C$ such that $b\subseteq c$.

22.9

If $T(X)$ is ccc then it preserves $G_\delta$-covers.

Assume this is not the case and fix a cover $C$, a name $\dot{y}$ for a countable subset of $2^\omega$ and some condition $p\in T(X)$ that forces that $\dot{y}$ is not covered by $C$.

Let $M$ be a countable elementary submodel of $H_\theta$. Since $C$ is a cover, we can find $c\in C$ such that $M\cap 2^\omega\subseteq c$. By our assumption there must be some $q\leq p$ and $n<\omega$ such that $q\Vdash \dot{y}(n)\not\in c$. Since $c$ is $G_\delta$, there must be some open $o\subseteq 2^\omega$, $c\subseteq o$ and $q\leq p$ such that $q\Vdash \dot{y}(n)\not\in o$. Let $a= q^\prime\cap M$. For $m<\omega$ let $S_m=\big\{t\in 2^m:\{r:r\Vdash t\not\sqsubseteq\dot{y}(n)\}\ \mbox{is not $a$-large}\big\}.$

Then, by the previous lemma, $S_m$ must be nonempty (otherwise $q$ would force that $\dot{y}(n)$ has no initial segment). Note also that $\mathcal S=\{S_m:m<\omega\}\in M$. It follows that there is $z\in M\cap 2^\omega$ a limit point of $\mathcal S$. By choice of $c$ we know that $z\in c$ so, in particular, $z\in o$. Since $o$ is open, there is $m_0$ such that $[z\upharpoonright m_0]\subseteq o$. Now find $m>m_0$ and $t\in S_m$ such that $t\upharpoonright m_0=z\upharpoonright m_0$. Now $q\Vdash \dot{y}(n)\not\in o$ so, in particular, $q\Vdash t\not\sqsubseteq\dot{y}(n_)$. Then $q\in\{r:r\Vdash t\sqsubseteq\dot{y}(n)\}=Q$. By a simple elementarity argument If follows that $Q$ is $a$-large. This is, of course, a contradiction.

22.10

A ccc $T(X)$ cannot add random reals since ground-model null sets form a $G_\delta$-cover which would have to be preserved.

### 22.1.3An Abstract Approach ?

22.11

We say that $RO(P)$ satisfies Axiom Y if for each countable elementary submodel of some $H_\theta$ and for each $q\in RO(P)$ there is a filter $\mathcal F\in M$ such that $\{p\in M\cap RO(P): q\leq p\}\subseteq\mathcal F.$

22.12

If $RO(P)$ satisfies Axiom Y then it preserves $G_\delta$-covers.

By contradiction fix a cover $C$ a name $\{\dot{y_n}:n<\omega\}$ and a condition $p\in P$ such that $p\Vdash (\forall c\in C)(\{\dot{y_n}:n<\omega\}\not\subseteq c).$ Again fix $c\in C$, an open $o\supseteq c$ and $q\leq p$ such that $q\Vdash \dot{y}\not\in o$. Use Axiom Y to find a filter $\mathcal F\in M$ for this $q$. Define $S_m=\{t\in 2^m:||t\not\sqsubseteq\dot{y_n}||\}.$ The same argument as before gives us $S_m\neq\emptyset$ and the rest of the proof follows as before.

22.13

If $X$ is ccc then $RO(T(X))$ satisfies Axiom Y.

## 22.2David's question for Jindra

22.14Forcing Idealized

If $P$ is a forcing such that each $\omega_1$-tree in the extension contains an $\omega_1$-tree in the ground model then it preserves Suslin Trees.

Let $T_0$ be a Suslin tree and suppose $P$ forces an uncountable antichain $A\subseteq T_0$. Let $T_1$ be the tree in the extension generated by $A$. This is an $\omega_1$-tree so it contains a ground model $\omega_1$-tree $T_2$. Let $B=\{t\in T_0:t\not\in T_2\ \&\ \mbox{all initial segments of}\ t\ \mbox{are in}\ T_2\}$. Then $B$ is an antichain and it must be uncountable: otherwise, since $T_2$ is an $\omega_1$-tree, there must be a $t\in T_2$ of height bigger than the height of elements of $B$. Then we can find an extension of $t$ in $T_0\setminus T_2$ and some initial segment of this extension of height bigger than $t$ will be in $B$ --- a contradiction.

## 22.3J. Zapletal: Iteration of ω-models

22.15Bartoszynski?

If a poset $P$ is sufficiently definable then the statement "$P$ is ccc" is absolute between generic extensions.

The original proof uses Keisler's compactness theorem. We will show a proof using iteration of models.

22.16

An $\omega$-model $M$ is a countable model of (a fragment of) ZFC with $\omega^M=\omega$.

22.17

Sufficiently definable in the above theorem means that an $\omega$-model correctly interprets everything that is needed for the argument.

Assume that there is an $\omega$-model $M$ such that $M\models "P\ \mbox{is not ccc}"$. Then $M$ is not ccc already in $V$. To show this we build a sequence of models $\langle M_\alpha:\alpha<\omega_1\rangle$ such that $M_{\alpha+1}$ is a generic ultrapower of $M_\alpha$ over $P(\omega_1)/\mathcal{NS}$ and a limits we take direct limits. Each $M_\alpha$ will be an $\omega$-model. Finally we take $M_{\omega_1}$. This will be an $\omega$-model which correctly interprets "being of size $\aleph_1$". Moreover it will see an uncountable antichain in $P$ (since each $M_\alpha$ sees that) and this antichain will be uncountable.

# 2318.6.2014

## 23.1J. Grebík: Ideals on products of Boolean algebras

23.1

Let $\mathbb B$ be a measure algebra and $\mathcal U\in\omega^*$. We consider $\mathbb B^\omega$ together with four ideals:

1. $\mathcal I_0=\{\overline{b}\in\mathbb B^\omega:\bigwedge\bigvee\overline{b}=\mathbf 0\}$
2. $\mathcal{I_U}=\{\overline{b}\in\mathbb B^\omega:\bigwedge_{U\in\mathcal U}\bigvee_{n\in U}b_n=\mathbf 0\}$
3. $\mathcal I^\mu=\{\overline{b}\in\mathbb B^\omega:\lim \mu(b_n) = 0\}$
4. $\mathcal I_{\mathcal U}^\mu=\{\overline{b}\in\mathbb B^\omega:\mathcal U-\lim \mu(b_n) = 0\}$

Recall that $\mathbb B^\omega/\mathcal I^\mu_{\mathcal U}$ is again some measure algebra.

23.2

When does the equality $\mathcal{I_U}=\mathcal I^\mu_{\mathcal U}$ hold?

The following can be found in Bartoszynski, Judah: Structure of the real line.

23.3

A sequence $\langle x_n:n<\omega\rangle\subseteq \mathbb B$ is measure independent if for each finite $A\in[\omega]^{<\omega}$ set of indices we have $\mu(\bigwedge_{n\in A} x_n) = \prod_{n\in A}\mu(x_n).$

23.4

Given a sequence or real numbers $\langle a_i:i<\omega\rangle$ in $[0,1]$ then there is a measure independent sequence $\langle x_n:n<\omega\rangle$ of elements of $\mathbb B(\omega)$ (the standard measure algebra of length $\omega$) with $\mu(x_n)=a_n$.

To have find a counterexample to question 23.2 we would need a sequence $\langle x_n:n<\omega\rangle$ such that $\mathcal U-\lim \mu(x_n)=0$ and $\sum \mu(x_n)=\infty$. However, we have the following theorem.

23.5Borel

If $\langle x_n:n<\omega\rangle$ is measure independent then $\sum_{n<\omega} \mu(x_n) = \infty \Rightarrow \bigwedge_{k<\omega}\bigvee_{n>k} x_n = 1$ and $\sum_{n<\omega} \mu(x_n) < \infty \Rightarrow \bigwedge_{k<\omega}\bigvee_{n>k} x_n = 0$

This gives the following conditions on $\mathcal U$: we need that $\mathcal U-\lim x_n=0$, while $\sum_{n\in U} x_n=\infty$ for each $U\in\mathcal U$. By the previous theorem we will also have $\bigwedge_{U\in\mathcal U}\bigvee_{n\in U}b_n=\mathbf 1$ for each $U\in\mathcal U$. This gives the following proposition.

23.6Grebík

The equality holds $\mathcal I^\mu_{\mathcal U} = \mathcal{I_U}$ iff $\mathcal U$ is a semiselective ultrafilter.

where

23.7Kunen

An ultrafilter is semiselective iff it is a rapid P-ultrafilter.

The proposition follows from the following characterization of semiselectivity.

23.8Grebík

$\mathcal U$ is semiselective iff for each sequence $\langle a_n:n<\omega\rangle$ of real numbers in $[0,1]$ such that $\mathcal U-\lim a_n = 0$ there is $U\in\mathcal U$ such that $\sum_{n\in U} a_n<\infty$.

23.9Jech

An ultrafilter $\mathcal U$ is semiselective iff for each Boolean algebra $\mathbb B$ and each sequence $\overline{b}=\langle b_n:n<\omega\rangle$ of elements of $\mathbb B$ such that $\overline{b}\in\mathcal I^\mu_{\mathcal U}$ there is an $U\in\mathcal U$ such that $\bigvee_{n\in U} b_n<\mathbf 1$.

## 23.2David Chodounský: Construction of ccc forcing using long diagonalization

The following is motivated by the question whether axiom Y (Y-cc) (see 22.11)
is productive.

Recall that

$\mbox{Knaster}\Rightarrow \mbox{productively ccc}\Rightarrow \mbox{powerfully ccc} \Rightarrow\mbox{ccc}$

and that, consistenly (e.g. under MA), all of the implications can be reversed. Moreover the implications cannot be separated by "definable" orderings.

23.10Thuemmel

The Todorčević's ordering, if ccc, is already productively ccc.

23.11

A Suslin tree is ccc but not powerfully ccc.

23.12

Take an indestructible gap $T$ and add a Cohen real $c$. Then $c\cap T$ and $T\setminus c$ are both destructible gaps. Let $P_1$ and $P_2$ be ccc forcings to, respectively, destroy $c\cap T$ and $T\setminus c$. Then $P_1\times P_2$ cannot be ccc, since it destroys $T$ but $T$ is indestructible. However $P_1,P_2$ are both even powerfully ccc.

23.13Todorčević

If $\mathfrak b=\aleph_1$ then $\mathbb{T(R)}$ is productively ccc but not Knaster.

23.14Baumgartner

The forcing to add an antichain to a Suslin tree by finite conditions is powerfully ccc but not productively ccc (as shown by the Suslin tree).

We will show that Y-cc does not fit into the above hierarchy. The following example shows that Y-cc is not productive.

23.15Galvin-Laver

Let $X$ be a space and $[X]^2= O_0\cup O_1$. Let $P_0 = \big\{ a\in [X]^{<\omega}:[a]^2\subseteq O_0\big\}$ ordered by reverse inclusion and $Q_0 = [X]^{<\omega}$ ordered as follows: $a\leq b \iff a\supseteq b\ \&\ b\times(a\setminus b)\subseteq O_0,$ i.e. a condition is stronger if each new vertex is connected to all old vertices (interpreting $(X,O_0)$ as a graph).

23.16

If $Q_0$ is ccc then $P_0$ is ccc.

23.17

If $Q_0$ is ccc then both $Q_0$ and $P_0$ are Y-cc.

So let $M$ be a countable elementary submodel of some $H_\theta$ and $p\in P_0$. Let $b=p\cap M$ and say that a set $S\subseteq P$ is $b$-big if for each $K\in[X\setminus b]^\omega$ there is a $a\in S$ such that $b\subseteq a$ and $a\cap K=\emptyset$.

Let $\mathcal F=\{\bigvee S:S\ \mbox{is}\ b-\mbox{big}\}$.

23.18

If $p\leq q\in RO(P_0)\cap M$ then $q\in \mathcal F$.

of Claim

Since $P_0$ is dense in $RO(P_0)$ we have $q=\bigvee\{a\in P_0:a\leq q\}$. To finish the proof of the claim we need only show that $S=\{a\in P_0:a\leq q\}$ is $b$-big. Since $S\in M$ we need only check this in $M$. Let $K\in M$. Since $M$ is transitive for countable sets, $K\subseteq M$. Then $p\in S$ and $p\cap K\subseteq b$. Then, by elementarity, there is $p^\prime\in M\cap S$ such that $p^\prime\cap K\subseteq b$.

23.19

$\mathcal F$ is centered.

of Claim

We will show that if $S_0,\ldots, S_n$ are $b$-big then there are $a_0\in S_0,\ldots, a_n\in S_n$ such that $\bigwedge_{i\leq n} a_i\neq\emptyset$. (At this point, I had to leave the seminar.)

23.20

It follows from the proof that both $Q_0$ and $P_0$ will be powerfully ccc.

# 243.9.2014

## 24.1D. Chodounský: Generalized Laver Forcing

24.1

Given a family $\mathcal A\subseteq[\omega]^\omega$ let $\mathbb L(\mathcal A)$ be a forcing consisting of trees $T$ having a stem $s$ such that $(\forall t\in T)(s\subseteq t\rightarrow succv_T(t)\in\mathcal A).$ where $succv_T(t)=\{s(|t|):s\in succ_T(t)\}$.

24.2

If $\mathcal A=[\omega]^\omega$ this is the classical Laver forcing. Laver forcing adds a dominating real. Moreover it has the Laver property, i.e. each bounded real in the extension is contained in some ground-model tunnel of width $2^n$.

24.3

If $\mathcal A$ is a filter then $L(\mathcal A)$ is $\sigma$-centered, in general $L(\mathcal A)$ is just proper (proof by fusion).

24.4

For $T,S\in\mathbb L(\mathcal A)$ with the same stem we define $T\leq_n S$ if $T\cap \omega^{|stem(T)|+n} = S\cap \omega^{|stem(T)|+n}$. A fusion sequence is a sequence of conditions $\langle T_n:n<\omega\rangle$ such that $T_{n+1}\leq_n T_n$ for each $n<\omega$.

24.5fusion

If $\langle T_n:n<\omega\rangle$ is a fusion sequence then there is $T\in\mathbb L(\mathcal A)$ which is a lower bound for all $T_n$.

24.6

A family $\mathcal A$ is hitting if for each $X\in[\omega]^\omega$ there is $A\in\mathcal A$ such that $|A\cap X|=\omega$, it is splitting if $X\in[\omega]^\omega$ there is $A\in\mathcal A$ such that $|A\cap X|=|A\setminus X|=\omega$. A family is $\omega$-hitting if for each sequence $\langle X_n:n<\omega\rangle\subseteq[\omega]^\omega$ there is an $A\in\mathcal A$ such that for each $n<\omega$ such that $|A\cap X_n|=\omega$ for each $n<\omega$. Similarly we define $\omega$-splitting.

24.7

In the $\omega$-variants it is sufficient to require the intersections to be non-empty.

24.8

Any P-ideal which is hitting is $\omega$-hitting.

24.9

Given a MAD family $\mathcal A$ the ideal $\mathcal I(\mathcal A)$ is hitting but not $\omega$-hitting.

The following theorem is from A. Dow: Two classes of Fréchet-Urysohn spaces, Proc. Ams. 108(1) 1990.

24.10A.Dow

Laver forcing preserves every $\omega$-hitting and $\omega$-splitting families.

24.11

A family $\mathcal X$ is $\mathcal F$-omega-hitting if for each sequence $\langle f_n:n<\omega$ of functions from $\omega$ to $\omega$ such that $f[\omega]\in\mathcal F$ there is $X\in\mathcal X$ such that $f[X]\in\mathcal F$ for each $n<\omega$.

24.12

In the above definition, $\mathcal F$ is usually a co-ideal.

24.13

If $\mathcal X$ is $\mathcal F$-omega-hitting then it is $\omega$-hitting.

Let $f_n$ be a bijection of $X_n$ onto $\omega$ and constant outside $X_n$.

24.14Hrušák, Chodounský, Guzmán

The following are equivalent for a filter $\mathcal F$ and a family $\mathcal X$:

1. $\mathcal X$ is $\mathcal F^+$-$\omega$-hitting.
2. $\mathbb L(\mathcal F^+)$ forces that $\mathcal X$ is $\omega$-hitting.

Suppose 1. fails and let $\langle f_n:n<\omega\rangle$ witness this. Let $\langle B_n:n<\omega\rangle$ be a partition of $\omega$ into infinite sets (eventually, $B_n$ will be a set of levels of a Laver tree). Let $T\in\mathbb L(\mathcal F^+)$ be a condition such that for each $t\in T$ if $|t|+1\in B_n$ then $succ_T(t)\subseteq f_n[\omega]$. Let $\dot{g}$ be the name for the generic real and let $\dot{X}^k_n=\check{f}_n^{-1}[\dot{g}[\check{B}_n\setminus k]].$ Then $T$ forces that $X^k_n$ is infinite. For each $X$ we can find $n<\omega$ such that $f_n[X]\in\mathcal F^*$. So we prune $T$ to a stronger $S$ such that for each $t\in S^\prime$ if $stem(S^\prime)\leq t$ then $succ_{S^\prime}(t)\cap f_n[X]=\emptyset$. This $S$ forces that $X\cap X^k_n$ is empty for some $k<\omega$ thereby showing that 2. fails.

On the other hand, assume that 1. holds. Let $T$ be a condition and $\dot{A}=\langle \dot{X}_n:n<\omega\rangle$ a name for a sequence of infinite sets. Let $M$ be an countable elementary submodel of some sufficiently large $H(\vartheta)$ such that $T,A\in M$. Let $\langle f_n:n<\omega\rangle$ enumerate all functions from $M$ with domain $\omega$ and range in $\mathcal F^+$. Use 1. to find $X\in\mathcal X$ such that $f_n[X]\in\mathcal F^+$ for each $n<\omega$. We fix this $X$ for the remainder of the proof.

24.15

Suppose $S,\dot{A}\in M$ such that $S\Vdash \dot{A}\in[\omega]^\omega$. Then there is a stronger $S^\prime\leq_0 S$ such that for each $T\leq S^\prime$ there is $t\in T$ such that $S^\prime_t \in M$ and $S^\prime_t\Vdash \check{X}\cap\dot{A}\neq\emptyset$

of Claim

Work in $M$ and, by recursion, construct a sequence $\langle t_n:n<\omega\rangle$ of nodes of $S$, a strictly increasing sequence of natural numbers $\langle k_n:n<\omega\rangle$ and a sequence of conditions $\langle R_n:n<\omega\rangle$ such that

1. $R_n\leq S_{t_n}$
2. $R_n\Vdash k_n\in \dot{A}$
3. The set $\{t_n:n<\omega\}$ is a maximal antichain in $S$.

Now let $S^\prime = \bigcup_{n\in X}R_n$ We need to show that $S^\prime \in\mathbb L(\mathcal F^+)$ and that it is as required. The second part is obvious, so we show that $S^\prime$ is a condition. For this it suffices to show that for each $t\in S^\prime$ we have $succv_{S^\prime}(t)\in \mathcal F^+$. By 3. there must be $n<\omega$ such that $t$ is compatible with $t_n$. If $t_n \subseteq t$ then we are done, as $succv_{S^\prime}(t)=succv_{R_n}(t) \in \mathcal F^+$. So suppose $t\subsetneq t_n$. Then, again by 3, for each $s\in succ_S(t)$ there is $t_{n(s)}$ compatible with $s$. Working in M, define $f:\omega\to\omega$ by letting $f(k_{n(s)})=s(|t|)$. Since $S$ is a condition, we have that the range of $f$ is in $\mathcal F^+$. It follows that $succv_{S^\prime}(t)=f[X]\in\mathcal F^+$.

We now use the claim to recursively construct a fusion sequence $\langle T_n:n<\omega\rangle$ with $T_0=T$ and such that:

1. For each $T^\prime\leq T_n$ there is $t\in T^\prime$ such that $T_n(t)\in M$ and $T_n(t)\Vdash A_n\cap\check{X}\neq\emptyset$.

The lower bound of the fusion seqence (see 24.5) is a condition which forces that $\check{X}$ hits each $\dot{A}_n$ and this finishes the proof.

# 2510.9.2014

## 25.1J. Verner: A simple question

25.1

Given a topological space $X$, its extent $e(X)$ is defined to be the smallest cardinality of an open subset of $X$.

25.2

Does there exist a space of size (and extent) $\aleph_1$ such that whenever it is partitioned into countably many pieces, at least one of the pieces contains a subspace of extent $\aleph_1$.

25.3

If $X$ is ccc and each of its subspaces has at most countable extent then $X$ is a union of a closed nowhere dense set and a countable set.

25.4

If $non(\mathcal M)=\omega_1$ then any nonmeager subset of $\mathbb R$ of size $\omega_1$ works.

## 25.2D. Chodounský: Axiom Y (joint work with J. Zapletal)

For motivation, recall 22.11.

25.5

$Y$-cc is a forcing-property, is preserved on regular subposets and each $\sigma$-centered poset has $Y$.

25.6

If $P$ is $Y$ then it is ccc.

For proof see 26.10

25.7

A forcing $P$ is $Y$-proper if for each countable elementary submodel of a sufficiently large $H(\theta)$ with $P\in M$ and for each $p\in P\cap M$ there is $q\leq p$ which is a $Y$-master condition, i.e. satisfies

1. q is a $M$-generic (i.e. is a master condition in the usual sense)
2. for each $r\leq q$ there is a filter $\mathcal F\subseteq RO(P)$, $\mathcal F\in M$ such that $\{ s\in RO(P)\cap M: r\leq s\} \subseteq \mathcal F.$
25.8

Being $Y$-proper is a forcing property, is preserved on regular subposets and each $Y$-cc forcing is $Y$-proper.

25.9

If $P$ is $\kappa$-cc, where $\kappa$ is smaller than the least measurable, then 2. already implies 1. in the above definition.

25.10

The Mathias-Prikry forcing on a measurable cardinal is an example of a forcing satsifying 2. but not 1. in the above definition.

25.11

Does $Y$-proper + ccc imply $Y$-cc.

25.12

A $Y$-proper forcing does not add random reals, adds unbounded reals and has many other properties of $Y$-cc forcings.

25.13

The forcing to kill all S-spaces, the forcing to force PID, the forcing to show that, under PFA, there are only 5 Tukey types of orderings, are all $Y$-proper.

25.14

A forcing which forces (an instance of) OCA cannot be proper (it adds anti-cliques to graphs).

### 25.2.1Laver forcings with axiom Y

Recall the definition 24.1 of $L(\mathcal I^+)$.

25.15

The following are equivalent for analytical P-ideals $\mathcal I$:

1. The ideal $\mathcal I$ is an intersection of $F_\sigma$-ideals.
2. $L(\mathcal I^+)$ is $Y$-proper.
3. For each compact Polish $X$, open $H\subseteq X^\omega$ and for each $G$ generic on $L(\mathcal I^+)$ we have $V[G]\models \Big(\forall Y\subseteq \check{X}, [Y]^\omega\cap H=\emptyset\Big) \Big(\exists \{Y_n:n<\omega\}\in V, [Y_n]^\omega\cap H=\emptyset\Big) \Big(Y_n\subseteq \bigcup_{n<\omega} Y_n\Big)$
25.16

If $\mathcal I$ is an analytical P-ideal then being an intersection of $F_\sigma$-ideals is equivalent to being an intersection of countably many $F_\sigma$-ideals. An example of an analytical P-ideal which does not satisfy 1. is the density zero ideal.

of Theorem

2.$\rightarrow$ 3.: This is true for all $Y$-proper forcings $P$. Let $X,H,G,\dot{Y}$ be given. Whenever $\mathcal F\subseteq RO(P)$ define $B(F,\dot{Y}) = \Big\{x\in X:(\forall O\in \mathcal U(x)^{V[G]}) (||\dot{O}\cap \dot{Y}^{V[G]}\neq\emptyset||\in\mathcal F) \Big\}$ (note that $B(G,\dot{Y})$ is the closure of $Y$ in $V[G]$ ).

25.17

$B(F,\dot{Y})$ is an $H$-anticlique, i.e. $[B(F,\dot{Y})]^\omega\cap H=\emptyset$.

of Claim

Assume, aiming for a contradiction, that there is $A=\{x_n:n<\omega\}\subseteq B(F,\dot{Y})$ such that $A\subseteq H$. Since $H$ is open, there are open neighbourhoods $O_0,\ldots,O_n$ of $x_0,\ldots,x_n$ such that $\prod_{i\leq n} O_i\times\prod_{i\geq n} X\subseteq H$. By the definition of $B$, we can find $p_i\in\mathcal F$ such that $p_i\Vdash O_i\cap\dot{Y}\neq\emptyset$. Let $p=\bigwedge_{i\leq n} p_i$. Then $p \Vdash \prod_{i\leq n} O_i \times X^{\omega\setminus n+1}\cap [\dot{Y}]^\omega\neq\emptyset$ However, $Y$ is an $H$-anticlique, a contradiction (the left side is contained in $H$).

Aiming for a contradiction, assume that 3. does not hold and that this is forced by some $p\in P$. Let $M$ be a countable elementary submodel of some sufficiently large $H(\theta)$ with $P,X,Y,p\in M$. By $Y$-properness there is a $Y$-master condition $q\leq p$. Let $\langle Y_n:n<\omega\rangle$ enumerate all sets of the form $B(F,\dot{Y})$ in $M$. Then no $r\leq q$ can force that $x\in\dot{Y}\setminus\bigcup_{n<\omega}Y_n$ for some $x\in X\cap V$ (since $X$ is second countable) a contradiction.

3.$\rightarrow$ 1.: Assume 1. fails, we will show that also 3. fails. Since 1. fails, there is $A\not\in\mathcal I$ which is in every $F_\sigma$ ideal containing $\mathcal I$. Let $P=\mathbb L(\mathcal I^+)$ and let $M$ be a countable elementary submodel of some sufficiently large $H(\theta)$ containing $\mathcal I, P, ...$. Let $Z=St(RO(P)\cap M)$. Let $X=K(Z)=\{K\subseteq Z:K\neq\emptyset, K\ \mbox{compact}\}$ with the hyperspace (Vietoris) topology. Define $H = \{\langle K_n:n<\omega\rangle: \bigcap_{n<\omega} K_n = \emptyset\}.$ Then $H$ is an open subset of $X$. Given $T\in P$ let $K_T = \Big\{F\in Z:\{p\in RO(P)\cap M:T\leq p\}\subseteq F\Big\}$ then $K_T\in X$. Let $G$ be a generic on $P$ and $Y=\{K_T:T\in G\}$. Now $Y$ is an $H$-anticlique. Given a sequence $\langle Y_n:n<\omega\rangle$ of anticliques and $T\in P$ it will be sufficient to find $S\leq T$ such that $K_S\not\in\bigcup_{n<\omega} Y_n$. Since being an anti-clique is equivalent to being a centered sequence of compact sets (i.e. filters), we may assume that each $Y_n$ is an ultrafilter on $RO(P)\cap M$.

It will be sufficient to find $S$ and a sequence $p_n\in Y_n$ of elements disjoint from $S$. Since $\mathcal I$ is an analytic P-ideal, we can fix a lscsm $\mu:[\omega]^{<\omega}\to\mathbb R^+_0$ with $I=Exh(\mu)$.

25.18

For each $k<\omega$, we can partition $A=A_0\cup \cdots\cup A_n$ into pieces in such that for each $i<\omega$ we have $|A_n|=1$ or $\mu(A_n)\leq 2^k$.

of Claim

Otherwise, let $\mathcal J$ be the ideal generated by sets which can be partitioned as in the claim. Then this is an $F_\sigma$-ideal containing $\mathcal I$ and not containing $A$, a contradiction.

Note that in the above claim we could have taken the partition to be in $M$.

25.19

For any $U\in P$ with $t=stem(U)$ and $n<\omega$ there is $V\leq_0 U$ (see 24.5) and $q_i\in Y_i$ such that

1. if $i then \(q_i$ is disjoint from $V$; and
2. $q_i$ is disjoint from $V$ after deleting finitely many successors of the stem of $V$.
of Claim

If the condition with stem $t$ and everywhere branching into $A$ is not in $Y_i$ then choose a condition $q_i\in Y_i$ disjoint with this condition and let $V=U$. Otherwise let $B=succv_U(t)\cap A$. Then $B\in\mathcal I^+$ so $\lim_{n\to\infty} \mu(B\setminus n) = \varepsilon > 0$. Choose $\langle k_i:i<\omega\rangle$ such that $\sum_{i<\omega}2^{-k_i} < \varepsilon/2$. Using the previous claim, fix a partition of $A$ into $\{A^i_j:j with each \(A^i_j$ of submeasure $\leq 2^{-k_i}$. Since $Y_n$ is an ultrafilter and since $A\in Y_n$ we can find $j(i) such that \(A^i_{j(i)}\in Y_n$. Let $B^\prime = B\setminus \Bigg(\bigcup_{i1} A^i_{j(i)}\Bigg).$ Finally let $V$ be constructed from $U$ by deleting the successors of $t$ in $B^\prime$. This works.

We now use the claim to construct a fusion sequence $S_n$ (applying the claim to each $S^n_t$ for $|t|=n, j=n$). By proposition 24.5 there is $S\leq S_n$. We construct $p_i$ as an intersection of a suitable finite set of $q_i$'s.

This finishes the proof of 3. $\rightarrow$ 1.

# 2624.9.2014

## 26.1J. Grebík: Graphs & Banach spaces

26.1

Let $P$ consists of finite graphs with vertices some countable ordinals. Ordered as follows $G\leq H$ if $V(G)\supseteq V(H)$ and $E_G\upharpoonright V(H)^2=E_H$. Let $P_\alpha$ consist of those graphs whose vertices are elements of $\alpha$.

26.2

$P_\alpha$ is Cohen and is a regular subposet of $P$ which is itself forcing equivalent to adding $\omega_1$ many reals. Also note that $P$ can be written as a two step iteration $P_\alpha * P_{\omega\setminus\alpha}$.

26.3

The generic graph added by $P$ does not have an automorphism.

Let us now try a similar forcing in a different setting.

26.4

A norm $||\cdot||:V\to\mathbb R$ on a space $V$ with basis $B$ is rational if it is generated by finitely many linear functionals $a_0,\ldots a_n:B\to\mathbb Q$ as follows: $||v|| = \max_{i\leq n} a_i(v),$ for each $v\in B$.

26.5

Each norm can be approximated by a rational norm with the identity being an $\varepsilon$-isomorphism.

26.6

Let $Q$ consist of finite-dimensional Banach spaces $p=(B_p,a_p)$, where $B_p\subseteq\omega_1$ is a finite basis for the space and $||\cdot||_p$ is a rational norm. A stronger condition must be a larger space and must preserve the norm. The generic space $V_G$ is a union of conditions in the generic. Take $W_G$ to be the completion of this space.

26.7

Again, we can define $Q_\alpha$ to consist of those conditions whose basis is a subset of $\alpha$. This will be a regular subposet of $Q$ which is forcing equivalent to Cohen forcing.

26.8

The space $W_G$ does not have an automorphism.

## 26.2D. Chodounský: News from Poland

### 26.2.1A new class of forcing notions

For motivation, recall 22.11. The following results can be found in D. Chodounský and J. Zapletal: Why Y-c.c., submitted.

26.9

Taking completions is necessary in the above definition. Consider collapse of $\omega_1$ to $\omega$, i.e. $P=\{p:n\to\omega_1\}$, then for each elementary submodel $M$ and each $p\in P$ we can take $F$ to be all conditions extending $"p\cap M"$ which is a filter on $P$ containing every condition in $M$ bigger than $p$. However, $P$ cannot be Y-cc (satisfy axiom Y), since it collapses $\omega_1$.

26.10

$\sigma$-centered $\Rightarrow$ Y-cc $\Rightarrow$ ccc.

The first implication is easy. Assume for a contradiction $P$ is Y-cc and $A\subseteq P$ is an uncountable antichain. Let $M$ be a countable elementary submodel of some sufficiently large $H(\theta)$. Let $A\in M$ and pick $q\in A\setminus M$. By Y-cc there is $\mathcal F\in M$ such that $\{p\in RO(P)\cap M:q\leq p\}\subseteq \mathcal F$. Let $G=\big\{B\subseteq A:\bigvee A\in\mathcal F\big\}.$ Note that for each $B\in M$ we have $B\in G$ iff $q\in B$, so $G$ is a non-principal $M$-ultrafilter on $A$. Now $G$ is $\sigma$-complete (in M) ultrafilter --- a contradiction ($\omega_1$ is not a measurable cardinal).

26.11

Forcing specializing an Aronszajn tree is Y-cc

26.12

Forcing freezing an $(\omega_1,\omega_1)$-gap is Y-cc.

26.13

The Todorčević ordering $T(X)$ is Y-cc if it is ccc. (cf. 21.4).

For basic properties of Y-cc reals, consult theorem 25.12.

26.14

Let $X$ be a second countable space and $H$ an open (in the product topology) on $X$, i.e. an element of $[X]^\omega$. then any anticlique in the extension is covered by a countable set of anticliques from the groundmodel. (cf. 3 of 25.15)

26.15

Y-cc forcings preserve $\omega_1$-covers by $G_\delta$-sets of a compact Polish spaces.

26.16

A finite support iteration of any length of Y-cc forcings is Y-cc.

We proof preservation under two-step iteration. So let $P$ be Y-cc and $P\Vdash \dot{Q}\ \mbox{is Y-cc}$. We need to show that $P*\dot{Q}$ is Y-cc. So let $M$ be an elementary submodel and $(p,\dot{q})\in P*\dot{Q}$ a condition. Observe that $RO(P*\dot{Q})=\{(r,\dot{s}):r\in RO(P), r\Vdash\in RO(\dot{Q})$). We know that there is $p_0\leq p$ and $\dot{\mathcal F_q}\in M$ such that $p_0$ forces that $\dot{\mathcal F_q}$ works for $\dot{q}$. Use Y-cc for $p_0\in P$ to find a filter $\mathcal F_p$. Finally let $\mathcal F=\{(r,s):r\in\mathcal F_p\ \&\ r\Vdash s\in\dot{\mathcal F_q}\}.$

The proof for limit stages is a little more complicated.

26.17

If $P,Q$ is Y-cc then $P\times Q$ need not be ccc. However it is open whether if $P\times Q$ is ccc for $P,Q$ Y-cc forcings then $P\times Q$ is Y-cc.

26.18

Let $\varphi$ be a property of complete Boolean algebras such that ZFC proves that $\varphi$ implies ccc, is preserved under iterations and complete subalgebras. Then if $\Diamond_{\kappa^+}(S_{\kappa})$ then there is a complete boolean algebra $B$ having $\varphi$ such that $B$ forces $MA_{\kappa}(\varphi)$.

Recall the definition 25.7 of Y-proper forcing notions.

26.19

Laver is Y-proper but not Y-cc.

26.20Balcar

Is Mathias forcing Y-proper?

The following notion was originally defined in W. Mitchell: Adding closed unbounded subsets of ω2 with finite forcing, Notre Dame J. Formal Logic 46(3), 2005, 357–371 (definition 2.3).

26.21Mitchell

Given a forcing $P$ and a countable elementary submodel $M\prec H(\theta)$ we say that $p\in P$ is a strong master condition if for each $r\leq p$ there is $r^\prime \in M$ such that for $q\in M$ we have $r\leq q\rightarrow r^\prime\leq q$.

26.22

A strong master condition $p$ forces that the generic filter is $V$-generic on $P\cap M$. Cf. this with the fact that a normal master condition forces the filter to be $M$-generic.

26.23

A strongly proper forcing is Y-proper.

26.24

Let $P$ be Y-proper, $p\in P$ and $\dot{f}$ a name for a function from $\kappa$ to $\kappa$ such that for each countable $a\in[\kappa]^\omega$ $p\Vdash \dot{f}\upharpoonright a\in V$ then $p\Vdash \dot{f}\in V$.

Let $p,\dot{f}\in M$ and find a Y-master condition $q$. Then, by assumption, there is $g\in V$ and $r\leq q$ such that $r\Vdash \dot{f}\upharpoonright (M\cap\kappa)=\check{g}$. By properness (i.e. since $q$ is a master condition) $q\Vdash\dot{f}[M]\subseteq M$ so $g[M]\subseteq M$. Y-master gives us a filter $\mathcal F\in M$. Given $\alpha,\beta\in\kappa$ let $b_{\alpha,\beta}=||\dot{f}(\alpha)=\beta||$ and define $h(\alpha)=\beta$ iff $h_{\alpha,\beta}\in\mathcal F$. Then $h$ is a (partial) function from $\kappa$ to $\kappa$ and is in $M$.

26.25

$h$ is a total function.

Moreover $h\upharpoonright M=g$.

# 2729.10.2014

## 27.1Šárka Stejskalová: Consistency of PFA

For more details one can consult:

27.1

A cardinal $\kappa$ is measurable if there is a normal measure on $\kappa$, i.e. a $\kappa$-complete ultrafilter extending the club filter.

27.2

If $\mathcal U$ is a normal measure on $\kappa$ then the mapping: $j_{\mathcal U}(x) \mapsto [c_x:\kappa\to V]_{\mathcal U}\in \prod_{\alpha\in\kappa} V/\mathcal U,$ assigning to each set $x$ the (equivalence class of the) constant function on $\kappa$ with value $x$ is an elementary embedding of $V$ into the ultraproduct $\prod_{\alpha\in\kappa} V/\mathcal U$.

27.3

If $j:V\to M$ is an nontrivial elementary embedding then we define $cp(j)$ to be the minimal ordinal $\alpha$ such that $\alpha. 27.4 If the GCH holds and \(\kappa$ is measurable, as witnessed by $\mathcal U$, then $\kappa^+ < j_{\mathcal U}(\kappa) < \kappa^{++}$

27.5

A cardinal $\kappa$ is $\lambda$-supercompact if there is an elementary embedding $j:V\to M$ with critical point $\kappa$ such that

1. $M$ is closed under $\lambda$-sequences, i.e. ${}^\lambda M\subseteq M$
2. $\lambda < j(\kappa)$

It is supercompact if it is $\lambda$-supercompact for each $\lambda$.

27.6

An ultrafilter $\mathcal U$ on $\mathcal P_{\kappa}(\lambda)=\{X\subseteq\kappa:|X|<\kappa\}$ is normal if for each $\langle X_\alpha:\alpha<\lambda\rangle\subseteq\mathcal U$ the diagonal intersection $\triangle_{\alpha<\lambda} X_\alpha = \{y\in\mathcal P_{\kappa}(\lambda):y\in\bigcap_{\alpha\in y}X_\alpha\}$ is an element of $\mathcal U$. An ultrafilter is fine if for each $\alpha<\lambda$ the set $\{y\in \mathcal P_{\kappa}(\lambda):\alpha\in y\}$ is an element of the ultrafilter.

27.7

$\kappa$ is $\lambda$-supercompact iff there is a $\kappa$-complete normal fine ultrafilter on $\mathcal P_{\kappa}(\lambda)$.

27.8

Let $\mathcal U$ be a $\kappa$-complete normal fine ultrafilter on $\mathcal P_{\kappa}(\lambda)$ and $j$ is its associated elementary embeding then

1. $[id] = j^{\prime\prime}\lambda$;
2. $\kappa$ is a critical point of $j$ and $\lambda< j(\kappa)$;
3. The range $M$ of $j$ is closed under $\lambda$-sequences;
4. $j^{\prime\prime}\alpha= [\langle y\cap\alpha:y\in \mathcal P_{\kappa}(\lambda)]\rangle$ for each $\alpha\leq\lambda$, in particular, $\alpha=[\langle otp(y\cap \alpha):y\in\mathcal P_{\kappa}(\lambda)\rangle]$;
5. ${}^{\lambda^{<\kappa}}M\subseteq M$ and $j^{\prime\prime}\mathcal P_{\kappa}(\lambda)\in M$
6. $2^{\lambda^{<\kappa}}\leq (2^{\lambda^{<\kappa}})^M 1.\(\supseteq$: take $\alpha<\lambda$, we show that $j(\alpha)\in[id]$, i.e. $\{y\in\mathcal P_{\kappa}(\lambda):j_(\alpha)(y)=id(y)$

1.$\subseteq$: Notice that $[f]\in[id]\equiv\{y\in\mathcal P_{\kappa}(\lambda): f(y)\in y\}\in\mathcal U$. So we want to find $\alpha<\lambda$ such that $\{y\in\mathcal P_{\kappa}(\lambda):f(y)=\alpha\}\in\mathcal U$. Assume, aiming towards a contradiction, that this is not the case. Then for each $\alpha<\lambda$ the set $X_\alpha=\{y\in\mathcal P_{\kappa}(\lambda):f(y)\neq\alpha\}$ is an element of $\mathcal U$. Since $\mathcal U$ is normal, the diagonal intersection $X$ of these $X_\alpha$s is in $\mathcal U$. But then $y\in\mathcal P_{\kappa}(\lambda):f(y)\in y$ is not in $\mathcal U$ --- a contradiction.

2.Let $\alpha=cp(j)$ be the minimal ordinal such that $[f]=\alpha. Then \(\{y\in\mathcal P_{\kappa}(\lambda):f(y) By \(\kappa$-completness, if $\alpha$ were below $\kappa$, there would be a $\beta<\alpha$ such that $\{y\in\mathcal P_{\kappa}(\lambda):f(y)=\beta\}\in\mathcal U$ contradicting the fact that $[f]=\alpha>\beta$. So, $\kappa\leq cp(j)$. To show the second part note that $\lambda = otp(j^{\prime\prime}\lambda)=otp([id])$ (by elementarity and 1.). It will be sufficient to show that $otp([id]), i.e., by \Lós theorem, that $\{y:\mathcal P_{\kappa}(\lambda):otp(id(x))=otp(x) 3.This follows directly from the following claim: 27.9 If \(j^{\prime\prime}X\in M=rng(j)$ and $|Y|\leq|X|$ for some $Y\subseteq M$, then $Y\in M$. The following theorem was proved by Laver in Laver, R.:Making the supercompactness of κ indestructible under κ-directed closed forcing, Israel Journal of Mathematics, 29(1978). 27.10Laver Let $\kappa$ be a supercompact cardinal. Then there is a function (Laver diamond) $f:\kappa\to V_\kappa$ such that for each $\lambda\geq\kappa$ and $x\in H(\lambda^+)$ there is a supercompact measure $\mathcal U$ on $\kappa$ such that $j_{\mathcal U}(f)(\kappa)=x$. Let $<$ be a wellordering of $V_\kappa$. Define $f$ as follows: \[ f(\alpha) = \min_{<} \{x: (\exists \alpha\leq\gamma<\kappa,x\in H(\gamma^+))(\not\exists \mathcal U)(j_{\mathcal U}(f\upharpoonright\alpha)(\alpha)=x) \}\cup\{\emptyset\}$ This $f$ works. Aiming towards a contradiction, assume not and choose a witness $x$ and $\lambda\geq\kappa$. Since $\kappa$ is $\lambda$-supercompact. Let $\rho=2^{2^\lambda}$ and $j:V\to N$ with $\rho. Then $N\models (\exists \nu,x\in H(\nu^+))(\mbox{there is no suitable supercompact measure})$ (Here, I got distracted. Nevertheless, I assume, Šárka did a good job of finishing the proof.) 27.11Baumgartner, 1979 Assume \(\kappa$ is a supercompact cardinal. Then there is a generic extension $M$ with $\kappa = \omega_2^M$ and $M\models PFA$.

Let $f$ be the laver function given by the previous theorem. We force with a countable support iteration of length $\kappa$: $P_\kappa=\langle \dot{Q}_\alpha:\alpha<\kappa\rangle$ such that if $f(\alpha)$ is a $P_\alpha$-name for a proper forcing, then $\dot{Q}_\alpha=f(\alpha)$ and $\dot{Q}_\alpha$ is trivial otherwise. $P_\kappa$ is $\kappa$-cc of size $\kappa$. Let $\dot{Q}$ be a $P_\kappa$ name for the Cohen forcing. By the properties of the Laver function there is a supercompactness measure $\mathcal U$ and an embedding $j_{\mathcal U}$ such that $j(f)(\kappa)=\dot{Q}$.

27.12

$j(P)_\kappa= P_\kappa$.

It follows that $Q_\alpha=\dot{Q}$ for $\mathcal U$-many $\alpha$'s so $P_\kappa$ adds $\kappa$-many subsets of $\omega$. Similarly one can show that $\kappa$ is collapsed to $\omega_2$: since this collapse is proper a name for it is small, so we can use the laver function to "capture it".

Now let $G$ be a generic filter on $P_\kappa$ and let $\dot{Q}$ be a name for a proper forcing in the extension. Find $\lambda\geq\kappa$ such that $\dot{Q}\in H(\lambda^+)$ and find a supercompact measure $\mathcal U$ on $\mathcal P_\kappa(2^{2^\lambda}=\rho)$ such that $j_{\mathcal U}(f)(\kappa)=\dot{Q}$.

27.13

If $j:V\to M$ is an elementary embedding with ${}^{<\kappa}M\subseteq M$ and if $P\in M$ is a $\kappa$-cc forcing and $G$ is a $V$-generic filter on $P$ then ${}^{<\kappa}M[G]\subseteq M[G]$.

of claim

Let $s\in V[G]$ be a sequence of length $<\kappa$ of elements from $M[G]$. Let $\dot{s}\in V$ be a name for this sequence. Since the forcing is $\kappa$-cc, we can assume $\dot{s}$ is small so that $\dot{s}\in M$ (by supercompactness of $\kappa$).

We want to show that $\dot{Q}$ is (a name for) a proper forcing in $M[G]$. The forcing $Q$ is proper in $V[G]$ and this is witnessed by a club $C\subseteq[H(\eta)]^\omega$ of countable elementary submodels of $H(\eta)$ for some $\eta<2^{2^\kappa}$. As $C\subseteq M[G]$, by the previous claim, $C\in M[G]$ and it will remain a witness showing that $Q$ is proper also in $M[G]$.

We now continue by showing that, given a sequence $\langle D_\alpha:\alpha<\omega_1\rangle$ of dense subsets of $Q$ in $M[G]$ there is a filter in $M[G]$ meeting all of them.

27.14Silver's lemma

Let $j:V\to M$ be an elementary embedding, $P\in M$ a forcing and $G$ a $V$-generic filter on $P$. Suppose, moreover, that $H$ is $M$-generic over $j(P)$. If $j[G]\subseteq H$ then there is an elementary embedding $j^*:V[G]\to M[H]$.

of claim

See, e.g., Cummings article referenced above.

Now let $g$ be a $V[G]$ generic filter on $Q$. We know that $j_U(P) = P*\dot{Q}*\dot{R}$ and let $H$ be $V[G*g]$-generic on $R$. Then $G*g*H$ is $V$-generic on $j(P)$ and $j[G]\subseteq G*g*H$ so we can use the previous claim to get an elementary embedding $j^*: V[G]\to M[G*g*H]$.

Now consider $j^*(Q)$ and the sequence $\langle j^*(D_\alpha):\alpha<\omega_1\rangle$. If we can show that $j^*[g]\in M[G*g*H]$, then the filter generated by $j^*[g]$ hits each $j^*(D_\alpha)$, so $M[G*g*H]$ has a generic for these sets so, by elementarity, so does $V[G]$. So it remains to show:

27.15

$j^*[g]\in M[G*g*H]$

of claim

It is sufficient to show that $j^*\upharpoonright Q\in M[G*g*H]$. Since $\dot{Q}$ is small, we know that $j^\upharpoonright\dot{Q}\in M$ (since $j^\upharpoonright\dot{Q}\subseteq M$). Now $j\upharpoonright\dot{Q} = \Big\{\big\langle\langle\dot{q},p\rangle, \langle j(\dot{q}),j(p)\rangle\big\rangle: \mbox{$\dot{q}$ is a $P$-name},p\in P \Big\}$ Define $\tau = \Big\{\big\langle\langle\dot{q},j(\dot{q})\rangle,j(p)\big\rangle: \mbox{$\dot{q},$ is a $P$-name}, p\in j(P)\Big\}.$ It is clear that $\tau\in M$ and some computation shows that $\tau_{G*g*H}=j^*\upharpoonright Q$.

The above claim finishes the proof of the theorem.

# 285.11.2014

## 28.1Š. Stejskalová: Laver's diamond (contd.)

We continued and finished the proof of the consistency of PFA (see theorem 27.11).

# 2919.11.2014

## 29.1P. Simon: F. Rothberger's paper: On families of real functions with a denumerable base

The following result is from Rothberger, F: On families of real functions with a denumerable base, Annals of Mathematics, 45 (1944).

29.1

$X$ has (*) if for each $M\subseteq X$ of size $\aleph_1$ there is a countable $S\subseteq X$ such that each $x\in M$ is a limit of a convergent sequence from $S$.

29.2

The following are equivalent: 1. $\mathbb R^{\omega_1}$ has (); and 2. $2^{\omega_1}$ has () 3. $\mathbb N^{\omega_1}$ has (*)

($2\rightarrow 1$): Since $\mathbb R\simeq (0,1)$ we will work with the open unit interval. Let $M\subseteq\mathbb (0,1)^{\omega_1}$ be of size $\aleph_1$. For each $f\in M$ choose $g_f^i\in 2^{\omega_1}$ such that $f(\alpha) = \sum_{i<\omega} 2^{-i}g^i_f(\alpha).$ By 2. there is an $S\subseteq2^{\omega_1}$ countable and $S_{g_f^i}\in S^\omega$ such that $S_{g_f^i}\rightarrow g_f^i$. Let $\bar{S} = \bigg\{\sum_{i=1}^{|s|} 2^{-i}\cdot s(i): s\in S^{<\omega}\bigg\}.$ Moreover let $h_{f,k,n}=\sum_{i=1}^k S_{g_f^i}(n).$ Clearly $h_{f,k,n}\in \bar{S}$ and $f = \lim_{n\to\infty} h_{f,n,n}.$ The other direction is trivial and other implications are similar.

29.3

$2^{\omega_1}$ has (*).

Let $M=\{f_\alpha:\alpha<\omega_1\}\subseteq 2^{\omega_1}$. Fix $S=\omega$. We will find an independent system $\{ A_{\alpha,i}:\alpha<\omega_1,i<2\}$ and convergent sequences $\{S_\alpha:\alpha<\omega_1\}$ with $S_\alpha\rightarrow f_\alpha$. We do this by transfinite induction. Let $A_{0,0},A_{0,1}$ be an arbirary partition of $\omega$ into two infinite sets. Moreover let $S_0\subseteq\omega$ be an infinite subset of $A_{0,f_0(0)}$ with an infinite complement in $A_{0,f_0(0)}$. Suppose we have now constructed $\mathcal A_\alpha=\langle A_{\beta,i}:i<2,\beta<\alpha\rangle$ and $\mathcal S_\alpha=\{S_\beta:\beta<\alpha\}$ satisfying the following:

1. $\mathcal A_\alpha$ is an independent family;
2. $\mathcal S_\alpha$ is an almost disjoint family;
3. $S_\beta\subseteq^* \bigcap_{\gamma\in K} A_{\gamma,f_\beta(\gamma)}$ for each $\beta<\alpha$ and $K\in[\alpha]^{<\omega}$;
4. for each $K\in[\alpha]^{<\omega}$ and $i:K\to2$ there is an infinite $B\subseteq\bigcap_{\gamma\in K} A_{\gamma,i(\gamma)}$ which is almost disjoint with each $S_\beta\subseteq^*\bigcap_{\gamma\in K} A_{\gamma,i(\gamma)}$.

For each $K\in[\alpha]^{<\omega}$ and $i:K\to2$ choose infinite disjoint subsets $M_{K,i}, N_{K,i}\subseteq \bigcap_{\gamma\in K} A_{\gamma,i(\gamma)}$ almost disjoint from $\mathcal S_\alpha$. By an old result of Bernstein the family $\{M_{K,i}, N_{K,i}:K\in[\alpha]^{<\omega},i:K\to2\}$ has an almost disjoint refinement by infinite sets so we may as well assume that they are already almost disjoint. Let $\mathcal S_i = \{S_\beta:\beta<\alpha\ \&\ f_\beta(\alpha) = i\},\quad i<2.$ Now there is a partition of $\omega$ into two infinite sets $A_{\alpha,0}, A_{\alpha,1}$ such that each $S_\beta\in\mathcal S_i$ is almost contained in $A_{\alpha,i}$ and each $M_{K,i}\subseteq^* A_{\alpha,0}$ and $N_{K,i}\subseteq^* A_{\alpha,1}$. It remains to construct $S_\alpha$. The family $\{A_{\gamma,f_\alpha(\gamma)}:\gamma\leq\alpha\}$ is a countable centered system of infinite sets. Let $S_\alpha$ be an infinite pseudointersection of this family almost dijsoint from $\mathcal S_\beta$. This finishes the construction. At the end we let $g_n(\alpha)=i\leftrightarrow n\in A_{\alpha,i}$.

## 29.2D. Chodounsky: Todorčević's ordering & Y-cc.

29.4

The following is probably true, but it is not what was actually on the board. The definition of Todorčević's ordering on the board took finite unions of sequences instead of sets of sequences. And in that case proposition 29.9 is probably false.

29.5

Given a topological space $Y$ let $Seq(Y)$ be the family of convergent sequences in $X$. The Todorčević's ordering on $Y$, denoted by $T(Y)$, consists of finite subsets of $Seq(Y)$ ordered as follows: $p\leq_T q$ if $p\supseteq q$ and $(\bigcup p)^\prime\cap (\bigcup q)=(\bigcup q)^\prime$.

Recall definition 22.11.

29.6Chodounský,Zapletal

$T(Y)$ is ccc iff it is $Y$-cc.

29.7

Let $X$ be a set together with a symetric relation $\pi$ on $X$. We define the ordering $Q_\pi(X)$ to consist of finite subsets of $X$ ordered as follows: $p\leq_\pi q$ if $p\supseteq q$ and $p\setminus q\times q\subseteq\pi$.

29.8Yorioka

If $T(Y)$ is ccc then $T(Y)$ is $Y$-cc.

29.9

Let $X=Seq(Y)$ and define $\pi$ on $X$ by letting $(p,q)\in\pi$ if $p$ and $q$ are compatible in $T(Y)$. Then $Q_\pi(X)$ is forcing equivalent to $T(Y)$.

The natural embedding works and is even onto the separative quotient of $Q_\pi(X)$: given any element of the quotient represented, say, by $s$. Then write $\bigcup s$ as a finite union of disjoint convergent sequences. This will be the preimage of $[s]$.

# 3026.11.2014

## 30.1R. Bonnet

30.1

A Boolean space is a compact zero-dimensional space topological space.

30.2

A meet semilattice is a pair $\langle S,\wedge\rangle$ where $\wedge$ is a symmetric associative operation on $S$. If the universe $S$ is a topological space, we also assume $\wedge$ to be continuous.

30.3

Any linear order of a space $X$ with a minimal element induces a (meet) semilattice operation by considering $x\wedge y = inf\{x,y\}$.

30.4

The Cantor set $2^\omega$ is linearly ordered by the lexicographic order and thus gives rise to a meet semilattice.

30.5

$(X,\leq)$ is a Priestley space if it is a Boolean space with a partial order such that whenever $y\not\geq x$ then there is a clopen $\leq$-final subset of $X$ containing $x$ but not y.

30.6

It follows from the definition that the partial order must be a closed relation in $X$.

30.7

A Boolean space with equality is a Priestley space.

30.8

Let $(X,\leq)$ be a Priestley space. The hyperspace of $X$, denoted by $H(X)$, consists of all nonempty compact final subsets of $X$. Given a clopen $U \in H(X)$ let $U^+=\{K\in H(X):K\subseteq U\}$. The $\{U^+, H(X)\setminus U^+:U\ \mbox{clopen}\ \&\ U\in H(X)\}$ is a basis of a zero-dimensional topology on $H(X)$.

30.9

If the order is equality, then the topology of $H(X)$ coincides with the Vietoris topology. This is also the case if $H(X)$ is closed in the standard hyperspace. In general, however, it is not clear whether the Vietoris topology cannot be stronger.

30.10

A Boolean LOTS $X$ is a Priestley space and $H(X)$ is homeomorphic to $X$.

30.11

The space $H(X)$ is compact, Hausdorff.

30.12

There is a natural continuous semilattice operation on $H(X)$: given $A,B\in H(X)$, we let $A\wedge B=A\cup B$.

30.13

For each Priestley space $X$ and each continuous increasing function $f:X\to Y$ into some Boolean semilattice there is a unique continuous, $\wedge$-preserving "lifting" $F:H(X)\to Y$ such that $f(x) = F(i(x))$, where $i(x)=[x,\rightarrow)$ is the natural embedding of $X$ into $H(X)$.

30.14

Any continuous increasing map $\varphi$ between two Priestley spaces can be uniquely and naturally extended to a map $\hat{\varphi}$ between their hyperspaces.

30.15

A Priestley space $X$ is a Skula space if each closed final segment is clopen.

30.16

A Boolean algebra $B$ (or its Stone space $S(B)$) has a clopen selector if there is a family $\mathcal U=\{U_x:x\in S(B)\}\subseteq B$ such that

1. $x\in U_x$ for each $x\in S(B)$;
2. $x\in U_y\rightarrow U_x\subseteq U_y$ for each $x\in S(B)$; and
3. for each $x\neq y$ either $x\not\in U_y$ or $y\not\in U_x$
30.17

The one-point compactification $\alpha D$ of a discrete space has a clopen selector: $U_d=\{d\}$ and $U_\alpha= D$.

30.18

If $S(B)$ has a clopen selector then it is scattered.

30.19

There is a scattered space (even one with Cantor-Bendixon rank 3) which does not have a clopen selector (or, equivalently, which is not a Skula space).

Fix $\mathcal A=\{A_\alpha:\alpha<\lambda\}$ an almost disjoint family on $\omega$ and $\{f_\alpha:\alpha<\lambda\}$ a dominating family of functions from $\omega$ to $\omega$. Consider $X = \omega\times(\omega+1)\cup\mathcal A \cup \{\infty\}$ with each $\{n\}\times\omega$ converging to $(n,\omega)$. Moreover, each $(n,m)\in\omega\times\omega$ is clopen. Then $g_\alpha= A_\alpha\times(\omega+1)\setminus graph(f_\alpha) \cup \{A_\alpha\}$ is a neighbourhood of $A_\alpha$.

Assume that it has a clopen selector. We may assume that each member of the selector does not contain $\infty$. Applying the Delta system lemma we continue to work a bit and arrive at a contradiction.

30.20

$S(B)$ has a clopen selector iff $B$ is generated (as a lattice) by a well-founded set iff it is a Skula space.

30.21

If $X$ is Skula then so is $H(X)$.

30.22

$cb(X)\leq wf(H(X))\leq\omega^{cb(X)}$.

30.23

Can we find an almost disjoint family $\mathcal A$ of size $\aleph_1$ such that the Mrówka space $\psi(\mathcal A)$ does not have a continuous semilattice operation?

# 313.12.2014

## 31.1J. Grebík: Questions around Suslin trees

31.1

Given a tree $T$ and a coloring $\chi:T\to2$ we say that a level $\alpha$ returns to $\beta<\alpha$ if there is a color $c\in 2$ and an increasing seqence $\langle\beta_n:n<\omega\rangle$ cofinal in $\alpha$ such that for each $t\in T_\beta$ the set $\{n:\chi(t\upharpoonright\beta_n)\neq c\}$ is finite.

31.2

If for each coloring $\chi:T\to2$ there is a level $\beta$ and an increasing sequence $\langle \alpha_\delta:\delta<\omega_1$ of levels returning to $\beta$ then $RO(T)$ with the sequential topology is compact.

# 327.1.2015

## 32.1Michael Hrušák: Weak Dimond

32.1Jensen

A $\Diamond$-sequence on $\omega_1$ is a sequence $\langle A_\alpha:\alpha<\omega_1\rangle$ of subsets of $\omega_1$ such that $A_\alpha\subseteq\alpha$ and for each $X\subseteq\omega_1$ the set $\{\alpha<\omega_1:X\cap\alpha=A_\alpha\}$ is stationary. The diamond principle is the statement that there is a $\Diamond$-sequence.

32.2Devlin-Shelah

The weak diamond is the statement that for each coloring $F:2^{<\omega_1}\to2$ there is a diamond sequence $g:\omega_1\to 2$ such that for each branch $f\in 2^{\omega_1}$ the set $\{\alpha<\omega_1:g(\alpha)=F(f\upharpoonright\alpha)\}$ is stationary.

32.3

Weak diamond is equivalent to $2^\omega<2^{<\omega_1}$.

32.4

Given sets $A,B$ and a relation $E\subseteq A\times B$ such that $E\subseteq A\times B$ satisfies $A\subseteq E^{-1}[B]$ and $A\not\subseteq E^{-1}(b)$ for each $b\in B$, define $\langle A,B,E\rangle = \min\{|B^\prime|:B^\prime\subseteq B\ \&\ (\forall a\in A) (\exists b\in B^\prime)aEb\}$

The following definition comes from Džamonja, M., Hrušák, M. and Moore, J.: Parametrized $\Diamond$ principles, Transactions of AMS, 356.6 (2004).

32.5

A cardinal invariant is Borel provided it can be written in the form $\langle A,B,E\rangle$ for some Borel $A,B,E$.

32.6

The parametrized diamond $\Phi(A,B,E)$ is the statement that for each coloring $F:2^{<\omega_1}\to A$ there is a guessing sequence $g:\omega_1\to B$ such that for each branch $f\in 2^{\omega_1}$ the set $\{\alpha<\omega_1: (F(f\upharpoonright\alpha),g(\alpha))\in E\}$ is stationary. The parametrized weak diamond $\Diamond(A,B,E)$ is the same statement where we restrict it to "definable colorings".

32.7

Typically, a coloring is definable means that each of its restrictions to each $2^\alpha$ is Borel.

32.8

The strongest weak diamond (denoted $\Phi(\omega_1)$ )is the principle $\Phi(\omega_1,\omega_1,=)$.

32.9

The (standard) $\Diamond$ principle is equivalent to CH plus the strongest weak diamond.

32.10

The strongest weak diamond and $\langle A,B,E\rangle\leq\omega_1$ implies $\Phi(A,B,E)$.

32.11

The strongest weak diamond does not imply CH.

Start with a model of GCH, add $\aleph_2$-many Cohen reals followed by $Fn(\omega_3,2,\omega_1)$.

Unfortunately, the strongest weak diamond implies $2^{\omega}<2^{\omega_1}$.

32.12

The strongest weak diamond + $non(\mathcal M)=\aleph_1$ implies the existence of a Suslin tree.

32.13

The diamond principle $\Diamond(\omega_1,\omega_1,=)$, abbreviated as $\Diamond(\omega_1)$, is $\Phi(\omega_1,\omega_1,=)$ restricted to colorings from $\mathbb L(\mathcal P(\omega_1))$

Most of the things true for $\Phi(\omega_1)$ are also true even for $\Diamond(\omega_1)$).

32.14

$\Diamond(\omega_1)$ + $non(\mathcal M)=\aleph_1$ implies the existence of a Suslin tree.

32.15

$\Diamond(\omega_1)$ plus CH is equivalent to $\Diamond$.

32.16

If a forcing $P$ is "sufficiently definable", proper of size $\leq\mathfrak c$ then the countable support iteration $P_{\omega_1}$ forces $\Diamond(\omega_1)$.

### 32.1.1Generalizations ?

32.17

$\Diamond(\omega_2)$ is the statement that for each coloring $F:2^{<\omega_2}\to\omega_2$ in $\mathbb L(\mathcal P(\omega_2))$ there is a guessing sequence $g:\omega_2\to\omega_2$ such that for each branch $f:\omega_2\to 2$ such that the set $\{\alpha:F(f\upharpoonright\alpha)=g(\alpha)\}$ is stationary.

32.18Hrušák

$\Diamond(\omega_2)$ is a theorem of ZFC.

Recall Baumgartner's result that, under PFA, all $\aleph_1$-dense sets of reals are order isomorphic. This motivates the question:

32.19

Is it consistent that all $\aleph_2$-dense sets of reals are order isomorphic.

32.20

Devlin-Shelah's weak diamond restricted to colorings from $\mathbb L(\mathcal P(\omega_1))$ implies that there are two non-isomorphic $\aleph_1$-dense subsets of $\mathbb R$.

of Theorem

Fix some set $X=\{x_\alpha:\alpha<\omega_1\}\subseteq\mathbb R$ and, without loss of generality, assume that $\mathbb Q\subseteq X$. By suitable coding (using the first $\omega$ levels of $2^{<\omega_1}$), we may assume that each $s\in 2^{<\omega_1}$ codes an order embedding $h_s$ of $\mathbb Q$ into $\mathbb R$ and some $x^s_\alpha$. Each $h_s$ can be uniquely extended to an order isomorphism $\overline{h}_s:\mathbb R\hookrightarrow \mathbb R$. Define $F(s) = 0 \iff \overline{h}_s(x^s_\alpha)\in X.$ Clearly $F$ is in $\mathbb L(X)$. Now, using the weak diamond, let $Y=\{\alpha:g(\alpha)=1\}$. Then $Y$ clearly cannot be order isomorphic to $X$. $Y$ need not be $\aleph_1$ dense, but this can be mended in the obvious way by, e.g., using odd levels to guarantee this and even levels as we defined $F$ above.

32.21

The proof works also for $\aleph_2$. In particular, $\Diamond(\omega_1)$ implies this. This means that if $\Diamond(\omega_1)$ is a theorem of ZFC then Baumgartner's result could not be extended to $\aleph_2$.

### 32.1.2Applications in topology

Recall that

32.22

A topological space $X$ is sequential if for each $A\subseteq X$ which is not closed there is a converging sequence $\langle x_n:n<\omega\rangle$ in $A$ whose limit is not in $A$. The order $ord(X)$ of a sequential space is the smallest ordinal $\alpha$ such that iterating sequential closure $\alpha$-many times always gives the full closure.

32.23Bashkirov,70s

CH implies that there is a compact sequential space of order $\omega_1$.

32.24

There is a compact sequential space of order 2 (e.g. a one-point compactification of a $\Psi$-space). If we assume $\mathfrak b = \mathfrak c$, Dow showed that we can get one of order 4. Under MA, such a space of order 5 exists.

There is also Balogh's solution to the Moore-Mrówka problem:

32.25Balogh,80s

Under PFA every compact countably tight space is sequential.

32.26

A space $X$ is super-sequential if it is sequential, scattered and the CB-rank coincides with the sequential order.

32.27Dow,Shelah

Under PFA each compact super sequential space has sequential order at most $\omega$.

32.28Hrušák

$\Diamond(\omega^\omega,\omega^\omega,\not\geq^*)$ implies that there is a supersequential compact space $X$ with order $\omega_1$.

32.29

If a space $X$ is locally countable than any order 2 sequential space contains a $\Psi$-space.

32.30Hrušák

$\Diamond(\omega^\omega,\omega^\omega,\not\geq^*)$ (also abbreviated $\Diamond(\mathfrak b)$) implies that there is a supersequential compact space $X$ with order 3.

We need a MAD $\mathcal A$ family on $\omega$ and a family $\mathcal B$ of infinite subsets of $\omega$ such that

1. for each $B\in \mathcal B, A\in\mathcal A$ either $A\subseteq^* B$ or $A\cap B=^*\emptyset$;
2. $\mathcal B$ is $\mathcal I(\mathcal A)$-almost disjoint.
3. for each sequence $\langle A_n:n<\omega\rangle$ there is a $B\in\mathcal B$ almost containing infinitely many $A_n$'s

Once we have them, we take the space $\Psi(\mathcal A)$ add $\mathcal B$ on top with the natural topology and do a one-point compactification (formally, take the Stone space of the subalgebra of $\mathcal P(\omega)$ generated by $\mathcal A\cup\mathcal B\cup\omega$). This space will have sequential order 3. To use the diamond we use coding to associate with each $s\in 2^{<\omega_1}$ a triple $(X_s,\langle B^s_\beta:\beta<\alpha_s\rangle,\langle A^s_\beta:\beta<\alpha_s\rangle)$. Moreover, fix a borel function $\varphi$ which transforms this triple so that the $B$s form a partition of $\omega$ and $A$s are partitions of the $B$s. If $X_s\subseteq\omega$, then $F(s)$ is the function $d_s=\max A^s_n\cap X_s$ for $A^s_n$s partitioning either all $B^s$s or the least $B^s$ which hits $X_s$ infinitely often (this will guarantee that $\mathcal A$ and $\mathcal B$ are MAD). If $X_s$ is a sequence of infinite sets, we similarly guarantee 3. Then, recursively construct $\mathcal B,\mathcal A$ at each step using the guessing function $g$ to go to the next step.

32.31

A similar proof shows that $\Diamond(\mathfrak s)$ gives a counterexample to the Scarborough-Stone problem.

# 3314.1.2015

## 33.1E. Thuemmel

### 33.1.1Forcing with homogeneous sets

33.1Chodounský

Given a coloring $\pi:[A]^\to 2$ we define the set $P_\pi = \{ F\in[A]^{<\omega}:F\ \mbox{is 0-homogeneous}\}$ and order it by reverse extension.

33.2

If a subset of $P_\pi$ is linked then it is already centered.

33.3Chodounský

If $P$ is a forcing whose linked subsets are centered then there is a set $A$ and a coloring $\pi:[A]^2\to2$ such that $P$ is forcing equivalent to $P_\pi$.

Let $A=P$ and define $\pi(p,q) =0\iff p||q$. Then $\varphi(p)=\{p\}$ is an embedding of $P$ into $P_\pi$ onto a dense subset (of the separative quotient).

### 33.1.2Hrušák's question

In the following all ideals are assumed to be tall ideals on $\omega$.

33.4

Recall the following arrow notation:

1. $X\rightarrow (Y)^n_m$ is the statement that for each set $A\in X$ and a coloring $\chi:[A]^n\to m$ there is a homogeneous $B\subseteq A$ in $Y$.

2. $X\rightarrow (Z,Y)^n$ is the statement that for each set $A\in X$ and a coloring $\chi:[A]^n\to 2$ there is either a set $B\subseteq A$ in $Y$ which is homogeneous in color $0$ or there is a set $B\subseteq A$ in $Z$ which is homogeneous in color $1$. $X\rightarrow (\omega,Y)^n$ is a shorthand for $X\rightarrow (\{B:|B|=\omega\},Y)^n$.

33.5Hrušák 6.6

Is there a (Borel) ideal $\mathcal I$ such that $\mathcal I^+\not\rightarrow (\mathcal I^+)^2_2,$ while $\mathcal I^+\rightarrow (\omega,\mathcal I^+)^2.$

33.6E. Thuemmel

The ideal $nwd(\mathbb Q)$ answers the previous question.

Enumerate $\mathbb Q=\{q_n:n<\omega$). To show that the first arrow does not hold, consider the coloring $\chi(q_i,q_j)=0\iff q_i where \(i. A homogeneous set is a monotone sequence, hence nowhere dense. To show the second arrow, consider a set \(A\in nwd^+$ and a coloring $\chi:[A]^2\to2$. Fix $\{U_i:i<\omega\}$ a topological basis of $\mathbb Q$. Recursively try to pick $\{(q_i,A_i):i<\omega\}$ so that $q_{i+1}\in A_i\setminus A_{i+1}$, $A_{i+1}\subseteq A_i$ and $\chi(q_i,A_i)=0$. If we are successful, then the $q_i$'s form an infinite $0$-homogeneous set. Otherwise we stop at some $i_0<\omega$. Then for each $q\in A_i$ the set $\{q^\prime\in A_{i+1}:\chi(q,q^\prime)=0\}$ is in $nwd$. We now pick $p_i, B_i\in nwd^+$ such that $p_{i+1}\in (B_i\cap U_i)\setminus B_{i+1}$, $B_{i+1}\subseteq B_i$, $A_{i_0}\setminus B_i\in nwd$ and $\chi(p_i,B_i)=1$. This process cannot stop and in the end gives a $nwd$-positive 1-homogeneous set.

33.7

It follows from the proof that $\mathcal I^+\not\rightarrow (\mathcal I^+)^2_2$ for every ideal $\mathcal I$ which is not $(\omega,2)$-distributive, i.e. for which $\mathcal P(\omega)/\mathcal I$ is an $(\omega,2)$-distributive algebra. The idea is to take a witnessing matrix $\{(A_n,\omega\setminus A_n):n<\omega\}$ of $\mathcal I$-positive sets with no refinement. The bad coloring is $\chi(i,j)=0$, $i, iff either \((\forall n<\omega)(i\in A_n\leftrightarrow j\in A_n)$ or $(\exists n)((\forall m 33.8E.Thuemmel If \(\mathcal I$ is $(\omega,2)$-distributive then $\mathcal I^+\not\rightarrow (\mathcal I^+)^2_2$ iff $\mathcal I^+\not\rightarrow (\omega,\mathcal I^+)^2$.

One implication is clear. We show that if $\mathcal I^+\not\rightarrow \mathcal (I^+)^2_2$ then $\mathcal I^+\not\rightarrow (\omega,\mathcal I^+)^2$. So fix $A\in\mathcal I^+$ and a coloring $\chi$ witnessing the failure of the stronger arrow. Let $A_i=\{j\in A(>i):\chi(i,j)=0\}$. Consider the partitions $\{(A_i,A\setminus A_i):i\in A\}$. By distributivity there is a positive $B$ such that, wlog, $B\setminus A_i\in\mathcal I$ for each $i\in B$. The proof now splits according to two cases.

Case 1. There is a positive $C\subseteq B$ such that $|C\setminus A_i|<\omega$ for each $i<\omega$. Then $1-\chi\upharpoonright C$ witnesses the failure of the weaker arrow since a $1$-homogeneous set is in $\mathcal I$ and a $0$-homogeneous set $D$ must have $D(>i)\subseteq A\setminus A_i$ for each $i$ so $|D(>i)|<\omega$.

Case 2. Each subset of $B$ which is a pseudointersection of the $A_i$'s is in $\mathcal I$. Disjointify the $A_i$'s, i.e. let $R_i=((A\setminus (A_i\cup \bigcup_{j. Then \(R_i\in\mathcal I$, as $B\setminus A_i\in\mathcal I$. Let $D=\bigcup_{i<\omega} R_i$. Then $B\setminus D = B\setminus(\bigcup_{i<\omega} A\setminus A_i) = B\cap\bigcap_{i<\omega}A_i\in\mathcal I$ the last $\in$ following from the fact that $\bigcap_{i<\omega} A_i$ is
$0$-homogeneous. It follows that $D\in\mathcal I^+$. The rest of the proof is based on the proof that $Fin\times Fin^+\not\rightarrow(\omega,Fin\times Fin)^2$ as witnessed by the coloring $\chi((u,v),(x,y))=0$ iff the line connecting the points $(u,v)$ and $(x,y)$ is strictly decreasing. A set homogeneous in color $1$ is dominated by some function (or contained in a vertical column), a set homogeneous in $0$ is finite.

Consider the $R_i$'s as vertial lines (in $\omega\times\omega\simeq D$).

33.9

The ideal $conv(\mathbb Q)$ is not $(\omega,2)$-distributive however, considering the positive set $\{1/n+1/k:n,k<\omega\}$ one again shows, in the same way as with $Fin\times Fin$, that $conv^+\not\rightarrow(\omega,conv^+)^2$. Similarly for any $F_\sigma$-ideal.

Recall that

33.10

An ideal $\mathcal I$ is semiselective if it is $\omega$-distributive and $\mathcal I^+\rightarrow(\mathcal I^+)^2_2$.

33.11

No tall semiselective ideal can be analytic.

33.12Farah

There is an ideal $\mathcal I$ which is not $\omega$-distributive and $\mathcal I^+\rightarrow(\mathcal I^+)^2_2$. Therefore, by the previous results, it must be $(\omega,2)$-distributive.

We construct a labeled tree $\{A_s:s\in {}^{<\omega}\mathbb R\}$ with $A_r\in[\omega]^\omega$ such that $\{A_{s^{\smallfrown} x}:x\in\mathbb R\}$ is a MAD system on $A_s$ and such that for each $\chi:[A_s]^2\to2$ there is $x\in\mathbb R$ such that $A_{s^{\smallfrown} x}$ is homogeneous. The co-ideal will consist of sets containing some $A_s$. It is clear that it satisfies the required arrow. It is a co-ideal since if some $A_s$ is partitioned into $B_0,B_1$ then we color pairs in $[A_s]^2$ by $0$ if they are contained in the same part of the partition and by $1$ otherwise. A $1$-homogeneous set has size at most $2$ a $0$-homogeneous set is contained in one member of the partition. It is clearly not $\omega$-distributive since the labeled tree is a name for a collapse of $\mathfrak c$ to $\omega$.

# 3428.1.2015

## 34.1E. Thuemmel: Ideals on ω

Recall the definition 33.4 due to M. Hrušák.

34.1Hrušák

Is there an ideal $\mathcal I$ such that $\mathcal I^+\not\rightarrow(\omega,\mathcal I^+)^2$ while $\mathcal I^+\not\rightarrow(\mathcal I^+)^2_2$.

Last week we have shown (33.6) that, in particular,

34.2E. Thuemmel

The ideal $\mathbf{nwd}$ of nowhere dense subsets of $\mathbb Q$ is such an idal.

In fact, we have shown even more, see 33.8. Today we aim to show that the ideal $\overline{ctble}$ of subsets of $\mathbb Q$ with countable closure (in $\mathbb R$).

34.3W. Kubiś

The ideal $\overline{ctble}$ consists precisely of the scattered linear suborders of $\mathbb Q$.

Also recall

34.4

An ideal $\mathcal I$ is selective if it is $\omega$-closed (i.e. the quotient algebra is $\sigma$-closed) and $\mathcal I^+\rightarrow(\mathcal I^+)^2_2$.

and

34.5

If $\mathcal I$ is $\omega$-closed then $\emptyset\times\mathcal I$ is $\omega$-distributive and $\mathcal I^+\not\rightarrow(\mathcal I^+)^2_2$. {} Proof: Distributivity is clear, since $\emptyset\times\mathcal I$ is locally isomorphic to $\mathcal I$. It is not $\omega$-closed (consider the sets $(\omega\setminus n)\times\omega$).

We will also show the following example.

34.6

The ideal $\mathbf{nwd}\times Fin\times Fin$ is not $(\omega,2)$-distributive and does not satisfy the weaker arrow.

34.7

The ideal $\overline{ctble}$ is not $(\omega,2)$-distributive.

The sets $D_n = \bigcup_{i<2^{n-1}}\left[\frac{2i}{2^n},\frac{2i+1}{2^n}\right)$ witness nondistributivity.

34.8

$\overline{ctble}^+\rightarrow(\omega,\overline{ctble}^+)^2$.

Let $\chi$ be a coloring of the pairs of a positive set $A$. We first try to find an infinite set homogeneous in color $0$. Recursively choose $a_n,A_n$ such that

1. $\chi(a_n,A_n)=0$
2. $A_n$ is positive
3. $a_{n+1}\in A_n$, $A_{n+1}\subseteq A_n$


If we can choose ininitely many $a_n$s then they form an infinite set homogeneous in color $0$. Otherwise there is an $A$ such that for each $a\in A$ the set $\{b\in A:\chi(a,b)=0\}$ is small. Again consider the sets $D_n = \bigcup_{i<2^{n-1}}\left[\frac{2i}{2^n},\frac{2i+1}{2^n}\right)$ and for $s\in{}^{<\omega}\{-1,1\}$ let $D_s = A\cap \bigcap_{i<|s|} s(i)D_i,$ where $-1D_i=[0,1]\setminus D_i$ and $1D_i=D_i$. Let $S=\{s\in{}^{<\omega}\{-1,1\}:D_s\in\overline{ctble}^+\}$. Then $S$ is a tree without finite branches.

34.9

In fact, $S$ is a perfect tree.

of Claim

Assume to the contrary that for some branch $f\in[S]$ there is $s\in S$ such that $s\subseteq f$ and $S$ does not split above $s$. Then, by the choice of $D_s$ and the fact that $S$ does not split above $s$ we have that $D_s\setminus D_{f\upharpoonright n}\in\mathcal I$ for each $n\geq|s|$. Since the $D_n$s and hence, a fortiori, the $D_t$s were witnessing non-distributivity we have that $D_s$ is small --- a contradiction.

Enumerate the tree $S$ as $\{s_n:n<\omega\}$. Recursively choose $a^n\in D_{s_n}$ such that $\chi(a^i,a^n)=1$ for each $i. This can be done since \(D_s\subseteq A$ and the set $\{b\in A:\chi(a,b)=0\}$ is small for each $a\in A$. Then $\{a^n:n<\omega\}$ is homogeneous in color $1$ and positive, since $S$ has perfect closure.

Recall the definition of the Katětov order for ideals:

34.10Katětov

We say that an ideal $\mathcal I$ is Katětov below $\mathcal J$, written as $\mathcal I\leq_K\mathcal J$, if there is a function $f:\omega\to\omega$ such that the preimages of $\mathcal I$-small sets are $\mathcal J$-small.

M. Hrušák has a program of finding "Katětov critical" ideals, i.e. given a property $P$ he wants to find an ideal $\mathcal I_P$ such that $\mathcal J$ has $P$ iff it is Katětov above $\mathcal I_P$. Since $(\omega,2)$-distributivity is not upwards closed in the Katětov order, we cannot hope for a critical ideal. However, we have:

34.11

If an ideal $\mathcal I$ is nowhere $(\omega,2)$-distributive then it is Katětov above $\overline{ctble}$.

34.12

If $\omega\not\rightarrow(\omega,\mathcal J^+)^2$ and $\mathcal I$ is nowhere $(\omega,2)$-distributive then $\omega\not\rightarrow(\omega,(\mathcal I\times\mathcal J)^+)^2$, and the ideal $\mathcal I\times\mathcal J$ is nowhere $(\omega,2)$-distributive.

Let $\chi:[\omega]^2\to 2$ witness the failure of the arrow for $\mathcal J$ and $\{A_n:n<\omega\}$ witness the non-distributivity of $\mathcal I$. We claim that the "lift" of $\chi$, i.e. $\chi^\prime:[\omega\times\omega]^2\to 2$ which depends on the second coordinates and is 1 if they are equal, witnesses the failure of the arrow for the product ideal.

Further, the sets $\{A_n\times\omega:n<\omega\}$ witness the non-distributivity.

We now turn to descriptive complexity. Recall that Farah, motivated by the Galvin-Prikry theorem, proved

34.13Farah

If a family of $\mathcal M$ subsets of $\omega$ is analytic and $\mathcal I$ is semiselective then there is an $\mathcal I$ positive set $A$ such that either $[A]^\omega\subseteq \mathcal M$ or $[A]^\omega\cap \mathcal M=\emptyset$.

For the definition of semiselective see 33.10.

34.14

If a semiselective ideal is tall then it is not analytic.

34.15

If $\mathcal I$ is tall, analytic and $\omega$-distributive then $\mathcal I^+\not\rightarrow(\mathcal I^+)^2_2$

34.16

Given an ideal $\mathcal I$ we say that a sequence $\langle A_n:n\in A$ is an $\mathcal I$-tower if $A\in\mathcal I^+$ and for each $n\in A$ we have $A\setminus A_n\in\mathcal I$ and, optionally, the sets $A_n$ are decreasing. A set $D$ is a diagonal of this tower if for each $n\in D$ the set $D\setminus (n+1)\subseteq A_n$.

34.17

Each tall, $\omega$-distributive analytic ideal $\mathcal I$ contains all diagonals of some $\mathcal I$-tower.

We know that $\mathcal I^+\not\rightarrow(\mathcal I^+)^2_2$ so there is a positive set $A$ and a coloring $\chi$ of the pairs of $A$ with two colors which has no positive homogeneous subset. Define $B_n^i=\{m\in A:b\setminus(n+1):\chi(n,m) = i\}.$ This is a (mod-Fin) partition of $A$. By distributivity, there is an $f:\omega\to 2$ and a positive $B$ such that $B\subseteq B_n^{f(n)}$. Since $\mathcal I$ is an ideal, we can wlog assume that $f$ is constant, e.g. 0. We let $B_n = B\cap\bigcap_{i. Then the \(B_n$s form an $\mathcal I$-tower. All diagonals of this tower are small.

We can now generalize Farah's theorem:

34.18

If a family of $\mathcal M$ subsets of $\omega$ is analytic and $\mathcal I$ is $\omega$-distributive then there is an $\mathcal I$-tower such that either all diagonals are in $\mathcal M$ or all diagonals are not in $\mathcal M$.

Thus we have the following picture:

# 3511.2.2015

## 35.1O. Guzmán: Generalized diamonds

35.1

The notation $\Diamond^\mu_\kappa(2,=)$ abbreviates the following statement: For each coloring $F:2^{<\kappa}\to 2$ of the binary tree of height $\kappa$ by two colors such that for each $\alpha<\kappa$ the restriction $F\upharpoonright\alpha\in \mathbb L(OR^\mu)$ there is a guessing function $g:\kappa\to2$ such that for each $R\in2^\kappa$ the set $\{\alpha:F(R\upharpoonright\alpha)=g(\alpha)\}$ is stationary.

35.2

$\Diamond_{\omega_1}^\omega$ implies that $\mathfrak p=\omega_1$, there are no Q-sets, and that the Borel version of weak diamond holds.

It is clear that $\Diamond_{\omega_1}^{\omega_1}\rightarrow\Diamond_{\omega_1}^\omega$. We aim to show that the implication cannot be reversed.

35.3

Let $T$ be an everywhere branching Suslin tree. Then $T$ forces $\Diamond_{\omega_1}^\omega$.

Let $\dot{F}$ be a name for a suitably definable coloring. Then, by a result of Chang, choose for each $\alpha<\omega_1$ a name $\dot{S}_\alpha$ for a countable set such that $T\Vdash \dot{F}\upharpoonright\alpha\ \mbox{is coded by}\ \dot{S}_\alpha$. Since $T$ is Suslin, for every $\alpha$ there is $\gamma_\alpha>\alpha$ such that each $t\in T_{\gamma_\alpha}$ decides $\dot{S}_\alpha$. For each such $t\in T_{\gamma_\alpha}$ let $\{A_t^i:i<2\}$ be a partition of $succ_T(t)$ into two (nonempty) sets. Let $R\subseteq T$ be a generic branch. In $V[R]$ we define the guessing function $g$ as follows: $g(\alpha)= i\iff R\upharpoonright{\gamma_\alpha+1}\in A^i_{R\upharpoonright\gamma_\alpha}.$ We will show that $g$ guesses $F$. Let $p\in T$ be a condition,$\dot{R}$ a name for the generic and $\dot{C}$ a name for a club. Let $M$ be a countable elementary submodel of some sufficiently large $H(\theta)$ containing $p,\dot{R},\dot{C},T$ etc. Let $\delta=M\cap\omega_1$. Pick $q\in T_\delta$ extending $p$ such that $q$ is an $M$-generic branch for $T$. Note that $q$ decides $\dot{R}\upharpoonright\delta$. Also $q\Vdash \delta\in\dot{C}$ (since $M[R]$ is a generic extension of $M$ so its height is $\delta$ and since if a club is in a model, the height of the model is in the club). Let $q^\prime\in T_{\gamma_\delta}$ extend $q$. Now there is an $i<2$ such that $q^\prime\Vdash\dot{F}(\dot{R}\upharpoonright\delta)=i$. Let $q^{\prime\prime}$ extend $q^\prime$ such that $q^{\prime\prime}\in A^i_{q^\prime}$. Then $q^{\prime\prime}$ forces that $\delta\in\dot{C}\ \&\ \dot{F}(\dot{R}\upharpoonright\delta)=\dot{g}(\delta)$.

35.4

Given two sets $a,b$ we say that $a\preceq b$ if there is an onto function $f:b\to a$.

35.5

$\neg \Diamond^{\kappa^+}_{\kappa^+}(2,=)\iff \mathbb L(OR^{\kappa^+})\models 2^{\kappa^+}\preceq 2^\kappa$

35.6

Devlin and Shelah proved that $\neg \Phi_{\kappa^+}(2,=)\iff 2^{\kappa^+}\preceq 2^\kappa$, where $\Phi$ is the same as $\Diamond$ with the definability assumption dropped. (See Devlin, K. Shelah, S.: A weak version of $\diamondsuit$ which follows from $2^{\aleph _{0}}<2^{\aleph _{1}}$, Israel J. Math 29 (1978) 239-247)

35.7

If $V$ fails $\Diamond_{\omega_1}^{\omega_1}$ then it also fails after forcing with a Suslin tree.

Let $N=\{\dot{A}:\dot{A}\ \mbox{is a (nice) name for a subset of}\ \omega_1\}$ It is easy to see that $\mathbb L(OR^{\omega_1})\models|N|=2^{\omega_1}$. So, by 35.5 there is, in $\mathbb L(OR^{\omega_1})$, a surjection $H:2^\omega\to N$. This lifts to a surjection in the extension from $2^\omega$ onto $2^{\omega_1}$.

35.8

$\Diamond_{\omega_1}^\omega(2,=)\not\rightarrow\Diamond_{\omega_1}^{\omega_1}$.

The following proposition disproves the conjecture 32.18 of M. Hrušák (see also definition 32.17).

35.9

It is consistent that $\Diamond_{\omega_2}^{\omega_2}(2,=)$ fails.

35.10

Let $\mathcal A$ be an AD family. We say that $X$ separates $\mathcal B\subseteq\mathcal A$ if $B\subseteq^* X$ for each $B\in \mathcal B$ and $A\cap X=^*\emptyset$ for each $A\in\mathcal A\setminus\mathcal B$. An AD family is normal if each of its subfamilies are separated.

First notice that if there is a normal AD family of size $\omega_2$ then $\Diamond_{\omega_2}^{\omega_2}(2,=)$ fails since $F:[\omega]^\omega\to\mathcal P(\mathcal A)$ defined: $F(X)=\{B\in\mathcal A:B\subseteq^* X\} \in\mathbb L(OR^{\omega_2}).$ is a surjection, since $\mathcal A$ is normal, from $2^\omega$ onto $2^{\omega_2}$. We now show that $\mathfrak p>\omega_2$ implies there is a normal AD family of size $\omega_2$. For an AD family $\mathcal A$ let $P(\mathcal A,\mathcal B)$ be the natural forcing adding a separator for $\mathcal B$ with finite conditions. Note that, e.g., for a Luzin gap this forcing will always collapse. However, if $\mathcal A$ is an AD family consisting of branches of a tree then the forcing will be $\sigma$-centered and we can use $MA_{<\mathfrak p}(\sigma\mbox{-centered})$.

35.11David

Can the above be reversed, i.e. does $\mathfrak p\leq\omega_2$ imply $\Diamond_{\omega_2}^{\omega_2}(2,=)$.

35.12

The above is not clear, since pseudointersections of $<\mathfrak p$-sized families exist but need not be "definable".

## 35.2J. Cancino-Manríquez: Ideal independent families

(joint work with O. Guzman and A. Miller)

35.13

A family $\mathcal I\subseteq[\omega]^\omega$ is ideal independent if no element $A\in\mathcal I$ is almost covered by a finite set $\mathcal F\in[\mathcal I]^{<\omega}$ of elements of $\mathcal I$ not containing $A$.

35.14

If $\mathcal I$ is ideal independent it generates a proper ideal.

35.15

Any AD family as well as an independent family is ideal independent.

35.16

If $\mathcal I$ is an infinite ideal independent family, the the ideal generated by $\mathcal I$ is not maximal and is not a P-ideal.

35.17D. Monk

The spread generating number ($\mathfrak s_{mm}$) is defined to be the minimal cardinality of a maximal ideal independent family of infinite subsets of $\omega$.

35.18D. Monk

Does $\mathfrak s_{mm}=\mathfrak u$?

35.19

$\mathfrak r\leq\mathfrak s_{mm}$.

Let $\mathcal I$ witness $\mathfrak s_{mm}$. Then $\mathcal R = \left\{A\setminus\bigcup F: A\in\mathcal I\ \&\ F\in[\mathcal I]^{<\omega}\ \&\ A\not\in F\right\}$ is a reaping family.

35.20

$\mathfrak d\leq \mathfrak s_{mm}$.

Assume to the contrary that $\mathfrak s_{mm}<\mathfrak d$ and let $\mathcal I$ witness this. Note that $\bigcup\mathcal I=^*\omega$ (otherwise the infinite part not covered could be added to $\mathcal I$ contradicting maximality). WLOG let $\{A_n:n<\omega\}\subseteq\mathcal I$ be elements such that $\bigcup_{n<\omega} A_n=\omega$. For $n<\omega$ define, by recursion, $B_0=A_0$ and $B_n = A_n\setminus\bigcup_{i. For a set \(A\in\mathcal I\setminus\{A_n:n<\omega\}$ and a finite $F\in[\mathcal I]^{<\omega}$ not containing $A$ or the define $f_{A,F}(n+1) = \min\{k<\omega: k>f_{A,F}(n)\ \&\ (\exists m>n+1)(k\cap B_m\cap(A\setminus\bigcup F)\neq\emptyset)\}$ By our assumption there is an increasing $h_0:\omega\to\omega$ not dominated by any $f_{A,F}$. For a nonempty finite $F\in[\mathcal I]^{<\omega}$ define $g_F(n+1) = \min\{ k<\omega: k>g_F(n)\ \&\ (\exists m>n+1)(k\cap (B_m\setminus\bigcup F))\setminus h_0(m)\neq\emptyset) \}.$ Again, by our assumption, there is an increasing $h_1:\omega\to\omega$ dominating $h_0$ which is not dominated by any $g_{F}$. Let $X=\bigcup_{n<\omega}[h_0(n),h_1(n))$. Then $\mathcal I\cup\{X\}$ is ideal independent contradicting the maximality of $\mathcal I$. This finishes the proof.

35.21

In the Miller model $\mathfrak u<\mathfrak s_{mm}$.

35.22

It is consistent that $\mathfrak s_{mm}<\mathfrak i$ and $\max\{r_\sigma,\mathfrak d\}<\mathfrak s_{mm}$. Also, $\Diamond_{\omega_2}^\omega(r_\sigma,\mathfrak d)$ implies that $\mathfrak s_{mm}=\omega_2$.

35.23

Shelah has a model where $\mathfrak i < \mathfrak u$.

## 35.3J. Cancino-Manríquez: Irresolvable spaces

(joint work with M. Hrušák and D. Meza-Alcántara)

In the following all spaces are assumed to be regular and crowded.

35.24

A topological space $X$ is irresolvable if it does not contain disjoint dense subsets.

35.25

A space is irresolvable if all dense sets have nonempty interior.

35.26M. Scheepers

The irresolvability number $(\mathfrak{irr}$) is defined to be the minimal $\pi$-weight of a countable irresolvable space.

35.27

$\mathfrak r\leq \mathfrak{irr}\leq \mathfrak i$.

35.28M. Scheepers

Does $\mathfrak r=\mathfrak{irr}$?

35.29

$\mathfrak d\leq\mathfrak{irr}$.

35.30

The $\pi$-weight in the definition of $\mathfrak{irr}$ can be replaced by weight, i.e. $\mathfrak{irr}=\min\{w(X):X\ \mbox{is countable irresolvable}\}.$

Let $\tau$ be a topology on $\omega$ witnessing $\mathfrak{irr}$ and $\mathcal B$ its minimal $\pi$-base. Let $X=\bigcup\mathcal B$. Then $X$ is open dense in $\omega$. For each $U\in\mathcal B$ and $x,y\in U$ choose disjoint clopen (every countable regular space is zerodimensional) neighbourhoods $V(U,x,y), W(U,x,y)\subseteq U$ of $x,y$. Then $\{V(U,x,y),W(U,x,y):x,y,\in\omega, U\in\mathcal B\}$ is a subbase of clopen sets of a regular irresolvable topology on $X$.

of theorem

Assume to the contrary that $\mathfrak{irr}<\mathfrak d$ and let $\tau$ be a topology on $\omega$ witnessing $\mathfrak{irr}$. By recursion construct $\langle A_n:n<\omega\rangle,\langle B_n:n<\omega\rangle$ increasing sequences of disjoint nowheredense subsets of $\omega$ such that $n\in\overline{A_{n+1}}\cap\overline{B_{n+1}}$. Suppose $A_n,B_n$ are constructed. If $n\in\overline{A_n}\cap\overline{B_n}$ let $A_{n+1}=A_n$ and $B_{n+1}=B_n$ Otherwise if $n\not\in\overline{A_n}\cap\overline{B_n}$ let $\{W_n:n<\omega\}$ be a partition of $\omega\setminus \overline{A_n}\cup\overline{B_n}\cup\{n\}$ into clopen sets. Let $\mathcal V_n$ be a local base at $n$ of cardinality $<\mathfrak d$. For $U\in\mathcal V_n$ define $f_U(m+1)=\min\{ k<\omega: f_U(m)\ \& (\exists l>m+1)(U\cap W_l\neq\emptyset) \}.$ Let $h_0:\omega\to\omega$ be a function not dominated by any $f_U$. Let $A_{n+1}=A_n\cup \bigcup_{n<\omega}W_n\cap h_0(n)$. Notice that $A_{n+1}$ is nowheredense and $n\in\overline{A_{n+1}}$. Similarly, for $U\in\mathcal V_n$ define $g_U(m+1)=\min\{ k<\omega: g_U(m)\ \& (\exists l>m+1)(U\cap W_l\setminus h_0\neq\emptyset) \}.$ Again, let $h_1:\omega\to\omega$ be a function not dominated by any $g_U$. Define $B_{n+1}=B_n\cup\bigcup_{n<\omega}W_n\cap[h_0(n),h_1(n))$.

Finally $\bigcup_{n<\omega} A_n$ and $\bigcup_{n<\omega} B_n$ are two disjoint dense subsets of $\omega$ contradicting irresolvability.

# 3625.2.2015

## 36.1D. Chodounský: Mathias like reals

36.1

Given a filter $\mathcal F\in V$ on $\omega$ we say that a real $r\subseteq\omega$ is Mathias-like for $\mathcal F$ if it satisfies:

1. $r$ is a pseudointersection of $\mathcal F$, and
2. For each $\mathcal F$-universal $H$ there is a finite $a\in[r]^{<\omega}$ which is also in $H$.

Recall that a set $H$ of finite subsets of $\omega$ is $\mathcal F$-universal (Laflamme) if it contains a finite subset of every element of $\mathcal F$.

36.2

Every Mathias-like real becomes a Mathias real possibly after intersecting it with a further Cohen real.

36.3

This is similar to the theorem that a dominating real is Hechler generic after adding it to a further Cohen real. A similar characterization exists for amoeba reals.

36.4

Assume $P,Q$ are forcings such that each generic extension over $Q$ contains a generic over $P$ then there is an $r\in RO(P)$ such that $RO(P)\upharpoonright r$ completely embeds into $RO(Q)$.

Fix a $Q$-name $\dot{G}_P$ for a $P$-generic ultrafilter. Define $i:RO(P)\to RO(Q)$ as follows: $i(p) = ||p\in\dot{G}_P||.$ Now let $r = \bigwedge\{p:i(p) = \mathbf 1\}.$ An easy density argument shows that $r>0$. We now show that $i$ is a regular embedding when restricted to $RO(P)\upharpoonright r$: but, by definition, this is clear: it preserves order, disjointness and maximality of antichains.

36.5

Assume a forcing $P$ adds a Mathias-like real for $\mathcal F$. Then $\mathbb M_{\mathcal F}$ completely embeds into $P\times Q$. for any forcing $Q$ adding a Cohen real.

Proof (idea): Use the previous proposition, the fact that if a real is Mathias-like for $\mathcal F$ then it is Mathias-like for any restriction of $\mathcal F$ and the fact that a direct sum of countably many copies of $P\times Cohen$ is isomorphic to $P\times Cohen = P*Cohen$ and the fact that $P\times Cohen$ regularly embeds into $P\times Q$ for any $Q$ adding a Cohen real.

## 36.2D. Chodounský: Eventually different reals

Recall the relation $=^\infty$ on $\omega^\omega$. This is a symmetric, reflexive relation which, however, is not transitive. We can, nevertheless, define the bounding and dominating cardinal invariants $\mathfrak b_\infty$ and $\mathfrak d_\infty$ for this relation. The following is well known:

36.6Bartoszynski ?

$\mathfrak b_\infty=cov(\mathcal M)$ and $\mathfrak d_\infty=non(\mathcal M)$.

thus we can write (a part of) Cinchoń's diagram as follows:

36.7Bartoszynski

After adding two infinitely equal reals we get a Cohen real.

The previous theorem motivates the definition:

36.8

A forcing is half Cohen iff it adds infinitely equal reals.

36.9Shelah

Suslin ccc forcings add half Cohen reals iff they add Cohen reals.

36.10Zapletal

There is a forcing adding a half Cohen real which does not add a Cohen real.

36.11?

An model $M$ has an eventually different real if the groundmodel reals are meager.

36.12?

If a Suslin ccc forcing adds a real number which is not in any Cohen generic extension of the ground model, then it adds an eventually different real.

36.13

Given a filter $\mathcal F$ on $\omega$ the following are equivalent 1. $\mathbb M_{\mathcal F}$ does not add an eventually different real, 2. $\mathcal F^{<\omega}$ is $+$-selective, and 3. $\mathcal F^{<\omega}$ is $+$-Ramsey.

where a filter is $+$-selective if any sequence of positive sets has a positive selector while a filter is $+$-Ramsey if in the game where the second person is building a selector (picking a single point in each move) one has a winning strategy.

of Theorem

To show that 1 implies 3. we show something stronger:

36.14

If a model with a Mathias-like real does not contain an eventually different real then the filter is $+$-Ramsey.

of Claim

Let $r$ be the Mathias-like real and let $T$ be an $\omega$-branching tree labeled by positive sets which gives a strategy for the first player. We need to find a positive branch. This will show that $T$ is not a winning strategy. Working in the extension let $O_n=\{a\in[T]:(\exists m>n)(a(m)\subseteq r\setminus n)\}.$ Each $O_n$ is an open set and, since $r$ is Mathias-like, it is dense. By the Baire Category theorem $O=\bigcap_{n<\omega} O_n$ is co-meager. Any branch $a\in O$ is positive and, since no e.d. reals were added, the ground model reals are not meager, so there is a ground model branch $a\in O$.

Next 3. immediately implies 2. so it remains to show that 2. implies 1. This is a normal forcing argument: let $\dot{X}$ be a name for a function in from $\{(n,m):m>n\}$ into $\omega$. Enumerate $[\omega]^{<\omega}=\{s_i:i<\omega\}$ such that $s_i. Define $X_k = \{t\in[\omega\setminus k]^{<\omega}: (\forall i 36.15 \(X_k$ is $\mathcal F^{<\omega}$-positive. of Claim . Now use $+$-selectivity to get a selector $Y$. Define $Y(i,k) = h^t_i(k)$ iff $t\in Y\cap X_k$. Then $Y$ is forced to be infinitely equal to $X$. # 3718.3.2015 ## 37.1J. Grebík: Free sequences in Boolean algebras. 37.1 Let $B$ be a Boolean algebra. A sequence of elements $\langle a_\alpha:\alpha<\kappa\rangle$ is a free sequence if for each finite number of distinct indices $\{\delta_0<\cdots<\delta_n<\gamma_0<\cdots<\gamma_n\}$ the intersection \[ \bigwedge_{i=0} a_i \wedge \bigwedge_{i=0}^n -a_i$ is nonzero.

37.2

A strictly decreasing sequence is free but not independent.

37.3

The cardinal invariant $\mathfrak f$ is defined to be the minimal cardinality of a non-extendible free sequence in $\mathcal P(\omega)/fin$ of this length.

Recall that for an ultrafilter $\mathcal U$, $\chi(\mathcal U)$ denotes the character of $\mathcal U$, i.e. the minimal cardinality of a basis of $\mathcal U$. Similarly, $\pi\chi(\mathcal U)$ is the $\pi$-character of $\mathcal U$, i.e. the minimal cardinality of a $\pi$-basis of $\mathcal U$. (A $\pi$-basis of $\mathcal U$ is a family of infinite sets such that each element of $\mathcal U$ contains an element of the $\pi$-base). Also recall that $\mathfrak u$ is the minimal cardinality of an ultrafilter base while $\mathfrak r$ (the reaping number) is the minimal cardinality of an ultrafilter $\pi$-base.

The following cardinal invariant is less known.

37.4

The cardinal invariant $\mathfrak u^*$ is defined: $\mathfrak u^* = \min\{\kappa:(\exists\mathcal U)(\chi(\mathcal U)=\pi\chi(\mathcal U)=\kappa)\}\cup\{\mathfrak c\}$

37.5Kunen

If $\mathfrak c$ is regular then there is an ultrafilter $\mathcal U$ such that $\chi(\mathcal U)=\pi\chi(\mathcal U)(=\mathfrak c)$.

Proof (Hint): Construct $\mathcal U$ by induction to $\mathfrak c$ using an $\mathfrak c\times\mathfrak c$ independent matrix.

37.6

If $\mathfrak r=\mathfrak u$ then $\mathfrak u^*=\mathfrak u$.

37.7

$\mathfrak r\leq\mathfrak f\leq\mathfrak u^*$

Let $\mathcal A=\langle A_\alpha:\alpha<\kappa\rangle$ be a free sequence of length $\kappa<\mathfrak r$. This sequence generates some filter $\mathcal F$. Define $comb(\mathcal A)=\Big\{\bigwedge_{\delta\in\Delta} A_\delta\wedge \bigwedge_{\gamma\in\Gamma}-A_\gamma: \Delta<\Gamma\in[\kappa]^{<\omega}\Big\}.$

37.8

For each $b\in\mathcal F^+$ if $b$ is not above any $a\in comb(\mathcal A)$ then $\mathcal A$ can be extended by $b$.

Now extend $\mathcal F$ to an ultrafilter $\mathcal U$ and notice that $comb(\mathcal A)$ is not a $\pi$-base of $\mathcal U$. Now use the above claim.

The second inequality is proved similarly: Let $\mathcal U$ witness $\mathfrak u^*$. Let $\{U_\beta:\beta<\mathfrak u^*\}$ be a basis of $\mathcal U$. Recursively construct a free sequence $\langle A_\alpha:\alpha<\mathfrak u^*\rangle$ such that $A_\alpha\leq U_\alpha$ and $A_\alpha\in\mathcal U$. If $A_\alpha$ is constructed then $\mathcal U\upharpoonright U_{\alpha+1}$ is an ultrafilter on $U_{\alpha+1}$. To construct $A_{\alpha+1}$ notice that $comb(\mathcal A_\alpha)\upharpoonright U_{\alpha+1}$ is not a $\pi$-base of the restriction $\mathcal U\upharpoonright U_{\alpha+1}$. The sequence $\mathcal A$ is a basis of $\mathcal U$ so it cannot be extended.

37.9

If $\mathfrak r=\mathfrak u$ then $\mathfrak f=\mathfrak u$.

37.10

The cardinal invariant $\mathfrak r^*$ is defined: $\mathfrak r^*=\min\{\kappa:(\exists\mathcal F)(\chi(\mathcal F)=\kappa\ \&\ (\exists \mathcal B) (|\mathcal B|\leq\kappa\ \&\ (\forall\mathcal U\supseteq\mathcal F) (\mathcal B\ \mbox{is a}\ \pi\mbox{-base of}\ \mathcal U))\}$

So we have the following picture:

37.11

Is it consistent that $\mathfrak f>\mathfrak u$?

Note that countable support iteration does not work, since in the required model we need to have $\mathfrak u>\mathfrak r$ so we'd need $\mathfrak c\geq\omega_3$.

The following question might be simpler:

37.12

Is it consistent that $\mathfrak f>\mathfrak i$?

37.13

Is $\mathfrak f=\mathfrak r^*$?

37.14

Is $\mathfrak i\geq\mathfrak r^*$?

37.15

Is $\mathfrak f$ a regular cardinal? Does it at least have an uncountable cofinality? Can $\mathfrak f$ always be realized by a sequence of cardinal length? What about the cofinality of the witnessing sequence?

37.16

It is consistent that $\mathfrak f<\mathfrak i$.

37.17

If $B$ is a complete ccc Boolean algebra, is it consistent that $\mathfrak r_B<\mathfrak u_B$?

# 3825.3.2015

## 38.1J. Verner: Anti-P-points

38.1

A point $x\in X$ is an anti-P-point if there is a local base $\mathcal B$ at $x$ such that for any (nontrivial) countable sequence $\langle B_n:n<\omega\rangle\subseteq\mathcal B$ the interior of the intersection $\bigcap_{n<\omega}B_n$ does not contain $x$.

38.2

Does $\omega^*$ contain an anti-P-point?

## 38.2D. Chodounský

### 38.2.1A Dynamical question

38.3

$FIN=[\omega]^{<\omega}\setminus\{\emptyset\}$. We will consider $FIN$ with operation $\cup$ restricted to disjoint sets.

38.4

Given a disjoint sequence $A=\langle a_n:n<\omega\rangle\subseteq FIN$ we define $FU(A) = \Big\{\bigcup_{i\in x} a_i:x\in FIN\Big\}$

38.5Hindman

Whenever $FIN=A_0\cup A_1$ there is a disjoint sequence $D$ and an $i<2$ such that $FU(D)\subseteq A_i$.

Fix an interval partition $\mathcal I=\langle I_n:n<\omega\rangle$. For $C\subseteq FIN$ we say that $C$ is $\mathcal I$-big if $\bigcup C$ intersects all but finitely many intervals from $\mathcal I$.

38.6

Given an interval partition $\mathcal I$ can we extend Hindman's theorem so that the $D$ is $\mathcal I$-big.

38.7

If $\mathcal I$ contains infinitely many partition of the same length then the answer is no.

### 38.2.2Indestructible gaps

38.8

It is consistent that there is an indestructible non-Hausdorff gap.

38.9

An $(\omega_1,\omega_1)$-pregap is a sequence $\langle L_\alpha,R_\alpha:\alpha<\omega_1\rangle$ such that

1. $L_\alpha\cap R_\alpha=\emptyset$ for each $\alpha<\omega_1$
2. $L_\alpha\subseteq^* L_\beta, R_\alpha\subseteq^* R_\beta$ for each $\alpha<\beta$

It is a gap if there is no separator, i.e. an $S$ such that $L_\alpha\subseteq^* S, R_\alpha\cap S=^*\emptyset$ for each $\alpha<\omega_1$. A gap satisfies the $(K)$ ($(O)$ respectively) condition if for each $\alpha<\beta\in\omega_1$ the set $L_\alpha\cap R_\beta \cup R_\alpha\cap L_\beta$ ($L_\alpha\cap R_\beta$ respectively) is nonempty. It satisfies the $(H)$ condition if $(\forall n)(\forall\beta<\omega_1) (|\{\alpha<\beta:L_\alpha\cap R_\beta\subseteq\}|<\omega)$ We say that a gap is special if it has a subgap satisfying $(K)$, oriented if it has a subgap satisfying $(O)$ and Hausdorff if it has a subgap satisfying $(H)$.

38.10

Each Hausdorff gap is oriented and each oriented gap is special.

38.11

A special gap is indestructible by forcings preserving $\omega_1$.

38.12Hausdorff

A Hausdorff gap exists.

38.13

The restriction of any gap to a Cohen real is not special, in particular it is consistent that there are nonspecial gaps. On the other hand, $MA(\sigma-linked)$ or PID implies that each gap is Hausdorff and OCA implies that each gap is special.

38.14

Is it consistent that there are special and not oriented or oriented and not Hausdorff gaps?

38.15Hirshorn, unpublished

There is a special non-Hausdorff gap.

After adding $\omega_1$-many Cohen reals there are expamples showing that there are special and not oriented and oriented and not Hausdorff gaps.

We will show how to force an oriented non-Hausdorff gap. Recall that the definition of condition $(H)$ can also be used for towers (9.13).

It should be evident that if a gap satisfies $(H)$ then its left part satisfies $(H)$.

38.17

Is there a tower which generates a meager ideal?

38.18

If there is a tall meager $\omega_1$-generated P-ideal then there is a meager tower.

38.19

It is consistent that there is an oriented non-Hausdorff gap.

We will add a gap which is oriented but its left side does not satisfy $(H)$. The forcing will consist of conditions of the form $\langle F, n, \langle L_\alpha, R_\alpha:\alpha\in F\rangle\rangle$ where $F\in[\omega]^{<\omega_1}$,$L_\alpha,R_\alpha\subseteq n$ are disjoint.and for $\alpha<\beta\in F$ the intersection $L_\alpha\cap R_\beta$ is nonempty. The ordering is the natural choice.

38.20

The forcing is Knaster.

Let $\{p_\alpha:p<\omega_1\}$ be conditions. We may assume that the first coordinates form a delta system with root $F$ (which is below the the tails), that the second coordinates are equal to some fixed $n$ and the third coordinates are "isomorphic" (i.e. have the same size, same restriction to $F$, etc.). In this case all such conditions are compatible.

We now show by contradiction that the left part does not satisfy $(H)$. So assume there is a condition $a$ and a name $\dot{I}$ for a cofinal subset of $\omega_1$ such that $a$ forces that $\{L_\alpha:\alpha\in I\}$ satisfies $(H)$. Let $X=\{\alpha:(\exists p_\alpha\leq a)(p_\alpha\Vdash\alpha\in\dot{I}\ \&\ \alpha\in F_{p_\alpha}).$ Then $X$ is uncountable. Again, as in the proof of the claim, we may assume that the $\{p_\alpha:\alpha\in X\}$ form a "delta"-system. Let $p_0$ be the first condition "in" $X$ and $p_\omega$ the $\omega$-th condition. Construct $q_0\leq p_0,p_\omega$ as in the proof of the above claim. By our assumption on $a$ there is a $q_1\leq q_0$ and $k<\omega$ such that $q_1\Vdash |\{\alpha\in I: \alpha_\omega\ \&\ L_\alpha\setminus L_{\alpha_\omega}\subseteq n\}|\leq k.$ Extend $q_1$ to $q_2$ by adding $k+1$-many blocks between $F_{\alpha_0}$ and $F_{\alpha_\omega}$ and copying the respective conditions $P_\alpha$. Moreover increase the $n_{q_1}$ by $k$ and use the space above $n_{q_1}$ to guarantee the failure of $(H)$ and arrive at a contradiction. (There are some technical details to work out.)

38.21

It is not hard to see that forcing used in the proof is equivalent to adding $\omega_1$-many Cohen reals.

# 3915.4.2015

## 39.1News from Singapore

39.1

Let $\mathcal I$ be a tall ideal on $\omega$. $\mbox{cov}^*(\mathcal I) =\min\big\{ |\mathcal H|:\mathcal H\subseteq\mathcal I\ \&\ (\forall A\in[\omega]^\omega)(\exists H\in\mathcal H)(|H\cap A|=\omega)\big\}$

39.2Dilip, Shelah

$\mbox{cov}^*(\mathcal Z_0)\leq\mathfrak d$

39.3Dilip

$\mbox{cov}^*(\mathcal Z_0)\leq\max\{\mathfrak{b,s}\}$

39.4Dilip

$\mbox{cov}^*(\mathcal Z_0)\leq\mathfrak b$

39.5Todorčević

Y-PFA implies $\mathfrak c=\aleph_2$.

## 39.2J. Wódka: Monotone spaces

### 39.2.1What I am working on...

39.6

A function $f:\mathbb R\to\mathbb R$ is a swiatkowski function if $(\forall a where $C(f)$ is the set of points where $f$ is continuous and $I(u,v)$ is the open interval between $u,v$. A function $f$ is quasi-continuous if \[ (\forall x\in\mathbb R)(\forall\varepsilon>0)(\forall I\in\mathcal U(x)) (\exists J\subseteq I)(\mbox{diam} f[\{x\}\cup J]<\varepsilon).$

### 39.2.2Monotone spaces

39.7

A metric space $(X,d)$ is $c$-monotone if there is a linear order $\leq$ on $X$ such that $d(x,y)\leq c\cdot d(x,z)$ for all $x\leq y\leq z$. It is monotone if it is $c$-monotone for some $c$.

39.8

A metrizable space is a GO-space iff there is a matric which makes it a monotone space.

39.9

At this moment I was forbidden by the speaker to make notes.

# 406.5.2015

## 40.1A. Enayat: Leibnizian motives in set theory

The following are some references:

1. J. Mycielski: New set-theoretical axioms derived from a lean metamathematics, JSL 1995
2. A. Enayat: On the Leibniz-Mycielski axiom in set theory, Fund. Math. 181 (2004)
3. A. Enayat: Leibnizian models of set theorey, JSL 69 (3), 2004
4. A. Enayat: Models of set theory with definable ordinals, Arch. Math. Logic 44 (2005)
5. V. Kanovei and V. Lyubetsky: A countable definable set of reals with no definable element, arXiv:1408.3901 [math.LO], 2014

We start with some definitions. In the following will have a language $\mathcal L$ and a structure $\mathcal M$ for this language

40.1

Let $\mathcal M$ be a structure for some language $\mathcal L$. We say that $\mathcal M$ is pointwise definable if every singleton is definable by a formula without parameters.

40.2

It is clear that if $\mathcal M$ is pointwise definable it has cardinality at most $|\mathcal L|\cdot\aleph_0$.

40.3

$(\mathbb N,<)$, $(V_\omega,\in)$, $(\mathbb A,+,\cdot)$, where $\mathbb A$ is the set of real algebraic numbers.

40.4

An $\mathcal L$ structure $\mathcal M$ is Leibnizian if for each pair $a,b$ of distinct elements of $M$ there is an $\mathcal L$-formula $\varphi$ with one free variable such that $\mathcal M\models \varphi(a)\ \&\ \neg\varphi(b)$

40.5

$(\mathbb R,+,\cdot)$ is Leibnizian (since rationals are definable and dense) but not pointwise definable.

40.6

$(V_{\omega+1},\in)$ is Leibnizian (any two reals differ in a natural number) but not pointwise definable.

40.7

If $\mathcal M$ is Leibnizian then $|M|\leq 2^{\aleph_0\cdot|\mathcal L|}$.

40.8

An ultrafilter $\mathcal U$ is Hausdorff iff for each $f,g\in{}^\omega\omega$ if $f(\mathcal U)=g(\mathcal U)$ then $f=_{\mathcal U}g$, where $f(\mathcal U)=\{X:f^{-1}[X]\in\mathcal U\}$.

40.9

A selective ultrafilter is Hausdorff. Bartoszynski and Shelah have a paper where they claim to show that it is consistent that there are no Hausdorff ultrafilters. However, there might be a gap in that paper.

40.10

Let $\mathcal L$ have a unary predicate $X$ for each $X\subseteq\omega$. Let $\mathcal U$ be an ultrafilter on $\omega$. Then the ultraproduct $\prod \omega/\mathcal U$ is Leibnizian iff $\mathcal U$ is Hausdorff.

40.11

An $\mathcal L$ structure $\mathcal M$ is weakly Leibnizian if for each pair $a,b$ of distinct elements of $M$ there is a $\mathcal L$-formula $\varphi$ with two free variables such that $\mathcal M\models \neg( \varphi(a,b)\leftrightarrow\varphi(b,a))$

40.12

Any structure with a definable linear order is weakly Leibnizian.

40.13

It follows that there are arbitrarily large weakly Leibnizian structures (independent of the size of the language).

40.14

$(\mathbb R,<)$ is weakly Leibnizian and not Leibnizian. If $V=L$ then so is $(V_{\omega+2},\in)$ since in that case the universe can be definably linearly ordered.

40.15

An $\mathcal L$ structure $\mathcal M$ is barely Leibnizian if for each pair $a,b$ of distinct elements of $M$ there is a $\mathcal L$-formula $\varphi$ with two free variables such that $\mathcal M\models \varphi(a,b)\ \&\ \neg\varphi(a,a)$

40.16

If equality is definable then the structure is barely Leibnizian. In particular any model of set theory is barely Leibnizian due to the axiom of extensionality.

40.17

From the definitions it immediately follows that $\mathcal M$ is pointwise definable $\rightarrow$ $\mathcal M$ is Leibnizian $\rightarrow$ $\mathcal M$ is weakly Leibnizian $\rightarrow$ $\mathcal M$ is barely Leibnizian. The examples, on the other hand, show that none of the implications can be reversed.

40.18classical

A consistent extension $T$ of $ZF$ has a pointwise definable model if $T$ is consistent with $V=OD$. Or, to put it differently, a complete consistent extension of $ZF$ has a pointwise definable model if it proves $V=OD$.

Recall that $OD = \Big\{ a\in V: (\exists \alpha\in Ord) (\exists \overline{\beta}\subseteq Ord) (\exists \varphi(x,\overline{y})) (a\ \mbox{is definable in}\ (V_\alpha,\overline{\beta})) \Big\}$ and $HOD=\{a\in V:\mbox{tcl}(\{a\})\subseteq OD\}$. Also recall that $ZF \vdash "HOD\models ZF+\mbox{definable global choice}".$

The proof of the above theorem uses the following standard theorem of Model theory.

40.19

If $T$ has definable Skolem functions and $\mathcal M$ is a model of $T$ then the pointwise definable elements of $\mathcal M$ form an elementary substructure of $\mathcal M$.

40.20Vopěnka

The universe is set-generic over $HOD$ if $V=L[A]$ for some set $A$ of ordinals.

40.21

Grigorieff showed that the forcing going from $HOD$ to $V$ in the above theorem is homogeneous.

In this context recall

40.22Balcar, Vopěnka

$ZFC\vdash(\forall x)(\exists y\subseteq Ord)(x\in L[y])$

also, I recall that about 20 years ago a master's student of Sy Friedman proved that

40.23

The universe is class-generic over $HOD$.

However, it was probably not written up.

40.24Roguski

If $M$ is a countable model of ZFC then there is a generic extension $M[G]$ such that $M$ is the HOD of $M[G]$.

40.25

The following are equivalent

1. $M\models V=OD$ (note this is the same as $V=HOD$)
2. $M$ has a definable (without parameters) global well ordering
3. The theory of $M$ has a pointwise definable model.
40.26Paris

A model $M$ of $ZF$ is a Paris model (or a D.O. model) if every ordinal of $M$ is pointwise definable.

40.27Paris

Every consistent extension of $ZF$ has a Paris model. Moreover, this model is unique iff the extension proves that $V=OD$.

40.28

If a consistent extension of $ZF$ does not prove that $V=OD$ then there are continuum many pairwise non-equivalent Paris models.

40.29

A Paris model has countably many ordinals. However, it may be arbitrarily wide.

40.30Henkin-Orey

Let $T$ be a theory in a countable language and $p$ a non-principle 1-type. Then $T$ has a model omitting $p$.

40.31

Mostowski and Ryll-Nardzewski have a proof using the Baire category theorem.

Paris' theorem (the first part) can be proved rather easily using the omitting types theorem. The second part also follows from the omitting part theorem if we additionally use the Balcar-Vopěnka theorem:

40.32Balcar-Vopěnka

If two models of $ZF$ have the same height, the same sets of ordinals and one of them satisfies $AC$ then they are isomorphic.

40.33

The Leibniz-Mycielski principle ($LM$) says that for distinct $a,b$ there is some $V_\alpha$ such that $a,b\in V_\alpha$ and there is a formula $\varphi(x)$ without parameters such that $V_\alpha\models \varphi(a)\ \&\ \neg\varphi(b)$.

This is motivated by the classical Leibniz principle of the identity of indiscernibles:

If two objects are distinct they differ in some property

40.34Mycielski

The following are equivalent for a complete consistent extension $T$ of $ZF$:

1. $T$ has a model with no indiscernibles (i.e. a Leibnizian model).
2. $T$ proves the Leibniz-Mycielski principle.
40.35

The proof uses the reflection theorem and Paris models.

40.36

The Kinna-Wagner selection principle ($GKW_1$) says that every family $X$ of sets, each of which has size at least 2, can be split by a set, i.e. there is an $S$ such that $\emptyset\neq S\cap x\neq x$ for each $x\in X$.

40.37

The Kinna-Wagner selection principle is equivalent to the statement ($GKW_2$) that every set can be injectively mapped into the power set of an ordinal.

40.38

The Kinna-Wagner selection principle is weaker than the Axiom of Choice.

40.39

The Kinna-Wagner selection principle has a global version (similar to the global version of $AC$): there is a definable class function which to each set of size at least 2 assigns a nonempty proper subset.

40.40

The following are equivalent:

1. This global version of the Kinna-Wagner selection principle
2. There is a definable injective class function $H$ from $V$ into subsets of $2^{ 3. The Leibniz-Mycielski principle holds. \(1.\rightarrow 2.$. Basically, use the splitting function to assign a binary tree to each $X$. The set of branches of this tree will work.

$2.\rightarrow 3.$. Given $a,b$ they are both assigned a set of binary sequences. Pick the first $\alpha$ where a difference in the sequences appears.

$3.\rightarrow 1.$. Given $a,b$ find the $\alpha$ and $\varphi$ given by the Leibniz-Mycielski principle. Then the set $\{a,b\}$ can be split by $\{x:V_\alpha\models \varphi(a)\}$.

We thus have the following picture where none of the implications can be reversed:

40.41

$\mathbb L$ of a Cohen real fails $LM$. Adding $\aleph_0$ side-by-side Jensen reals one can build a model $M$ of $LM$ and $\neg AC$.

40.42

Is there a model of $AC+LM$ where $V\neq OD$?

This is connected to the following theorem of Kanovei:

40.43

There is a countable set of reals none of which are definable.

40.44

Can the above set be of the form $[a]$, where $[a]=\{b:a=^*b\}$.

40.45Kanovei-Lyubetsky

Yes!

# 4113.5.2015

## 41.1D. Chodounský: Y-PFA and the continuum.

Recall that last time we were proving that PID (9.8) implies $\mathfrak b\leq\aleph_2$. This went somewhat along the following lines

41.1

A pair $G=(\langle A_\alpha:\alpha<\kappa\rangle, \langle T_\alpha:\alpha<\lambda\rangle)$ is called a $(\kappa,\lambda)$-gap if

1. $T_\alpha\subseteq^* T_\beta$ for each $\alpha<\beta<\lambda$;
2. $A_\alpha\cap T_\beta=^*\emptyset$ for each $\alpha<\kappa,\beta<\lambda$

If $\kappa=\lambda=\omega_1$ We say that $G$ satisfies $(H)$ if, moreover,

1. $(\forall\beta<\omega_1)(\forall n<\omega) (|\{\alpha<\beta:A_\alpha\cap T_\beta\subseteq n\}|<\omega)$

Given a $(\kappa,\lambda)$-gap $G$, we define a P-ideal $I(G)$ on $\kappa$ as follows $\mathcal I(G) = \{X:(\forall n)(\exists\beta<\lambda) (|\{\alpha\in X:A_\alpha\cap T_\beta\subseteq n\}|<\omega) \}$

Assume PID. If the first alternative holds for $I(G)$ then $G$ must be a Hausdorff gap (i.e. has a cofinal subgap satisfying condition H). Otherwise the second alternative shows that $G$ must contain an $(\omega,\lambda)$-gap. By Rothberger, $\mathfrak b$ is the minimal $\lambda$ such that there is an $(\omega,\lambda)$-gap. If $\mathfrak b>\omega_2$ then there must be a gap of the form $(\langle T_\alpha\rangle^\perp,T_\alpha)_{\alpha<\omega_2}$ (take an $<^*$-unbounded family of functions of size $\mathfrak b$, and take the first $\omega_2$-functions and everything below their graphs in $\omega\times\omega$). Then the first alternative of PID applied to $I(G)$ cannot hold so the second alternative gives an $(\omega,\omega_2)$-gap so $\mathfrak b\leq\aleph_2$ --- a contradiction.

41.2

$Y-PFA$ implies that continuum is $\aleph_2$

The strategy is as follows: $Y-PFA$ implies that $\mathfrak b\geq\omega_2$ (apply the axiom to Hechler's forcing). Next we need to show that $Y-PFA$ implies PID. Finally we show that $Y-PFA$ implies that $\mathfrak b=\mathfrak c$.

41.3

$Y-PFA$ implies PID.

Let $\mathcal I$ be a P-ideal consisting of countable subsets of $\kappa$ and assume that $\kappa$ cannot be written as a countable union of $\mathcal I^\perp$ sets. Then we need to force the first alternative. Let $\mathcal K$ be the $\sigma$-ideal $\sigma$-generated by $\mathcal I^\perp$. Fix a large enough $H(\theta)$. We say that $\langle M,I,\alpha\rangle$ is a suitable triple (an s-triple) if

1. $M$ is a countable elementary submodel of $H(\theta)$;
2. $I$ almost covers every element of $\mathcal I$ in $M$; and
3. $\alpha<\kappa$ is not contained in any $K\in\mathcal K\cap M$.

We order suitable triples as follows: $\langle M,I,\alpha\rangle< \langle N,J,\beta\rangle,$ if $\langle M, I,\alpha\rangle\in N$. The forcing $P$ we will apply $Y-PFA$ consists of finite sets of suitable triples each of which is linearly ordered by $<$. The conditions are ordered as follows: for $p,q\in P$ we write $p\leq q$ if

1. $p\supseteq q$; and
2. for each $\langle M,I,\alpha\rangle\in p\setminus q$ which is below some $\langle N,J,\beta\rangle\in q)$ the $\alpha$ is an element of $J$.

Given a generic for $P$ the union of the $\alpha$'s will be the required $\omega_1$-sized subset of $\kappa$ such that all countable subsets will be in $\mathcal I$. The fact that it has size $\omega_1$ is immediate. Before we show the second claim, we show that $P$ is $Y$-proper (25.7). So let $M$ be a countable elementary submodel of some sufficiently large $H(\lambda)$ ($\lambda>\theta$) and $p\in M\cap P$. Let $N=M\cap H(\theta)$. It is clear that $p\in N$. Extend $p$ by adding an arbitrary suitable triple of the form $\langle N,I,\alpha\rangle$. We claim that this condition is $Y$-master. Given an element $r\in RO(P)\cap M$ choose a downwards closed $A_r\in M$ of conditions in $P$ such that $r=\bigwedge A_r$.

41.4

The game $G(p,A)$ is a game where the first player plays elements of $\mathcal K$ and the second player responds with an $\alpha$ outside of the move of the first player. The second player wins if there is a condition $r\in A$ stronger than $p$ such that the last coordinates of the triples in $r\setminus p$ are exactly the first $n$ moves of the second player for some $n<\omega$.

41.5

This game is closed and hence determined. We say that a set $A$ is $p$-big if the second player has a winning strategy in $G(p,A)$.

Given $r\leq p^{\smallfrown}\langle N,I,\alpha\rangle$ and $s\geq r$ in $M$ we claim that $A_s$ is $r\cap M$-big. Notice that the game $G(r\cap M, A_s)$ is an element of $M$. It follows that I. cannot have a winning strategy (II. beats it by enumerating $r\setminus r\cap M$). So II. has a winning strategy (which is an element of $M$). Now let $O_0$ be the set of all responses of the winning strategy to moves of I. Then $O_0\in M$ and, by definition, $O_0\not\in\mathcal K$ so there is an infinite $I_0\subseteq O_0$ which is in $\mathcal I$. Moreover this $I_0$ can be chosen from $M$. Then, by definition of the ordering of triples, $I_0$ is covered by the second coordinates of each triple in $r\setminus M$. And here I got lost :-)

# 4220.5.2015

## 42.1J. Grebík: Oscillations of real numbers

The motivation for the following is the Prikry problem:

42.1Prikry

Does there exist an atomless ccc forcing which does not add a Cohen real and does not add a Random real.

There are two strategies for trying to answer this question. The first strategy is to look at, e.g., weakly distributive posets, which cannot add a Cohen real, and try to find one which does not add Random reals. The other strategy is to look at, e.g. Y-cc posets, which cannot add Random reals and look for one which does not add Cohen.

42.2Shelah

If a Suslin ccc forcing adds an unbounded real then it adds a Cohen real.

42.3

The above was also proved by Kamburelis. Note that it also follows from the above result that a yes answer to Prikry's question is necessarily non Suslin.

42.4Blaszcyk, Shelah

There is a $\sigma$-centered forcing not adding a Cohen real iff there is a nowhere dense ultrafilter.

42.5

The above forcing is, in fact, the Laver forcing with a nowhere dense ultrafilter.

42.6

It is consistent that the answer to Prikry's question is yes.

Recall Theorem 26.10. According to the second strategy, we will try to construct a Y-cc forcing which does not add Cohen reals. The idea is to take an unbounded sequence of functions $\{f_\alpha:\alpha<\mathfrak b\}$ and let $P_\pi = \{A\in[\mathfrak b]^{<\omega}:\pi[A^2] = 0\},$ where, e.g., $\pi(\alpha,\beta) = o(f_\alpha,f_\beta),$ for a suitable oscillation function. We will also consider the forcing $Q_\pi(\mathfrak b)$ (recall Definition 29.7). Here we have a nice theorem

42.7

$Q_\pi$ is disjoint-ccc if there is no uncountable antichain of disjoint conditions. (Recall that conditions in $Q_\pi$ are finite sets.)

42.8Chodounský, Zapletal

If $Q_\pi$ is disjoint-ccc then $P_\pi$ is Y-cc.

42.9Todorčević

Given two functions $f,g\in\omega^\omega$ we define their oscillation, $osc(f,g)$, as follows: $osc(f,g) = |\{n<\omega:f(n)\leq g(n)\ \&\ f(n+1)>g(n+1)\}|.$

42.10

The following is easy.

1. $osc(f,g)<\omega\iff f\leq^* g\vee g\leq^* f$.
2. $osc$, as defined above, is not symmetric, i.e. there are $f,g$ with $osc(f,g)\neq osc(g,f)$.

We can extend the definition of oscillation to partial functions.

42.11

Let $P$ consist of strictly increasing functions and let $P^k$ consist of $k$-element subsets of $P$ which are strictly $\leq^*$ increasing. Let $R=\bigcup_{k<\omega} P^k$. For each $x\in R$ fix its enumeration $\{x(0),\ldots,x(k-1)\}$ in $\leq^*$ increasing order. Naturally order $R$ as follows: $x if each \(f\in x$ is $\leq^*$-strictly below each $g\in y$. We can now extend the oscillation to $(R)^2$: $osc(x,y) = (osc(x(i),y(j)))_{i=0,j=0}^{k,l},$ where $x\in P^k,y\in P^l$.

42.12

PFA implies $\mathfrak b=2^{\aleph_0}$.

The proof will use the following notion of coding:

42.13

Given $r\in \omega^\omega$ and $O:[\mathfrak b]^2\to\omega$ we say that $\delta <\mathfrak b$ of cofinality $\omega_1$ codes $r$, if there is a club $C\subseteq\delta$ such that for all $n<\omega$ there is a partition of $C$ into sets $\{C^n_i:i<\omega\}$ such that for all $\alpha,\beta\in C^n_i$ we have $O(\alpha+n,\beta+n)=r(n)$.

42.14

A given $\delta$ cannot code two distinct $r_0,r_1$.

Otherwise let $\delta$ code $r_0\neq r_1$. Without loss of generality $r_0(0)\neq r_1(0)$. By coding, there are witnessing clubs $C_0,C_1$. Let $C=C_0\cap C_1$. By coding there are partitions $\{C^0_i:i<\omega\}, \{D^0_i:i<\omega\}$ of $C$ such that $O(\alpha,\beta)=r_0(0)$ for $\alpha,\beta\in C^0_i$ and $O(\alpha,\beta)=r_1(0)$ for $\alpha,\beta\in D^0_i$. Let $\{C_i:i<\omega\}$ be a common refinement of these partitions. Then at least one element has to contain at least two ordinals which immediately leads to a contradiction.

# 4327.5.2015

## 43.1Filip Šedivý: Variants of Hechler forcing

The following talk is a presentation of the paper: Palumbo, J.: Unbounded and Dominating Reals in Hechler Extensions

43.1

Given two reals $x,y\in\omega^\omega$ we write $x\leq^* y$ if $x(n)\leq y(n)$ for all but finitely many $n<\omega$.

43.2

A real $x\in V[G]$ is dominating (over $V$) if $y\leq^* x$ for all $y\in V$. It is unbounded if there is no $y\in V$ such that $x\leq^*y$.

43.3

A forcing $P$ is forcing equivalent to $Q$ if for each generic filter $G$ on $P$ there is, in $V[G]$, a generic filter $H$ on $Q$ such that $V[H]=V[G]$ and vice versa.

We will now define three variants of the Hechler forcing.

43.4

The first variant of Hechler forcing, $D$, consists of pairs $(s,f)$ where $s\in\omega^{<\omega}$ and $f\in\omega^\omega$ ordered as follows: $p \leq q \iff s_p\supseteq s_q\ \&\ (\forall n)(f_p(n)\geq f_q(n))\ \&\ (\forall n\in\mbox{dom}(s_p)\setminus\mbox{dom}(s_q)) (s_p(n)\geq f(n)).$ The second variant $D_{nd}$ is a suborder of $D$ consisting of those pairs such that the first coordinate is non-decreasing. The third variant $D_{tree}$ consists of those trees $T\subseteq\omega^{<\omega}$ such that each $t\in T$ extending the stem has co-finitely many successors.

We aim to show that $D$ is forcing equivalent to $D_{nd}$ but not to $D_{tree}$.

43.5

$D$ is forcing equivalent to $D_{nd}$. {} Proof: We first show that $D_{nd}*C$ is forcing equivalent to $D$, where $C$ is the Cohen forcing. So let $d$ be a $D$-generic real. Define $d_{nd}=\min\{d(k):k\geq n\}$ Then $d_{nd}$ is generic on $D_{nd}$. Let $d^\prime=d-d_{nd}$. Then $d^\prime$ is Cohen over $V$. Let $A=\{n:d_{nd}(n)=d_{nd}(n+1)\}$. Then $d^\prime\upharpoonright A$ is Cohen over $V[d_{nd}]$. On the other hand let $d_{nd}*c$ be generic over $D_{nd}*C$. Let $A=\{n:d_{nd}(n)=d_{nd}(n+1)\}$. Define $d(n)=d_{nd}(n)+c(n)$ when $n\in A$ and $d(n)=d_{nd}(n)$ for $n\not\in A$.

43.6

Let $d$ be a $D_{tree}$-generic real. Then for any unbounded $x\in V[d]\cap\omega^\omega$ there is a dominating $y\in V[d]\cap\omega^\omega$ such that $x\not\leq^*y$.

43.7

Let $d$ be a $D$-generic real then there is an unbounded $x\in V[d]\cap\omega^\omega$ such that $x\leq^* y$ for each dominating $y\in V[d]\cap\omega^\omega$.

# 443.6.2015

## 44.1Jan Grebík: Oscillations of real numbers

We continue with the proof of theorem 42.12. For the proof we will need the notion of coding defined in 42.13 and the definition 42.9 of oscillation.

of Theorem 42.12

Given a function $O:[\mathfrak b]^2\to\omega$ consider the forcing $P=Col(\omega_1,2^\omega)$. This forcing adds a club $C\subseteq\mathfrak b^V$ of type $\omega_1$. We let $Q_n^{i\prime}$ be a $P$-name for the poset consisting of $i$-homogeneous (wrt $O$ finite subsets of $C$, i.e. if $p\in Q_n^{i\prime}$ and $\alpha,\beta\in p$ then $O(\alpha+n,\beta+n)=i$. Let $Q_n^i$ consist of finite sequences of elements of $O_n^{i\prime}$. We will choose the function $O$ so that, in $V^P$ for each $r\in\omega^\omega$ the finite support product $\prod_{n<\omega}Q_n^{r(n)}$ is ccc. The choice will use several technical lemmas. We will only show one which is relatively simple to give a taste of the flavour:

44.1

If $Z\subseteq P^k$ is unbounded and countably directed then there is a countable subset $D\subseteq Z$ and a $k$ by $k$ matrix $h:k\times k\to\omega$ having zeroes on the diagonal such that for each $l<\omega$ there is $d\in D$ and $z\in Z$ such that $osc(d,z) = h+l,$ i.e. $osc(d(i),z(j)) = h(i,j)+l,$ for each $i. 44.2 Taking \(k=1$ and $Z$ a witnesses to $\mathfrak b$ the above lemma shows that $Z$ realizes all oscillations.

44.3

Given a strictly $<^*$-increasing witness $\{a_\alpha:\alpha<\mathfrak b\}$ to $\mathfrak b$ there area $\alpha,\beta$ such that $a_\alpha\leq a_\beta$ (i.e. $osc(a_\alpha,a_\beta)=0$).

of Lemma

Since $\omega^\omega$ is separable there is $\gamma<\mathfrak b$ such that $\{a_\alpha:\alpha<\gamma\}$ is dense in $\{a_\alpha:\alpha<\mathfrak b\}$. Now there is a cofinal $D\subseteq\mathfrak b$ and $n_0$ such that $a_\gamma(n) for all $d\in D$ and $n>n_0$. We may also assume that there is a single $t_0$ such that $d\upharpoonright n_0 = t_0$ for each $d\in D$. Let $n_1$ be the minimal number such that \[ \{ d(n_1):d\in D\}$ is unbounded. We may again assume that there is a single $t_1$ such that $d\upharpoonright n_1 = t_0^{\smallfrown} t_1$. Fix a countable set $\{d_i:i<\omega\}\subseteq D$ such that $\{ d_i(n_1):i<\omega\}$ is unbounded. Since $\{a_\alpha:\alpha<\gamma\}$ is dense there is $\beta<\gamma$ and $n_2<\omega$ such that $a_\beta\upharpoonright n_1 = t_0^{\smallfrown} t_1$ and $a_\beta(n) for each \(n>n_2$. It easily follows that $\beta,\gamma$ is the required pair.

# 4517.6.2015

## 45.1E. Thuemmel: The Prikry problem

The idea presented during the last seminars was to attack the Prikry problem, i.e. the problem of the existence of a ccc forcing which doesn't add neither a Cohen real nor a Random real, was to find such a forcing in the form $P_\pi$. We know that if $Q_\pi$ (recall Definitions 29.7 and 33.1) is ccc then $P_\pi$ is Y-cc so, in particular, it is ccc and does not add a Random real. However, we aim to show that this cannot work:

45.1

If $Q_\pi$ is atomless and ccc and $P_\pi$ is atomless then $P_\pi$ adds a Cohen real.

45.2

$\pi^i(x)=\{y\in X:\pi(x,y)=i\}$.

45.3

The forcing $Q_\pi$ is ccc iff $|\pi^1(x)|\leq\omega$ for each $x\in X$.

$\rightarrow$. Aiming towards a contradiction assume that $Q_\pi$ is ccc and there is an $x\in X$ and a set $Y=\{y_\alpha:\alpha<\omega_1\}$ such that $pi[\{x\}\times Y]=1$. Then $\{\{x,y_\alpha\}:\alpha<\omega_1\}$ is an antichain giving us the desired contradiction.

$\leftarrow$. Let $D=\{p_\alpha:\alpha<\omega_1\}\subseteq Q_\pi$. We will show that it is not an antichain. We may assume that $D$ forms a delta system with root $R$, $|p_\alpha|=n$ for all $\alpha$ and some $n$ and that $\pi$ behaves uniformly on each $p_\alpha$. By our assumption $\pi[R\times(p_\alpha\setminus R]=0$ for each $\alpha<\omega_1$. Using the assumption again there is some $\alpha<\omega_1$ such that $\pi[p_0\setminus R\times p_\alpha\setminus R]=0$. Then $p_0$ and $p_\alpha$ are compatible.

of Theorem

Let $C=\{ A\in[X]^\omega:\pi[A\times X\setminus A]=0\}$.

45.4

The family $C$ is a club.

of Claim

The fact that $C$ is closed is immediate. Unboundedness follows from the previous lemma.

Now let $D=\{A\in[X]^\omega: P_\pi\upharpoonright A\ \mbox{is atomless}\}$.

45.5

The family $D$ is a club.

45.6

For any $A\in C$ the forcing $P_\pi(A)$ is a regular subforcing of $P_\pi$. (Hence also $P_\pi\simeq P_\pi(A)\times P_\pi(X\setminus A)$).

Now pick any $A\in D\cap C$. Then $P_\pi(A)$ is regularly embedded in $P_\pi$. But $P_\pi(A)$ is countable and atomless so it is, in fact, Cohen forcing.

45.7

If $P_\pi$ has uniform density. Consider the Boolean algebra $B_\pi$ (finitely) generated by $P_\pi$ in its completion. Given an $A\in C$ we know that $P_\pi(A)$ is regularly embedded in $B_\pi$ and so we can consider it to be a countable subset of $B_\pi$. It is easy to see that the mapping $\Phi:C\to[B_\pi]^\omega$ given by $\Phi(A)=\langle P_\pi(A)\rangle_{B_\pi}$ is continuous. This allows us to argue that $\Phi[C]$ is a club. So $B_\pi$ has a club of regularly embedded countable subalgebras so it is Semi-Cohen (originally defined by Jech and Fucino, see also the paper by Balcar, Jech and Zapletal or Koppelberg's article in the Handbook). The following theorem even allows us to conclude that $P_\pi$ is, in fact, a long Cohen forcing:

45.8Bandlowsee Balcar, Jech, Zapletal

An algebra $B$ is (a possibly long) Cohen iff $\{C\in[B]^\omega:C\ \mbox{is a regular subalgebra of}\ B\}$ contains a club $C$ such that the Boolean algebra generated by $D_1,D_2$ is in $C$ for each $D_1,D_2\in C$.

Hence we have the following theorem:

45.9

If $Q_\pi$ is ccc and has uniform density $\kappa$ then it is isomorphic to $C(\kappa)$.

It is still not clear, however, whether $Q_\pi$ which are disjoint-ccc could not, in theory, work.

## 45.2D. Chodounský: The new version of the Y-cc paper.

45.10

A forcing $P$ is suitable if there is a function $w$ assigning to each condition of $P$ a finite set such that

1. if $p\leq q$ then $w(p)\supseteq w(q)$,
2. if $p$ is compatible with $q$ then there is $r\leq p,q$ such that $w(r)=w(p)\cup w(q)$, and
3. if $\langle p^i_\alpha:\alpha<\omega_1,i<2\rangle$ are two sequences of conditions such that $\{w(p_\alpha^i):\alpha<\omega_1,i<2\}$ forms a delta system, then there are $\alpha,\beta<\omega_1$ such that $p_\alpha^0$ is compatible with $p_\beta^1$.
45.11

A suitable forcing is Y-cc.

The obvious question now is whether a suitable forcing adds a Cohen real.

## 45.3D. Chodounský: Raghavan's and Shelah's recent paper

The paper contains the following interesting theorem.

45.12

If $\kappa$ is regular uncountable then $\mathfrak s_\kappa\leq\mathfrak b_\kappa$.

However we will be more interested in a different theorem from the paper:

45.13

$cov^*(\mathcal Z_0)\leq\mathfrak d$

Recall that, when $\mathcal I$ is a tall ideal, $cov^*(I)$ is the size of the smallest hitting subfamily of $\mathcal I$.

45.14

Dilip claims he can replace $\mathfrak d$ with $\max\{\mathfrak b,\mathfrak s\}$. Hrušák can replace $\mathfrak d$ with $\max\{\mathfrak b,\mbox{non}(\mathcal N)\}$.

of Theorem

In the following a suitable partition of $\omega$ will be an interbal partition such that each interval has size some power of 2 and that their sizes strictly increase. Suppose $A\subseteq\omega$ satisfies:

1. $\lim_{n\to\infty}\frac{|A\cap I_n|}{|I_n|}=0$
2. $\lim_{n\to\infty} \max\{|a-b|:a,b\in A\cap I_n\}=\infty$ for some suitable partition. Then $A\in\mathcal Z_0$.
45.15

Given an interval $I$ of size $2^k$ and a function $\sigma\in {}^{ we let $I_\sigma=\{n\in I:n-\min I =\sigma\ (\mbox{mod}\ 2^{|s|})\}$ Given \(f:\omega\to2^{<\omega}$ and an interval partition we let $I_f = \bigcup_{n<\omega} I_{f(n)\upharpoonright\log |I_n|}$

45.16

If $|f(n)|\rightarrow\infty$ then $I_f$ satisfies 1. and 2.

45.17

Given two interval partitions $\mathcal I,\mathcal J$ we say that $\mathcal I$ dominates $\mathcal J$, $\mathcal J \leq \mathcal I$, if each interval in $\mathcal I$ contains an interval in $\mathcal J$.

45.18

Given a suitable interval parition there are functions $S=\{f_\alpha:\alpha<\omega_1\}$ such that each $f_\alpha:\omega\to2^{<\omega}$ has the property that $|f_\alpha(n)|=\log |I_n|$ and, moreover, for each $F\in[\omega_1]^{2^m}$, there are infinitely many $n$'s for which $I_n = \bigcup_{\alpha \in F} (I_n)_{f_{\alpha}(n)\upharpoonright m}$

If $\mathcal J$ is a suitable partition and $f:\omega\to2^{<\omega}$ define $f^{\mathcal J}$ as follows: $f^{\mathcal J}(n) = f(n)\upharpoonright k\iff n\in J_k.$

For each suitable partition $\mathcal I$ fix $\{f_{\alpha,\mathcal I}:\alpha<\omega_1\}$ as given by the lemma. Let $\mathcal D=\{\mathcal I_\beta:\beta<\mathfrak d\}$ be a dominating family of suitable partitions. Then the family consisting of the sets $\mathcal I_{f_{\alpha,\mathcal I}^{\mathcal J}}$ where both $\mathcal I$ and $\mathcal J$ are from $\mathcal D$ is hitting. Otherwise fix some $A$ which is not hit. Then for each $\mathcal I\in\mathcal D$ and $\alpha<\omega_1$ there must be $i_{\mathcal I,\alpha}<\omega$ such that $|\mathcal I_{f_{\alpha,\mathcal J}\upharpoonright i}\cap A|<\omega$ Fix a suitable partition $\mathcal I$ such that each element of $I$ contains a member of $A$. Find $m<\omega$ and $F\in[\omega_1]^{2^m}$ such that $i_{\mathcal I,\alpha}=m$ for each $\alpha\in F$. This however contradicts the properties of $f_{\alpha,\mathcal I}$ guaranteed by the lemma.

It remains to prove the lemma. It is sufficient to find suitable functions $\{f_\alpha:\alpha<\omega_1\}$ such that for each $m\leq 2^i, F\in[\omega_1]^m$ there is an infinite set $B\in[\omega]^\omega$ such that $f_\alpha(n)\upharpoonright i\neq f_\beta(n)\upharpoonright i$ for each $n\in B$ and $\alpha\neq\beta\in F$. This is a straightforward, although somewhat involved, induction.

# 4615.7.2015

## 46.1J. Zapletal: Kσ-cliques

46.1

$A\subseteq X$ is a $G$-anticlique (where $G\subseteq X^n$) if $A^n\cap G=\emptyset$.

46.2

Define the following cardinal invariant $\mathfrak{gr}_n$ to be the minimal size of a $G$-anticlique for some $F_\sigma$-graph in $X^n$ ($X$ a Polish space) which cannot be covered by a countable union of closed anticliques.

46.3

$\mathfrak{gr}_{n+1}\leq \mathfrak{gr}_n$ and the inequality is consistently strict.

46.4Zapletal

If $G$ is an $F_\sigma$-graph and $A$ a $G$-anticlique, then there is a $\sigma$-centered forcing which forces that it can be covered by a countable union of closed anticliques.

46.5

It follows that $\mathfrak p\leq \mathfrak{gr}_{<\omega}$.

46.6

Is it consistent that $\mathfrak p< \mathfrak{gr}_{<\omega}$.

## 46.2M. Hrušák: Incomparable families and trees in P(ω)/fin

46.7Geschke, Bouler

Does there exists a family $\mathcal I\subseteq[\omega]^\omega$ such that

1. $\mathcal I$ is closed under complements;
2. if $A,B\in\mathcal I$ then $A\not\subseteq^* B|$; and
3. for any two disjoint $I,J\subseteq\omega$ there is an $A\mathcal I$ such that $A\subseteq I$ or $A\subseteq J$ or it separates $I$ from $J$.

These families are called uniform self-dual matroids.

46.8

If $\mathfrak r=\mathfrak c$ then such a family exists.

Recall that $\mathfrak r$ is the size of the smallest family $\mathcal R$ such that for each infinite $A\in[\omega]^\omega$ there is $R\in\mathcal R$ such that $R\subseteq A$ or $R\cap A=\emptyset$.

46.9Monk

Is there a family of infinite subsets of $\omega$ which is maximal with respect to 2. of size $<\mathfrak c$ (i.e. a maximal family of incomparable elements of $\mathcal P(\omega)/fin$).

46.10

Each maximal family of incomparable (non $1$) elements of $\mathcal P(\omega)/fin$ has size $\mathfrak c$; so the answer to Monk's question is no.

Let $\mathcal I$ be an incomparable family of size $<\mathfrak c$. Pick $A\in\mathcal I$. Find $A_0\subseteq A$ such that no $B$ in the algebra $\langle\mathcal I\rangle$ splits $A_0$ (this can be done since $|\mathcal I|<\mathfrak c$). Find a similar $A_1\subseteq A\setminus A_0$. Then $\overline{A}=(A\setminus A_0)\cup A_1$ can be added to $\mathcal I$.

46.11

We will call $T\subseteq[\omega]^\omega$ a tree, if $(T,\supseteq^*)$ is a tree. We say that a tree $T$ end extends a tree $S$ if $T\subseteq S$ and each node in $S\setminus T$ is either above a leaf of $T$ or is only comparable with $\omega$ in $T$. A tree is maximal if it has no proper end-extension.

46.12

Disjoint nodes in $T$ need not be almost disjoint, just $\subseteq^*$ incomparable.

46.13Monk

Let $\mathfrak{tr}$ be the minimal size of a maximal tree.

46.14Monk

Is it consistent that $\mathfrak{tr}<\mathfrak c$?

46.15

It is consistent that $\mathfrak{tr}<\mathfrak c$.

46.16

If there is a completely separable mad family then there is a maximal tree of height $2$.

46.17

A base tree is a maximal tree.

46.18

Is $\mathfrak d\leq\mathfrak{tr}$?

46.19

It is easy to see that $\mathfrak{tr}\leq\mathfrak r$.

46.20Hrušák, Brendle

$\Diamond(\max\{\mathfrak{d,r}\})$ implies that there is a maximal tree of size $\omega_1$.

46.21

If $T$ contains an infinite A.D. family then $|T|\geq\mathfrak d$.

46.22

There is no maximal tree all of whose levels would be countable.

46.23

Is (it consistent that) there is a maximal $T$ of width $\omega_1$ and height $\omega_2$?

## 46.3Balcar's question

46.24Simon

There is an independent splitting family.

The following can be found in J. Brendle: There is an independent splitting family.

46.25Brendle

There is an independent splitting family of size $\mathfrak s$.

Let $\mathcal S=\{S_{\alpha,\beta}:\alpha<\mathfrak s,\beta<\mathfrak t\}$ be a splitting family and $T=\{T_\beta:\beta<\mathfrak t\}$ a tower and $\mathcal I=\{I_{\alpha,\beta}:\alpha,\beta<\mathfrak s\}$ an independent family. Let $J_{\alpha,\beta} = \Big((T_\alpha\times I_{\alpha,\beta}) \cup(S_\alpha\times\omega)\Big) \cap \bigtriangleup,$ where $\bigtriangleup=\{(n,m):m\leq n\}$.

## 46.4Hrušák: Miscellanea (Interlude)

46.26

Consider the ideal $\mathcal W^\prime$ of sets which do not contain, for any $n$, arbitrarily long arithmetic sequences with difference $n$.

46.27

Is $\mathcal P(\omega)/\mathcal W^\prime$ proper?

46.28Laflamme

Each $F_\sigma$-ideal can be destroyed by an $\omega^\omega$-bounding forcing.

46.29

Can every $F_{\sigma\delta}$ ideal be destroyed by a forcing not adding a dominating real?

46.30

First add $\omega_1$-many Cohen reals and then force with $F_\sigma(\mathcal I)$ and then force with Mathias of the generic ultrafilter, where $F_\sigma(\mathcal I) = \{\mathcal J: \mathcal J\ \mbox{is}\ F_\sigma\ \langle\mathcal J\cup\mathcal I\rangle\cap \mathcal I^*=\emptyset\}$

46.31

Without first adding $\omega_1$-many Cohens the above forcing adds a dominating real.

46.32Raghavan, Shelah

Killing density zero always adds an unbounded real.

46.33Folklore

It is consistent that all towers generate a meager filter.

Force with, e.g., Hechler; having $\mathfrak b>\omega_1$ and all towers of cofinality $\omega_1$ is sufficient.

46.34

It is consistent that every tower generates a non-meager filter. {} Proof: Start with CH and force $MA(\sigma-\mbox{centered})$ by a finite support iteration of $\sigma$-centered partial ordered sets.

46.35Chodounsky

Does there exist a meager tower and $\mathfrak t=\omega_1$?

46.36

Does $\mathfrak t=\omega_1$ imply that there is a tight gap?

Recall that a tight gap is a gap consisting of functions which has no partial separator, i.e. a system $\{(f_\alpha,g_\alpha):\alpha<\omega_1\}$ such that the first coordinates are $\leq^*$-increasing, the second are $\leq^*$-decreasing and there is no partial function $s:A\to\omega$ such that $f_\alpha\upharpoonright A\leq^*s\leq^*G_\alpha\upharpoonright A$.

# 475.8.2015

## 47.1M. Doucha: Continuous Scott Analysis

Joint work with A. Nies, I. Ben Yaacov and T. Tsankov. We consider the logic $L_{\omega_1,\omega}$ allowing finitely many quantifiers and countably many conjunctions. A classical result of Dana Scott says that every countable structure can be described by just a single sentence in this logic.

47.1

Given a formula $\varphi$ in $L_{\omega_1,\omega}$ we define its quantifier rank, $qr(\varphi)$, of $\varphi$ as follows: the rank of atomic formulas and their negations is $0$; the rank of a disjunction is the supremum of the ranks of the disjuncts; and the rank of $(\exists x)\varphi$ is one plus the rank of $\varphi$.

47.2

Given a countable language $\mathcal L$, a countable $\mathcal L$-structure $M$ and an ordinal $\alpha<\omega_1$ we say that two tuples $\overline{a},\overline{b}\in M$ are $\alpha$-equivalent, denoted by $\overline{a}\equiv_\alpha\overline{b}$, if they satisfy the same formulas of rank at most $\alpha$.

47.3

Given $M$ as above and $\overline{a}\in M^n$ we let $\varphi^{M,\overline{a}}_0$ be the infinite conjunction of rank 0 formulas satisfied by $\overline{a}$. If $\alpha<\omega_1$ is limit we let $\varphi^{M,\overline{a}}_\alpha$ be the conjunction of formulas $\varphi^{M,\overline{a}}_\beta$ for $\beta<\alpha$. Finally if $\alpha=\beta+1$ we let $\varphi^{M,\overline{a}}_\alpha = \varphi^{M,\overline{a}}_\beta \wedge \bigwedge_{b\in M}(\exists v)\varphi^{M,\overline{a}b}_\beta(v) \wedge (\forall v)\bigvee_{b\in M}\varphi^{M,\overline{a}b}_\beta(v).$

47.4

Two tuples $\overline{a},\overline{b}$ are $\alpha$-equivalent iff $\varphi^{M,\overline{a}}_\alpha(\overline{b})$.

47.5

Two tuples $\overline{a},\overline{b}\in M$ are $\alpha$-equivalent iff player II. has a winning strategy in the Ehrenfeucht-Fraïssé game $EF_\alpha((M,\overline{a}),(M,\overline{b}))$.

47.6

The Scott rank of $M$, denoted by $sr(M)$, is $\min\{\alpha<\omega_1: \overline{a}\equiv_\alpha\overline{b}\rightarrow \overline{a}\equiv_{\alpha+1}\overline{b}, \overline{a},\overline{b}\in M\}$ and the Scott sentence of $M$ is the formula $\varphi^{M,\emptyset}_{sr(M)}\wedge \bigwedge_{\overline{b}\in M} \big( (\forall v)( \varphi^{M,\overline{b}}_{sr(M)}(\overline{b})\rightarrow \varphi^{M,\overline{b}}_{sr(M)+1}(\overline{b}) ) \big)$

47.7Scott

If $N$ satisfies the Scott sentence of $M$ then it is isomorphic to $M$.

Our result is a similar theorem for continuous logic.

## 47.2M. Doucha: Uniform topological groups

(Sometimes also called balanced groups or SIN, i.e. small invariant neighbourhoods, groups.)

47.8

A topological group $G$ is called uniform if there is a compatible uniformity on $G$ such that the group operations are uniformly continuous.

Recall that a topological group $G$ carries two natural uniformities: the left and right uniformity $u_L$ and $u_R$ generated, respectively, by the entourages $\{(g,h):g^{-1}h\in\mathcal O\}$, $\mathcal O$ a symmetric neighbourhood of the identity; and $\{(g,h):gh^{-1}\in\mathcal O\}$, $\mathcal O$ a symmetric neighbourhood of the identity.

47.9

A topological group $G$ is uniform if one of the following equivalent conditions is satisfied:

1. $u_L=u_R$; or
2. the uniform completion w.r.t. $u_L$ (or $u_R$) is a group; or
3. there is a system $\mathcal O_\alpha$ of neighbourhoods of the identity forming a basis at the identity such that $(\forall \alpha)(\forall g\in G) (a^{-1}\mathcal O_\alpha a=\mathcal O_\alpha)$.

If, moreover, $G$ is metrizable, then the above are equivalent to

1. there is a bi-invariant compatible metric on G, i.e. $(\forall x,y,a,b\in G)d(x,y)=d(axb,ayb)$.
47.10

The following are straightforward examples:

1. All Abelian groups are uniform.
2. All compact groups are uniform.
47.11

A Heisenberg group, i.e. a group of 3-by-3 matrices with 1s on the diagonal, arbitrary reals above and zeros below. This group is metrizable so to show that it is not uniform it is sufficient to show that there cannot be a bi-invariant compatible metric.

47.12

If $G$ is a group and $(M,d)$ a bounded metric space such that $G$ faithfully acts on $M$ by isometries then we can define on $G$ a metric $D(g,h)=\sup\{d(gm,hm):m\in M\}.$ This is a bi-invariant metric on $G$ so $(G,D)$ is uniform. Note that if $(G,d)$ is a group with a bi-invariant metric then left-multiplication is a faithful action of $G$ on $G$ giving rise to the same metric.

47.13

There is a universal separable Abelian group with a bi-invariant metric. I.e. each separable Abelian group with a bi-invariant metric is isometric to a subgroup of this universal group.

47.14

There is a separable group with a bi-invariant metric bounded by $1$ that is universal for all separable groups with 1-bounded bi-invariant metrics. This universal group is a second countable uniform group.

On the other hand there is no separable universal group for groups with a uniform unbounded bi-invariant metrics.

47.15

A uniform Banach group is a Banach space $X$ with an additional group structure with uniformly continuous group operations (w.r.t. the norm).

47.16

There is a uniform Banach group which has $F_\infty$ as dense subgroup, where $F_\infty$ is the free group on countably many generators. In particular it is not Abelian.

## 47.4D. Chodounský: Work with S. Geschke

47.17

What is the relationship between the two right and left-shift dynamical systems on $\omega^*$?

47.18

Consistently, e.g. under PFA, the above are different, i.e. there is no automorphism $\varphi:\omega^*\to\omega^*$ such that $\varphi^{-1}\circ sh_L\circ \varphi=sh_R$. It is not too hard to see that an automorphism satisfying this condition cannot be trivial. However, what about under CH? What if there is a $P$-point with character $\omega_1$?

# 4812.8.2015

## 48.1B. Farkas: Towers in filters

The following is joint work with Joerg Brendle.

48.1

Given an ideal $\mathcal I$ the star-uniformity of $\mathcal I$ is defined to be: $\mbox{non}^*(\mathcal I)= \min\{|\mathcal X|:\mathcal X\subseteq[\omega]^\omega\ \&\ (\forall A\in\mathcal I)(\exists X\in \mathcal X) (|A\cap X|<\omega) \}$

48.2

If $\mathfrak t\leq\mbox{non}^*$ then $\mathcal I$ cannot contain a tower.

48.3

The Random graph is a countable graph $(\omega,E)$ such that for all finite $F,H\in[\omega]^{<\omega}$ there is a vertex $v\in\omega$ such that each $w\in F$ is connected to $v$ and each $w\in H$ is not connected to $v$. The ideal $\mathbf{Ran}$ is the ideal on $\omega$ generated by homogeneous subsets of the Random graph. The star-uniformity of the ideal Ran is $\omega$: since the Random graph is universal, it contains a copy of $\omega\times\omega$ where each column $\{n\}\times\omega$ is a clique and there are no edges between different columns.

48.4

The Solecki's ideal $\mathcal S$ has countable star-uniformity.

48.5

The ideal $\mathbf{Nwd} = \{A\subseteq[0,1]\cap\mathbb Q:int(\overline{A})=\emptyset\}$ which is $F_{\sigma\delta}$ of nowhere dense subsets of rationals and the ideal $\mathbf{Conv}$ generated by convergent sequences of rationals, which is $F_{\sigma\delta\sigma}$, have countable star-uniformity.

48.6

The ideal $\mathcal{ED}$ is defined to be $\mathcal{ED} = \{A\subseteq\omega\times\omega:\lim\sup |A_n|<\omega\},$ where $A_n=\{k:(n,k)\in A\}$ is the vertical section of $A$ at $n$. This ideal has countable star-uniformity.

48.7

If $\mathcal I\leq_{KB}\mathcal J$ then $\mbox{non}^*(\mathcal I)\leq\mbox{non}^*(\mathcal J)$.

The following diagram shows various $KB$-connections between different ideals:

48.8

We know that $\mathcal W\not\leq_{KB}\mathcal G_{fc}$ but we do not know whether $\mathcal W\leq_{K} G_{fc}$.

48.9

M. Hrušák and his students have been able to reduce the question whether $\mathbf{Ran}\leq\mathcal S$ to the following statement: $(\forall k\in\omega)(\exists n\in\omega) (\forall \chi:[2n]^n\to 2)(\exists c\in 2) (\forall x\in[2n]^k)(\exists a\in[2n]^n) (\chi(a)=c\ \&\ x\subseteq a)$

48.10

The ideal $\mathcal{ED}_{fin}$ is the ideal $\mathcal{ED}$ restricted to $\Delta=\{(k,n):n\leq k\}$.

48.11

The following are equivalent:

1. $\mathcal{ED}_{fin}\leq_{KB}\mathcal I$;
2. $\mathcal I$ is not weak Q, (i.e. if for each partition of; and $\omega$ into finite sets there is a $\mathcal I$-positive selector)
3. (if $\mathcal I$ is Borel) $\mbox{non}^*(\mathcal I)\geq\mathfrak{t}$.
48.12

If $\mathcal I\leq_{KB}\mathcal J$ and $\mathcal I$ contains a tower then $\mathcal J$ contains a tower.

48.13

The van der Waerden ideal $\mathcal W$ does not contain a tower.

Assume, aiming towards a contradiction, that $\langle A_\alpha:\alpha<\kappa\rangle$ is a tower in $\mathcal W$. Let $ap(A)=\sup\{|P|:P\subseteq A\ \&\ P\ \mbox{is an arithmetic progression}\}$. Next let $ap^*(A)=\min\{ ap(A\setminus n):n<\omega\}$. It is easy to see that $ap^*(A)\leq ap^*(B)$ for $A\subseteq^* B$. It follows that, wlog, there is $N<\omega$ such that $ap^*(A_\alpha)=N$ for all $\alpha<\kappa$. Recursively choose $N$-long arithmetic progressions $A_n$ in $A_0$ and $b_n$'s such that $A_n^{\smallfrown} b_n$ is an arithmetic progression and $b_n. If we let \(B=\{b_n:n<\omega\}$ then $\omega\setminus B$ is a pseudo-union of the tower --- a contradiction.

48.14

Given $A\subseteq A^{<\omega}$ its $G_\delta$ closure is $[A]=\{x\in2^\omega:(\exists^\infty n)(x\upharpoonright n\in A)\}.$ The ideal $\mbox{tr}(\mathcal N)$ is $\mbox{tr}(\mathcal N) = \{A\subseteq 2^{<\omega} [A]\in\mathcal N\},$ where $\mathcal N$ is the $\sigma$-ideal of Lebesgue measure zero subsets of $2^\omega$.

Recall Mazur's and Solecki's characterization of definable ideals:

48.15Mazur, Solecki

Let $\mathcal I$ be an ideal then:

1. $\mathcal I$ is $F_\sigma$ iff there is a lscsm $\varphi$ such that $\mathcal I=Fin(\varphi)$ (Mazur)
2. $\mathcal I$ is an analytic P-ideal iff there is a lscsm $\varphi$ such that $\mathcal I = Exh(\varphi)$ (Solecki),

where $\varphi$ is a lscsm (lower semicontinuous submeasure) if

1. $\varphi(\emptyset)=\emptyset$;
2. $\varphi(A\cup B)\leq\varphi(A)+\varphi(B)$;
3. $\varphi(A)\leq\varphi(B)$ for $A\subseteq B$ and
4. $\varphi(A)=\sup\{\varphi(A\cap n):n<\omega\}$

and $Fin(\varphi)=\{A\subseteq\omega:\varphi(A)<\infty\}$ and $Exh(\varphi)=\{A\subseteq\omega:\varphi(A\setminus n)\rightarrow 0\}$.

48.16

Even more is true: an ideal $\mathcal I$ is an $F_\sigma$ P-ideal iff there is a lscsm $\varphi$ such that $\mathcal I=Fin(\varphi)=Exh(\varphi)$.

48.17

$\mbox{tr}(\mathcal N)=Exh(\varphi)$ where $\varphi(A)=\sum\{2^{-s}:s\in A,\ s\ \mbox{is}\ \subseteq-\mbox{minimal}\}.$

It is easy to see that under CH all P-ideals contain a tower (they are even generated by a tower). This can be generalized:

48.18

After adding $\omega_1$-many Cohen reals every tall analytic P-ideal contains a tower.

Let $\mathcal I=Exh(\varphi)$ be a tall analytic P-ideal. Wlog assume $\varphi$ is in the ground model (it has to appear at some stage since it is coded by a real --- the submeasure $\varphi$).

48.19

After adding a Cohen real there is $A\in\mathcal I$ such that for each ground-model $X\in[\omega]^\omega$ the intersection $|X\cap A|$ is infinite.

(Note that for the claim it is sufficient that $\mathcal I$ is (co)analytic and is KB-above $\mathcal{ED}_{fin}$; note that all tall analytic P-ideals are KB-above $\mathcal{ED}_{fin}$.)

Recursively use the claim to construct a $\subseteq^*$-increasing sequence $X_\alpha\in V_\alpha$ in $\mathcal I$ such that $X^\prime_{\alpha}$ is a pseudo-union of $\langle X_\beta:\beta<\alpha\rangle$ and $X_\alpha= X^\prime_{\alpha}\cup A_\alpha$ where $A_\alpha$ is given by the claim.

The above proof does not work for longer extensions. So we now show how to build long towers in tall analytic P-ideals.

48.20

A forcing notion $P$ destroys an ideal $\mathcal I$ if it forces an infinite set disjoint from each $X\in\mathcal I\cap V$. It dominates a Borel ideal $\mathcal I$ if it forces a pseudounion of $\mathcal I\cap V$ into $\mathcal I$.

48.21

If $\mathcal I$ is not a P-ideal, it cannot be dominated.

48.22

If a forcing dominates an ideal, a simple iteration will force a tower into the ideal.

48.23

The Localization forcing (Loc) dominates all tall analytic P-ideals.

48.24

The proof of the lemma is essentially proving that $\mbox{cof}(\mathcal I)\leq\mbox{cof}(\mathcal N)$ for all tall analytic P-ideals $\mathcal I$. All that is needed for the proof is that Loc is ccc and adds a slalom which almost catches all groundmodel functions.

Let $S$ be the slalom added by Loc. Enumerate $Fin=\{F_k:k<\omega\}$ and let $\mathcal I=Exh(\varphi)$ in the ground model. Work in the extension and let $A_S = \bigcup_{n<\omega}\bigcup\{F_k:k\in S(n)\ \&\ \varphi(F_k)<2^{-n}\}.$ It is easy to see that $A_s\in\mathcal I$. Moreover, if $B\in\mathcal I\cap V$, then $B\subseteq^* A_S$: let $d(n)=\min\{k:\varphi(B\setminus k)<2^{-n}\}$ and $I_n= B\cap[d(n),d(n+1)]$. Then $B$ is almost included in the union of the $I_n$s. However $I_n$ is finite so there is $c:\omega\to\omega$ such that $I_n=F_{c(n)}$. By assumption the measure of $F_{c(n)}$ is small. The proof is finished by noting that $c$ is caught by the slalom.

48.25

If $R$ is a set of regular cardinals then towers of length in $R$ can be simultaneously forced into any tall analytic P-ideal.

48.26

In the above theorem it is not clear what happens outside of $R$.

48.27

Does the tower-spectrum $S_{tr}(\mathcal I)$, i.e. the set of regular lengths of towers contained in $\mathcal I$, have some closure properties?

The main question, however, is:

48.28Hrušák's Conjecture

Assume $\mathcal I$ is Borel ($F_\sigma$) and $\mathcal I$ contains a tower. Does $\mathcal I$ contain a tall analytic P-ideal?

48.29

Assume $V\models CH$. Then the $\omega_2$-stage FS iteration of Mathias or Laver forcings corresponding to $\mathcal I^*$ forces that there are no towers in $\mathcal I$.

48.30

If $P$ is $\sigma$-centered and $\mathcal I$ is $F_\sigma$ then a $\subseteq^*$-decreasing sequence in $\mathcal I^*$ with no pseudointersection in $\mathcal I^*$ still has no pseudointersection in $\mathcal I^*$ after forcing with $P$. If $P$ is an analytic P-ideal then any $\subseteq^*$-decreasing sequence $\mathcal I^*$ with no pseudointersection $X$ such that $||\omega\setminus X||_\varphi<||\omega||_\varphi$, where $||X||_\varphi = \lim_{n\to\infty} \varphi(X\setminus n),$ still has no such pseudointersection after forcing with $P$.

We prove only the second part. Let $\langle A_\alpha:\alpha<\kappa\rangle$ be a $\subseteq^*$-descending sequence with the above property. Let $P=\bigcup_{n<\omega}P_n$ with each $P_n$ centered and let $\dot{X}$ be a name such that $P\Vdash||\omega\setminus\dot{X}||_\varphi<||\omega||_\varphi$. So there is $N<\omega$ and $\varepsilon >0$ such that $P\Vdash \varphi(\omega\setminus(\dot{X}\cup N)<||\omega||_\varphi-\varepsilon.$ Let $X_n=\{m:(\forall p\in P_n)(p\not\Vdash m\not\in \dot{X})\}.$ It is not hard to show that $||\omega\setminus X_n||_\varphi<||\omega||_\varphi$. Then $X_n$ cannot be a pseudointersection of the sequence, so we can find $\alpha_n$ such that $X_n\not\subseteq A_{\alpha_n}$. Since $\mathcal I^*$ P-ideal the supremum $\alpha=\sup_{\alpha_n:n<\omega}$ is shorter than the length of the sequence. It is not hard to show that $P$ forces that $\dot{X}$ is not almost contained in $A_\alpha$.

48.31

Is it possible that $\mathcal Z$ contains a tower but $\mathcal I_{1/n}$ does not?

48.32

Assume $\mathcal{I,J}$ are two tall analytic P-ideals and $\mathcal I\not\leq_KB\mathcal J\upharpoonright X$ for any $\mathcal J$-positive $X$ then Laver forcing with $\mathcal J^*$ does not add a pseudointersection to $\mathcal I^*$.

The problem with the above theorem is that we do not know how to preserve towers in $\mathcal I^*$ (we do not add pseudointersections to the whole ideal, but that is not sufficient).

# 4911.11.2015

## 49.1Victor Torres-Perez: Shelah's Semi-Stationary Reflection Principle

49.1

Suppose we have a family of intervals of some linear order. Under what conditions can we color these intervals by two colors such that same-colored intervals are disjoint?

49.2

A family $\mathcal F$ of intervals is $n$-disjoint if there is a $n$-coloring of $F$ such that intervals of the same color are disjoint.

A family $\mathcal F$ of intervals is $n$-disjoint iff every subfamily of size $n+1$ is two-disjoint.

A family $\mathcal F$ of intervals is $\aleph_0$-disjoint iff every subfamily of size at most $\aleph_1$ is $\aleph_0$-disjoint.

49.5Todorčević,84

Rado's conjecture is independent of ZFC modulo a supercompact cardinal.

49.6Todorčević,94

Rado's conjecture implies the SCH (singular cardinal hypothesis), $2^\omega\leq\aleph_2$, the strong Chang Conjecture, $\neg\square(\kappa)$ for $\kappa\geq\aleph_2$;

Recall, that

49.7

For an uncountable cardinal $\kappa$ the sequence $\langle C_\alpha:\alpha\in\mbox{Lim}(\kappa^+)\rangle$ is a $\square_\kappa$-sequence if $C_\alpha$ is club in $\alpha$ of ordertype at most $\kappa$ and the $C_\alpha$'s cohere in the sense that $C_\alpha\cap\beta=C_\beta$ for $\beta\in\mbox{Lim}(C_\alpha)$. It is a $\square(\kappa^+)$-sequence if instead of the condition on ordertype we require that it is not a trace of some club on $\kappa^+$. The principles $\square(\kappa^+)$ and $\square_\kappa$ assert, respectively, that a $\square(\kappa^+)$ and a $\square_\kappa$ sequence exists.

49.8

$\square_\kappa\Rightarrow\square(\kappa^+)$

49.9

Strong Chang's Conjecture says that there are arbitrarily large $\theta$ such that for all $\lambda\geq\omega_2$, $M\preceq\langle H_\theta,\ldots\rangle$ countable, $a\in[\lambda]^\omega$, there is a countable $M^*\preceq\langle H_\theta,\ldots\rangle$...

### 49.1.1Rado's conjecture vs. Martin's Maximum

Recall that Martin's Maximum (MM) says that for every stationary set-preserving forcing and $\aleph_1$-many dense-sets there is a filter meeting all of them. MM also implies all of the consequences in the above theorem (and also the weak reflection principle). However, Rado's conjecture implies the negation of even $\mbox{MA}_{\omega_1}$.

49.10Schimerling

A $\square_{\kappa,\mu}$-sequence is a sequence $\langle \mathcal C_\alpha:\alpha\in\mbox{Lim}(\kappa^+)\rangle$ such that each $\mathcal C_\alpha$ consists of club subsets of $\alpha$ of ordertype at most $\kappa$, $|\mathcal C_\alpha|\leq\mu$ and for each $C\in\mathcal C_\alpha$ and $\beta\in\mbox{Lim}(C)$ we have $C\cap\beta\in\mathcal C_\beta$.

49.11

Martin's Maximum implies 1. $\neg\square_{\lambda,<\mbox{cf}(\lambda)}$, $\lambda\geq\omega_1$; 2. $\neg\square_{\omega_1,\omega_1}$ (due to Baumgartner); 3. $\neg\square_{\lambda,\lambda}$, for $\lambda$ singular of cofinality $\omega$; and 4. $\neg\square_{\lambda,<\lambda}$, for $\lambda$ of cofinality $\omega_1$.

49.12

The following are consistent, modulo a supercompact cardinal: 1. MM and $\square_{\lambda,\mbox{cf}(\lambda)}$ for all $\lambda$ of cofinality $>\omega_1$; and 2. MM and $\square_{\lambda,\lambda}$ for all $\lambda$ of cofinality $\omega_1$.

For the following recall that CH implies $\square_{\omega_1,\omega_1}$ and that Rado's Conjecture implies $2^\omega\leq\omega_2$. In particular if we want to negate the square under Rado's Conjecture, we need $2^\omega=\omega_2$.

49.13Todorčević,Torres

Assume that Rado's Conjecture is true then 1. $\neg\square_{\lambda,\mbox{cf}(\lambda)}$ for $\lambda\geq\omega_1$; 2. $\neg\square_{\lambda,\lambda}$ for $\lambda$ of countable cofinality; and 3. $\neg\mbox{CH}$ implies $\neg\square_{\omega_1,\omega_1}$.

49.14Döbler, Schimmerling

Strong Chang's Conjecture is equivalent to the Semi-Stationary Reflection Principle (SSR).

Note that MM implies WRP which implies SSR.

49.15Sakai

SSR implies the missing negations of $\square$ (as well as 1--3).

49.16Sakai,Velickovic,2014

SSR implies $\neg\square(\theta)$ if $\theta\geq\omega_2$ is regular.

49.17Torres,Wu

$\neg\square(\theta,\omega)$ if $\theta\geq\omega_2$.

49.18Torres,Wu

PID implies $\neg\square(\theta,\omega)$ if $\theta\geq\omega_2$.

49.19

Given two sets of ordinals $x,y$ we write $x\sqsubseteq y$ if $x\subseteq y$ and $x\cap\omega_1=y\cap\omega_1$.

49.20

Given $\lambda\geq\omega_2$ we call a set $S\subseteq[\lambda]^\omega$ semi-stationary if $\{y\in[\lambda]^\omega:(exists x\in S)(x\sqsubseteq y)\}$ is stationary.

49.21

The Semi-Stationary Reflection Principle (SSR) principle says that for every semi-stationary $S\subseteq[\lambda]^\omega$ there is a $W\in[\lambda]^{\omega_1}$ containing $\omega_1$ such that $[W]^\omega\cap S$ is semi-stationary.

49.22

For $\lambda=\omega_2$ SSR is equivalent to WRP and for larger $\lambda$ it is implied by WRP but does not imply it (Sakai).

49.23Torres,Wu

SSR for $\lambda\geq\omega_2$ imples $\neg\square(\lambda,\omega$).

Assume that there is a square sequence $\langle \mathcal C_\alpha:\alpha\in\mbox{Lim}(\lambda)$. We use the following lemma due to Sakai and Velickovic:

49.24

If there is a stationary, "weakly full" $X\subseteq[\lambda]^\omega$ such that $(\forall I\in[\lambda]^{\omega_1}) (\exists J\supseteq I, J\in[\lambda]^{\omega_1},\mbox{sup}(I)=\mbox{sup}(J), X\cap[J]^\omega\ \mbox{is nonstationary})$ then SSR does not hold at $\lambda$.

We use the square sequence to define the set $X$ required by the lemma: $X=\{ x\in[\lambda]^\omega: \mbox{sup}(x)\in\mbox{Lim}(\lambda)\ \&\ (\forall C\in\mathcal C_{\mbox{sup}(x)})(\xi^x_C=\mbox{sup}(x\cap C)<\mbox{sup}(x)\}\ \&\ (\forall\beta\in C_{\mbox{sup}(x)}\setminus \xi^x_C)(\mbox{cf}(\min(\lambda\setminus\beta))=\omega_1) \}$

49.25

$X$ is stationary.

of Claim

Take $F:[\lambda]^{<\omega}\to\lambda$. We need to find $x\in X$ closed under $F$. To do this we play the following game where the players alternate in playing triples of ordinals as follows. Player one plays $\alpha$, player two plays $\beta$ and then player one plays $\gamma$ of cofinality $\omega_1$ above $\alpha,\beta$. Player one wins iff the closure points of the $\gamma$'s under $F$ do not intersect the interval $[\alpha,\beta)$. By a lemma due to Sakai and Velickovic, player one has a winning strategy. Let $\tau$ be a winning strategy for player one and consider $C\subseteq\lambda$ a set of ordinals closed under $F$ and $\tau$. We will now play according to $\tau$, choosing suitable $\beta$s. We use the following key lemma

49.26

There is a sufficiently large $\theta$ and a countable elementary submodel $M\preceq\langle H_\theta,\ldots\rangle$ such that $\delta=M\cap\lambda$ is an ordinal of cofinality $\omega$ and there is an increasing sequence of ordinals $\{\delta_n:n<\omega\}$ cofinal in $\delta$ such that $\delta_n\in C\setminus\bigcup_{i\leq n} \mbox{Lim}(C^\delta_i)$ (recall, that $\mathcal C_\delta=\{C^\alpha_i:i<\omega\}$).

Given $\delta_n$ there is $\beta_n<\delta_n$ such that $C^\delta_n\cap[\beta,\delta_n)=\emptyset$. Then player 1 responds by playing $\gamma$ and next $\alpha$. At the end we let $x$ consist of the closure points of the $\gamma$'s under $F$.

The fact that $X$ is weakly full is rather easy and finishes the proof.

# 5213.1.2016

## 52.1Š. Stejskalová: Itay Neeman: Forcing with sequences of models of two types. (II)

Recall that we are working with conditions which are sequences of models, i.e. we have $H(\lambda)$,$\mathcal S$ a system of countable elementary submodels of $H(\lambda)$ and $\mathcal T$ a system of transitive elementary submodels of $H(\lambda)$. Moreover given $W\in\mathcal T$ and $Q\in\mathcal S$ with $W\in S$ we have $W\cap Q\in\mathcal S$. In the following, unless otherwise stated, the letter $W$ will denote models in $\mathcal T$ and the letter $Q$ models in $\mathcal S$. A condition in our forcing $I(\mathcal S,\mathcal T)$ is a sequence $p=\{ M^p_i:i such that \(M^p_i\in M^p_{i+1}$ which is closed under intersection, i.e. $M^p_i\cap M^p_j\in p$. We order the conditions by reverse inclusion.

52.1

Given a condition $s$ and a model $Q\in s$ we define the residuum $res_Q(s)=s\cap Q$. Given $W\in s\cap \mathcal T$ and $Q\in\mathcal S$ we define the residual gap of $Q$ in $s$, denoted by $[Q\cap W,W)^s$ to consist of models $M\in s$ such that $Q\cap W\in M$ and $M\in W$.

Last time we proved the following facts.

52.2

The residuum $res_Q(s)$ is a condition.

52.3

If $t\in W$ for some $W\in\mathcal T$ and $s$ is a condition and $t\leq s$ then $t\cup s$ is a condition. If $W\in\mathcal S$ then $t\cup s$ need not be a condition but closing it under intersections will give a condition.

52.4

A set $s$ of models "linearly ordered" by $\in$ is a condition iff it is closed under intersection of the form $Q\cap W$ where $Q\in\mathcal S$, $W\in\mathcal T$ and $W\in Q$.

Note that $\in$ is not transitive, so "linearly ordered" in the above fact means that there is a linear ordering of $s$ such that the immediate successor of any $M\in s$ contains $M$ as an element.

I (Jonathan) am not quite sure the following lemma was actually stated in this way...

52.5

If $t$ and $s$ are conditions and $t\cup s$ is incresing in $\in$ then $\langle t\cup s\rangle$, i.e. closing $t\cup s$ under intersections of the form given in the previous fact, is a condition.

Suppose that $W\in t$ and, in $s$, there is no transitive model between $W$ and $Q\in\mathcal S$. Let $W^\prime$ be the first transitive model in $s\cup t$ above $Q$. Then define $E_W$ to be the interval $[Q,W^\prime)$, in $s$. Otherwise let $W^\prime$ be the first transitive model in $[W,Q)$ in $s$ and define $E_W=[Q\cap W^\prime,W^\prime)$. Finally let $F_W=\big\{M\cap W:M\in E_W\big\}.$ We claim that $r=s\cup t\cup \bigcup_{W\in t\cap\mathcal T} F_W$ is a condition. (The rest was too fast for me to follow.)

We aim to show that the forcing we are talking about is strongly-proper w.r.t. some models. First the relevant definitions.

52.6

Given a forcing $P$, an elementary submodel $M\preceq H(\theta)$ with $P\in M$ we say that a condition $q\in P$ is a master condition for $M$ if for every $D\in M$ dense in $P$ and $p\leq q$ the condition $p$ is compatible with some element of $D\cap M$ (i.e. $D\cap M$ is predense below $q$). The forcing $P$ is $M$-proper if for every $p\in M\cap P$ there is a stronger $q\leq p$ which is a master condition for $M$. The forcing is proper if it is $M$-proper for every countable elementary submodel of a sufficiently large $H(\theta)$. Similarly, given a family $\mathcal S$ of models we say that $P$ is $\mathcal S$-proper if it is $M$-proper for every $M\in\mathcal S$.

52.7

Given $P$, $M$ then an $M$-master condition forces that $\dot{G}\cap M$ is $M$-generic over $M\cap P$, equivalently that that there are no new ordinals in $M[G]$ equivalently there are no new $V$-sets in $M[G]$.

52.8

Given a forcing $P$ a condition $p\in P$ is a strong master condition for $M$ if for every $D$ dense in $P$ the intersection $D\cap M$ is pre-dense below $p$, i.e. if $p$ forces that $\dot{G}\cap M$ is (fully) generic over $M\cap P$. Similarly as above we define when $P$ is $M$-strongly proper and $\mathcal S$-strongly proper and strongly proper.

52.9

The relationship between proper and strongly proper is the same relation as between ccc and semi-Cohen.

The following, while probably not literally true, is nevertheless true in spirit.

52.10

A forcing $P$ is $M$-strongly proper iff $P\cap M$ is a regular subforcing of $P$.

52.11

Suppose $P$ is strongly $M$-proper and $G$ is a generic over $P$. Then

1. $M[G]\preceq H(\theta)[G]$ and $M[G]\cap V=M$; and
2. given a $P$ name $\dot{f}\in M$ for a function defined on some ordinal then $(\dot{f}\cap M)^G = \dot{f}^G\upharpoonright M$.
52.12

A set $C\subseteq [\lambda]^{<\kappa}$ is closed unbounded (or club) if it is closed under increasing unions of length $<\kappa$ and is unbounded. A set is stationary in $[\lambda]^{<\kappa}$ if it intersects every club in $[\lambda]^{<\kappa}$.

52.13

A set $S$ is stationary in $[\lambda]^{<\omega_1}$ iff for every function $f:\lambda^{<\omega}\to\lambda$ there is $x\in S$ which is closed under $f$. A set $S$ is stationary in $[\lambda]^{<\kappa}$ iff for every function $f:\lambda^{<\omega}\to\lambda$ there is $x\in S$ which is closed under $f$ and $x\cap Ord\in\kappa$.

52.14

If $P$ is $\mathcal S$-proper for some $\mathcal S$ stationary (unbounded should be sufficient) in $[H(\theta)]^{<\delta}$ then $\delta$ remains a cardinal in the extension.

52.15

Let $M\in\mathcal S\cup \mathcal T$ be a model and $t\in I(\mathcal S,\mathcal T)$ with $Q\in t$. Then $t$ is a strong master condition for $Q$. In particular $I(\mathcal S,\mathcal T)$ is $\mathcal S\cup\mathcal T$-strongly proper. {} Proof: Write $P=I(\mathcal S,\mathcal T$. Let $Q\in t$ and, aiming towards a contradiction, assume that there is $s\leq t$ and $D$ dense in $P\cap Q$ such that $s\Vdash D\cap \dot{G}\cap Q=\emptyset.$ Consider $res_Q(s)$. Since $D\cap Q$ is dense, there is $r\in D$ extending $res_Q(s)$. Let $r^*$ extend both $r$ and $s$ (by Lemma 52.5) but this is a contradiction.

52.16

If $\mathcal S\cup\mathcal T$ is sufficiently large then $I(\mathcal S,\mathcal T)$ is strongly proper.

# 5410.2.2016

## 54.1José de Jesús Pelayo-Gómez

54.1Zapletal

The cut-and-choose game for an ideal $\mathcal I$ (denoted by $G(\mathcal I)$ is a game where player I starts by cutting $\omega$ into two pieces and player II chooses one of the pieces and an element from the piece. Player one then cuts the chosen piece again and the game continues. Player I wins the game if the set of elements chosen by player II is in the ideal.

54.2

Player I has a winning strategy in $G(\mathcal{ED})$.

54.3

If $\mathcal I$ is not tall then II has a winning strategy in $G(\mathcal I)$.

54.4

If $\mathcal I$ is a maximal ideal then I has a winning strategy iff it is not selective.

54.5

$\omega\rightarrow (\mathcal I^+)^2_2$ iff $\mathcal R\not\leq_K\mathcal I$, where $\mathcal R$ is the Random-graph ideal, i.e. the ideal generated by cliques and anticliques of the Random graph, and $\leq_K$ is the Katětov order (see 12.16).

54.6

Player I has a winning strategy in $G(\mathcal R)$.

Let $c:[\omega]^2\to2$ be the random coloring. In the first step Player I partitions $\omega$ as $A_0^0\cup A_0^1$ with $A_0^0=\emptyset$ and $A_0^1=\omega$. Player II has to choose $n_0\in A_0^1$. Player I partitions $A_0^1$ into $A_1^0$ and $A_1^1$ such that $m\in A_1^i\iff c(n_0,m)=i$. Define $f:\{n_k:k<\omega\}\to 2$ as follows: $f(n_k)=c(n_k,n_{k+1})$. Then $f^{-1}(0)$ is a clique and $f^{-1}(1)$ is an anticlique. (Note that it is just the regular proof of the Ramsey theorem.)

54.7

If Player I does not have a winning strategy in $G(\mathcal I)$ then $\omega\rightarrow(\mathcal I^+)^2_2$.

The corollary directly follows from the following observation:

54.8

If player I has a winning strategy in $G(\mathcal I)$ and $\mathcal J\geq_K\mathcal I$ then player I also has a winning strategy in $G(\mathcal J)$.

54.9

The Solecki ideal is an ideal on the countable set $\Omega=\{A\subseteq 2^\omega:A\ \mbox{is clopen and}\ \lambda(A)=1/2\}$ consisting of sets $A\subseteq\Omega$ such that there is $F_A\in[2^\omega]^{<\omega}$ such that each open $U\in A$ intersects $F$, formally: $\mathcal S=\big\{A\subseteq\Omega: (\exists F\in[2^\omega]^{<\omega})(\forall U\in A)(U\cap F\neq\emptyset) \big\}$

The ideal is generated by clopen sets containing a branch. Note

54.10

Does player II have a winning strategy in $G(\mathcal S)$?

(This is, of course, motivated by the question whether $\mathcal R\leq_K\mathcal S$.)

54.11

If the player II has a winning strategy in $G(\mathcal I)$ then $\mathcal I$ is not tall.

The above conjecture is false, even for $F_\sigma$-ideals:

54.12

There is an $F_\sigma$-tall ideal $\mathcal I$ such that player II has a winnning strategy in $G(\mathcal I)$.

Fix $\{A_s:s\in\omega^{<\omega}\}$ such that $A_\emptyset=\omega$, $\{A_{s^{\smallfrown} n}:n<\omega\}$ is a partition of $A_s$ for every $s\in\omega^{<\omega}$ and for each $n,m<\omega$ there is $k<\omega$ and $s,t\in\omega^k$ such that $n\in A_s$ and $m\in A_t$. Say that $X\subseteq\omega$ is big if it contains an infinite set $A$ such that whenever $|A\cap A_s|=\omega$ then there are infinitely many $n$'s such that $|A\cap A_{s^{\smallfrown} n}|=\omega$. Let $\mathcal H$ consist of sets which are not big.

54.13

$\mathcal H$ is an ideal.

of Claim

The proof is technical and will be skipped.

Similarly we define that $X$ is big below $s\in\omega^{<\omega}$ if it is big in the tree rooted at $s$. Note that if a set is big below some $s$ then it is big below infinitely many of its successors. We say that $T\subseteq\omega^{<\omega}$ is a small-branching tree if $|\mbox{succ}_T(s)|\leq|s|+1$ for each $s\in T$. Given $A\subseteq\omega$ let $T_A=\{s\in\omega{^<\omega}:A\cap A_s\neq\emptyset$. Let $D_0$ consist of those $A$ such that $T_A$ is a small-branching tree; and $D_1$ consist of those $A$ such that there is $s\in\omega^{<\omega}$ such that $A\subseteq A_s$ and $(\forall n)(|A\cap A_{s^{\smallfrown} n}|=1)$. Let $PC$ be the ideal generated by $D_0\cup D_1$.

Note that $PC$ is tall by definition and is clearly $F_\sigma$. We will show that player II has a winning strategy in $G(PC)$. Player II will alwasy choose the set $A_n^i$ which is big (i.e. not in $\mathcal H$) and chooses the numbers so that the resulting $A=\{n_k:k<\omega\}$ cannot be covered by finitely many small-branching trees.

54.14

Is there a Borel ideal $\mathcal I$ such that $\mathcal I^+\rightarrow(\mathcal I^+)^2_2$.

54.15Thuemmel

For each analytic ideal $\mathcal I$ player I has a winning strategy in $G^\prime(\mathcal I)$, where in the modified game player I first chooses an infinite subset and then the game is played restricted to this set. Moreover, there should be a co-analytic ideal for which player I does not have a winning strategy.

## 54.2Oswaldo Guzmán: Sierpińsky's principle and meager sets

54.16Sierpińsky

The principle (*) is the statement that there is a family $\{\varphi_n:n<\omega\}\subseteq\omega_1^\omega$ such that for each uncountable $A\subseteq\omega_1$ there is $n<\omega$ such that $\varphi_n[A]=\omega_1$.

54.17Sierpińsky

CH implies (*).

54.18

A family $\{f_\alpha:\alpha<\omega_1\}\subseteq\omega^\omega$ is a weak Luzin-set if for each $g\in\omega^\omega$ there is $\alpha<\omega_1$ such that $|f_\beta\cap g|=\omega$ for each $\beta\geq\alpha$.

54.19

If $g\in\omega^\omega$ let $ED(g)=\{f:|f\cap g|<\omega$. This set is meager. A weak-Luzin set has countable intersection with each $ED(g)$. In particular a Luzin set is a weak-Luzin set.

54.20Miller

(*) iff there is a Luzin set.

54.21

If there is a Luzin set then () and () implies that $\mbox{non}(\mathcal M)=\omega_1$.

54.22

Can one of these implications be reversed?

Miller showed that in the Miller model (*) holds; Judah and Shelah showed that there are no Luzin sets in the Miller model. In particular the first implication cannot be reversed.

54.23O. Guzmán

$\mbox{non}(\mathcal M)=\omega_1$ implies that there is a weak-Luzin set (in particular, that (*)).

In fact, the proof gives a meager weak-Luzin set. So one can ask:

54.24

Does $\mbox{non}(\mathcal M)=\omega_1$ imply that there is a non-meager weak-Luzin set?

54.25O. Guzmán

No!

We will show a sketch of the proof of the theorem. First we need a lemma.

54.26

If $\mbox{non}(\mathcal M)=\omega_1$ then there is a family $X=\{f_\alpha:\alpha<\omega_1\}$ such that

1. each $f_\alpha$ is a partial function from $\omega$ to $\omega$;
2. the domains of the functions form an A.D. family;
3. for every $g:\omega\to\omega$ there is an $\alpha<\omega_1$ such that $|g\cap f_\alpha|=\omega$

Enumerate $\omega^{<\omega}$ as $\{s_n:n<\omega\}$ such that $s_0=\emptyset$ and if $s_n\subseteq s_m$ then $n\leq m$. Define a function $F$ which assigns to each $f\in\omega^\omega$ an infinite partial function from $\omega$ to $\omega$ as follows: for $f\in\omega^\omega$ let $\mbox{dom}(F(f))=\{n:s_n\subseteq f\}$ and if $s_n\subseteq f$ then $F(f)(n)=f(|s_n|)$ (i.e. let $i$ be the inverse of the enumeration of the domain and let $F(f)$ be $f\circ i$). Let $Y\subseteq\omega^\omega$ be non-meager. Then $F[Y]$ is the required family. Conditions 1. and 2. are clear. We show 3: Let $g:\omega\to\omega$. Let $N(g)=\{f:|F(f)\cap g|<\omega\}$. Then $N(g)$ is a meager set, since $N(g)=\bigcup_{k<\omega} N_k(g)$, where $N_k(g)=\{f:|F(f)\cap g|\leq k\}$ and the latter is nowhere dense. Since $Y$ is nonmeager, for every $g$ there is $f\in F[Y]\setminus N(g)$, i.e. $|F(f)\cap g|=\infty$.

of Theorem

Let $X$ be as in the above lemma. Without loss of generality assume that $\{\mbox{dom}(f_n):n<\omega\}$ is a partition of $\omega$. We our weak-Luzin set $Y=\{h_\alpha:\alpha<\omega_1\}$ as follows. If $n<\omega$ then let $h_n$ be arbitrary. Assume $\alpha\geq\omega$ and let $\alpha=\{\alpha_n:n<\omega\}$. Then $\{f_{\alpha_n}:n<\omega\}$ is an A.D. family covering $\omega$. Let $B_0=\mbox{dom}(f_{\alpha_0})$ and $B_{n+1}=\mbox{dom}(f_{\alpha_{n+1}})\setminus\bigcup_{i\leq n}B_i$. Then $\{B_n:n<\omega\}$ is a partition of $\omega$. Let $h_\alpha=\bigcup_{n<\omega} f_{\alpha_n}\upharpoonright B_n.$

54.27

The family $Y$ is a weak-Luzin set.

of Claim

Let $g:\omega\to\omega$. By the lemma there is $\alpha<\omega_1$ such that $f_\alpha$ has infinite intersection with $g$. However, $f_\alpha$ is almost contained in $h_\beta$ for each $\beta\geq\alpha$.

54.28

The family $Y$ is meager.

of Claim

Clear.

We now sketch the No answer to the question 54.24. We define a forcing $P_{cat}$ to consist of conditions $p=(s_p,M_p,F_p)$ which are triples satisfying:

1. $s_p\in\omega^{<\omega}$;
2. $M_p=\langle M^p_0,\ldots,M^p_n\rangle$ is an $\in$-chain of countable elementary submodels of $H((2^{\mathfrak c}))^+$;
3. $F_p:M_p\to\omega^{\omega}$ is a function such that $F_p(M^p_i)$ is Cohen over $M_i$ and, if $i, then \(F_p(M^p_i)\in M^p_{i+1}$; and
4. $s_p\cap F_p(M^p_i)=\emptyset$ for each $i\leq n$.

Ordering is the reverse end extension in the first coordinate and reverse inclusion inclusion in the second and third coordinates.

54.29

The forcing $P_{cat}$ has the following properties:

1. it is proper (but it cannot be strongly proper because of the next property);
2. if $X$ is a non-meager weak-Luzin set then it forces that $X$ is not weak-Luzin; and
3. it does not destroy category, i.e. it forces that the old reals are not meager in the extension; in fact, even
4. the countable support iteration of $P_{cat}$ (of any length) does not destroy category
54.30

Assuming the consistency of an inaccessible cardinal it is consistent that $\mbox{non}(\mathcal M)=\omega_1$ and each weak-Luzin set is meager.

Do a countable support iteration of $P_{cat}$ of an inaccessible length. Since the length is inaccessible, the iteration will have a suitable chain condition. By the chain condition every non-meager weak-Luzin set will appear at an intermediate step and we use 2.

of Proposition

We first prove properness.

54.31

Let $M$ be an elementary submodel of $H((2^{\mathfrak c})^{++})$ and $p\in M\cap P_{cat}$. Then there is $f\in\omega^\omega$ such that if $N=M\cap H((2^{\mathfrak c})^+)$ then $p^\prime=(s_p,M_p\cup \{N\},F_p\cup\{(N,f)\})$ is a condition.

Let $M$ be an elementary submodel of $H((2^{\mathfrak c})^{++})$ and $p\in M\cap P_{cat}$. We will show that $p^\prime$ is $(M,P_{cat})$-generic. So let $D\in M$ be an open dense set and $q\leq p^\prime$. Wlog we may assume that $q\in D$ (since $D$ is open dense). Let $q_0$ be the "restriction of $q$ to $M$", i.e. $s_{q_0}=s_q$, $M_{q_0}=M\cap M_q$ and $F_{q_0}=F_q\upharpoonright M$. Then $q_0\in M$. Since $q\leq q_0$, $q\in D$ and has the same stem as $q_0$, by elementarity there is $r\in D\cap M$ such that $r\leq q_0$ and has the same stem as $q_0$ (and as $q$). Then $r$ is compatible with $q$ because $(s_r,M_r\cup M_q,F_r\cup F_q)$ is a condition.

Next we prove 3.

54.32Folklore

$P$ does not destroy category iff $P$ does not add an eventually different real.

54.33Kuratowski-Ulam

If $X$ and $Y$ are Polish spaces and $N\subseteq X\times Y$ is nowheredense then $\{x:N_x\ \mbox{is nowheredense in}\ Y\}$ is co-meager in $X$ (where $N_x$ is the vertical section of $N$ at $x$).

Let $p\in P_{cat}$ be a condition, $i\leq |M_p|$, $g_j=F_p(M_{i+j})$ and $m=|M_p|-i$. Assume $D\subseteq(\omega^\omega)^{m+1}$ is nowheredense and $D\in M^p_i$. Then $(g_0,\ldots,g_m)\not\in D$. (Since $g_j$ is Cohen above $M^p_i$.) We show that $P_{cat}$ does not add an eventually different real (this will finish the proof of 3). So assume $p$ is a condition and $\dot{g}$ is a name for a function. Let $\{M_n:n<\omega\}$ be an $\in$-chain of countable elementary submodels such that $p,\dot{g}\in M_0$. Let $\mathcal A=\{A_n:n<\omega\}$ be a partition of $\omega$ in $M_0$. For simplicity assume that $A_n\cap n=\emptyset$. For every $n<\omega$ find $h_n:A_n\to\omega$ such that $h_n\in M_{n+1}$ and if $f\in M_n\cap\omega^\omega$ then $|h_n\cap f|=\omega$. Let $M=\bigcup_{n<\omega} M_n$ and $h=\bigcup_{n<\omega}h_n$. Let $p^\prime=(s_p,M_p\cup\{M\},F_p\cup\{(M,c)\})$ for some Cohen $c$ over $M$.

54.34

The condition $p^\prime$ forces that $|\dot{g}\cap h|=\omega$.

of Claim

Let $q\leq p^\prime$ and $k<\omega$. We extend $q$ to a condition which forces that $\dot{g}$ intersects $h$ above $k$. Let $q^\prime=(s_q,M_q\cap M,F_q\upharpoonright M)\in M$. There is $n>k$ such that $q^\prime\in M_n$. Define $D=\{t\in\omega^{<\omega}:(\exists l\in A_n,r\in M_n)(s_r=t\ \&\ r\leq q^\prime\ \& \ r\Vdash\dot{g}(l)=h_n(l))\}$ Now $D\in M_{n+1}$. Let $m=|M_q\setminus M_{q^\prime}|$. Define $N(D)$ to be $\big\{ (f_0,\ldots,f_m): (\forall t\in D)(t\cap(f_0\cup\cdots\cup f_m)\not\subseteq s_q)\big\}$ It is clear that $N(D)\in M_{n+1}$. We will show that $N(D)$ is nowheredense. This is enough to finish the proof of th claim: let $\mbox{Im}(F_q)\setminus M=(f_0,\ldots,f_m)$. By Kuratowski-Ulam this sequence is not in $N(D)$. But, by the definition of $N(D)$ and $D$ this is enough.We now show that $N(D)$ is nowhere dense. (At this point, my battery gave up.)

# 5517.2.2016

## 55.1M. Olšák: Random Reals and Suslin Trees (R. Laver, 1987)

55.1

If $MA_\kappa$ holds and $B$ is a measure algebra then every tree of size $\leq\kappa$ with no uncountable chain is special, i.e. can be written as a countable union of antichains.

55.2

In contrast with this a single Cohen real adds a Suslin tree.

Let $T=(\kappa,<_T)$ be a tree in the extension (i.e. $<_T$ is a name for the tree ordering). Recursively Fix a sequence of countably generated complete subalgebras $\langle B_\alpha:\alpha<\kappa\rangle$ of $B$ as follows. Let $B_0$ be an arbitrary, countably generated complete subalgebra of $B$ such that the measures of elements of $B_0$ are dense in $[0,1]$. Assume we have constructed $B_\beta$ for all $\beta<\alpha$. Let $B_\alpha$ be the algebra completely generated by $\bigcup_{\delta<\alpha, ||\delta<_T\alpha||>0} B_\alpha\cup\{||\delta<_T\alpha||\}.$

55.3

For every $\alpha<\kappa$ there are at most countably many $\delta<\alpha$ such that $||\delta<_{T}\alpha||>0$.

of Claim

Since $B$ is ccc, there is $\gamma<\omega_1$ such that $\alpha$ is forced by $B$ to be on a level of height at most $\gamma$. For a fixed $\gamma_0<\gamma$ and two distinct $\delta_0,\delta_1$ the values $||\delta <_{T}\alpha\ \&\ \mbox{lvl}(\delta)=\gamma_0||$ are disjoint. So, again by ccc, there are at most countably many possibilities for $\delta$.

Moreover, fix a countable $C_\alpha\subseteq B_\alpha^+$ which is dense in $B_\alpha$ in the sense of the metric given by the measure, i.e. for any $b\in B_\alpha$ and $\varepsilon > 0$ there is $c\in C_\alpha$ such that $\mu(c\Delta b)<\varepsilon$.

Consider the forcing $P$ consisting of functions $f: E\times W\to B$ with $W\in[\omega]^{<\omega},E\in[\kappa]^{<\omega}$ such that

1. $f(\alpha,n)\wedge f(\beta,n)\Vdash \alpha\perp_{<_T}\beta$;
2. $\mu(f(\alpha,n)) = 0$ or $\mu(f(\alpha,n)) > 1/2$; and
3. $f(\alpha,n)\in B_\alpha$.

The order on $P$ is defined as follows: $f\leq g$ iff

1. $\mbox{dom}(f)\supseteq\mbox{dom}(g)$;
2. $(\forall (\alpha,n)\in\mbox{dom}(g))f(\alpha,n)\leq g(\alpha,n)$; and
3. $(\forall (\alpha,n)\in\mbox{dom}(g))g(\alpha,n) > 0 \rightarrow f(\alpha,n) > 0)$.
55.4

$P$ is ccc.

of Claim

Let $\{f_\alpha:\alpha<\omega_1\}\subseteq P$. WLOG there is $W\in[\omega]^{<\omega}$ such that $\mbox{dom}(f_\alpha)=E_\alpha\times W$ and the $E_\alpha$'s form a $\Delta$-system with root $E$ and constant $|E_\alpha|$. We may also assume that there is $\varepsilon > 0$ such that if $\mu(f_\alpha(\beta,n))>1/2$ then $\mu(f_\alpha(\beta,n))>1/2+\varepsilon$. So we can find $c_{\beta,n}\in C_\beta$ such that, without loss of generality for all $\alpha<\omega_1$, $\mu(f_\alpha(\beta,n)\Delta c_{\beta,n})<\varepsilon/4$. It is easy to see that $\{f_\alpha\upharpoonright E\times W\}$ are pairwise compatible, so we may as well assume that $E=\emptyset$. Next we may assume that for all $\gamma<\delta$ and $\alpha\in E_\gamma,\beta\in E_\delta$ we have $||\alpha<_T\beta||=0$. Fix $\mathcal U$ a uniform ultrafilter on $\omega_1$. Let $I(\alpha,\beta)<\omega$ be a witness for the incompatibility of $f_\alpha$ and $f_\beta$. For each $\beta<\omega_1$ fix $n_\beta<\omega$ such that $\{\alpha:I(\alpha,\beta)=n\}\in \mathcal U$. Next fix $n<\omega$ such that $n_\beta=n$ for $\mathcal U$-many $\beta$. Write $E_\gamma=\{\sigma_{\gamma,i}:i<|E_\gamma|\}$. Let $\gamma,\delta<\omega_1$ be such that $I(\gamma,\delta)=n$. Then there are $i,j<|E_\gamma|=|E_\delta|$ such that $\mu(||\sigma_{\gamma,i}<_T\sigma_{\gamma,j}||)\geq\varepsilon/|E_\gamma|$. Again fix $i,j$ which occurs $\mathcal U\times\mathcal U$-often. We can now apply the following lemma to $b_{\alpha,\beta}=||\sigma_{\alpha,i}<_T\sigma_{\beta,j}||$.

55.5

Suppose $\varepsilon > 0$, $X\in[\omega_1]^{\omega_1}$, $X_\alpha\in \mathcal U$ for each $\alpha\in X$ and $\mu(b_{\alpha,\beta})\geq\varepsilon$ for $\alpha\in X,b\in X_\alpha$. Then there is a generic filter $G$ on $B$ such that, in the extension by $G$, if we define $R(\alpha,\alpha^\prime)\iff (\exists \beta)(b_{\alpha,\beta}\wedge b_{\alpha^\prime,\beta}\in G)$ then there is $Y\in[X]^{\omega_1}$ such that $R(\alpha,\alpha^\prime)$ for each $\alpha,\alpha^\prime\in Y$.

The lemma gives us a generic $G$ and $Y\in V[G]$ such that $\sigma_{\gamma,i}$ and $\sigma_{\gamma,j}$ are compatible but this contradicts the fact that $<_T$ does not contain uncountable chains thus finishing the proof of claim.

Given $\alpha<\kappa$ and $c\in C_\alpha$ with $\mu(c)<1$ let $D_{\alpha,c} = \Big\{f\in P_{E,W}: (\exists n)(\mu(f(\alpha,n))>1/2\ \&\ \mu(f(\alpha,n)-c)<1/2(1-\mu(c)))\Big\}.$ This set is clearly dense in $P$. We now apply $MA_\kappa$ to get a generic filter $G$ and let $a_{\alpha,n} = \bigwedge_{f\in G}f(\alpha,n).$ Then $\{a_{\alpha,n}:\alpha<\kappa,n<\omega\}$ is a name for the decomposition of $(\kappa,<_T)$ into countably many antichains: Fix a condition $b\in B$ and $\alpha<\kappa$. Without loss of generality we may assume $b\in B_\alpha$ (since $\{a_{\alpha,n}:n<\omega\}\subseteq B_\alpha$). Fix $c\in C_\alpha$ such that $\mu(c-b)<1/4\mu(c)$. Choose $f\in G\cap D_{\alpha,c}$. Then there is $n$ such that $a_{\alpha,n}\leq f(\alpha,n)$ and $\mu(a_{\alpha,n})\geq 1/2$. So, in particular, $b\wedge a_{\alpha,n}$ has non-zero measure and forces that $\alpha\in A_n$ ($A_n$ is the $n$-th antichain).

We still need to prove Lemma 55.5. First, however, observe:

55.6

Fix $0<\delta<\varepsilon$. Given $X\in\mathcal U$, $\{b_\beta\in B:\beta\in X\}$ with each $b_\beta$ of measure at least $\varepsilon$ there is $c\in B$ such that for each $c^\prime\leq c$ the set $\Big\{\beta:\mu(b_\beta\wedge c^\prime)\geq\delta\mu(c^\prime)\Big\}$ is an element of $\mathcal U$

of Observation

This is clear (use $\sigma$-additivity of $\mu$).

of Lemma 55.5

Fix $0<\delta<\varepsilon$ and apply the above lemma to each row $\{b_{\alpha,\beta}:\beta<\omega_1\}$ to get a $c_\alpha$. Since $B$ is ccc find $c\in B$ such that $c\Vdash W=\{\alpha:c_\alpha\in \dot{G}\}\ \mbox{is uncountable}.$ Then any generic $G$ containing $c$ will do: assume that it doesn't. Work in $V[G]$ and using the following special case of Dushnik-Miller's theorem $W\rightarrow(\aleph_1,N)^2$ with $N\geq 1/\delta$ find $\alpha_0,\ldots,\alpha_{N-1}\in W$ such that $R(\alpha_i,\alpha_j)$ does not hold for any $i\neq j. This immediately gives a contradiction. 55.7 Does Laver, Miller or Sacks add a Suslin tree? 55.8Carlson-Laver Under \(PFA$ Sacks forcing does not add a Suslin Tree.

55.9

The random forcing adds a poset which is ccc but not powerfully ccc so random forcing destroys $MA$.

# 572.3.2016

## 57.2E. Thuemmel: Remarks on Pelayo's lecture

Recall the cut and choose game from Definition 54.1. and Corollary 54.7. What about the converse to this Corollary. Can it be proved? Moreover, the local version of the corollary can also be proved, i.e.:

57.1

Assume that $\mathcal I^+\not\rightarrow(\mathcal I^+)2_2$. Then I has a winning strategy for the local version (i.e. we start on a postiive set) of the cut and choose game.

57.2

There is no analytic ideal with $\mathcal I^+\rightarrow(\mathcal I^+)^2_2$.

57.3

An ideal $\mathcal I$ is weakly homogeneous if for each positive $A\in\mathcal I^+$ there is a further positive subset $B\subseteq A$ such that $\mathcal I\simeq\mathcal I\upharpoonright B$. It is homogeneous if for each positive $A\in\mathcal I^+$ we have $\mathcal I\simeq\mathcal I\upharpoonright A$.

57.4ED

The ideal $\mathcal{ED}$ is generated by columns and graphs of functions in $\omega\times\omega$.

57.5

The ideal $\mathcal{ED}$ is not weakly homogeneous.

57.6

If $\mathcal I$ is weakly homogeneous then $\omega\rightarrow(\mathcal I^+)^2_2$ iff $\mathcal I^+\rightarrow(\mathcal I^+)^2_2$

57.7

There is a co-analytic example $\mathcal I$, due to Farah, such that $\mathcal I^+\rightarrow(\mathcal I^+)^2_2$. It is not clear what one can say about the cut and choose game for this ideal.

## 57.3J. Grebík: A Fraïssé class without a Katětov functor

This talk is motivated by the problem:

57.8

Is there a Fraïssé class without a Katětov functor?

which appeared in Kubiś, Mašulović:Katetov functors, published as preprint 1412.1850 on the ArXive.

57.9

Let $\mathcal L$ be a countable language with equality and consider the category $Fin(\mathcal L)$ of finitely generated $\mathcal L$ structures with embeddings (iff for relations [including equality]).

57.10

A category $\mathcal C\subseteq Fin(\mathcal L)$ is a Fraïssé class, if

1. it has the joint embedding property, i.e. for each $X,Y\in\mathcal C$ there is a $Z\in\mathcal C$ which embeds both $X$ and $Y$;
2. it has amalgamation, i.e. for all $X\rightarrow X_0$ and $X\rightarrow X_1$ there is a unique $Z$ completing the diagram, i.e. $X_0,X_1\rightarrow Z$;
3. it has the hereditary property, i.e. for all $X\in\mathcal C$ every finitely generated substructure of $X$ is an object of $\mathcal C$; and
4. it is "essentially" countable, i.e. it has only countably many pairwise non-isomorphic objects.
57.11

Finite graphs, finite linear orders, finite groups, finite Boolean algebras and finite dimensional vector spaces all form a Fraïssé class.

57.12

Given a Fraïssé class $\mathcal C$ let $\sigma\mathcal C$ denote the category of all co-limits of countable sequences.

57.13

Given $C\in\mathcal C$ let $Age(C)$ denote the category of all subobjects of $C$.

57.14

An object $C\in\mathcal C$ is homogeneous if any isomorphism between subobjects of $C$ can be extended to an isomorphism of $C$.

57.15

If $\mathcal C$ is a Fraïssé class then there is a unique (up to isomorphism) homogeneous $C\in\sigma\mathcal C$ with $Age(C)=\mathcal C$. This is called the Fraïssé limit of $\mathcal C$.

57.16

A Fraïssé class $\mathcal C\subseteq Fin(\mathcal L)$ has a Katětov functor $\mathcal K$ if $\mathcal K:\mathcal C\rightarrow\sigma\mathcal C$ is a covariant functor with a natural transformation $\eta$ from the identity functor to $\mathcal K$ such that every one-point extension of $X\in\mathcal C$, i.e. a $Y\in\mathcal C$ generated by $X$ and a single element, is realized in $\mathcal K(X)$, i.e. the embedding $\eta_X$ extends to all one-point extensions of $X$.

57.17

Given a finite metric space $(X,\rho)$ Katětov considered the set $K(X)=\{f:X\to\mathbb R^+:f(x)+f(y)\geq\rho(x,y)\},$ i.e. all one-point extensions of the metric space $X$, with the metric $d(f,g)=\mbox{sup}\{|f(x)-g(x)|:x\in X\}.$ In this case $(K(X),d)$ is a separable metric space and $K$ is in fact a Katětov functor.

57.18

A Katětov functor on $\mathcal C$ can be extended to $\sigma\mathcal C$.

The answer to problem 57.8 is yes.

57.19Grebík

There is a Fraïssé class $\mathcal C$ without a Katětov functor but which nevertheless has a faithful functor from $\sigma\mathcal C$ into the Fraïssé limit of $\mathcal C$.

Let $\mathcal C$ be the class of all finite linear orders with colorings of pairs by countably many colors without monochromatic triangles.

# 589.3.2016

## 58.1W. Bielas: Cantorval

58.1

Given a sequence $x=\langle x_n:n<\omega\rangle$ let $\sum(x) = \{\sum_{n\in A}x_n:A\subseteq\omega\}.$ Let $x=\langle 3/4,2/4,3/16,2/16,3/16,2/64,3/64,3/256,\ldots$. Then $\sum(x)$ is an example of a Cantorval (see 60.6)

## 58.2P. Simon

58.2

There is a Cantor set $A$ (compact, such that for each non-zero $x$ we have $|A+x\cap A|\leq 1$.

The construction giving the cantor set $A$ is relatively easy. In the standard construction of $A=\bigcap_{n<\omega} A_n$ at step $n<\omega$ we just need to split the intervals into more than 3 subintervals so that $|A_n+x\cap A_n|\leq 1$. The question is what is the least number of intervals we need to split it into.

## 58.3J. Verner: A simple lemma supposedly proved by my uncle

58.3

Given $z\in\mathbb R$, $|z|>1$, a natural number $k<\omega$ a finite $K\subseteq\mathbb R$ and a sequence $x\in K^\omega$ define $z(x,k) = \sum_{n=1}^\infty \frac{x(n+k)}{z^n}.$

58.4

Assume that $x\in K^\omega$ is a sequence such that $(\exists^\infty k)(z(x,k)=z(x,0)).$ Then it follows that the set $\big\{ z(x,k):k<\omega\big\}$ is finite!

58.5

Is the above lemma true?

## 58.4J. Grebík: A Fraïssé class without a Katětov functor

Recall the Definition 57.10 of a Fraïssé class, the Definition 57.16 of a Katětov functor and the Notation 57.12 $\sigma\mathcal C$. Note that if the Fraïssé class has pushouts then it has a Katětov functor.

58.6

The class of all finite groups has no pushouts, but it is a Fraïssé class (this is difficult). Does it have a Katětov functor?

58.7

Given a Fraïssé class with a Katětov functor if we iterate the functor on any object in the class the limit of this sequence is the Fraïssé limit of the class. In particular we can then define a different Katětov functor which assigns to each object the Fraïssé limit of the class. It then follows, from the definition of a Katětov functor, that the automorphism group of every object embeds into the automorphism group of the Fraïssé limit.

58.8

Let $\mathcal C$ be a Fraïssé class, $x\subseteq a\in C$ and $A,X\in\sigma\mathcal C$ such that the inclusions of $x$ into $a,X$ and $a,X$ into $A$ commute. Then we say that $A$ is a homogeneous extension of $X$ over $x\subseteq y$ if for all $\alpha\in Aut(x)$ automorphisms whose restriction to $x$ is the identity on $x$ there is an automorphism $\beta\in Aut(A)$ extending $\alpha$ which restricts to the identity on $a$. If $a$ is a one-point extension of $x$ and $A$ a one-point extension of $X$ we say that $A$ is a one-point homogeneous extension of $X$.

58.9

If there is a Katětov functor $K$ for $\mathcal C$ then for all $X\in\mathcal C$ and all embeddings $e$ of $x$ into the Fraïssé limit $F(\mathcal C)$ of $\mathcal C$ and all one-point extensions $\langle x,t\rangle$ of $x$ the embedding $e$ can be extended to an embedding of $\langle x,t\rangle$ into $\langle F(\mathcal C),t\rangle$.

Simple three-dimensional arrow chasing using the defining condition of the Katětov functor (i.e. that there is a natural transformation from the identity + one-point extensions).

58.10

Assume that $A$ is the Fraïssé limit of $\mathcal C$ and that there is an embedding $e:A\to A$ (different from the identity) and $E:Aut(A)\to Aut(A)$ such that for each automorphism $\alpha\in Aut(A)$ the diagram $eE(\alpha)=\alpha e$ commutes. Given a finite $x\subseteq A$ consider a one-point extension $\langle x,t\rangle\subseteq A$ with $t\in A\setminus e[A]$. Then there exists an $x^\prime\in\mathcal C$ a superstructure of $x$ such that $\langle A,t\rangle$ is a one-point homogeneous extension of $\langle x^\prime,t\rangle$.

58.11

If $\mathcal C$ has a Katětov functor then $e$ in the above lemma can be taken to be the natural transformation from the identity to $K$ and $E$ is $K$.

We now proceed to prove Theorem 57.19. We just give the definition of $\mathcal C$ and show that there is no Katětov functor:

of Theorem 57.19

Let $Q$ be a countable set of colors and let $\mathcal C$ consist of finite linear orders with a fixed coloring without monochromatic triangles. It is not hard to see that $\mathcal C$ is a Fraïssé class (it is enough to proof amalgamation). We show that there are no one-point homogeneous extensios of the Fraïssé limit $F(\mathcal C)$ of $\mathcal C$ (this is enough by Lemma 58.10). Pick some nonempty $(x,c_x)\in\mathcal C$ and let $\langle x,t\rangle$ be a one-point extension of $x$. If there were a one-point homogeneous extension then we would force a monochromatic triangle into it.

The talk is based on the notes J. Grebík:An Example of a Fraïssé class without a Katětov functor.

# 5916.3.2016

## 59.1J. Grebík: Katětov functors

We continue the proof of 57.19 by showing that there is, in fact, a faithful functor. Again, first recall the Definition 57.10 of a Fraïssé class, the Definition 57.16 of a Katětov functor and the Notation 57.12 $\sigma\mathcal C$. Now we continue the proof:

of Theorem 57.19

Fix an $\subseteq$-increasing sequence $\langle Q_n:n\leq\omega\rangle$ of colors with $|Q_{n+1}\setminus Q_n|=\omega$ and, given $n\leq\omega$ let $\mathcal C_n$ be the Fraïssé class of finite linear orders colored by colors in $\mathcal C_n$ and having no monochromatic triangles. We now define functors $F_n:\mathcal C_n\to\sigma\mathcal C_{n+1}$ such that

1. There is a natural transformation $\eta$ from the inclusion to $F_n$; and
2. every one-point extension $\langle x,t\rangle\in\mathcal C_n$ embeds into $F(x)$ (in $\mathcal C_{n+1}$).

Then we can extend $F_n$ from $\mathcal C_n$ to $\sigma\mathcal C_n$. Recursively we build $F:\sigma\mathcal C_0\to\sigma\mathcal C_\omega$. Moreover all $x\in\sigma\mathcal C_0$ will be mapped by $F$ into the Fraïssé limit (see 57.15) of $\sigma\mathcal C_\omega$. We define $F_0$. Given $x\in\mathcal C_0$ let $O_x$ be the set of all proper, partial one-point extensions of $x$. We will let $F_0(x)=x\cup O_x$. So we now need to put an ordering on $F_0(x)$ and a coloring on the pairs. Given $\psi\in O_x$ we write it as $\psi=\langle supp(\psi),t_\psi\rangle$ with $supp(\psi)\subseteq x$. Fix a color $p\in Q_1\setminus Q_0$ and $<$ an ordering of $Q_1$ in order-type $\omega$. Given $\psi\in O_x$ and $v\in x\setminus supp(\psi)$: we color the pair $(t_\psi,v)$ by $p$; we let $t_\psi \leq v$ iff $\forall w\in supp(\psi))\neg(v. Given \(\xi\not=\psi$ we let $t_\xi if 1. either there is \(v\in x$ with $t_\xi; or 2. \(|supp(\xi)|<|supp(\psi)|$; or
3. the biggest $v\in supp(\xi)\Delta supp(\psi)$ is in $\xi$; or
4. $supp(\xi)=supp(\psi)$ and the biggest $v$ in the supports such that the color of $(v,t_\xi)$ differs from the color of $(v,t_\psi)$ and the color of the first pair is less than the color of the second pair.

It remains to define the coloring on pairs of elements of $O_x$ (the map $F_0$ acts on the arrows in a natural way). We first define an equivalence relation on $[O_x]^2$: two pairs $\xi_0<\psi_0$ and $\xi_1<\psi_1$ are equivalent if there is an isomorphism $e:supp(\psi_0)\cup supp(\psi_0)\to supp(\psi_1)\cup supp(\psi_1)$ such that $F_0(e)(t_{\xi_0})) = t_{\xi_1}$ and $F_0(e)(t_{\psi_0})) = t_{\psi_1}$. Finally we inductively define the coloring on $O_x$'s by assigning each equivalence class a distinct color such that the embeddings are respected. The proof is finished by showing that there will be no chromatic triangles in $O_x$.

# 6023.3.2016

## 60.1W. Bielas: On the center of distances of a given metric space

\

(This is joint work with S. Plewik and M. Walczyńska.)

Let $X$ be a compact metric space and assume we are given two sequences $\langle a_n:n<\omega\rangle$ and $\langle b_n:n<\omega\rangle$ such that $d(a_n,b_n)\to0$. Observe that the two sequences have the same set of cluster points. The following theorem of von Neumann, proved in 1935, gives a sort of converse:

60.1von Neumann

If two sequences $\langle a_n:n<\omega\rangle$ and $\langle b_n:n<\omega\rangle$ in a compact metric space have the same set of cluster points then there is a permutation of the indices $\pi$ such that $d(a_{\pi(n)},b_n)\to0$.

Assume now that $d(a_n,b_n)\to\alpha$ and the sequences again have the same set of cluster points. Then for any cluster point $x$ there is a different cluster point $y$ such that $d(x,y)=\alpha$. This leads to the following definition.

60.2

Given a compact metric space we define the center of distances of X as follows: $S(X) = \{\alpha:(\forall x\in X)(\exists y\in X)(d(x,y)=\alpha)\}.$

60.3

If two sequences $\langle a_n:n<\omega\rangle$ and $\langle b_n:n<\omega\rangle$ in a compact metric space $X$ have the same set of cluster points $C$ and $\alpha\in C$ then there is a permutation $\pi$ of the indices such that $d(a_{\pi(n)},b_n)\to\alpha$.

60.4

Given a sequence $\langle a_n:n<\omega\rangle$ let $\sum(a_n) = \Big\{\sum_{n\in A}:a_n:A\subseteq\omega\Big\}.$ be the set of subsums (or the subsum set) of the sequence. We will tacitly assume that the terms of the sequnce are non-negative and that the sequence has a sum.

The set of subsums is symmetric, i.e. if $\{a_n:n<\omega\}$ and $a=\sum a_n$ then $x\mapsto a-x$ is a symmetry of $\sum(a_n)$.

60.5Guthrie-Nymann

The sumbsum set of a positive summable sequence is either:

1. a finite union of closed intervals
2. a Cantor set
3. a symmetric Cantorval
60.6

A set $C\subset R$ of reals is a Cantorval if $C$ is a nonempty compact regular closed subset such that both endpoints of any nontrivial component are accumulation points of trivial components.

60.7

Let $C$ consist of real numbers in $[0,1]$ written in base $3$ such that the first place where digit $1$ occurs is even. Then $C$ is a Cantorval.

60.8

If $\langle a_n:n<\omega\rangle$ is positive and summable then $\{a_n:n<\omega\}\subseteq S(\sum(a_n))$.

60.9

Let $X=\sum(2/4^n,3/4^n)$. Then $S(X)=\{3/4^n:1\leq n<\omega\}\cup\{2/4^n:1\leq n<\omega\}\cup\{0\},$ and $S(\mbox{bd}X) = \{ 1/4^n:n<\omega\}\cup\{0\}.$ Moreover,

1. $X\subseteq[0,5/3]$ and $[2/3,1]\subseteq X$ is a component of $X$;
2. $x\mapsto 5/3-x$ is a symmetry of $X$;
3. $[0,1/2)\cap X=1/4X$ and $[1/2,2/3]\cap X=X\cap[0,1/6] + 1/2$;
4. $\lambda(X)=1$;
5. the set $[0,2/3]\setminus X$ is closed under the functions $1/4x, 1/2+1/4x$ and $1/4(5/3-x)$ whose ranges are disjoint;
60.10

If a sequence $\{a_n:n<\omega\}\neq\{2/4^n,3/4^n:n<\omega\}$ then $\sum(a_n)\neq\sum(2/4^n,3/4^n)$.

60.11

$S(\sum(a/q^n)) = \{ a/q^n:0 whenever $q>2$ and $a>0$. # 6130.3.2016 ## 61.1Jerzy Król: Connections between forcing and physics ### 61.1.1History 1. P. Benioff: Models of Zermelo Frankel set theory as carriers for the mathematics of physics. I Journal of mathematical Physics, 17, 618 (1976) 2. G. Takeuti: Two Applications of Logic to Mathematics Princeton, 1976 IS and PUP 3. R. A. van Wesep: Hidden variables in quantum mechanics: Generic models, set-theoretic forcing, and the appearance of probability Annals of Physics, vol. 321(10) 2006 # 6213.4.2016 ## 62.1M. Doucha: Automatic continuity of homomorphisms from locally compact groups 62.1 Every locally compact group carries a left-invariant $\sigma$-additive measure which is finite on compact sets and is approximated by closed sets from below. 62.2J. Kuznetsova Assume $G,H$ are topological groups and $\varphi:G\to H$ is a homomorphism. Then 1. If $G$ is locally compact, $\varphi$ measurablem and $add(\mathcal N)=\mathfrak c$ then $\varphi$ is continuous. 2. If $G$ is a Baire group, $\varphi$ Baire-measurable and $add(\mathcal M)=\mathfrak c$ then $varphi$ is continuous. 62.3 It is known (Banach, Otis ?) that if $H$ is separable, then the additivity assumptions are not necessary. 62.4 If $G$ is locally compact and $\mu(A)>0$ or $G$ is Baire and $A$ has the Baire property and is not meager then $A^{-1}\cdot A$ contains an open neighbourhood of the unit element. We will first show how to prove the result due to Banach for separable $H$. of 62.3 Let $U$ be a neighbourhood of $1_H$. Fix $V$ an open neighbourhood of $1_H$ such tat $V\cdot V^{-1}\subseteq U$. Since $H$ is separable, let $D=\{d_n:n<\omega\}$ be a countable dense set in $H$. Then $H=\bigcup_{n<\omega} h_nV$. It follows that $G=\bigcup_{n<\omega}\varphi^{-1}(h_nV)$. Since $\varphi$ is measurable and $\mu G>0$ there must be $n<\omega$ such that $\mu \varphi^{-1}(h_nV)>0$. Now use the above fact to fix an open neighbourhood $W$ of $1_G$ such that $W\subseteq \varphi^{-1}(h_nV)^{-1}\cdot (\varphi^{-1}(h_nV))$. Now $\varphi(W)\subseteq V^{-1}h^{-1}_n h_nV\subseteq V^{-1}V\subseteq U$. 62.5 A nonempty subset $A\subseteq G$ of a locally compact topological group is extra-measurable if $S\cdot A$ is measurable for all $S\subseteq G$. 62.6 Open sets are extra-measurable while singletons are not. 62.7 Does there exist an extra-measurable null set if $add(\mathcal N)<\mathfrak c$? 62.8 Assume that $G$ is locally compact. Then the following are equivalent: 1. There is a non-continuous measurable homomorphisms $\varphi:G\to H$. 2. There is a countable family $\mathcal A=\{A_n:n<\omega\}$ of extre-measurable null subsets of $G$ with the following properties (which are necessary for $\mathcal A$ to be a system of neighbourhoods of unity in some topology): 1.1 $1_G\in A_n$ for each $n<\omega$; 1.2 $A_n=A_n^{-1}$ for each $n<\omega$; 1.3 $(\exists m)(A_m^2\subseteq A_n)$ for each $n<\omega$; 1.4 (not needed for commutative groups) Assume $2$ holds. Let $\tau$ be the topology on $G$ generated by the family $\mathcal A$ of neighbourhoods of $1_G$. Then the identity is a non-continuous homomorphism into $(G,\tau)$. Assume now that $1$ holds. Let $U\subseteq H$ be an open neighbourhood of $1_H$ whose preimage $A=\varphi^{-1}(U)$ is not open. We may assume that $\mu A=0$ (otherwise, by Banach's theorem, $A$ would be open). Now for every $S\subseteq G$ we have $S\cdot A=\varphi^{-1}(\varphi(S)\cdot A)$ (one has to do some computations to see this). It follows that $A$ is extra-measurable. We finish the proof by finding a system of subsets of $U$ as in (2) and its preimage $\mathcal A$ will be as required. 62.9 If $G$ has no extra-measurable null set then all measurable homomorphisms from $G$ are continuous. 62.10 Assume $add(\mathcal N)=\mathfrak c$ and let $G$ be a locally compact group $G$. Then there are no extra-measurable null sets in $G$. We will only show the proof for Polish groups where additivity of null is the same. We may assume that $G$ perfect (otherwise it would be countable and hence discrete). Let $A$ be a non-empty null set. We show that it is not extra-measurable. The proof follows the construction of a Bernstein set. Let $\{P_\alpha:\alpha<\mathfrak c$) be an enumeration of all compact perfect subsets of $G$ with positive measure. We construct disjoint $S=\{s_\alpha:\alpha<\mathfrak c\}$ and $D=\{d_\alpha:\alpha<\mathfrak c\}$ such that $S\cdot A\cap P_\alpha\neq \emptyset\neq D\cdot A^{-1}\cap P_\alpha$. This is easily done using the additivity assumption. Then $S\cdot A$ is not measurable. # 6320.4.2016 ## 63.1D. Chodounský: Towers and gaps Recall that a tower (or a chain) is a sequence $\mathcal T=\langle T_\alpha:\alpha<\omega_1\rangle$ of infinite subsets of $\omega$ such that $T_\alpha\subseteq^*T_\beta\iff\alpha\leq\beta$. We say that the tower $\mathcal T$ has property S if $T_\alpha\not\subseteq T_\beta$ whenever $\alpha\neq\beta$. The tower is special if it has a cofinal subsequence with property S. If $T_\alpha\not\subseteq T_\beta$ for $\alpha\leq\beta$ we shall write $\alpha\otimes\beta$. The tower can be naturally considered as a subspace of $2^\omega$ and $\otimes$ can be thought of as a symmetric open relation on $T$ as a subspace of $2^\omega$. Thus an easy consequence of OCA is that every tower is special. Also recall that a pregap is a pair of towers $(\mathcal A,\mathcal B)$ such that $A_\alpha\cap B_\alpha=\emptyset$ for every $\alpha<\omega_1$. A pregap $(\mathcal A,\mathcal B)$ is a gap if there is no $C$ separating $\mathcal A$ from $\mathcal B$, i.e. almost-containing each $A_\alpha$ and almost disjoint from each $B_\alpha$. Given a gap we define $\alpha\boxtimes\beta$ if $(A_\alpha\cap B_\beta)\cup (A_\beta\cap B_\alpha)\neq\emptyset$. The relation $\boxtimes$ is again an open relation (on $2^\omega\times2^\omega$). 63.1 A pregap is a gap iff there is no $\neg\boxtimes$-homogeneous set of size $\omega_1$. Recall that a gap is called special (or indestructible) if it is a gap in every bigger model which preserves $\omega_1$. 63.2 A gap is special iff it is indestructible by ccc forcing. 63.3 A gap is special iff there is a $\boxtimes$-homogeneous set of size $\omega_1$. Note, the PID axiom implies that every gap is special. On the other hand, PID is consistent with CH, in particular with $\mathfrak b=\omega_1$ which implies that there is a non-meager tower and so is necessarily destructible. Also note that while both PID and OCA are consequences of PFA, they are independent of each other and the consistency of PID needs some large cardinals while the consistency of OCA does not. Also PID implies that there is no Suslin Tree. It is not clear what the situation with OCA and Suslin trees is. We will no try to show a model where every tower is special while there are non-special gaps. We start with a simple lemma. 63.4 Assume $G\subseteq\omega_1$ is an uncountable set and $R$ is a symmetric relation on $G$ with no uncountable set homogeneous in $\neg R$. Given $\alpha\in G$ let $X_\alpha=\{\beta\in G:\alpha\otimes\beta\}$. Then 1. The set $|\{\alpha:|X_\alpha|\leq\omega\}|$ is at most countable; and 2. There is a countable set $D\subseteq G$ such that $G=\bigcup_{\alpha\in D}X_\alpha$. We will use the above lemma in a situation where $M$ is a countable elementary submodel of some $H(\kappa)$ with $R,G\in M$. Then from the lemma it follows that for each $\alpha\in G\setminus M$ the set $X_\alpha$ is uncountable (1) and, moreover, $\alpha$ is in relation with some $\beta\in M$ (2). Given a non-special tower $\mathcal T$ there is a simple ccc forcing $P_{\mathcal T}$ which specializes $\mathcal T$: 63.5 The forcing $P_{\mathcal T}$ consists of finite subsets which are $\otimes$-homogeneous ordered by reverse inclusion. We now start with a model of GCH (and, e.g., $\Diamond$, so that there are destructible gaps). By a bookkeepting argument we will do an $\omega_2$-stage finite support iteration of the $P_{\mathcal T}$ forcing to specialize all towers. We have to show that both $P_{\mathcal T}$ and its iteration preserves destructibility of gaps. Note in the proof we will consider a natural topology on $P_{\mathcal T}$ inherited from the topological sum $\bigoplus_{n<\omega}(2^\omega)^n$. The compatibility relation is open in this topology. 63.6 $P_{\mathcal T}$ preserves the destructibility of gaps. Assume to the contrary and let $\mathcal A,\mathcal B$ be a destructible gap and let $\dot{J}$ be a name for an uncountable subset of $\omega_1$ which is homogeneous in $\boxtimes$ or $\neg\boxtimes$. Assume first that it is homogeneous in $\boxtimes$. Then there must be an uncountable $G\subseteq\omega_1$ such \[ (\forall \alpha\in G)(\exists p_\alpha\in P)(p_\alpha\Vdash\alpha\in\dot{J}).$ Notice that if $p_\alpha||p_\beta$ then $\alpha\boxtimes\beta$. Let $M$ be a countable elementary submodel of some $H(\kappa)$ containing all the relevant things. Use the $\Delta$-system lemma in $M$ to find an uncountable $G^\prime\subseteq G$ such that for $\alpha,\beta\in G$ $p_\alpha\cap p_\beta=\Delta$ and $p_\alpha\setminus \Delta whenever \(\alpha<\beta$. Fix $\alpha\in G\setminus M$. Applying the above lemma to $G^\prime$ and $\boxtimes$ we find $\beta>\alpha$ such that $\alpha\boxtimes\beta$. Fix an open neighbourhood $O\in M$ of $\alpha$ such that $\gamma\boxtimes\alpha$ for all $\gamma\in O$ and an open neighbourhood $U\in M$ of $p_\beta$ such that $T_\delta\not\subseteq T_\gamma$ for each $p\in U$ and all $\delta\in p,\gamma\in p_\alpha\setminus\Delta$. (The fact that the neighbourhoods can be chosen in $M$ is by second countability). Since $\beta\in O$ and $p_\beta\in U$ by elementarity there is $\xi \in M$ with $x\in O$ and $p_\xi\in U$. Then $p_\xi || p_\alpha$ (since $p_\xi\in U$) and $\xi\boxtimes\alpha$.

If it were homogeneous in $\neg\boxtimes$ everything would still go through except for the choice of the open neighbourhood $O$, since $\neg\boxtimes$ is not open. Instead of the $O$ we consider the set $G^{\prime\prime}=\{\xi\in G^\prime:A_\beta\cap n=A_\xi\cap n\ \&\ B_\beta\cap n= B_\xi\cap n\},$ where $n<\omega$ is chosen so that $A_\alpha\setminus n \subseteq A_\beta$ and $B_\alpha\setminus n\subseteq B_\beta$. We now use the second part of the previous lemma to $G^{\prime\prime}$ and $\neg\boxtimes$, so we find $\xi\in M$ such that $\xi\neg\boxtimes\beta$. The rest of the proof is finished similarly as above by showing that $\xi\neg\boxtimes\alpha$ and $p_\xi||p_\alpha$.

63.7

Finite support iteration of $P_{\mathcal T}$ preserves destructibility of gaps.

Using the delta system lemma on supports.

# 6427.4.2016

## 64.2M. Hrušák: CDH spaces

64.1

A separable metric space $X$ is CDH (countable dense homogeneous) if for every two countable dense subsets $D,E\subseteq X$ there is a autohomeomorphism $h:X\to X$ such that $h[D]=E$.

64.2

Does there exist a connected meager in itself (meager for short) CDH $X$?

64.3

A space $X$ is a $\lambda$-set if every countable subset of $X$ is $G_\delta$.

64.4

If a CDH $X$ is meager then it is a $\lambda$-set.

64.5

In the Cohen model every $|lambda$ set has size $\omega_1$.

64.6

In the Cohen model there is no connected meager CDH space.

64.7M. Hrušák, J. van Mill

CH implies that there is a connected meager CDH space $\subseteq[0,1]^\omega$.

64.8

Can such a space be constructed in the plane?

64.9

There is a connected $\lambda$-set iff there is a $\lambda$-set of size $\mathfrak c$.

First notice that if there is a $\lambda$-set of size $\mathfrak c$ then there is a $\lambda$-set $X\subseteq 2^\omega$ of size $\mathfrak c$. We can also assume that $|X\cap U|=\mathfrak c$ for each nonempty open $U\subseteq 2^\omega$. It is easy to construct a function $g:X\to[0,1]$ such that for each open $U\subseteq 2^\omega\times[0,1)\cup\{\infty\}$ there is $x\in X$ such that $(x,g(x))\in\mbox{bd}(U)$. Now $Y=\{\infty\}\cup\{(x,g(x)):x\in X\}$ is a connected $\lambda$-set.

64.10

Under CH there is a connected $\lambda$-set in the plane.

## 64.3M. Hrušák: Gruff ultrafilters

64.11

An ultrafilter $p$ on $\mathbb Q$ is gruff if it is generated by "perfect" sets (i.e. closed without isolated points).

64.12van Douwen

Do gruff ultrafilters exist?

64.13van Douwen

Yes, if $\mbox{cov}(\mathcal M)=\mathfrak c$.

64.14K.P. + E. Copláková

Yes, if $\mathfrak b=\mathfrak c$.

64.15M. Hrušák + D. Fernandéz

Yes if either

• $\mathfrak d=\mathfrak c$
• V is the Random model
• $\Diamond(r(\mathcal P(\mathbb Q)/nwd))$

## 64.4M. Hrušák: Raving MAD families

64.16

A MAD family $\mathcal A$ is called Shelah-Steprans if for each family $X$ of nonempty finite subsets of $\omega$ either

1. There is $I\in\mathcal I(\mathcal A)$ which hits each $a\in X$ or
2. There is $Y\in[\omega]^\omega$ such that $\bigcup Y\in\mathcal I(\mathcal A)$.
64.17D. Raghavan

It is consistent that no such families exist.

64.18

Given an ideal $\mathcal I$ we write $\mathcal I^{<\omega} = \{ X\subseteq[\omega]^{<\omega}:(\exists I\in\mathcal I) (\forall a\in X)(a\cap I\neq\emptyset)\}.$

64.19

A MAD family $\mathcal A$ has property (+) (is raving) if for each $\{X_n:n<\omega\}\subseteq\mathcal I(\mathcal A)^{<\omega}$ there is a sequence $\langle a_n:n<\omega\rangle$ such that $a_n\in X_n$ and $\bigcup_{n<\omega} a_n\in\mathcal I(\mathcal A)$.

64.20

The property (+) is very strong. Moreover (+) MAD families cannot be destroyed by definable forcings. Also property (+) implies the Shelah-Steprans property.

64.21

Is it consistent that there are raving MAD families with of size $>\omega_1$?

64.22

A Borel ideal $\mathcal I$ has property (+) iff it is Katětov above $Fin\times Fin$.

64.23

An non-meager ideal is Shelah-Steprans (even raving).

## 64.5M. Hrušák: Sequential order of compact spaces

64.24Archangelskii-Franklin

Does there exist a compact sequential space $X$ with sequential order $>2$?

64.25Bashkirov,1970

There is one of order $\omega_1$.

64.26A. Dow

If $\mathfrak b=\mathfrak c$ then there is a space of order 4. Under MA there is one of order 5.

64.27A. Down

Under PFA if the sequential order coincides with the Cantor-Bendixon rank then the order is at most $\omega$.

64.28M. Hrušák

Under $\Diamond(\mathfrak b)$ there is one of size $\omega$. Under $\Diamond(\max\{\mathfrak s,\mathfrak b\})$ then there is one of size $\omega_1$.

# 654.5.2016

## 65.1D. Chodounský: Independent families

We will show Shelah's result that the following is consistent: $\mathfrak i <\mathfrak u$ J. Grebík subsequently showed that in the model $\mathfrak f = \mathfrak i$.

65.1

The independence number $\mathfrak i$ is the smallest size of a maximal independent family on $\omega$, i.e. a family $\{A_\alpha:\alpha<\kappa\}$ of infinite subsets of $\omega$ such that for any disjoint finite $F,H\subseteq\kappa$ the intersection $\bigcap_{\alpha\in F}A_\alpha\cap\bigcap_{\alpha\in H}\omega\setminus A_\alpha$ is infinite while any larger family does not satisfy this property.

65.2

For a set $A\subseteq\omega$ we write $A^0=A$ and $A^1=\omega\setminus A$. For a family of sets $\mathcal A$ we write $Fn(\mathcal A)$ for the set of finite partial functions from $\mathcal A$ to $2$. Given $h\in Fn(\mathcal A)$ we write $\mathcal A^h = \bigcap_{A\in\mbox{dom}(h)}A^{h(A)}$

In this notation a system $\mathcal A$ is independent iff $\mathcal A^h$ is infinite for each $h\in Fn(\mathcal A)$. It is maximal independent if for each $X\in[\omega]^\omega$ there is a function $h_X\in Fn(\mathcal A)$ such that $\mathcal A^{h_X}\subseteq^* X$ or $\mathcal A^{h_X}\cap X=^*\emptyset$. In the first case we say that $X$ hits $\mathcal A$, in the second it misses $\mathcal A$. If either of the cases happens we say that $X$ reaps $\mathcal A$.

65.3

The ultrafilter number $\mathfrak u$ is the smallest size of an ultrafilter base.

Note the consistency of $\mathfrak u<\mathfrak i$ is shown, e.g., in the Miller model, where $\mathfrak a=\mathfrak i=\mathfrak d=\mathfrak g=\mathfrak c=\omega_2$, while $\mathfrak u=\mathfrak r= \mathfrak b=\mathfrak p=\mathfrak s=\omega_1$. The proof of the consistency of $\mathfrak i<\mathfrak u$ is a standard iteration argument which destroys each ultrafilter with a small basis while preseving a specially chosen independent system.

65.4

An independent system $\mathcal A$ is dense if for each $X\in[\omega]^\omega$ and for each $h\in Fn(\mathcal A)$ there is an extension $h^\prime\supseteq h,h^\prime\in Fn(\mathcal A)$ such that $\mathcal A^h\cap X=^*\emptyset$ or $\mathcal A^h\cap X\subseteq^*\emptyset$. A family $D\subseteq Fn(\mathcal A)$ is open dense if every function in $Fn(\mathcal A)$ has an extension in $D$ and $D$ is closed under extension. We will denote by $\mathcal D(\mathcal A)$ the family of all open dense subsets of $Fn(\mathcal A)$.

65.5

A dense independent system is a maximal independent system. Moreover the mod-finite equality and inclusion in the above definition can be replaced by strict equality and inclusion without changing the notion.

65.6

Given an independent family $\mathcal A$ let $\mathcal F_{\mathcal A}$ be the filter generated by sets of the form $F_D = \bigcup_{h\in D}\mathcal A^h,$ for some $D\in\mathcal D(\mathcal A)$. Moreover let $\mathcal C_\mathcal A$ be the downwards closure of $\{ \omega\setminus \mathcal A^h:h\in Fn(\mathcal A)\}.$

65.7

An independent family $\mathcal A$ is dense iff $\mathcal F_{\mathcal A}\cup\mathcal C_{\mathcal A}=\mathcal P(\omega)$.

65.8

A forcing is Cohen preserving if each open dense subset of $Fn(\mathcal A)$ in the extension contains a ground-model open dense subset of $Fn(\mathcal A)$.

65.9

A forcing is Cohen preserving if it is Cohen preserving for a single countable $\mathcal A$ since the Cohen forcing is ccc.

65.10

The definition of $\mathcal C_{\mathcal A}$ is absolute (up to downwards closure) and the definition of $\mathcal F_{\mathcal A}$ is absolute for Cohen preserving extensions.

65.11

An independent family is selective if it is dense and the filter $\mathcal F_{\mathcal A}$ is a rare P-filter.

65.12

A filter $\mathcal F$ is P if countable subfamilies of $\mathcal F$ have pseudointersections in $\mathcal F$. It is non-meager if the family $\{e_F:F\in\mathcal F\}$ is unbounded. It is rare (or a Q-filter) if the family $\{d_F:F\in\mathcal F\}$ is dominating, where $d_F(n) = e_F(n+1)-e_F(n).$

65.13

A filter $\mathcal F$ is rare if it has a selector for every partition of $\omega$ into finite pieces. A filter is non-meager if it is non-meager as a subset of $2^\omega$.

65.14

A filter $\mathcal F$ is a non-meager P-filter iff every sequence $\langle F_n:n<\omega\rangle\subseteq\mathcal F$ has a pseudointersection $F\in\mathcal F$ such that $(\exists^\infty n)(F\setminus (n+1)\subseteq F_n)$ (or, equivalently, $(\exists^\infty n\in F)(F\setminus (n+1)\subseteq F_n)$)

65.15

A filter $\mathcal F$ is a rare P-filter iff every sequence $\langle F_n:n<\omega\rangle\subseteq\mathcal F$ has a pseudointersection $F\in\mathcal F$ such that $(\forall n\in F)(F\setminus (n+1)\subseteq F_n)$.

We now construct one step of our iteration. Given an ultrafilter $\mathcal U$ we construct a proper Sacks property Cohen-preserving forcing $P_{\mathcal U}$. Recall that a forcing $P$ has the Sacks property if every new function is contained in a ground-model slalom of size $2^n$, i.e. for every new $f:\omega\to\omega$ there is $F:\omega\to[\omega]^{<\omega}$ such that $|F(n)|<2^n$ and $f(n)\in F(n)$ for all $n<\omega$.

65.16

Sacks property probably implies Cohen-preserving.

65.17

For our purposes we shall call as sequence $E=\langle e_n:n<\omega\rangle$ a partition of $\omega$ if it is a sequence of pairwise disjoint sets such that each $e_n$ is infinite and $\min(e_n)<\min(e_m)$ for $n. We shall write $e_* = \omega\setminus\bigcup_{n<\omega}e_n.$ Given an ideal \(\mathcal I$ we say that a partition is an $\mathcal I$-partition if $e_n\in\mathcal I$ for each $n<\omega$ as well as $e_*\in\mathcal I$.

65.18

Given a tree $T$ (without finite branches) we say that $T_n$ is a branching level of $T$ if each $s\in T_n$ is a splitting node. A tree is uniformly branching if each splitting node is on a branching level. Given two levels $n\leq m$ we say that $m$ depends on $n$ if $n$ is a branching level and $s(m)+s(n)\equiv_2 t(m)+t(n)$ for each $s,t\in T_{m+1}$.

65.19

Oswaldo interprets the dependence by a party simile: Natural numbers are deciding whether to go to a party. Natural numbers corresponding to branching levels are undecided. Natural numbers depending on $n$ base their decisions only on $n$ and other numbers are decided, but their decisions are more complicated.

65.20

Given a uniformly branching tree $T$ we let $e_n(T)$ consist of all $m$ which depend on the $n$-th branching level of $T$. Given an ideal $\mathcal I$ we say that a uniformly branching tree $T$ is $\mathcal I$-suitable if $e(T)$ is an $\mathcal I$-partition. The forcing $P_{\mathcal U}$ consists of $\mathcal U^*$-suitable trees.

65.21

If $\bigcup_{m\in M} e_m(T)\in\mathcal I$ then removing the branching levels from $M$ maintains suitability. If two sets of branching levels are disjoint we can always remove one of them while preserving suitability.

65.22

Given $f:n\to 2$ we let $T_f$ be the tree consisting of $s\in T$ such that if $T_m$ is the $i$-th branching level of $T$ for $i and \(m<|s|$ then $s(m)=f(i)$. Given $a\subseteq \omega$ we write $S\leq_a T$ if the $n$-th branching level of $T$ is a branching level of $S$ for $n\in a$.

65.23

If $a\cup b=\omega$ and $T\in P_{\mathcal U}$ then there is $S\in P_{\mathcal U}$ such that $S\leq_a T$ or $S\leq_b T$.

65.24

The forcing $P_{\mathcal U}$ forces that $\mathcal U$ does not generate an ultrafilter in the extension.

# 6611.5.2016

## 66.1D. Chodounský: Independent families II.

Recall that last time we defined, given an ideal $\mathcal I$, $\mathcal I$-suitable trees (Definition 65.20), $\mathcal I$-partitions (Notation 65.17) and the $\leq_a$ ordering on $\mathcal I$-suitable trees for $a\subseteq\omega$ (Notation 65.22).

66.1

An $\mathcal I$-partition $E$ is coarser than an $\mathcal I$-partition $G$ if each element of $E$ is a union of elements of $G$.

66.2

Given an ideal $\mathcal I$ we define the following game of two players. Player I. starts the game by playing an $\mathcal I$-partition $E_0$. Player II. continues by playing a coarser partition $G_0$ and then player I. respons by playing a coarser partition $E_1$ and so on. Given a game $\langle E_n,G_n:n<\omega\rangle$ we define $\Delta_n=g_n^*\setminus e_n^*$. Player II wins if $\Delta=\bigcup_{n<\omega}\Delta_n$ is $\mathcal I$-positive.

66.3

If $\mathcal I$ is maximal then Player I. does not have a winning strategy.

If I had a winning strategy then he would have a winning strategy such that each $i\in\omega$ is either in $\Delta$ or in some $e^*_n$. So, aimimng towards a contradiction, assume $\sigma$ is such a strategy. We play two games: player II will always play the move played by player I in the other game. Then clearly player I lost in one of these games.

66.4

If $T\in P_{\mathcal U}$ and $\dot{x}$ is a name for set in $V$ then for every $n$ there is $S\leq_n T$ such that $S_f$ decides $\dot{x}$ for each $f:n\to 2$.

Enumerate ${}^n2$ as $f_i:i<2^n$. Let $T_0^\prime\leq T_{f_0}$ decide $\dot{x}$. Find $T_0\supseteq T_0^\prime$ such that $(T_0)_{f_0}=T_0^\prime$ and $T_0\leq_n T$. Continue this process inductively to arrive at $S$.

66.5

The forcing has the Sacks property.

Let $\dot{f}$ be a name for a function from $\omega$ to $\omega$ and $T\in P_{\mathcal U}$. We will extend $T$ to a condition $R$ such that for each $g\in{}^{<\omega}\omega$ the condition $R_g$ decides $\dot{f}\upharpoonright|g|$. We play the above game: start with $T$ and apply the previous proposition to get $T^\prime_0\leq_0 T$. Player I plays $e(T^\prime_0)$ and player II responds by some $E^\prime$; $\Delta_0=e^\prime_*\setminus e_*(T^\prime_0)$. We construct a tree $T_0$ such that

1. $e_0(T^\prime_0)\cup\Delta_0\subseteq e_0(T_0)$; and
2. $e_j(T_0)=e^\prime_j$.

Continue by induction. Since this is not a winning strategy for player I we can assume that he lost. Then $R=\bigcap_{n<\omega}T_n$ is the required condition.

# 6710.5.2016

## 67.2D. Chodounský

Recall Definition (12.16) of the Katětov order. Also recall that there is, up to graph isomorphism, a unique graph on $\omega$ such that for any two disjoint finite $A,B\in[\omega]^{<\omega}$ there is $n\in\omega$ which is connected to every vertex in $A$ and not connected to any vertex in $B$. This graph is called th random graph.

67.1

The random ideal $Ran$ is generated by homogeneous subsets of the random graph on $\omega$, i.e. subsets of $\omega$ which can be covered by finitely many cliques and anti-cliques.

Note that the random ideal is an $F_\sigma$ ideal and is one of the lowest Borel ideals in the Katětov order.

67.2

An ideal $\mathcal I$ is Katětov above $Ran$ iff $\omega\not\rightarrow(\mathcal I^+)^2_2$.

67.3

The Solecki ideal is the ideal on clopen subsets of $2^\omega$ having measure $1/2$ which is generated by sets of the form $I_x=\{A:x\in A\}$ for some $x\in 2^\omega$.

The Solecki ideal is an $F_\sigma$ ideal (the submeasure is given by the number of $I_x$ covering the set). The following is still an open question:

67.4

Is the Solecki ideal Katětov above the random ideal?

The following diagram, taken from J. Brendle, B. Farkas and J. Verner: Towers in filters, cardinal invariants, and Luzin type families, is known:

Note that most of the connections already appeared in papers by M.~Hrušák, J.~Brendle and J.~Flašková.

## 67.3P. Vojtáš:

Consider the space $c_0^+$ of sequences of positive reals converging to 0 together with the $l^1$ norm given by: $||a||_1 = \sum_{n=0}^\infty a_n.$ Let $l^1$ be the space of sequences with finite norm (together with the $l^1$ norm). The $l^2$ space is given by the norm: $||a||_2 = \sqrt{\sum_{n=0}^\infty a^2_n}.$

67.5

For each $a\in l^1$ the set $\{b:b\leq^*a\}$ is dense and of first category (meager).

67.6

For each $a\in l^2\setminus l^1$ the set $\{b:a\leq^*b\}$ is dense of first category in $l^2$.

67.7

Given $a,b\in l^1$ define $b\preceq^* a$ if $(\forall^\infty n)\left(\frac{a_{n+1}}{a_n} \geq \frac{b_{n+1}}{b_n}\right)$

67.8Vojtáš, 93

Under MA the algebras $RO(l^1,\succeq^*)$ and $RO(c_0^+\setminus l^1,\geq^*)$ are isomorphic to $RO(\mathcal P(\omega)/fin,\subseteq^*)$.

(Vojtáš, P.: Boolean isomorphism between partial orderings of convergent and divergent series and infinite subsets of $\mathbb N$, Proc. Amer. Math. Soc. 117 (1993), 235-242)

67.9Fuchino, Mildenberger, Shelah, Vojtáš, 99

It is consistent that the above algebras are not isomorphic.

The proof works by showing that the $\mathfrak h$ of the Boolean algebras are different. It can be found in:

Fuchino S., Mildenberger H., Shelah S. and Vojtáš P.: On absolutely divergent series, Fund. Math. 160 (1999)

67.10

Given $a,b\in c_0^+\setminus l^1$ we say $a\leq^{c*} b$ if $\sum\{a_n-b_n:a_n> b_n\}<\infty$.

# 688.6.2016

## 68.1P. Vojtáš: Divergent series

We consider the structure $S=(c_0^+\setminus l^1,<^*)$ of (absolutely) divergent series converging to $0$ and the algebras $RO(S)$ and $RO(\mathcal P(\omega)/fin)$. We will show that, starting with CH, any $\aleph_2$-length iteration of Laver-property forcings which makes $\mathfrak h=\aleph_2$ will make $h(S)=\aleph_1$.

Let $N_{\omega_2}$ be the final extension and $b\in S\cap N$. Find the stage $\delta$ such that $b\in N_\delta=M$. Working in $M$, CH holds, so we can enumerate $M\cap[\omega]^\omega=\{H_\nu:\nu<\omega_1\}$. In $N$ define $D_\nu = \{a\in S:a\ \mbox{diverges on exactly one of}\ H_\nu\ \mbox{or} (\omega\setminus H_\nu)\}.$ It is not hard to see that $D_\nu$ is dense in $S$. We will show that $D=\bigcap_{\nu<\omega_1} D_\nu$ is empty. Assume, towards a contradiction, that there is $c\in D$. Work in $M$. Construct a sequence $\langle m_i:i<\omega\rangle$ such that $b(x)<1/2^{m_i}$ for each $x\in [m_{i+1},m_{i+2})=I_{i+1}$. Using the Laver property, find $y_i$ of size $\leq 2^{i^2}$ such that $c\upharpoonright I_{i+1}\in y_i$. Moreover let $w_i = \{s\in y_i:\sum_{x\in I_i}s(x) > 1/i^2\}.$ Then if $l\in S$ and fits into the slalom $y$ then there is an infinite $A$ such that $l\upharpoonright I_n\in w_n$ for each $n\in A$ and $l$ diverges on $\bigcup_{n\in A}I_n$. We now use the following theorem, which can probably be found in a book by Allon, Spencer, Erdös:

68.1

For each $n<\omega$ we can partition $I_n$ into two pieces $I_n =J^0_n\cup J^1_n$ such that for each $d\in w_n$ and $i<2$ we have $1/3 \leq \frac{\sum_{x\in J^i_n} d(x)}{\sum_{x\in I_n}d(x)}\leq 2/3.$

## 68.2M. Hrušák: Strong measure zero

68.2

If $X$ is a metric space a set $A\subseteq X$ is SMZ (strong measure zero) if for each sequence $\langle \varepsilon_n:n<\omega\rangle$ of positive non-zero real numbers there is a sequence of points $\langle x_n:n<\omega\rangle$ such that $A\subseteq\bigcup_{n<\omega}B(x_n,\varepsilon_n).$

68.3

For topological groups we can equivalently define that $A$ has (left) strong measure zero if for every sequence $\{U_n:n<\omega\}$ of neighbourhoods of unity there is a sequence of points $\langle x_n:n<\omega\rangle$ such that $A\subseteq\bigcup_{n<\omega} x_nU_n.$

68.4Galvin, Mycielski, Solovay

A set $A\subseteq\mathbb R$ is SMZ iff $A+M\neq\mathbb R$ for each meager $M\subseteq R$.

68.5

The following is known:

1. If $G$ is a locally compact group then a set has SMZ iff $A+M\neq\mathbb R$ for each meager $M\subseteq G$ (we say that $G$ has the GMS-property)
2. Under the Borel-Conjecture (SMZ consists precisely of countable sets) every group has the GMS property.
68.6M.,Wohofsky, Zindulka

Under CH the group $\mathbb Z^\omega$ does not have the GMS property.

68.7Zapletal

Under CH (or some other reasonable condition for counteraxamples to GMS) if $G$ is Abelian or if G has an invariant metric then $G$ has the GMS property iff $G$ is locally compact. Moreover, $S_\infty$ does not have the GMS property. In fact, a closed subgroup $H\subseteq S_\infty$ has the GMS iff it is locally compact.

68.8

A group $G$ does not have the GMS property iff there is a meager set $M\subseteq G$ such that for each sequence $\{U_n:n<\omega\}$ of neighbourhoods of unity there is a sequence of points $\{x_n:n<\omega\}\subseteq G$ such that $g\cdot(\bigcup_{n<\omega}x_n U_n)\cap M$ is dense in $M$ for each $g\in G$.

# 6929.6.2016

## 69.1M. Doležal: The ideal of σ-porous sets

This is joint work with Preiss and Zelený.

69.1

A set $P\subseteq X\times 2^X$ is a porosity relation on $X$ if

1. If $B\subseteq A$ and $P(x,A)$ then $P(x,B)$;
2. $P(x,A)\iff (\exists r>0)(P(x,B_r(x)\cap A))$; and
3. If $P(x,A)$ then $P(x,\overline{A})$.

The intended meaning of $P(x,A)$ is that the set $A$ is small around $x$. We say that $A$ is $P$-porous at $x$ if $P(x,A)$. A set is $P$-porous if it is $P$-porous at each of its points. A set is $\sigma$-P-porous if it is a countable union of $P$-porous sets.

69.2

The standard porosity relation $P(x,A)$ holds if $\lim\sup_{R\rightarrow 0} \frac{\sup\{ r>0:(\exists z)(d(x,z)0$

69.3

The strong porosity relation $P(x,A)$ holds if for each $R>0$ there is $z$ such that $B_{d(x,z)}(z)\cap A=\emptyset$.

69.4

Let $f:X\to\mathbb R$ be continuous. Define $P(x,A)$ if $f$ is differentiable at $x$ on $A$.

69.5

A porosity relation $P$ satisfies condition $(\alpha)$ if there is a nondecreasing function $\psi:[0,\infty)\to[0,\infty)$ satisfying:

1. $\psi(0)=0$;
2. $0<\psi(r)\leq r$ for $r>0$; and
3. if $P(x,C)$ then $P(x,\{y:d(y,C)\leq\psi(d(x,y))\})$.

It satisfies condition $(\beta)$ if for every Suslin (analytic) set $A$ the set $\{x\in A:\neg P(x,A)\}$ is Suslin.

69.6

Let $X$ be a compact metric space and $P$ porosity relation on $X$ which satisfies conditions $(\alpha),(\beta)$. If $A\subseteq X$ is a Souslin (analytic) non-$\sigma$-P-porous set. Then there is a compact $K\subseteq A$ which is still not $\sigma$-P-porous.

of Theorem

(idea) Characterize $\sigma$-P-porosity using a game $G_{fin}(A)$ such that for Suslin non-$\sigma$-porous $A$ the first player will have a winning strategy and the second player will have a winning strategy iff $A$ is $\sigma$-P-porous. Then for a Suslin non-$\sigma$-P-porous set $A$ we will find a compact $K\subseteq A$ such that the same strategy for player one as in $G_{fin}(A)$ also works for $G_{fin}(K)$.

Fix two sequences $\langle R_n:n<\omega\rangle$ and $\langle r_n:n<\omega$ such that $R_{n+1} < r_n < R_n/n$. Given a set $A$ we first define an "infinite" game $G(A)$ which is played as follows: The first player plays a sequence of points (centers of balls) $\langle x_n:n<\omega\rangle$ and the second player responds by playing open sets $\langle S_n^i:i\leq n,n<\omega\rangle$ such that

1. $x_{n+1}\in B(x_n,(R_n-r_n))$
2. $S_{n+1}^i\subseteq B(x_{n+1},R_{n+1})$
69.7

$\overline{B(x_{n+1},R_{n+1})}\subseteq B(x_n,R_n)$ and there is a unique $x\in\bigcap_{n<\omega}B(x_n,R_n)$. This $x$ is the result of the play.

Player two wins if either the result of the game $x$ is outside of $A$ or there is $i<\omega$ such that $x\not\in\bigcup_{i\leq n<\omega}S_n^i$ and $P(x,X\setminus \bigcup_{i\leq n<\omega} S_n^i\cap B(x,r_n)).$

69.8

If $A$ is $\sigma$-P-porous then the second player has a winning strategy in $G(A)$.

of Lemma

Write $A=\bigcup_{n<\omega} A_n$ with each $A_n$ being $P$-porous. The second player plays $S_n^i$ as holes in $A_i$, in the $n$-th move: $S_n^i=B(X_n,R_n)\setminus \overline{A_i}.$ This is a winning stragey for player II.

69.9

If $A$ is Suslin and not $\sigma$-P-porous then the first player has a winning strategy.

of Lemma

(hint) There is a $P$-Foran-Zajíček schema, i.e. a system $\mathcal F=\{F(s):s\in\omega^{<\omega}\}$ of nonempty subsets of $A$ satisfying conditions

1. The union $\bigcup_{n<\omega} F(t^{\smallfrown} n)$ is dense in $F(t)$ for every $t\in\omega^{<\omega}$ and $F(t)\subseteq F(s)$ for each $s\subseteq t$;
2. The set $F(t)$ is not $P$-porous in any point of $F(t^{\smallfrown} k)$ for every $k<\omega$ and $t\in\omega^{<\omega}$; and
3. The intersection $\bigcap_{n<\omega}F(f\upharpoonright n)\cap G_n\neq\emptyset$ for each $f\in\omega^\omega$ and each sequence $\langle G_n:n<\omega\rangle$ of open sets, closure-descending with diameters tending to zero and satisfying $F(f\upharpoonright n)\cap G_n\neq\emptyset$.

The strategy of player I is as follows. He plays a sequence $\langle s_n:n<\omega\rangle\subseteq\omega^{<\omega}$ on the side such that $|s_n|\leq n$, the sequence $s_n$ is compatible with $s_{n+1}$ and then plays a point $x_n\in F(s_n)$. He plays the sequences $s_n$ so that they converge to a branch $f\in\omega^\omega$ This will guarantee that $x\in\bigcap_{n<\omega}F(f\upharpoonright n)\subseteq A$. In the first move he plays arbitrarily, say $s_0=\emptyset$ and $x_0\in F(\emptyset)$. Assume we are to play in the $n$-th move, i.e. $\langle x_m:m\leq n\rangle$, $\langle s_m:m\leq n\rangle$ and $\{S^i_m:m\leq n\}$ have played. There are three possible cases.

69.10

If There is $i\leq |s_n|$ such that $B(x_n,R)\not\subseteq \bigcup_{m=i}^n S_m^i$ and $Z=\bigcup_{m=i}^n S_m^i\cap B(x_n,R_n-i\cdot r_n)\cap F(s_n\upharpoonright i)\neq\emptyset$, then we choose the smallest such $i$ and play $x_{n+1}\in Z$ and $s_{n+1}=s_n\upharpoonright i$.

69.11

If Case 1 fails and $x_n\in Z=\bigcup_{m=|s_n|}^n S_m^{|s_n|}\cap B(x_n,R_n-|s_n|\cdot r_n)\cap F(s_n\upharpoonright |s_n|)$ and $B(x_n,R_n)\not\subseteq\bigcup_{m=|s_n|}^n S_m^{|s_n|}$ then play $x_{n+1}=x_n$ and $s_{n+1}=s_n$.

69.12

If both previous cases fail then $F(s_n)\cap B(x_n,R_n-|s_n|\cdot r_n)\neq\emptyset$ and, using condition (1) of the Furan-Zajíček schema there is $k<\omega$ such that $F(s_n^{\smallfrown} k)\cap B(x_n,R_n-|s_n|\cdot r_n)\neq\emptyset$ and let $s_{n+1}=s_n^{\smallfrown} k$ and choose $x_{n+1}$ arbitrarily in the above intersection.

Finally we modify the game $G(A)$ to a game $G_{fin}(A)$ where player two is only allowed to play out of a finite set of options while preserving the winning strategies from the previous lemmas. This uses condition $(\alpha)$.

# 703.8.2016

## 70.1Samuel Goméz da Silva: Star covering properties

70.1Matveev

A space $X$ satisfies property (a) (absolutely countably compact) if for every open cover $\mathcal U$ and all dense sets $D$ there is a closed discrete in $X$ subset $F\subseteq D$ such that $st(F,\mathcal U)=X$, where $st(F,\mathcal U)= \bigcup\{U\in\mathcal U:U\cap F\neq\emptyset\}$

The motivation for the above definition is the following result of Flesichman:

70.2Flesichman,'70s

A Hausdorff space $X$ is countably compact iff for every open cover $\mathcal U$ of $X$ there is a finite $F\subseteq X$ such that $st(F,\mathcal U)=X$

Proof (Idea): Every open cover has a discrete kernel, i.e. a maximal set $A\subseteq X$ such that $x\neq y\in A\rightarrow x\not\in st(y,\mathcal U$. Clearly $A$ is closed discrete and, since $X$ is compact, has to be finite. For the other direction, let $A$ be countably infinite closed discrete. Write $A$ as a disjoint union $A=\bigcup_{n<\omega} A_n$ with $|A_n|=n$. By $T_2$ we can find open $\{U_a:a\in A\}$ such that $\{U_a:a\in A_n\}$ are pairwise disjoint and $U_a\cap A=\{a\}$. Let $\mathcal U=\{U_a:a\in A\}\cup\{X\setminus A\}$. This cover witnesses that the star-covering property fails.

70.3

Paracompact $T_1$ spaces are (a) (picking a point in the locally finite refinement).

70.4

The $\psi$-space $\psi(\mathcal A)$ is not (a) for any infinite MAD family $\mathcal A$.

By maximality closed discrete subsets of $\omega\subseteq\psi(\mathcal A)$ are finite. Pick a countable subfamily $\mathcal A^\prime=\{A_n:n<\omega\}\subseteq\mathcal A$ of $\mathcal A$. Then the cover $\{\{A_n\}\cup A_n\setminus n:n<\omega\}\cup \{\mathcal A\setminus\mathcal A^\prime\}$

70.5Matveev's version of Jones' lemma

If $X$ is (a) and $H\subseteq X$ is closed discrete then $|H|<2^{d(X)}$.

(by contradiction) Assume $H$ is closed discrete of size $\geq 2^{d(X)}$ and let $D\subseteq X$ be dense of size $d(X)$. Without loss of generality $H\cap D=\emptyset$. Pick $H^\prime\subseteq H$ indexed by closed discrete subsets of $D$, i.e. $H^\prime=\{x_G:G\subseteq D\ \mbox{is closed discrete}\}$. By discreteness pick for each $G$ a neighbourhood $U_x$ of $x$ such that $U_x\cap H=\{x\}$. Let $\mathcal U=\{X\setminus H\}\cup\{U_{x_G}\setminus G\}$

Since, if $2^{\aleph_0}<2^{\aleph_1}$, then Jones' lemma actually gives countable extent for separable normal spaces. This motivates the following question:

70.6

Does $2^{\aleph_0}<2^{\aleph_1}$ imply that separable (a) spaces have countable extent (i.e. closed discrete spaces are at most countable).

70.7dS.

If $|\mathcal A|=\aleph_1$ is an AD family and $\psi(\mathcal A)$ has property (a) then $\mbox{cf}({}^{\omega_1}\omega,\leq)\leq 2^{\aleph_0}$

Together with the following

70.8Jech, Prikry

If $2^{\aleph_0}$ is regular, $2^{\aleph_0}<2^{\aleph_1}$ and $\mbox{cf}({}^{\omega_1}\omega,\leq)\leq 2^{\aleph_0}$ then there is an inner model with a measurable cardinal.

this shows that $\psi$-spaces cannot be counterexamples to the question unless there are large cardinals.

of 70.7

Let $\mathcal =\{A_\alpha:\alpha<\omega_1\}$. For all closed discrete $G\subseteq\omega$ define $F_G:\omega_1\to\omega$ as follows: $F_G(\alpha)=\mbox{sup}(A_\alpha\cap G)+1.$ Then $\{F_G:G\subseteq\omega\ \mbox{is closed discrete}\}$ is a dominating family in ${}^{\omega_1}\omega$: given a function $g:\omega_1\to\omega$ consider the following cover: $\mathcal U=\{\{A_\alpha\}\cup A_\alpha\setminus g(\alpha):\alpha<\omega_1\}\cup\{\omega\}.$ Since $\psi\mathcal(\mathcal A)$ is (a) there is a closed discrete $G^\prime$ such that $st(G^\prime,U)$ covers $\psi(\mathcal A)$. But then $F_{G^\prime}$ dominates $g$.

70.9C. Morgan, dS., 2009

If $P$ is normality, (a) or countable paracompactness we define ${\mathfrak m}_{P}=\min\{|\mathcal A|:\psi(\mathcal A)\ \mbox{is not}\ P\}$ and ${\mathfrak n}_{P}=\min\{\kappa:|\mathcal A|=\kappa\rightarrow\mathcal A\ \mbox{is not}\ P\}$

70.10

$\mathfrak m_{normal}=\aleph_1$ (Luzing gaps), and if $2^{\aleph_0}<2^{\aleph_1}$ then $\mathfrak n_{normal}=\aleph_1$ (Jones' lemma).

Since normal $\psi$-spaces are countably paracompact we also have $\mathfrak m_{normal}\leq\mathfrak n_{c-paracompact}$.

70.11

An AD family $\mathcal A$ is soft (sa) if there is $P\subseteq\omega$ such that $(\forall A\in\mathcal A)(0<|P\cap A|<\omega)$.

70.12Szeptycki

The space $\psi(\mathcal A)$ is (a) iff all finite modifications of $\mathcal A$ are soft.

It follows that $\mathfrak m_{sa}=\mathfrak m_{(a)}$. Szeptycki and Vaughan proved that $\mathfrak p\leq\mathfrak m_{sa}$ and Brendle, Yatabe and Szeptycki proved that the upper bound is $\mathfrak b$. However, there is not much known about $\mathfrak n$. E.g., the following question is open:

70.13

Does $2^{\aleph_0}<2^{\aleph_1}$ imply that $\mathfrak n_P =\aleph_1$ for $P\in\{\mbox{countably paracompact, (a)}\}$.

70.14

The diamond principle $\Diamond(\omega,\omega<)$ for Borel functions implies $\mathfrak n_{sa}=\mathfrak n_{c-paracompact}=\aleph_1$.

Recall that the diamond principle gives for all Borel $F:{}^{<\omega_1}2\to\omega$ there is $g:\omega_2\to\omega$ such that $\{\alpha:F(f\upharpoonright\alpha) is stationary for each branch \(f$.

### 70.1.1The selective version of (a)

The following was introduced by Casarta, di Maio, Kočinac in 2011:

70.15

A space $X$ is selectively (a) if for every sequence $\langle \mathcal U_n:n<\omega\rangle$ of open covers and for all dense $D\subseteq X$ there is a sequence $\langle A_n:n<\omega\rangle$ of subsets of $D$ which are closed discrete in $X$ such that $\{ st(A_n,\mathcal U_n):n<\omega\},$ is a cover of $X$.

70.16dS.,2014

If $X$ is selectively (a) and $H\subseteq X$ is closed discrete then $|H|<2^{d(X)}$.

Note that under CH, (a) and selectively (a) are equivalent for $\psi$-spaces, since, under CH, only countable $\psi$-spaces can be selectively (a). This equivalence is also true under $\mathfrak p=\mathfrak c$.

70.17dS.,2014

Let $\mathcal A$ be almost disjoint and $X=\psi(\mathcal A)$. Then

1. If $|\mathcal A|<\mathfrak p$ then $X$ is (a)
2. If $|\mathcal A|<\mathfrak d$ then $X$ is selectively (a)
3. If $\mathcal A$ is MAD then it is selectively (a) iff $|\mathcal A|<\mathfrak d$.

Proof (idea): 2. Let $\mathcal A$ have size $<\mathfrak d$. For each $n<\omega$ and $A\in'mathcal A$ fix $U_{A,n}\in\mathcal U_n$ containing $A$. Let $f_A(n)=\min\{U_{A,n}\cap\omega$. Let $f$ be unboundaded. Then for each $A$ there is $m$ such that $f_A(m). Let $P_n=\{k:0\leq k\leq f(n)\}\cup\{n\}.$ 1. Aiming towards a contradiction assume that \(\mathcal A$ is MAD of size $\geq\mathfrak d$. Wlog $\mathcal A=\{A_\alpha:\alpha<\mathfrak d\}$. Let $\{f_\alpha:\alpha<\mathfrak d\}$ be a dominating family. Define $\mathcal U_n=\Big\{\{A_\alpha\}\cup A_\alpha\setminus f_\alpha(n):\alpha<\mathfrak d\Big\} \cup \Big\{X\setminus\mathcal A\Big\}$ If $\langle P_n:n<\omega\rangle$ is an arbitrary sequence of closed discrete subsets of $\omega$ by maximality of $\mathcal A$ they are finite. Let $g(n)=\mbox{max}P_n+1$ Now $g$ is dominated by some $f_\alpha$, wlog everywhere. But then $A_\alpha\not\in st(P_n,\mathcal U_n)$ contradicting that the space $X$ is selectively (a).
70.18

It is consistent that there is a selectively (a) non (a) $\psi$-space. Moreover it is consistent that $2^{\aleph_0}<2^{\aleph_1}$ and there is a separable selectively (a) space with an uncountable closed discrete subset (adding $\aleph_{\omega_1}$-many Cohen reals to a model of CH).

70.19

$2^{\aleph_0}<2^{\aleph_1}$ does not imply $\mathfrak n_{s(a)}=\aleph_1$.

70.20dS., 2015

Separable, $T_1$, locally compact, selectively (a)-spaces have countable extent assuming the Borel diamond $\Diamond({}^\omega\omega,<)$.

Proof (by contradiction): Fix a countable dense $D\subseteq X$, wlog $D=\omega$, and let $H=\omega_1\setminus\omega$ an uncountable closed discrete subset of $X$. We will show that $X$ is not selectively (a). For $\beta\in\omega_1\setminus\omega$ fix open $U_\beta$ such that $U_\beta\cap H=\{\beta\}$. By local compactness we may assume $U_\beta\subseteq K_\beta$ for some compact $K_\beta$. Let $A_\beta=U_\beta\cap\omega$. Let $X_f:f\in 2^\omega$ be a Borel enumeration of all sequences of subsets of $\omega$, say $X_f=\langle X_{f,n}:n<\omega\rangle$. For $h\in{}^{<\omega_1}2$ with $\mbox{dom}(h)\geq\omega$ and for $n<\omega$ let $Y_{h,n} = A_{\mbox{dom}(h)}\cap X_{h\upharpoonright\omega,n}$ Define "the project" $F:{}^{<\omega_1}2\to{}^\omega\omega$ as follows: $F(h)(n) = \mbox{sup}(Y_{h,n}+1\ \mbox{if}\ |h|=\aleph_0\ \mbox{and}\ |Y_{h,n}|<\omega)$ and $0$ otherwise. Then $F$ is Borel. Let $g:\omega_1\to{}^\omega\omega$ be the "funding" (oracle) given by the diamond. We define covers witnessing that $X$ is not selectively (a) as follows: $\mathcal U_n=\{X\setminus(\omega_1\setminus\omega)\}\cup \{U_\beta\setminus g_\beta(n):\beta\in\omega_1\setminus\omega.$ Now if $P=\langle P_n:n<\omega\rangle$ is a sequence of closed discrete subsets of $\omega$ then pick a branch $f:\omega_1\to2$ such that $P=X_{f\upharpoonright\omega}$. Then $S = \{\beta<\omega_1:F(f\upharpoonright\beta) is stationary. Then clearly any $\beta\in S\setminus\omega$ is not covered by any $st(P_n,\mathcal U_n)$. {} # 7113.10.2016 ## 71.1E. Thuemmel: Extremally disconnected groups We present the following result of Reznichenko and Sipacheva: 71.1Reznichenko,Sipacheva It is consistent that there are no countable, nontrivial (i.e. T_1 and nondiscrete) extremally disconnected groups. 71.2 The uncountable case is still open. 71.3Sirota There is a countable dense subset of the Stone space of Mathias forcing with a selective ultrafilter which is an extremally disconnected group. The same works for Prikry forcing with a measure. 71.4Malychin There is an extremally disconnected group under $\mathfrak p=\mathfrak c$. We show that if there is no no rapid filter then there is no nontrivial countable extremally disconnected group. 71.5 A filter $\mathcal F$ is not rapid if for each infinite $\{n_k:k<\omega\}$ there is a partition of $\omega$ into finite sets $Y_k$ such that for each $F\in\mathcal F$ there is $k<\omega$ such that $|F\cap Y_k|>n_k$. The proof strategy will be to construct two disjoint sets $\xi_0,\xi_1\subseteq G$ which both have $e$ as the single common point in their closures. This is enough since for each $x\in \xi_i$ we find a clopen neighbourhood which is disjoint from $\xi_{1-i}$. We disjointify these neighbourhoods and get disjoint open $U_i$ with $e$ in their closures which contradicts extremal disconnectedness. We also use the following theorem of Malychin. 71.6Malychin If there is an extremally disconnected group then it contains an extremally disconnected Boolean subgroup. In the following definitions, propositions and examples, we assume, for simplicity, that the groups are Boolean, because that is sufficient for our applications. In fact, they hold for arbitrary groups. 71.7 Call $M\subseteq G$ thick if there is $n$ such that for each $P\in[G]^n$ there are distinct $a,b\in P$ such that $ab\in M$. If $M$ is thick, let $j_M$ be the least $n$ as in the definition. 71.8 If $W\subseteq G$ is disjoint from its square (i.e. $W\cap W^2=\emptyset$) and $G$ is Boolean, then $G\setminus W$ is thick. We show that $j_{G\setminus W}\leq 3$. Choose $\{a,b,c\}\subseteq G$. Assume that $ab,ac\in W$ (otherwise we are done). But, since $G$ is Boolean, $bc=abac$ and, as $ab,ac\in W$ we know that $bc\in W^2\subseteq G\setminus W$. 71.9 If $H$ is a subgroup of $G$ with finite index $i_H$ then it is thick. We show that $j_H\leq i_H+1$. If $P\in[G]^{i_H+1}$ then there are two distinct $a,b\in P\cap gH$ for some $g$. But then $ab\in gHgH=H$. 71.10 Call $M\subseteq G$ very thick if for each $m$ there is $n(m)$ such that for each $P\in[G]^{n(m)}$ there is $Q\in[P]^n$ such that $Q^2\subseteq M\cup\{e\}$. 71.11 If $M$ is thick iff it is very thick. {} Proof: The implication from right to left is clear. So let $M$ be thick and choose $m<\omega$. We may assume $m>j_{M}$. Now by Ramsey's theorem there is $R_m$ such that for each $P\in[G]^{R_m}$ there is $Q\in[P]^m$ with $Q^2\subseteq M\cup\{e\}$ or $Q^2\cap M=\{e\}$. However, the second choice is ruled out since $m>j_{M}$. 71.12 The thick subsets of $G$ form a filter. Let $M_1,M_2$ be thick and assume $j_{M_1}\leq j_{M_2}$. Since $M_1$ is very thick, there is $n=n(j_{M_2})$ such that for every $P\in[G]^n$ there is $Q\in[P]^{j_{M_2}}$ such that $Q^2\subseteq M\cup\{e\}$. But then for distinct $a,b\in Q$ we have $ab\in M$ showing that $M_1\cap M_2$ is thick. From now on we assume there is no rapid filter. 71.13 If $G$ is a countable Boolean group then there are a finite-to-one onto function $f:G\to X$, some filter $\mathcal F$ on $G$ which converges to $e$ (i.e. $e\in\overline{F}$ for each $F\in\mathcal F$) and a countable set $\{M_n:n<\omega\}$ of thick subsets of $G$. Then there is $\xi\subseteq G$ which is a pseudointersection of the thick subsets and for each $F\in\mathcal F$ there are $a,b\in F$ with distinct $f$-images $f(a)\neq f(b)$ and such that $ab\in\xi$. 71.14 It is sufficient to assume, in the above lemma, that $f[\mathcal F]$ is not rapid. First we may assume that the $M_n$'s are decreasing. Consider $f[\mathcal F]$. It is not rapid so there is a partition $X=\bigcup Y_n$ into finite sets such that for each $F\in\mathcal F$ there is $n$ with $|f[F]\cap Y_n|>j_{M_n}$. Since the $M_n$'s are thick we can find, for each $F\in\mathcal F$, distinct $a,b\in F$ with $f(a)\neq f(b)$ and $ab\in M_n$. Now consider the set $\xi_n=\{ab\in M_n:f(a)\neq f(b)\in Y_n\}$. Let $\xi =\bigcup_{n<\omega}\xi_n$. Since $f$ is finite to one and $Y_n$ is finite we easily have that $\xi$ is a pseudointersection of the $M_n$'s. We now apply the lemma to get $\xi_1,\xi_2$. Let $X=G$, $f=id_G$ and $\mathcal F$ the filter of neighbourhoods of $e$. First notice that $G$ has a descending sequence $\langle U_n:n<\omega\rangle$ of neighbourhoods of $e$ with $\bigcap_{n<\omega} U_n=\{e\}$ and $U_{n+1}^3\subseteq U_n$. Now for every $g\not\in U_n$ we have $gU_{n+1}\cap (gU_{n+1})^2=\emptyset$. In particular $G\setminus gU_{n+1}$ is thick for each $g\not\in U_n$. Since all Boolean groups are of the form $([|G|]^{<\omega},\triangle)$ and since $G$ is countable we may assume that $G=[\omega]^{<\omega}$. Let $H_n=\{g\in G:\min g > n\}$. Then $H_n$ is a subgroup of $G$ with finite index (i_{H_n}=n+1) so, in particular, it is thick. Let \[ \{M_n:n<\omega\}=\{H_n:n<\omega\}\cup\{G\setminus gU_{n+1}:g\not\in U_n\}$ By the previous lemma there is $\xi^\prime i_1\subseteq G$ which is a pseudointersection of the $M_n$'s and has the desired property w.r.t. $\mathcal F$, $f$. In particular, we have

1. for each $g\not\in U_n$ we have $|\xi^\prime_1\cap gU_{n+1}|<\omega$; and
2. $\lim_{g\in\xi^\prime_1}(\min g)=\infty$.
3. for every $F$ neighbourhood of $e$ there are distinct $a,b\in F$ with $ab\in\xi^\prime_1$

If $g\neq q$ then there is $n$ with $g\not\in U_n$ then $|\xi_1\cap gU_{n+1}|<\omega$ and, in particular, $g\not in\overline{\xi^\prime_1}$. So the only limit point of $\xi^\prime_1$ can be $e$. We now show that it is, in fact, a limit point. Let $U$ be a neighbourhood of $e$. Find $V$ neighbourhood of $e$ with $V^2\subseteq U$. Then $V\in\mathcal F$. So there are distinct $a,b\in V$ with $ab\in\xi^\prime_1\cap V^2\subseteq U$.

Let $y_0=0$. Inductively define $y_{n+1}=\max(\{\max g:\min g Now partition $\xi_1^\prime$ into two sets: \[ \xi_1^\prime= \{g\in\xi^\prime_1:\min g\in\bigcup_{n<\omega}[y_{2n},y_{2n+1})\} \cup \{g\in\xi^\prime_1:\min g\in\bigcup_{n<\omega}[y_{2n+1},y_{2n+2})\}$ one of these sets must have $e$ in its closure and let $\xi_1$ be this set. Now define $f(g) = \left\{ \begin{array}{ll} n & g\in\xi_1\ \&\min g\in[y_n,y_{n+1})\cr g & g\not\in\xi_1\cr \end{array} \right.$ and let $X=f[G]$. Then $f$ is finite-to-one. Note also that if $a,b\in\xi_1$ with $f(a)\neq f(b)$ then $ab\not\in\xi_1$. Let $\{M_n:n<\omega\} = \{ G\setminus gU_{n+1}:g\not\in U_n\}$ and let $\mathcal F$ be the filter generated by neighbourhoods of $e$ and $\xi_1$. Applying the lemma again we get a $\xi^\prime_2$. Let $\xi_2=\xi^\prime_2\setminus \xi_1$. Using the same argument we have that there is no other limit point of $\xi_2$ except, possibly, $e$. We now need to show that $e$ is, in fact, a limit point of $\xi_2$: Let $U$ be a neighbourhood of $e$. Find $V$ neighbourhood of $e$ with $V^2\subseteq U$. By (3) above, since $\xi_1\cap V\in\mathcal F$ there are distinct $a,b\in V\cap \xi_1$ with $f(a)\neq f(b)$ and $ab\in\xi^\prime_2$. But since $ab\in V^2\subseteq U$ we know that $ab\not\in\xi_1$ so $ab\in\xi_2$. This finishes the proof that $e$ is a limit point of $\xi_2$.

It is clear that the $\xi_1,\xi_2$ are as required.

# 723.11.2016

## 72.1M. Pawliuk: Fraïssé classes & the Hrushovski property

72.1

Let $X$ be a structure with a "lot of homeomorphisms". What dynamical properties does $Aut(X)$ have? E.g. how does it act on compact spaces, in particular, does it have fixed points when it acts on compact spaces?

72.2

$Age(X)$ is the family of all finite substructures of $X$.

72.3Kechris, Pestov, Todorčević, 2005

If $X$ is countable and homogeneous then actions of $Aut(X)$ are determined by the finite combinatorics of $Age(X)$.

The result intuitively says that topological dynamics of $X$ is "dual" to finite combinatorics of the $Age(X)$ with fixed points corresponding to Ramsey properties. Amenability turns out to correspond to finite measures.

### 72.1.1Three examples

72.4

Consider $(\mathbb Q,\leq)$, the unique countable linear dense in itself linear order without endpoints.

The characterization of $\mathbb Q$ can be reformulated as follows: For all $a_0 and any consistent position of an \(x$ which gives a linear order is realized by some $q\in\mathbb Q$. I.e. any type over a finite set of parameters is realized.

72.5

Consider the Random graph, the unique countable graph such that for any disjoint finite sets of vertices $A,B$ there is a vertex $v$ connected to every vertex in $A$ and not connected to any vertex in $B$.

The characterization is again equivalent to saying that every type over a finite set of parameters is realized.

72.6

Consider the rational Urysohn space $\mathbb (U,d)$, the unique countable metric space with rational distances which has the "1-point extension property", i.e. for every finite $A\subseteq U$ and prescribed distances to points in $A$ respecting the triangle inequality, there is a point in $\mathbb U$ having these prescribed distances.

Again, the characterization is equivalent to realizing types over finite sets of parameters.

We now consider directed graphs.

72.7

A directed graph is a set $V$ with an irreflexive, antisymmetric binary relation.

72.8Cherlin, 1998

If $X$ is a countable, homogeneous digraph then it is one of the following:

1. $(\mathbb Q,<)$
2. The random complete digraph (tournament)
3. The random complete $n$-partite digraph
4. ... 15.
72.9

What are the dynamical properties of $Aut(D)$ where $D$ is a countable homogeneous digraph. In particular, which are amenable? If amenable, is the mean/measure unique?

72.10Sotic,Pawliuk

The answer for the above questions is known for all $D$ with one exception.

72.11Hrushovski,1992

Let $\{(A_i,B_i):i be a finite set of pairs of induced subgraphs of a finite graph \(C$ and graph isomorphisms $f_i:A_i\to B_i$. Then there is a finite graph $G\supseteq C$ and automorphisms $g_i:G\to G$ such that $f_i\subseteq g_i$.

72.12

This theorem is interesting even for $k=1$ as well as if $A,B$ are single points. Also, the case $k=2$ is as difficult as the general case.

The following is a big open question:

72.13

Does the theorem still hold if we replace graph by tournament, i.e. a complete digraph? (Yes for $k=1$)

The following is a little less known but still quite big.

72.14

Is there any assymetric class (e.g. digraphs) with the Hrushovski property?

72.15Solecki, circa 2009

Finite metric spaces have the Hrushovski property.

### 72.1.2Amenability

The motivating example will be the Random graph. Its associated object (the universal minimal flow) $M(G)$ is the set of all dense linear orders on $\omega$ which has the product topology inherited from $2^{\omega^2}$.

72.16Kechris, Pestov, Todorčević

Suppose $\mathcal K$ is a Fraïssé class and "some other assumptions". Then the following are equivalent:

1. The universal minimal flow $M(G)$ is equal to $X_{\mathcal LO(\mathcal K)}$ the collection $LO(\mathcal K)$ of all dense linear orders on $\mbox{dom}(\mathcal K)$ with the product topology.
2. $LO(\mathcal K)$, the is a Ramsey class
72.17Nguyen van Thé

Suppose $\mathcal K$ is a Fraïssé class and $\mathcal K^*$ is a class of (precompact) expansions of $\mathcal K$. Then

1. $M(G)=X_{\mathcal K^*}$
2. $\mathcal K^*$ is a Ramsey class.
72.18

We say that $G$ is amenable if $M(G)$ has a $G$-invariant measure. It is uniquely ergodic if $M(G)$ has a unique $G$-invariant measure. It is extremely amenable if $|M(G)|=1$.

72.19Angels, Kechris, Lyov

If $\mathcal K$ is a Fraïssé class with precompact expansions ($\mathcal K^*$) and $G=Aut(\mathcal K)$ then $G$ is amenable if there are measures $\mu_A$ for each $A\in\mathcal K$ such that

1. $\mu_A(A^*)\in[0,1]$ for each $A^*\in\mathcal K^*$ expansion of $A$.
2. $\mu_A$ is a probability measure.
3. For each $A\leq B\in\mathcal K$ and for all expansions $A^*\in\mathcal K^*$ of $A$ $\mu_A(A^*) = \sum_{B^*}\mu_B(B^*)$

See also L. Nguyen van Thé, J. Jasinski, C. Laflamme, R. Woodrow: Ramsey precompact expansions of homogeneous directed graphs, Electron. J. Combin., 21 (4), 31pp, 2014

# 744.1.2017

## 74.1D. Chodounský: Preserving ultrafilters

74.1

Whenever we add an unbounded real all non P-point ultrafilters are destroyed. Whenever we add a real we destroy at least one ultrafilter.

The first part is easy (consider a decreasing sequence sets from the ultrafilter with no pseudointersection in the ultrafilter and visualize it as the columns of $\omega\times\omega$---then the graph of an unbounded real splits $\omega\times\omega$ into two parts not decided by the ultrafilter).

For the second part first construct a tree $T$ such that

1. At every level all nodes either split or don't split
2. The distance between splitting levels is large enough so that for every subset $A$ of the splitting nodes on a level there is a level below, above the previous splitting level $n$, such that $A=\{s:s(n)=1\}$.
3. For every level $n$ there are $s,t$ on the level such that $s(n)\neq t(n)$

Given a perfect tree $S\subseteq T$ let $X_S=\bigcap[S]$ and $Y_S=\omega\setminus\bigcup[S]$. Consider the ideal $\mathcal I$ generated by $\{X_S,Y_S:S\subseteq T\ \mbox{is perfect}\}$. First notice that all singletons are in $\mathcal I$. It is also not hard to show, using (2), that $\omega\not\in\mathcal I$: Given $S_0,\ldots,S_n$ find a splitting level $k$ and $A\subseteq T^k$ such that $A\cap S_i\neq\emptyset=\neq A\setminus S_i$. Then $n$ given by (2) is not contained in any $X_{S_i}$ or $Y_{S_i}$. We show that the co-ideal $\mathcal I^+$ is not reaping in the extension. Since we added a new real there is a new branch $r\subseteq\omega$ through $T$. This branch is not reaped by $\mathcal I^+$: Suppose, aiming towards a contradiction, that there is $x\in\mathcal I^+\cap V$ so that, wlog, $a\subseteq r$. The set $\{x\in [T]:a\subseteq x\}$ is a (family of branches of a) subtree $T^\prime$ of $T$. Now $r$ is a branch through $T^\prime$. It follows that $T^\prime$ must be perfect (any tree with a new branch must be perfect). However $X_{T^\prime}\in\mathcal I^+$---a contradiction.

74.2

A similar argument shows that after adding a real at least one MAD family is destroyed.

74.3

A definable (Suslin) ccc forcing either contains a Cohen real or a Random real (Judah-Shelah) which are splitting, so no definable ccc forcing preserves ultrafilters.

74.4

After any iteration of definable forcings of length $\omega_2$ there are P-points.

A typical argument showing that P-points exist either uses some cardinal invariants (e.g. $\mathfrak d=\mathfrak c$) or preservation of some Borel diamonds. These arguments don't work for the Random model. A different argument by Paul Cohen (not the one from forcing) was thought to show the existence of P-points in the Random model. However a gap was found in this proof so it is an open question whether P-points exists in Random forcing. It is also open whether P-points exist in the Silver model. Since both the Random and Silver forcings add splitting reals, no ground-model P-points are preserved. Also, if there are P-points at the end, they must reflect to some earlier stage (i.e. P-points cannot be added at the final step).

74.5

After adding a Silver real the filter generated by a ground-model P-point is no-longer $P^+$,

74.6

Assume $P$ is a definable and bounding/Sacks-property forcing destroying a P-point $\mathcal U$. Moreover assume $X$ in the extension is not reaped by $\mathcal U$. Does it follow that there is a $U\in\mathcal U$ such that $U\cap X$ is a splitting real (on $[U]^\omega$).

74.7

If $P$ is a definable and bounding/Sacks-property forcing destroying a P-point then it adds a splitting real.

74.8Zapletal

Assume CH. Then for all definable proper $P$ the following are equivalent:

1. $P$ preserves $P$-points.
2. $P$ does not add splitting reals and has the weak Laver property
74.9

Definability means that, either:

1. $P$ is of the form $Borel(X)\setminus \mathcal I$ for some $\Pi^1_1$ on $\Sigma^1_1$ $\sigma$-ideal $\mathcal I$ on $X$.
2. LC (determinacy for universally Baire sets) + $P$ is of the form $Borel(X)\setminus \mathcal I$ for some universally Baire $\sigma$-ideal $\mathcal I$ on $X$.
74.10

A forcing $P$ has the Laver property if for each bounded real $x$ in the extension there is an $F:\omega\to[\omega]^{<\omega}$ in the ground model with $|F(n)|\leq n$ such that $x(n)\in F(n)$ for all $n$. It has the weak Laver property if $x(n)\in F(n)$ for all $n$ in some infinite ground-model set.

74.11

If $P$ has the weak Laver property it does not add a bounded eventually different real.

The combinatorial core of Zapletal's theorem is the essence of the following proposition:

74.12

A forcing $P$ satisfies (2) iff for each $F_\sigma$ ideal $\mathcal I$ the forcing $P$ forces that $(\mathcal I^+)^V$ is reaping.

Suppose $P$ does not have the weak Laver property (if it adds splitting reals it clearly cannot preserve any reaping family). Let $f$ be a function bounded by some $b\in V$ witnessing the failure of the weak Laver property. Let $X=\bigcup\mathcal P(b(n))$. We define let $\mu_n(A)= \max\{|\bigcap A\cap\mathcal P(b(n))|,|b(n)\setminus \bigcup A\cap\mathcal P(b(n))|\}.$ Let $\mathcal I$ be the ideal generated by sets $A$ satisfying $\mu_n(A)\geq 2^n$ for all $n<\omega$. Then $[\omega]^{<\omega}\subseteq I$ and $\mathcal I$ is an $F_\sigma$-ideal. Moreover $\omega\not\in\mathcal I$.

Now consider the set $A\subseteq X$ defined as follows: $A=\{x\in X:(\forall n<\omega)(\forall a\in x\cap \mathcal P(b(n)))(f(n)\in a)\}.$ Then $A$ is not reaped by any ground-model $\mathcal I$-positive set: given a positive $C\in\mathcal I^+\cap V$ there must be infinitely many $n's$ such that $\mu_n(C)<2^n$. This allows us to define a function $F_C$ such that $|F_C(n)|<2^n$ and $f(n)\in F_C(n)$ for infinitely many $n$s.

On the other hand assume that $\mathcal I=Fin(\mu)$ is an $F_\sigma$-ideal. Partition $\omega=\bigcup_{n<\omega}I_n$ such that $\mu(I_n)\geq n\cdot 2^{2^n}$. Let $G\subseteq\omega$ be a new set. Consider $f_g(n)=I_n\cap G$. Then $f_g$ is a bounded function. By the weak Laver property there is an infinite $A\in[\omega]^\omega$ and $F:A\to[\omega]^{<\omega}$ such that $f_g(n)\in F(n)$ for all $n\in A$ $\ldots$

# 7525.1.2017

## 75.1E. Thuemmel: The P-point game for co-ideals

75.1

A Boolean algebra $B$ is $(\omega,2)$-distributive iff for any sequence $\langle a_i:i<\omega\rangle$ and any $a\in B^+$ there is $b\in B^+$ such that $b\leq a$ and for each $i<\omega$ either $b\leq a_i$ or $b\leq -a_i$. We say that an ideal is $(\omega,2)$-distributive iff the algebra $\mathcal P(\omega)/\mathcal I$ is $(\omega,2)$-distributive.

75.2

An ideal $\mathcal I$ is $P^-$ if for each positive $X\in \mathcal I^+$ and for each partition $X=\bigcup_{n<\omega}R_n$ of $X$ into pieces $R_n\in \mathcal I$ there is a positive $Y\in\mathcal I^+\upharpoonright X$ such that $|Y\cap R_n|<\omega$.

75.3

Given an ideal $\mathcal I$ we consider the following game $\mathcal G_P(\mathcal I)$: the first player starts by choosing an $X\in\mathcal I^+$ and then they alternate moving as follows: The first player plays a partition $A^0_n\cup A^1_n=X$ and the second player chooses one of the pieces, $i_n<2$, and a finite $F_n\in[\bigcap_{m\leq n} A^{i_m}_n]$ set coming from the intersection of all his previous choices. The second player wins if the union of the finite sets is $\mathcal I$-positive.

75.4

Player I has a winning strategy iff either $\mathcal I$ is not $P^{-1}$ or $\mathcal I$ is not $(\omega,2)$-distributive.

The next proposition says that a strenghtening of non-distributivity (*) is in fact equivalent to the condition in the above theorem. It is rather easy to see and we omit the proof.

75.5

An ideal is either not distributive or not $P^-$ iff there is a positive $X\in\mathcal I^+$ and a sequence $\langle A_n^i:i<2,n<\omega\rangle$ of partitions of $X$ such that there is no $\mathcal I$-positive $Y$ below $X$ which would be almost included in one of the pieces of the all partitions.

75.6

An ideal has (*) iff there is an $\mathcal I$-positive $X$ such that $\mathcal I\upharpoonright X$ is Katětov-above $conv$.

of the Theorem

We use ideas of Galving and Shelah. Proposition 75.5 immediately gives a winning strategy for Player I in case $\mathcal I$ is either non-distributive or not $P^-$. On the other hand if $\sigma$ is a winning strategy for the first player we show that $\mathcal I$ has (). The $X$ guaranteed by () is the first move of Player I. If $g$ is a game played according to $\sigma$, let $A^i_{gn}$ be the $n$-the move of player I. Consider the set $\{ A^i_{gn}:n<\omega,g\ \mbox{is a game played according to}\ \sigma,i<2\}.$ This set is countable set of partitions witnessing (*)

75.7

If $\mathcal I$ is contained in an $F_\sigma$ ideal then II has a winning strategy in the game where player I is not allowed to choose $X$.

Easy.

75.8Hrušák, 3.15 in joint article

If $\mathcal I$ is analytic then either $\mathcal I$ has (*) or it has an $F_\sigma$ extension.

75.9

An ideal $\mathcal I$ is Ramsey iff $\mathcal I^+\rightarrow(\mathcal I^+)^2_2$

75.10

Is there an Analytic Ramsey ideal?

75.11

If $\mathcal I$ has (*) then it is not Ramsey.

We use a trick due to S. Sierpińsky. Let $X$ and $A^i_n$ be given by (). Define $\varrho:\omega\to{}^\omega2$ as follows: $\varrho(k)(n) = \left\{ \begin{array}{ll} 0 & k\in A_n^0\\ 1 & k\in A_n^1 \end{array} \right.$ and, for $n let $\chi(k,l) = \left\{ \begin{array}{ll} 0 & \varrho(k)<_{LEX}\varrho(l)\\ 1 & \varrho(k)>_{LEX}\varrho(l)\\ 2 & \varrho(k)=\varrho(l) \end{array} \right.$ This coloring witnesses that \(\mathcal I$ is not Ramsey: homogeneity in $2$ is ruled out by (). If there was a big homegeneous set in 0 or 1 then there would be a strictly increasing or strictly decreasing sequence in $({}^\omega2,\leq_{LEX})$. But its limit would be a big pseudointersection contradicting (*).

75.12

If $\mathcal I$ is a tall analytic and is contained in an $F_\sigma$-ideal then it is not Ramsey.

75.13

If the above conjecture is true then there is no analytic tall Ramsey ideal.

## 75.2Saeed Ghasemi: Scattered C-star algebras

75.14

Let $H$ be a Hilbert space (i.e. a complete normed vector space over $\mathbb C$ with the norm coming from an inner product). We let $B(H)$ be the set of all bounded linear operators. Given an operator $T\in B(H)$ there is a unique $T^*:H\to H$ such that $\langle Tx,y\rangle = \langle x,T^*y\rangle$ for all $x,y\in H$. The norm of $T$ is defined to be $||T||=\sup\{||Tx||:||x||\leq 1\}.\leqno(*)$ It is not hard to see that $||TT^*||=||T||^2$.

75.15

An concrete C-star algebra is a norm-closed subspace of $B(H)$ for some hilbert space $H$.

75.16

An abstract C-star algebra is a $*$-Banach algebra satisfying $(*)$.

75.17Gelfland, von Neumann

Every abstract C-star algebra is a concrete C-star algebra.

75.18

The finite-dimensional C-star algebras are exactly the matrix algebras over $\mathbb C$.

For the rest of this talk assume $H$ is separable (even isomorphic to $l_2)$.

75.19

Let $K(H)$ be the ideal of all compact operators on $H$. Then we have a projection $\pi:B(H)\to B(H)/K(H)$. The quotient $B(l_2)/H(l_2)$ is called the Calkin algebra.

For the following fix a basis $\{e_n:n<\omega\}$ for $l_2$, e.g. $e_n(m) = \left\{ \begin{array}{ll} 1 & n=m\\ 0 & n\neq m \end{array} \right.$

75.20

An element $p\in A$ is a projection if $p=p^2=p^*$.

75.21

The Boolean algebra $\mathcal P(\omega)\rightarrow Proj(B(H))$ (the second part is a complete Lattice given by the order $p\leq q\iff pq=p$.

75.22

A family $\{P_\xi:\xi<\kappa\}$ of non-compact projections is an almost orthogonal family if $P_\xi P_\eta$ is compact for every $\xi\neq\eta$.

75.23

$\mathfrak a^*$ is the minimal size of an inrinite maximal almost orthogonal family of projections.

75.24Wolfey

MA implies that $\mathfrak a^*=2^{\aleph_0}$ and it is consistent that $\mathfrak a^*<2^{\aleph_0}$.

75.25

It is not known whether $\mathfrak a\leq \mathfrak a^*$, $\mathfrak a\geq \mathfrak a^*$ or $\mathfrak a=\mathfrak a^*$ can be proved in ZFC.

Recall that given an AD family $\mathcal A$ on $\omega$ we can define the $\psi$-space $\psi(\mathcal A)$ on the set $\omega\cup\mathcal A$ where each $n\in\omega$ is isolated and the neighbourhoods of $A\in\mathcal A$ are of the form $A\setminus F$ for some finite $F\subseteq\omega$. Then $\psi(\mathcal A)$ is a scattered locally compact Hausdorff space of Cantor-Bendixon rank $2$.

Note that if $X$ is a locally compact Hausdorff space then the space $C_0(X)$ of complex-functions on $X$ converging to 0 forms a commutative C-star algebra.

75.26Gelfand

Every commutative C-star algebra is of the form $C_0(X)$ for some locally compact Hausdorff space $X$.

This allows us to generalize topological notions to C-star algebras, e.g.

1. A C-star algebra is connected iff it has no nontrivial projections
2. A compactification of space corresponds to the unitization of a C-star algebra
3. An algebra is "compact" if it is unital
4. There is a Czech-Stone compactification (universal unitization, given by the multiplier algebra of A), one-point-compactification (a minimal unitization)
5. etc.
75.27

To define the multiplier of $A$ consider a faithful -homomorphism $\sigma: A\to B(H)$. Then we can let $\mathcal M(A) = \{T\in B(H): (\forall a\in A)(T\sigma(a)\in\sigma[A],\sigma(a)T\in\sigma[A])\}$ The algebra $\mathcal M(A)/A$ is called the corona algebra of $A$*.

Note that whenever $\mathcal A$ is an AD family of size $\kappa$ then we have the following short exacet sequence: $0\rightarrow C_0(\omega)\rightarrow C_0(\psi(\mathcal A))\rightarrow C_0(\kappa)\rightarrow 0$

Unfortunately, although the most interesting part of the talk followed, I was too confused with C-star algebras to continue recording it.

# 768.1.2017

## 76.1S. Navarro-Flores -- Ramsey spaces

76.1

Recall that we write $[a,A]=\{B\in[\omega]^\omega:a\sqsubseteq B\subseteq A\}$ for $a\in[\omega]^{<\omega}$ and $A\in[\omega]^\omega$. These sets form the basis of a topology on $[\omega]^\omega$. This topology is called the Ellentuck topology.

76.2

A family $X\subseteq[\omega]^\omega$ is called Ramsey if for each nonempty $[a,A]$ there is $[a,B]\leq [a,A]$ such that $[a,B]\subseteq X$ or $[a,B]\cap X=\emptyset$.

76.3Ellentuck

A family $X$ is Ramsey iff X has the Baire property.

76.4

Let $(\mathcal F,\sqsubseteq)$ be a partial order and $T\subseteq F$ a tree of height $\omega$ in this order. We define the following order $\leq$ on the branches of $T$ as follows: moreover, assume that $\mathcal F$ admits a partial order $\leq$. We write

76.5Solecki, Todorcević

Every analytic ideal contains a cofinal (w.r.t. inclusion) $G_\delta$ set.

76.6

Does every co-analytic co-ideal contain a $G_\delta$ dense set (w.r.t. inclusion)

# 7715.2.2017

## 77.1O. Guzmán: Desctructibility of MAD families

77.1

Let $P$ be a forcing and $\mathcal A$ a MAD family. We say that $P$ destroys $\mathcal A$ if $\P\Vdash\mathcal A\ \mbox{is not maximal}$. $P$ destroys an ideal $\mathcal I$ iff $P$ adds a set almost disjoint with every element of $\mathcal I$.

77.2

Kunen showed that there is a Cohen indestructible family under CH (to show that $\mathfrak a=\omega_1$ in the Cohen model). Later Steprans showed that after adding an arbitrary number of Cohen reals to a model of CH there will still be a Cohen indestructible family. This led to the following question:

77.3

Does there always exist a Cohen indestructible MAD family?

77.4

There is a MAD family $\mathcal A$ that is destroyed by every forcing that adds reals (the Sierpińsky construction of a MAD family --- e.g. an arbitrary family extending the branches of the binary tree.)

77.5

Let $\mathcal I,J$ be ideals on $\omega$. We say that $f:\omega\to\omega$ is a Katětov function from $(\omega,\mathcal I)$ to $(\omega,\mathcal J$) if $f^{-1}[A]\in\mathcal I$ for every $A\in\mathcal J$. We write $\mathcal I\leq_K\mathcal J$ if there is a Katětov function from $\mathcal J$ to $\mathcal I$.

77.6

If $\mathcal I\leq_K\mathcal J$ and $P$ destroys $\mathcal J$ then it also destroys $\mathcal I$.

77.7

Define $\pi:\mathcal P(\omega^{<\omega})\to\mathcal P({}^\omega\omega)$ as follows: $\pi(a) = \{f:(\exists^\infty n)(f\upharpoonright a\in a)\}.$ Given a $\sigma$-ideal $K$ on ${}^\omega\omega$ let $tr(K)=\{a\subseteq\omega^{<\omega}:\pi(a)\in K\}$

77.8Hrušák, Zapletal

Let $K$ be a $\sigma$-ideal on $C=2^\omega$ (or $C={}^\omega\omega$) and $\mathcal J$ an ideal on $\omega$. Then the following are equivalent:

1. There is a $B\in Borel(C)/K$ such that $K\Vdash \mathcal J\mbox{is not tall}$
2. There is $X\in tr(K)$ such that $\mathcal J\leq_K tr(K)\upharpoonright X$,


It follows that Cohen indestructible implies Miller indestructible implies Sacks intedtructible. Random indestructible implies Sacks indestructible. The reason is the inclusion between the respective ideals: countable, $\sigma$-compact, meager and null.

77.9

Adding a real destroys $tr(countable)$ so that if Sacks destroys an ideal then every forcing adding a real destroys it.

77.10Hrušák

Is there, in ZFC, a Sacks indestructible MAD family?

Equivalently, can there be a model such that adding any real destroys all MAD families. Note that the same question for ultrafilters is also open.

77.11

It is known that $\omega_2$-length iteration of definable proper forcings has a Cohen indestructible MAD family. A regular-length finite-support iteration of ccc forcings has a Sacks indestructible family.

77.12

Under CH one can construct MAD families survirving different combinations of forcings. Also one can construct a Sacks indestructible MAD family which is destroyed by the two step iteration of Sacks forcing.

77.13Miller

Is there an ultrafilter which survives one Sacks but not two?

The usual proof of $\mathfrak b\leq\mathfrak a$ in fact gives:

77.14
1. If $\mathcal A$ is a MAD family then $\mathcal I(\mathcal A)\leq_{K}Fin\times Fin$.
2. A forcing $P$ destroys $Fin\times Fin$ iff $P$ adds a dominating

77.15

A MAD family $\mathcal A$ is $Fin\times Fin$-like if for every definable $\mathcal I$ on $\omega$ we have that $\mathcal I(\mathcal A)\leq_K\mathcal I$ implies $Fin\times Fin\leq_K\mathcal I$.

77.16Hrušák, Zapletal

If a $\sigma$-ideal $K$ is not ccc then $tr(K)$ is not Borel.

77.17

If $P$ is a definable forcing (i.e. of the form $Borel/K$) that does not add dominating reals and $\mathcal A$ is $Fin\times Fin$-like then $P$ does not destroy $\mathcal A$.

77.18

Forcing with countable AD families is $\sigma$-closed and adds a generic MAD family which is $Fin\times Fin$-like ($\sigma$-closed implies that no new definable ideals are added).

It is interesting that the above generic MAD family can be destroyed without adding dominating or splitting reals.

77.19

Given an ideal $\mathcal I$ on $\omega$. We define $\mathcal (I^{<\omega})^+=\{X\subseteq[\omega]^{<\omega}:(\forall A\in\mathcal I) (\exists a\in X)(a\cap A=\emptyset)\}.$ An ideal $\mathcal I$ is Shelah-Steprans if for each $X\in\mathcal (I^{<\omega})^+$ there is an infinite $W\subseteq X$ such that $\bigcup X\in\mathcal I$. Equivalently if for every $X\subseteq[\omega]^{<\omega}$ there is $A\in\mathcal I$ hitting every $x\in X$ or there is $A\in\mathcal I$ containing infinitely many $x\in X$.

77.20

If $\mathcal I$ is non-meager then it is Shelah-Steprans. The ideal $Fin\times Fin$ is also Shelah-Steprans.

77.21Brendle, Guzmán, Hrušák

Let $\mathcal I$ be a Borel ideal then $\mathcal I$ is Shelah-Steprans iff $Fin\times Fin\leq_K\mathcal I$.

Under suitable LC assumptions this is true for all definable ideals.

77.22

A MAD family is Shelah-Steprans if $\mathcal I(\mathcal A)$ is.

77.23

If $\mathcal I$ is Shelah-Steprans and $\mathcal I\leq_K\mathcal J$ then $\mathcal J$ is Shelah-Steprans.

77.24LC

If $\mathcal A$ is Shelah-Steprans then it is $Fin\times Fin$-like.

77.25

Mathias-Prikry forcing with $\mathcal I(\mathcal A_{gen})$ does not add unbounded or splitting reals.

77.26Brendle

Is it consistent that $\mathfrak b=\mathfrak s=\omega_1<\mathfrak a$.

77.27Chodounský, Zdomskyy

An ideal $\mathcal I$ is Hurewicz if it is a Hurewicz subset of $2^\omega$ or, equivalently, if for every sequence $\{X_n:n<\omega\}\subseteq(\mathcal I^{<\omega})^+$ there are finite $F_n\in[X_n]^{<\omega}$ such that for every infinite $A\in[\omega]^\omega$ the union $\bigcup_{n\in A}F_n\in(\mathcal I^{<\omega})^+$.

77.28

A MAD family $\mathcal A$ is Zdomskyy if $\mathcal I(\mathcal A)$ cannot be extended to a Hurewicz ideal.

77.29Zdomskyy

Are there Zdomskyy MAD families, say under CH?

A strategy to answer Brendle's question, assuming there are no Zdomskyy families under CH: Start with a model of GCH and use a bookeeping argument to kill all Hurewicz MAD families in $\omega_2$-many steps. In the resulting model $\mathfrak b=\mathfrak s=\mathfrak \omega_1$.

77.30

It is consistent that there are no Shelah-Steprans MAD families.

77.31

If an ideal $\mathcal I$ is $F_\sigma$ or an analytic P-ideal then it is not Shelah-Steprans.

We now aim to prove 77.21. So let $\mathcal I$ be a Borel ideal. First define the following game $G(\mathcal I)$: The first player chooses sets from $\mathcal I^*$, missing $n$ in the $n$-th move, and the second player chooses finite subsets of these. He wins if the union of his moves is positive.

77.32

An ideal $\mathcal I$ is a weak P-ideal if every collection of sets in $\mathcal I^*$ has a positive pseudointersection.

77.33

An ideal is not a weak P-ideal iff it is Katětov above $Fin\times Fin$.

77.34Laflamme

The first player has a winning strategy iff $\mathcal I$ is not a weak P-ideal. The second player has a winning strategy iff there is a sequence of $\mathcal I^{<\omega}$-positive sets such that for each $A\in\mathcal I^*$ there is a member of this sequence such that each element of this member hits $A$.

77.35

If $\mathcal I$ is not Shelah-Steprans then the second player has a winning strategy.

77.36

If $\mathcal I$ is not Shelah-Steprans then the second player has a winning tactic (i.e. a strategy not depending on the previous moves of II).

77.37

This is easy for $F_\sigma$ or analytic P-ideals (just play a set of submeasure bigger than the minimum of player I's move).

77.38

II has a winning strategy iff it is not Shelah-Steprans.

77.39

II has a winning strategy iff he has a winning tactic.

# 7822.2.2017

## 78.1Bohuslav Balcar: 1943-2017

Bohuslav Balcar, a prominent Czech mathematician and a kind man, passed away last Friday. He will be remembered by his wife Maria, his daughters Veronika and Magdaléna, their families, and his many friends. The funeral will take place on Friday February 24th in Prague.

## 78.2E. Thuemmel: Ramsey ideals

The following was preceeded by approximately 40 minutes of interesting stuff which, unfortunately, I missed. It is a continuation of the talk sec33.1 given by E.~Thuemmel two years ago.

78.1

Given an ideal $\mathcal I$ we define the following game. Player I. plays partitions of $\omega$ into two parts while player II chooses one element of the partition and a finite subset of it. PLayer II wins if the union of the finite pieces is positive with respect to $\mathcal I$.

78.2

II has a centered winning strategy iff $\mathcal I$ has an $F_\sigma$-extension.

Assume first that $\mathcal I\subseteq\mathcal J$ with $\mathcal J$ an $F_\sigma$-ideal. Let $\mathcal J= Fin(\mu)$ and choose an ultrafilter $\mathcal J^*\subseteq\mathcal U$. Whenever player I plays a partition, player II chooses the set from the ultrafilter and chooses a finite subset such that, in the n-th move, it has submeasure at least $n$.

Assume, on the other hand, that $\sigma$ is a centered winning strategy for II and let $\mathcal U$ be the ultrafilter given by the strategy. The proof is based on an idea of J. Zapletal. First we will need the Kechris-Louveau-Woodin theorem:

78.3Kechris,Louveau,Woodin

If $\mathcal I$ is an analytic ideal disjoint from an ultrafilter $\mathcal U$ then either 1. there is an $F_\sigma$ set Z such that $\mathcal J\subseteq Z\subseteq 2^\omega\setminus\mathcal U$ or 2. there is a sequence $\{c_n:n<\omega\}\subseteq \mathcal U\}$ of sets from the ultrafilter such that the closure of this sequence is homeomorphic to $2^\omega$ and the (topological) border of this sequence is in the ideal $\mathcal I$

We apply this theorem to the ideal $\mathcal I$ and the ultrafilter $\mathcal U$. The first alternative immediately gives an $F_\sigma$-extension of $\mathcal I$. So assume that the second alternative applies. So we have the set $\{c_n:n<\omega\}$. We inductively construct a sequence $\{ n(k):k<\omega\}\subseteq\omega$ and a sequence $\{a_k:k<\omega\}$ of finite subsets of $\omega$ such that $a_k\sqsubseteq c_k,a_{k+1}$ and $c_{n(k)}\neq c_k$ by playing the following game: Player I plays $c_{n(k)}$ and $\omega\setminus c_{n(k)}$ (i.e. chooses $n(k)$) in the $k$-th step such that $c_k\not\in[a_k]$. Then the union of the $a_k$'s is in the closure of the $c_n$'s but not equal to any $c_n$ so it is in the ideal $\mathcal I$---a contradiction.

## 78.3O. Guzmán

78.4

Given an ideal $\mathcal I$ we say that a tree $T\subseteq\omega^{<\omega}$ is $\mathcal I^+$-branching if for every node $s\in T$ the set $suc_T(s)=\{n:s^{\smallfrown} n\in T\}$ is $\mathcal I$-positive.

78.5

A co-ideal $\mathcal I^+$ is Happy if for every $\mathcal I^+$-branching tree $T$ such that $\{suc_T(s):s\in T\}\cup\mathcal I^*$ is centered there is a branch $f\in [T]$ such that $f[\omega]\in\mathcal I^+$.

78.6Mathias

If $\mathcal A$ is a MAD family then $\mathcal I(\mathcal A)^+$ is Happy.

Let $\mathcal T$ be $\mathcal I(\mathcal A)^+$-branching satisfying the assumption of the definition of a Happy family. Enumerate $\{suc_T(s):s\in T\}=\{B_n:n<\omega\}$. Choose a pseudointersection $C_0$ of these sets such that there is $A_0\in\mathcal A$ containing $C_0$. Subtract $A_0$ for each $B_n$. The resulting family will still be centered so we can again choose $C_1,A_1$ as before. In the end we get distinct $\{A_n:n<\omega\}\subseteq\mathcal A$ with $|A_n\cap B_n|=\omega$. Find $f\in[T]$ such that $|f[\omega]\cap A_n|=\omega$ for every $n<\omega$. It follows that $f[\omega]$ is $\mathcal I$-positive.

78.7

An ideal $\mathcal I$ is +-Ramsey if for every $\mathcal I^+$ branching tree $T$ there is a branch $f\in[T]$ such that $f[\omega]\in\mathcal I^+$.

78.8

An ultrafilter $\mathcal U$ is Happy iff it is Selective iff it is +-Ramsey. Moreover, if $\mathcal I^+$ is Happy then $P(\omega)/\mathcal I$ adds a Ramsey ultrafilter disjoint from $\mathcal I$ so, if $\mathcal I$ is tall, it cannot be definable.

78.9

There is a MAD family $\mathcal A$ such that $\mathcal I(\mathcal A)$ is not +-Ramsey.

78.10Hrušák

Is there a +-Ramsey MAD family?

78.11

Yes, assuming either $cov(\mathcal M)=\mathfrak c$, $\mathfrak b=\mathfrak c$, $\Diamond(\mathfrak b)$.

In fact, the following is true!

78.12

There is a +-Ramsey MAD family (in ZFC).

The proof breaks down into two cases, depending on whether $\mathfrak a<\mathfrak s$. The case when $\mathfrak s\leq\mathfrak a$ is based on Shelah's construction of a completely separable MAD family. Note that in this case the family will have size $\mathfrak c$. In the other case the family will be smaller. It is not clear, whether one can always construct one of size $\mathfrak c$. We will only show the easier case $\mathfrak a<\mathfrak s$. First we need some preparations.

78.13Hrušák

The cardinal $cov(\mathcal M)$ is the minimal cofinality of an ideal which is not +-Ramsey.

78.14

This is similar to the characterization of $\mathfrak p$ as the minimal cofinality of a filter which is not P${}^+$.

Let $\kappa$ be the minimal cofinality of a non +-Ramsey ideal. Assume first that $cof(\mathcal I). We show that \(\mathcal I$ is not +-Ramsey. Let $T$ be an $\mathcal I^+$-branching tree. We force with $T$ (this is equivalent to forcing with Cohen). Let $g$ be the generic branch. It is not hard to see that $g[\omega]$ is positive and this is guaranteed by $\omega\cdot cof(\mathcal I)-many dense sets. On the other hand we show how to construct a non +-Ramsey ideal of cofinality \(cov(\mathcal M)$. Let $\{T_\alpha:\alpha be a family of trees such that \(\omega^\omega=\bigcup [T_\alpha]$ and $[T_\alpha]$ is nowhere dense. Let $\mathcal I$ be the ideal on $\omega^{<\omega}$ generated by these trees. We claim that $\mathcal I$ is not +-Ramsey: define the following tree $T=\{(s_0,\ldots,s_n):s_0\sqsubset s_1\sqsubset\cdots\sqsubset s_n\}$ on ${}^{<\omega}\omega$. Then $T$ is an $\mathcal I^+$ branching tree.

78.15

Let $\mathcal A$ be a MAD family. If it is Miller indestructible then it is +-Ramsey.

Let $T$ be an $\mathcal I(\mathcal A)^+$-branching tree. We first construct an infinitely branching subtree $S\subseteq T$ such that the successors of each node $s$ of $S$ are contained in some element of $\mathcal A$ and these elements are distinct for distinct nodes of $s$. We now consider $S$ as a Miller condition. Let $\dot{r}$ be a name for the generic raal. Then $S$ forces that $dot{r}[\omega]$ is $\mathcal I$-positive. Let $M$ be a countable elementary submodel of some large enough $H(\kappa)$ containing all the relevant things (e.g. $S$, $\mathcal A$, ...). Let $r\in[S]$ be a Miller real over $M$. Then $M[r]\models \mathcal A\ \mbox{is MAD}\ \&\ \dot{r}[\omega]\ \mbox{is positive}$. In particular $M[r]\models dot{r}[\omega]\in \mathcal I(\mathcal A)^{++}$ so, in $M[r]$, there are infinitely many $\{A_n:n<\omega\}\subseteq A$ such that $|A_n\cap r[\omega]|=\omega$. But this is absolute, so $r[\omega]$ is $\mathcal I(\mathcal A)$-positive.

78.16Folklore

If $\mathcal A$ has cardinality $<\mathfrak d$ then it is Miller indestructible.

Assume $|\mathcal A|<\mathfrak d$ and let $p$ be a Miller tree and $\dot{X}$ a name for an infinite subset of $\omega$. By the continuous reading of names there is an extension $q\leq p$ and a continuous function $F:[q]\to\omega$ (in V) such that $q\Vdash f[\dot{r}]=\dot{X}$.

Now for every $A\in\mathcal A$ let $K(A)$ be the set of all branches $f$ of $q$ such that $|A\cap F[f]|=\omega$. Since $\mathcal A$ is MAD we can write $[q]=\bigcup_{A\in\mathcal A}K(A).$ It is also easy to see that $K(A)$ are Borel.

78.17Kechris

A Borel subset of $\omega^\omega$ is either contained in a $\sigma$-compact set of contains the branches of a Miller tree.

However, since $|\mathcal A|<\mathfrak d$ there must be $A\in\mathcal A$ such that $K(A)$ is not $\sigma$-compact. Then, applying the previous proposition, $K(A)$ contains the branches of a Miller tree $q_1\leq q$. However, $q_1\Vdash |\dot{X}\cap A|=\omega$ which finishes the proof.

78.18

If $\mathfrak a<\mathfrak d$ then there is a +-Ramsey MAD family.

# 7922. 3. 2017

## 79.1J. Verner: Schritesser's & Shelah, Horowitz proof of the existence of definable MED families

The talk was a presentation of the article D. Schritesser: On Horowitz and Shelah's Borel maximal eventually different family

# 8029. 3. 2017

80.1

A cardinal $\kappa$ is $\lambda$-supercompact if there is an elementary embedding, i.e. a formula defining it, $j:V\to M$ with critical point $\kappa$ such that ${}^\lambda M\subseteq M$. A cardinal is supercompact if it is $\lambda$-supercompact for each $\lambda$.

80.2

If GCH holds below a measurable it holds at the measurable. If it holds below a supercompact then it holds everywhere.

80.3

A tree $T$ is special iff it can be covered by countably many antichains (or, equivalently, if there is a function $f:T\to\omega$ which is an injection on chains).

80.4

Rado Conjecture on $\omega_1$, $RC(\omega_1)$, claims that a tree $T$ of height $\omega_1$ is special iff each subtree $S\subseteq T$ of size $\omega_1$ is special.

80.5

In the above the width of the tree is not limited!

Note that in ZFC there is a special Aronszajn tree on $\omega_1$ but it is consistent (collapsing a super compact cardinal using Mitchell forcing) that there is no Aronszajn tree on $\omega_2$ (this is abbreviated as $TP(\omega_2)$, the Tree property at $\omega_2$) and CH implies that there is a special Aronszajn tree. Also $\kappa^{<\kappa}=\kappa$ implies that there is a $\kappa^+$-Aronszajn tree. A similar argument shows that $TP(\kappa^++)$ implies $2^\kappa>\kappa^+$. However, it is still unknown whether $TP(\aleph_{\omega+1})$ and $2^{\aleph_\omega}>\aleph_{\omega+1}$ is consistent.

The following theorem summarizes some facts about RC, proved by several authors including Todorcević.

80.6

$RC(\omega_1)$ implies $TP(\omega_2)+2^\omega\leq\omega_2$, stationary reflection at $\omega_2$ and SCH.

80.7

The consistency of a strongly compact cardinal implies the consistency of Rado Conjecture and Rado Conjecture implies the existence of Woodin cardinals in some inner models.

For simplicity we will only proof the following:

80.8

If there is a supercompact cardinal then GCH and $RC(\omega_1)$ is consistent as is $RC(\omega_1)$ and $\neg$CH.

Assume $\kappa$ is supercompact and $T$ is a tree of height $\omega_1$ in the future forcing extension. For simplicity assume $dom(T)\subseteq\lambda$. Since $\kappa$ is $\lambda$-supercompact, fix a witnessing embedding $j$. To get the first part (RC+GCH) it is sufficient to Lévy collapse $\kappa$ to $\omega_1$. To get the second, we will use Mitchell forcing.

The Lévy collapse $Coll(\omega_1,<\kappa)$ consists of countable partial functions from $\omega_1$ to $\kappa$. It can be written as $Coll(\omega_1,<\alpha)\times P^\alpha$, where $P^\alpha$ consists of partial functions $p:\kappa\times\omega_1\to\kappa$ such that $p(\alpha,\beta)<\alpha$ and $\xi,\beta)\in dom(p)\rightarrow \xi\geq\alpha$. The Lévy collapse is $\sigma$-closed and $\kappa$-Knaster.

Assume that $G$ is $P=Coll(\omega_1,\kappa)$ generic over $V$. We will lift our embedding $j:V\to M$ to an elementary embedding $j^*:V[G]\to M[H]$ for some $j(P)$-generic $H$ over $M$.

80.9Silver's lemma

Assume $j:V\to M$ is an elementary embedding, $G$ is a $P$-generic over $V$ and $H$ is a $j(P)$-generic filter over $M$. Then the following are equivalent:

1. There is an elementary embedding $j^*:V[G]\to M[H]$ with $j\upharpoonright V=j$
2. $j[G]\subseteq H$

By the above notes and elementarity $j(P)=Coll(\omega_1, and \(j(P)\upharpoonright\kappa=Coll(\omega_1,<\lambda)$. In particular, $j(P)=P\times j(P)^\lambda$. Let $H^*$ be $j(P)^\lambda$-generic over $V[G]$ and write $H=G\times H^*$. Since $j$ is the identity on $P$, we have $j[G]=G\subseteq G\times H^*$. Now apply Silver's lemma to get an elementary $j^*:V[G]\to M[H]$ (in $V[H]$). Since $P$ is Knaster, the name for $T$ is in $M$ so $T\in M[G]\subseteq M[H]$. It follows that $T$ is a subtree of $j(T)$. Assume $T$ is not special. We will show that $T$ is not special in $M[H]$.

80.10

If $T$ is a tree of height $\omega_1$ which is not special then it remains non-special in any $\sigma$-closed forcing extension. {} Proof: Let $p$ force that $f:T\to\omega_1$ is the specializing function. Inductively construct $g:T\to\omega_1$ and $p:T\to P$ such that $s implies that \(p(t)\leq p(s)$ and $p(s)\Vdash f(s)=g(s)$. At limit nodes we use the fact that $P$ is $\sigma$-closed.

By the above lemma, since $T$ is not special in $V[G]$ and so, afortiori, in $M[G]$ it is still not special in $M[G][H^*]=M[H]$. So, in $M[H]$, $j(T)$ has a subtree of size $and, by elementarity, \(T$ must also have a nonspecial subtree of size $\omega_1$.

To get the second part we use Mitchell forcing $M(\omega,\kappa)$ which consists of pairs $(p,q)$ such that $p\in Add(\omega,\kappa)$ and $q$ is a function with domain a countable subset of $\kappa$ and $q(\alpha)$ is a $Add(\omega,\alpha)$-name for a condition in $Add(\omega_1,1)$. The ordering is as follows $(p,q)\leq(p^\prime,q^\prime)$ if $p\leq p^\prime$, $dom(p)\supseteq dom(p^\prime)$ and for each $\alpha\in p^\prime$ we have that $p\upharpoonright\alpha\Vdash q(\alpha)\leq q^\prime(\alpha)$.

Mitchell forcing is $\lambda$-Knaster (a $\Delta$-system argument) so $\lambda$ is preserved and $2^\omega=\lambda$. Since if $2^\omega\geq \alpha$ then adding a single Cohen subset of $\omega_1$ forces $CH$ so forces that $|\alpha|=\omega_1$. Since $M$ preserves $\omega_1$, we have that in the extension $\lambda=\omega_2$. To show that $M$ preserves $\omega_1$ we show that there is a pseudoprojection $p:Add(\omega,\lambda)\times Q\to M$ where $Q$ is $\sigma$-closed. Since $Add(\omega,\lambda)$ is Knaster we know that the product cannot collapse $\omega_1$ so neither does $M$.

The rest of the proof is similar to the above using the fact that $j(M)=j(M)/M$, so we choose a $j(M)/M$ generic $H^*$ over $V[G]$.