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
Let
A filter
Every rapid filter is a measure centering filter.
If a sequence has a positive intersection number then there are infinitely many centered subsequences.
What about the weak intersection number?
320. 11. 2013
3.1E. Thuemmel: Pure decision property
This is equivalent to saying that for each
A coideal
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.
An ideal
An ideal is selective if it is Ramsey and
We will sometimes say that an ideal or a coideal is
If we add a Sacks real then groundmodel infinite reals form a semiselective coideal which is not selective.
selective
If
A semiselective coideal cannot be definable.
Can we have a definable Ramsey coideal?
The ideal conv is the ideal generated by Cauchy sequences of rationals
in the interval
If an ideal is not
Can we ommit properness in the above lemma?
No.
An ideal has
The following is a combinatorial reformulation of 3.14.
An ideal
It will be first instructive to notice the following reformulation of
An ideal
To see this, distributivity gives us a
"
"
We will now show the proof of Lemma 3.14.
Assume
Next we turn to give an example showing 3.16.
Farah has an ideal
In the above we could have everywhere replaced properness with
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.
A set
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.
A strong measure zero cannot contain a perfect set (as it is preserved under uniform continuous images).
Strong measure zero sets form a
A set
A set
If
Every meager shiftable has strong measure zero.
In
This motivates the following question.
The following is a partial answer.
If
The
(idea) Say that a group
Under CH there is an uncountable meagershiftable set in
The first
An ideal
Notice that if an ideal is
We say that a set is strongly meager if it is in
The Dual Borel Conjecture says that there are no uncountable strongly meager sets.
Bartoszynski showed that, in ZFC,
The meagershiftable ideal on a locally compact Polish group is
It might be that
If
The diameter of the Lebesgue neighbourhood is the Lebesgue number of the open cover.
4.2Honza Starý / BB
A submeasure
Let
Let
(Basically, an application of the Sikorski theorem) Consider
54. 12. 2013
5.1Honza Grebík: Asymptotic Density in Generic Extensions
Let
Given a finite sequence
For
\(d(P_2^1)=3/\pi^2\) \(\pi(x)\sim x/\log x\) \(\zeta(x)=0\rightarrow Re(x)\neq 1\)
5.2Balcar: Torturing Honza Starý
Recall ...
Let
And now for something entirely different...
A
Let
In case
First, since
The sequence
Note that the sequence is decreasing. It is sufficient to show that for each
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.
Does every ccc forcing add a random real or a Cohen real?
If the measure algebra cannot be embedded into the Talagrand algebra, it might be a candidate for a ZFC answer to the above problem.
Let
We will show that it is continuous (and hence
If
and the story continues...
5.3Balcar: Torturing Honza Grebík
Let
Let
611. 12. 2013
6.1E. Thuemmel: Exponent
6.1.1Pure decision
Recall the following question which motivates the subsequent.
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:
Mathias forcing
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, 73105.
Let us identify each condition
This map has the following property
For each
Let
We shall generalize this observation into a definition:
A monotone map
If a monotone map
If
We will now consider the extension of the
\(f[U_{s,A}] = U_{sh(s),A}\) \(f[U_{\emptyset, A}]\subseteq U_{\emptyset, A}\)
The set of fixed points of
The map
Aiming for a contradiction, assume that no
To summarize, starting from Mathias forcing, we have arrived to an extremally disconnected
compact space
All of this works in the more general setting with Mathias forcing replaced by
Given a regular cardinal
 a
\(\kappa\) complete homomorphism  uniform
(i..e. if
\(\{\alpha<\kappa:b_\alpha\neq\mathbf 0\}<\kappa\) then\(r(b)=\mathbf 0\) ).  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
Consider the characteristic function
Exponent has fusion in the following sense. Given
Exponent has pure decision in the following sense. Given a sentence
Given a general exponent
If
From now we will consider only
Given
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.
Matet forcing consists of pairs
The following can be found in the dissertation of Luz M. García Ávila.
Let
In this case, the shift function
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
Is there a separable Fréchet topological group which is not metrizable.
Consistently no.
(see Hrušák, Ariet: Malykhin's Problem, preprint)
Is it consistent that there is a countable Fréchet group of weight
A
Is it consistent that there is a
If
 A yes answer to the conjecture implies that there are no countable e.d. topological groups.
 A yes answer is equivalent to saying that
\(RO(G)\) does not add Cohen reals for any e.d. group.  If
\(h\) is a group homomorphism, then the answer is yes.
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
Is
7.2Countable Dense Homogeneity
Two countable sets
A typical example of a Polish space which has two types is the halfopen interval. It turns out
that, basically, this is the only possibility for finitely many types, i.e. all examples are of the
form
Is there a Polish space with
This is related to the Topological Vaughts conjecture. The following theorem can be found in the paper R. HernandezGutierrez, M. Hrušák, J. van Mill: Countable dense homogeneity and lambda sets, preprint.
If there is a
Is it consistent that there is a Baire CDH set of reals of size
7.3AD Families
The following is a recent result published in the paper Hrušák, Guzmán: nLuzin gaps, Top. Appl 160 (2013).
A system
\(A_\alpha\cap B_\alpha=\emptyset\) \(A_\alpha\cap B_\beta =^*\emptyset\) \((\forall \alpha\neq\beta)(A_\alpha\cap B_\beta\cup A_\beta\cap B_\alpha\neq\emptyset)\)
Assume PFA. If
Is MA enough in the above theorem?
7.4Cardinal invariants connected to AD families
It is consistent that
If
If
If
Given a tall ideal
The above theorem 7.23 follows from the following by choosing
If
Let
\(\mathcal A_\alpha\) is an AD on\(J_\alpha\) \(\bigcup_{\beta\leq\alpha}\mathcal A_\beta\) is AD\(\bigcup_{\beta\leq\alpha}\mathcal A_\beta\upharpoonright J_\alpha\) is MAD on\(J_\alpha\) .\(\bigcup_{\beta\leq\alpha}\mathcal A_\beta\) has size\(\omega_1\) .
7.5Generic existence of MAD families
A MAD family having property
\(\mathfrak a=\mathfrak c\) is equivalent to the fact that completely separable MAD families exist generically.\(\mathfrak b=\mathfrak c\) iff Cohen indestructible families exist generically. In known models Sacks indestructible families exist (either because
\(\mathfrak a<\mathfrak c\) or because they exist generically due to a cardinal invariant).
Do Sacks indestructible MAD families exist (in ZFC).
7.6Fsigmaideals
If
Does every tall Borel ideal contain a tall
An ideal
Is
Is there a tall Borel (Analytic) ideal
The above fails for
7.7Strong Measure Zero
If
The same proof works for any separable polish group with leftinvariant metric. He then asked about the reverse implication. A yes answer (on the reals) is due to GalvinMycielskiSolovay.
For which groups does the (Prikry)GalvinMycielskiSolovay theorem hold?
From now on
A set
A group is CLI if it carries a complete leftinvariant metric.
Each Polish group carries a complete metric as well as a leftinvariant metric. However
there are groups, e.g.
If
If
For which spaces/groups does meager imply uniformly meager. It should be exactly the locally compact but there is no proof yet.
A group has the strong GMS property if for every nowhere dense
strong GMS
The strong and weak GMS properties are absolute properties (i.e. with respect
to
A group
Compact
Suppose G is not Bergman and fix
A group
Bergman
GMS implies nonelastic.
So we have the following diagram:
88. 1. 2014
Let
There is a dense
\((\forall M\in[\omega]^\omega)(\bigcup_{k\in M} D_k\ \mbox{is dense})\)  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.)
Does each tall Borel ideal contain an
Let
We shall first show a proof by Hrušák. We will need some notation.
Given a set
There are two cases. Either there is an
Otherwise force with
Let us now show a proof via exponents.
An ideal
For the definition of selectivity and semiselectivity see the picture or take the following lemma as a definition.
Recall that
In the following we shall consider coideals instead of ideals.
Given a coideal
Given a coideal
\(t\sqsubseteq s\) ,\(A\subseteq B\) and\(A_i\subseteq B_i\) for all\(i\in A\) \(s\setminus t\) is a (finite) diagonal of\(\overline{B}\)
or, if we let
If
The embedding takes
Note that classical Ramsey theorems hold also in this setting.
For all
\((\forall B\in[A]^\omega)(\exists s\in\mathcal F)(s\sqsubseteq B)\) (i.e.\(\mathcal F\) is a barrier) or\([A]^{<\omega}\cap\mathcal F = \emptyset\)
Mathias generalized this to being able to select
If
\((\forall B\in diag_\omega(\overline{A}))(\exists s\in\mathcal F)(s\sqsubseteq B)\) or\(diag_{<\omega}(\overline{A})\cap\mathcal F = \emptyset\)
Consider next the Ellentuck theorem.
The Elentuck topology on
Note that a set
The Baire property is equivalent to being completely Ramsey
(in
Mathias and Farah generalized this taking some selective and semiselective, respectively,
A Marczewski pair is a pair
If
Assuming CH the
Using Marczewski's theorem it follows that, assuming CH, all analytic sets are
Let us now give an alternative proof of theorem 8.4:
Since
\([\emptyset,\overline{A}]\subseteq \mathcal I\) or\([\emptyset,\overline{A}]\cap\mathcal I=\emptyset\)
Since
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
Given a an uncountable regular cardinal
 We want the forcing to be
\(\kappa\) complete so that it does not add bounded subsets of\(\kappa\) .  Since we want the ideal to be
\(\kappa\) complete, it does not make sense to consider singular cardinals.  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.
Assume GCH and
The diagonal union is the dual notion to the diagonal intersection and is defined as follows:
Given a sequence
If an ideal is not closed under diagonal unions we first show that it is not a "generalized Pideal".
Then we just copy the standard proof that Grigorieff forcing with a nonPpoint collapses the
continuum. This direction does not need the GCH. For the other direction, since
Given two conditions
A sequence
\(p_{\alpha+1} \leq_{\alpha} p_\alpha\) for all\(\alpha<\kappa\) \(p_\beta = \bigcup_{\alpha<\beta} p_\alpha\) for limit\(\beta\)
It is routine to check that if the ideal
If
The assumption that
915. 1. 2014
9.1A. A. Gryzlov: On Dense subsets of Tichonoff products (preprint)
Let
 For each infinite
\(C\subseteq\omega\) the set\(\bigcup_{k\in C}Q_k\) is dense in D  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\(D\)  If
\(F\subseteq Q\) is bounded in\(Q_k\) 's (as in 2), then\(F\) is closed discrete in\(Q\) .
 condition 3. has no chance to hold in
\(\omega^\omega\) .  (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
The product of
There is an independent family
Let
To get a dense set of functions
For
Let
\((\forall k<\omega)(F\cap H_k\leq 1)\) \((\forall k\leq k_0)(F\cap H_k = \emptyset)\)
Then for each
Let
Let
We have
Case 1: If
Case 2: If
There is a set
Using the lemmas we obtain an independent matrix
9.2D. Chodounský: PID and Tukey Maximality
The following is based on the preprint D. Chodounský, P. BorodulinNadzeja: Hausdorff gaps and towers in
Recall that a poset
We have the following Tukeytypes of posets of size at most
Also recall that
PID is the following statement. Whenever
\( (\exists K\in[\omega_1]^{\omega_1})([K]^{\leq\omega}\subseteq\mathcal I)\) or\(\omega_1\) can be partitioned into countably many sets\(\{A_n:n<\omega\}\) such that for each\(I\in\mathcal I\) and\(n<\omega\) \(I\cap A_n=\emptyset\) .
The ideal
Under PFA these five types are the only possible types.
Assume PID. Then the existence of precisely 5 Tukey types
is equivalent to
David and Piotr BorodulinNadzeja were interested in Tukeytypes of (
Assume
They were, in fact, interested mainly in ideals generated by towers. The following definition is relevant in this context
An (
An ideal generated by a tower
Let
If
Consider any countable sequence
Now we are ready to apply PID to
There is
Otherwise
Consider now
The proof of Todorčević's and Raghavan's result is very similar however, they need to
guarantee that not only analytic Pideals are generated by
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)
For a Banach space
The span of the first (or second) coordinates need not be the full space (dual space).
Considering
Separable infinite dimensional Banach spaces have infinite biorthogonal systems. (in fact even satisfying the stronger condition mentioned in the above note).
it is natural to ask
Suppose
Assuming CH, there is a scattered, compact, Hausdorff, nonmetrizable space
Both of the above examples yield a nonseparable Banach space with no uncountable biorthogonal
system, viz
Assuming PID and
See 9.9 for the definition of the Pideal dichotomy (PID).
Assume PID. Is
An Asplund Banach space is a space whose separable spaces have separable duals.
If
For a Banach space
Assume PID and
HahnBanach extension theorem (14.2) means we can extend biorthogonal from subspaces. In particular we may assume
There is an uncountable subset
Apply PID to the ideal
There is an uncountable
Let
The following claim now finishes the proof.
There is a sequence
10.2J. LopezAbad: Families of finite sets
First a remark concerning Christina's talk.
An Auerbach base for a finite dimensional normed space
Every finite dimensional normed space has an Auerbach basis.
If
Now we will consider families of finite subsets of
Given
Let us consider when
The family
The family
Given
For any coloring
A family
For a family
Let
For every thin (Sperner) family
If
If
If
In particular, every precompact family has a
Suppose
For each
How do coloring of barriers behave?
For each barrier
Let
TFAE
1. There is an infinite
If
If
The set
A family
For a cardinal
A cardinal
The following is Fremlin's DUproblem.
Does there exist a
Taking the Cantor set
10.3A. Avilés: Tukey classification of orthogonals
The following theorem can be found in Avilés, A., Plebanek, G., Rodriguez, J.: Measurability in
Assume Analytic Determinacy. If
\(\{0\}\) ;\(\omega\) ;\(\omega^\omega\) ;\(K(\mathbb Q)\) (i.e.compact subsets of the rationals ordered by inclusion); or finite subsets of
\(\mathbb R\) .
Let
In the above theorem, analytic determinacy gives the same result for
Applying the theorem of Todorčević to the ideal
This theorem directly reduces 10.45 to Fremlin's characterization:
If
114. 2. 2014
11.1Arturo Antonio MartinezCelis Rodriguez: Porous Sets
The origin of the following definition goes back to the beginning of the 20th century.
Let
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
The following is from M. Repický: Cardinal invariants related to porous sets , Set theory of the reals (Ramat Gan, 1991), 433438, Israel Mathematical Conference Proceedings 6.
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
We will work with a different definition of the ideal
A subset
The additivity, covering, non and cofinality invariants for
For
For
Each
A forcing
The next lemma gives some examples for strong preservation of
If
Let
For every
By the above claim we can fix functions
Let
Let
We will need the following result:
Finite support iteration of forcings strongly preserving
It is consistent that
Start with a model of CH and let
Let
The following is also true
The above might not strictly be true, the definition of
11.2U. Ariet RamosGarcia: Extremal disconnectedness and ultrafilters
(see also his talk at the 2014 Winter School)
Our work with Michael is motivated by an old question of Arhangelskii:
Is there a nondiscrete extremally disconnected topological group?
Yes, under CH.
The following is probably (the notetakers guess) from A. Louveau: Sur un article de S. Sirota, Bull. Sci. Math, 1972.
The group
Given a point
1. If
If
Assume, aiming towards a contradiction, that some
Recall the following proposition, which follows from Stone duality.
IF
The previous proposition motivates the following conjecture.
if
The conjecture holds for countable Boolean ED topological groups and continuous homomorphisms.
Suppose that
Clear.
There is a partition
Recursively apply the previous lemma.
Again, using lemma 1, we can construct the
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,
For two sets
Given a partial order
Now we are ready to define the first two invariants:
The bounding number
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. 1124.
( where it was called simply
In forcing language, the above definition of
We will show another, equivalent, definition of
A family
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
Let us now start with some MAD family
Suppose we would like to continue and define a MAD family
Finally, we can let
Since MAD families naturally correspond to tall ideals, we can also rephrase the above theorem as follows:
(Recall, that an ideal
An AD family
An ideal
)
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
The idealized groupwise density number
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. 5056
A family
The groupwise density number
The connection between
An ideal
It is immediately clear that
It is consistent that
(see J. Brendle: Distinguishing groupwise density numbers, Monatshefte für Mathematik 152(3), 2007, pp. 207215)
It might not be clear, at first, that the cardinal invariant
Suppose now we have a function
A filter
The above ordering, when restricted to ultrafilters, is called the RudinKeisler
(and RudinBlass, respectively)
ordering and denoted
It is instructive to convince onself that the definitions can be repeated for ideals and (using the JalailiNainiTalagrand characterization) that an ideal is meager iff it is KatětovBlass above the ideal of finite sets.
We have shown above that ultrafilters cannot be meager. However, if a filter is nonmeager, it does not necessarily mean that it is an ultrafilter. How far are nonmeager filters from ultrafilters? The following principle says: not very much.
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:
The ultrafilter number
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. 93112:
The FD is equivalent to
12.2Honza Starý, inspired by the Winter School
Let
Recall that
A space
Also recall definition 11.24 of the spectrum of a point and observation 11.25 showing an equivalent definition for ED compac spaces.
For points
Next, recall
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
Given the stone space
A point
What is
For an ED space
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.
see S. Shelah: Groupwise density cannot be much bigger than the unbounded number, Math. Logic Quarterly 54(4), pp. 340344 2008
13.2J. Starý
For the following recall definition 12.27.
A topological space
Let
If
Infinite extremally disconnected compact spaces are not homogeneous.
Take the
For a complete Boolean algebra we define
Let
(see P. Simon: Points in extremally disconnected compact spaces, Rend. Circ. Math. di Palermo (2) 1990)
If
(see B. Balcar and P. Simon:
On minimal
Is there some
If
1412. 3. 2014
14.1J. Grebík: Extending asymptotic density to a measure
Given
Given a vector space
(Recall that
We will apply this theorem to the above situation. We let
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
The above is motivated by the following question:
All measures are elements of
14.2J. Verner: Zindulka/Hrušák's question
Given a real number
\(d(x,y)\leq c d(x,z)\) for each\(x\leq y\leq z\)  open intervals in the ordering are open in the metric
A metric space is
For the following, see Zindulka, O:
Is Every Metric on the Cantor Set
If a metric space
If a metric space
In particular,
Is there, in ZFC, a non
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:
Suppose that
14.4J. Starý
Recall the definition 12.22 of an extremally disconnected space.
A space is
A
A
A Boolean algebra
In this setting, recall Frolík's theorem 12.25.
The completion of a homogeneous Boolean algebra is always homogeneous.
14.5P. Simon, the history of Frolík's theorem
If
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.
Consider
Frolík's theorem 12.25 is a consequence of the following (due to Balcar and Franěk ?) and 14.16.
If
Let
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 3442), was different and used a result of K. Kunen that there are incomparable ultrafilters on
All of the above was motivated by the following considerations. The algebra
Under CH,
14.6J. Starý: Ultrapowers of Boolean algebras
Let
Consider, moreover, the set
Now let
Aiming to define an ultrafilter sum, for
An ultrafilter
If
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
Given a filter
Mathias forcing is ccc (no uncountable antichains), even
We will be interested in the following properties with respect to forcing extensions by Mathias forcing.
Given a model
The above definition of almost
A forcing
One part of the following equivalence is Lemma 3 in the referenced paper.
A forcing
Assume
Assume, on the other hand, that
Mathias forcing cannot be bounding.
The generic real for
In particular if the filter
The motivation for the following topological definitions which characterize
A topological space
Subsets of
Analytic subsets of
Preservation of these properties under products (or powers) is an area of active research.
The following theorem gives the standard definition of Menger property.
A space
To rephrase the Hurewicz property in a similar language we need some additional notions concerning covers.
An open cover
If
An open cover
A typical example of a
We are now ready to rephrase the Hurewicz property.
A space
We shall only show that a
Given
The following is core lemma for dealing with Hurewicz and Menger properties of filters.
Given a filter
We are given
Finally let
When checking the menger property for filters we need only check open covers
consisting of elements of the form
Due to lack of time, we will only recall the following definitions which Hrušák et al. use to deal with Mathias forcing.
Given a filter
A set
A filter
Given a filter
Given that the above definition may be reformulated as follows
A filter
this, together with Hrušák's and Minami's result, almost immediately gives a topological
reformulation of the property that
1626. 3. 2014
Tomorrow at 10:40 in Karlín, seminar room of the MÚ, Yasunao Hattori (Shimane University, Matsue, Japan) will talk about The Separation Dimension and InfiniteDimensional 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, 33573365
Let
A partition of
For the remainder of this talk fix
Use standard diagonalization to show that there is no
The (MAD) spectrum of
It is clear that
The above also follows from the fact that the Boolean algebra
When does
Is it consistent that
Does
Assume
Let
Every regular
First notice that
Let
Fix
Recall the definition of the bounding number for regular cardinals:
For singular cardinals it is convenient to modify the definition slightly:
Assume
Martin's axiom with
If
Can
Brendle proved that
A
There is a set
For each
If
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"
This has a connection to:
If
Shelah proved:
If
172. 4. 2014
17.1M. Fabián: Separable reductions and rich families in variational analysis
Let
Given a function
Fix
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 215221.
A family
 it is cofinal
 closed under closures of countable increasing unions (
\(\sigma\) complete)
This corresponds to the notion of a (
The intersection of countably many rich families is rich.
Let
The above observation can also be vaguely formulated saying: "continuity can be separably reduced".
Let
(see J. Lindenstrauss: On reflexive spaces having the metric approximation property, Israel J. Math, 3(4), 1965, pages 199204)
The following is a generalization due to W. Kubis in Banach spaces with projectional skeletons, J. Math. Anal. Appl. 350(2), 2009, pages 758776.
Given two linear functionals
There is
\(p\circ p^\prime = p\) for each\(p\preceq p^\prime\in\mathcal P\) ,\(p[X]\) is separable for each\(p\in\mathcal P\) \(\mathcal P\) is rich in\((L(X),\preceq)\) .
17.2J. Grebík: Ultrafilter Question
Is there an easy way to show that there is an ultrafilter
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 (
Also recall definition 15.26 and theorem 15.27.
Assume
Assume not. So let
A similar proof gives:
Assume
The above theorems can be reversed, the suitable unbounded (or dominating) family which
will be destroyed by
189.4.2014
Next week, Tue 15. 4. , there will be a workshop in Banach spaces organized by Wieslaw Kubiś and Jerzy Kakol at the Mathematical institute.
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
A dynamical system with discrete time is a space
The intention is that the map transforms the space
A point
If
Two points
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.
Given
Given a sequence
Given
For
The
 associative
 its restriction to
\(\omega\) is the standard addition  given
\(n<\omega\) then\(\mathcal U+n = n+\mathcal U\) .  is continuous in the right coordinate, i.e. the map
\(\mathcal U\mapsto\mathcal U+x\) is continuous for each\(\mathcal U\in\mathbb N\) .
Each
Let
There is no additive idempotent which would be, at the same time, a multiplicative idempotent.
(see his thesis)
Every compact dynamical system contains a minimal dynamical subsystem.
Every two different minimal subsystems are disjoint.
Suppose
Is there an interval partition
 If the answer is no then each union ultrafilter has a nonmeager core and is almost ordered.
 Even a consistent answer (either way) would be interesting.
 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
Let
If we take a different approach we can define, for
We let
Are the above inclusions strict?
Let
Is
Given
When is
An ultrafilter is
The measure
A set
If an ultrafilter
Let
An
There is a function
Assume
Consider
We shall show that
So let
The other implication follows along the lines of Fremlin's proof.
2030. 4. 2014
20.1J. Starý: Picharacters
Let
The following is proved in B. Balcar, P. Simon: On minimal picharacter of points in extremally disconnected compact spaces, Top. Appl. 41(1,2) 1991, pp. 133145.
Let
Is there a ccc, complete Boolean algebra for which
Mary Bell proved that the answer is yes if we drop the completeness requirement.
If
Let
If
Is there an EDC space
Yes,
(For the space see A. Dow, Gubbi, A. V. Gubbi, A. Szymański: Rigid Stone spaces within ZFC, Proc. Am. Math. Soc. 102(3), 1988.)
Let
20.2Torturing J. Grebík (bachelor thesis)
The outline of the thesis is as follows:
 Motivations (Riemann
\(\zeta\) function, squarefree numbers, ...)  Measure

Results
2.1 Density
2.2
\(\mathcal P(\omega)/\mathcal Z\)
Recall definition 19.12. Note that the ideal
Take a countable sequence
217. 5. 2014
21.1B. Balcar & P. Simon: Nonhomogeneity of Stone spaces
Consider the additive group of integers
Can we find an
21.2E. Thuemmel: Yorioka + Zapletal
Let
If
Assuming towards a contradiction assume that some
If
By recursion on
The set
Choose
By Königs theorem
2221.5.2014
22.1J. Zapletal: Yorioka's paper
We aim to show the proof of
Let
If
The proof is based on two tricks.
22.1.1Trick I.
Let
If
If
Using the largeness of
Then let
If
22.1.2Trick II.
A
If
Assume this is not the case and fix a cover
Let
Then, by the previous lemma,
A ccc
22.1.3An Abstract Approach ?
We say that
If
By contradiction fix a cover
If
22.2David's question for Jindra
If
Let
22.3J. Zapletal: Iteration of ωmodels
If a poset
The original proof uses Keisler's compactness theorem. We will show a proof using iteration of models.
An
Sufficiently definable in the above theorem means that an
Assume that there is an
2318.6.2014
23.1J. Grebík: Ideals on products of Boolean algebras
Let
\(\mathcal I_0=\{\overline{b}\in\mathbb B^\omega:\bigwedge\bigvee\overline{b}=\mathbf 0\}\) \(\mathcal{I_U}=\{\overline{b}\in\mathbb B^\omega:\bigwedge_{U\in\mathcal U}\bigvee_{n\in U}b_n=\mathbf 0\}\) \(\mathcal I^\mu=\{\overline{b}\in\mathbb B^\omega:\lim \mu(b_n) = 0\}\) \(\mathcal I_{\mathcal U}^\mu=\{\overline{b}\in\mathbb B^\omega:\mathcal U\lim \mu(b_n) = 0\}\)
Recall that
When does the equality
The following can be found in Bartoszynski, Judah: Structure of the real line.
A sequence
Given a sequence or real numbers
To have find a counterexample to question 23.2 we would need
a sequence
If
This gives the following conditions on
The equality holds
where
An ultrafilter is semiselective iff it is a rapid Pultrafilter.
The proposition follows from the following characterization of semiselectivity.
An ultrafilter
23.2David Chodounský: Construction of ccc forcing using long diagonalization
The following is motivated by the question whether axiom Y (Ycc) (see 22.11)
is productive.
Recall that
and that, consistenly (e.g. under MA), all of the implications can be reversed. Moreover the implications cannot be separated by "definable" orderings.
The Todorčević's ordering, if ccc, is already productively ccc.
A Suslin tree is ccc but not powerfully ccc.
Take an indestructible gap
If
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 Ycc does not fit into the above hierarchy. The following example shows that Ycc is not productive.
Let
If
If
So let
Let
If
Since
We will show that if
It follows from the proof that both
243.9.2014
24.1D. Chodounský: Generalized Laver Forcing
Given a family
If
If
For
If
A family
In the
Any Pideal which is hitting is
Given a MAD family
The following theorem is from A. Dow: Two classes of FréchetUrysohn spaces, Proc. Ams. 108(1) 1990.
Laver forcing preserves every
A family
In the above definition,
If
Let
The following are equivalent for a
filter
\(\mathcal X\) is\(\mathcal F^+\) \(\omega\) hitting.\(\mathbb L(\mathcal F^+)\) forces that\(\mathcal X\) is\(\omega\) hitting.
Suppose 1. fails and let
On the other hand, assume that 1. holds. Let
Suppose
Work in
\(R_n\leq S_{t_n}\) \(R_n\Vdash k_n\in \dot{A}\)  The set
\(\{t_n:n<\omega\}\) is a maximal antichain in\(S\) .
Now let
We now use the claim to recursively construct a fusion sequence
 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
2510.9.2014
25.1J. Verner: A simple question
Given a topological space
Does there exist a space of size (and extent)
If
If
25.2D. Chodounský: Axiom Y (joint work with J. Zapletal)
If
A forcing
 q is a
\(M\) generic (i.e. is a master condition in the usual sense)  for each
\(r\leq q\) there is a filter\(\mathcal F\subseteq RO(P)\) ,\(\mathcal F\in M\) such that\[ \{ s\in RO(P)\cap M: r\leq s\} \subseteq \mathcal F. \]
Being
If
The MathiasPrikry forcing on a measurable cardinal is an example of a forcing satsifying 2. but not 1. in the above definition.
Does
A
The forcing to kill all Sspaces, the forcing to force PID, the forcing
to show that, under PFA, there are only 5 Tukey types of orderings, are all
A forcing which forces (an instance of) OCA cannot be proper (it adds anticliques to graphs).
25.2.1Laver forcings with axiom Y
Recall the definition 24.1 of
The following are equivalent for analytical Pideals
 The ideal
\(\mathcal I\) is an intersection of\(F_\sigma\) ideals. \(L(\mathcal I^+)\) is\(Y\) proper. For each compact Polish
\(X\) , open\(H\subseteq X^\omega\) and for each\(G\) generic on\(L(\mathcal I^+)\) we have\[ V[G]\models \Big(\forall Y\subseteq \check{X}, [Y]^\omega\cap H=\emptyset\Big) \Big(\exists \{Y_n:n<\omega\}\in V, [Y_n]^\omega\cap H=\emptyset\Big) \Big(Y_n\subseteq \bigcup_{n<\omega} Y_n\Big) \]
If
2.
Assume, aiming for a contradiction, that there is
Aiming for a contradiction, assume that 3. does not hold and that this is
forced by some
3.
It will be sufficient to find
For each
Otherwise, let
Note that in the above claim we could have taken the partition to be in
For any
 if
\(i<n\) then\(q_i\) is disjoint from\(V\) ; and \(q_i\) is disjoint from\(V\) after deleting finitely many successors of the stem of\(V\) .
If the condition with stem
We now use the claim to construct a fusion sequence
This finishes the proof of 3.
2624.9.2014
26.1J. Grebík: Graphs & Banach spaces
Let
The generic graph added by
Let us now try a similar forcing in a different setting.
A norm
Each norm can be approximated by a rational norm with the identity
being an
Let
Again, we can define
The space
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 Yc.c., submitted.
Taking completions is necessary in the above definition. Consider
collapse of
The first implication is easy. Assume for a contradiction
Forcing specializing an Aronszajn tree is Ycc
Forcing freezing an
For basic properties of Ycc reals, consult theorem 25.12.
Let
Ycc forcings preserve
A finite support iteration of any length of Ycc forcings is Ycc.
We proof preservation under twostep iteration. So let
The proof for limit stages is a little more complicated.
If
Let
Recall the definition 25.7 of Yproper forcing notions.
Laver is Yproper but not Ycc.
Is Mathias forcing Yproper?
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).
Given a forcing
A strong master condition
A strongly proper forcing is Yproper.
Let
Let
Moreover
2729.10.2014
27.1Šárka Stejskalová: Consistency of PFA
For more details one can consult:
 Cummings, J: Iterated Forcing and Elementary embeddings, in Handbook of Set Theory, Springer 2010 (theorem 24.11)
 Devlin, K: Yorkshireman's guide to proper forcing, in Surveys in Set Theory, CUP 1983
 Jech, T: Multiple forcing, CUP 1986
 Jech, T: Set Theory (3rd edition), Springer 2006
A cardinal
If
If
If the GCH holds and
A cardinal
\(M\) is closed under\(\lambda\) sequences, i.e.\({}^\lambda M\subseteq M\) \(\lambda < j(\kappa)\)
It is supercompact if it is
An ultrafilter
Let
\([id] = j^{\prime\prime}\lambda\) ;\(\kappa\) is a critical point of\(j\) and\(\lambda< j(\kappa)\) ; The range
\(M\) of\(j\) is closed under\(\lambda\) sequences; \(j^{\prime\prime}\alpha= [\langle y\cap\alpha:y\in \mathcal P_{\kappa}(\lambda)]\rangle\) for each\(\alpha\leq\lambda\) , in particular,\(\alpha=[\langle otp(y\cap \alpha):y\in\mathcal P_{\kappa}(\lambda)\rangle]\) ;\({}^{\lambda^{<\kappa}}M\subseteq M\) and\(j^{\prime\prime}\mathcal P_{\kappa}(\lambda)\in M\) \(2^{\lambda^{<\kappa}}\leq (2^{\lambda^{<\kappa}})^M<j(\kappa)<(2^{\lambda^{<\kappa}})^+\)
1.
1.
2.Let
3.This follows directly from the following claim:
If
The following theorem was proved by Laver in Laver, R.:Making the supercompactness of κ indestructible under κdirected closed forcing, Israel Journal of Mathematics, 29(1978).
Let
Let
Assume
Let
It follows that
Now let
If
Let
We want to show that
We now continue by showing that, given a sequence
Let
See, e.g., Cummings article referenced above.
Now let
Now consider
It is sufficient to show that
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).
The following are equivalent:
1.
(
Let
\(\mathcal A_\alpha\) is an independent family;\(\mathcal S_\alpha\) is an almost disjoint family;\(S_\beta\subseteq^* \bigcap_{\gamma\in K} A_{\gamma,f_\beta(\gamma)}\) for each\(\beta<\alpha\) and\(K\in[\alpha]^{<\omega}\) ; 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
29.2D. Chodounsky: Todorčević's ordering & Ycc.
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.
Given a topological space
Let
If
Let
The natural embedding works and is even onto the separative
quotient of
3026.11.2014
30.1R. Bonnet
A Boolean space is a compact zerodimensional space topological space.
A meet semilattice is a pair
Any linear order of a space
The Cantor set
It follows from the definition that the partial order must be a closed
relation in
A Boolean space with equality is a Priestley space.
Let
If the order is equality, then the topology of
A Boolean LOTS
The space
There is a natural continuous semilattice operation on
For each Priestley space
Any continuous increasing map
A Priestley space
A Boolean algebra
\(x\in U_x\) for each\(x\in S(B)\) ;\(x\in U_y\rightarrow U_x\subseteq U_y\) for each\(x\in S(B)\) ; and for each
\(x\neq y\) either\(x\not\in U_y\) or\(y\not\in U_x\)
The onepoint compactification
If
There is a scattered space (even one with CantorBendixon rank 3) which does not have a clopen selector (or, equivalently, which is not a Skula space).
Fix
Assume that it has a clopen selector. We may assume that each member of the
selector does not contain
If
Can we find an almost disjoint family
313.12.2014
31.1J. Grebík: Questions around Suslin trees
Given a tree
If for each coloring
327.1.2015
32.1Michael Hrušák: Weak Dimond
A
The weak diamond is the statement that for each
coloring
Weak diamond is equivalent to
Given sets
The following definition comes from
Džamonja, M., Hrušák, M. and Moore, J.:
Parametrized
A cardinal invariant is Borel provided it can be written in the
form
The parametrized diamond
Typically, a coloring is definable means that each of its restrictions to
each
The strongest weak diamond (denoted
The (standard)
The strongest weak diamond and
The strongest weak diamond does not imply CH.
Start with a model of GCH, add
Unfortunately, the strongest weak diamond implies
The strongest weak diamond +
The diamond principle
Most of the things true for
If a forcing
32.1.1Generalizations ?
Recall Baumgartner's result that, under PFA, all
Is it consistent that all
DevlinShelah's weak diamond restricted to colorings from
Fix some set
The proof works also for
32.1.2Applications in topology
Recall that
A topological space
CH implies that there is a compact sequential space of order
There is a compact sequential space of order 2
(e.g. a onepoint compactification of a
There is also Balogh's solution to the MooreMrówka problem:
Under PFA every compact countably tight space is sequential.
A space
Under PFA each compact super sequential space has
sequential order at most
If a space
We need a MAD
 for each
\(B\in \mathcal B, A\in\mathcal A\) either\(A\subseteq^* B\) or\(A\cap B=^*\emptyset\) ; \(\mathcal B\) is\(\mathcal I(\mathcal A)\) almost disjoint. 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
A similar proof shows that
3314.1.2015
33.1E. Thuemmel
33.1.1Forcing with homogeneous sets
Given a coloring
If a subset of
If
Let
33.1.2Hrušák's question
In the following all ideals are assumed to be tall ideals on
Recall the following arrow notation:

\(X\rightarrow (Y)^n_m\) 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\(Y\) . 
\(X\rightarrow (Z,Y)^n\) 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\(X\rightarrow (\{B:B=\omega\},Y)^n\) .
Is there a (Borel) ideal
The ideal
Enumerate
To show the second arrow, consider a set
It follows from the proof that
If
One implication is clear. We show that if
Case 1. There is a positive
Case 2. Each subset of
Consider the
The ideal
Recall that
An ideal
No tall semiselective ideal can be analytic.
There is an ideal
We construct a labeled tree
3428.1.2015
34.1E. Thuemmel: Ideals on ω
Recall the definition 33.4 due to M. Hrušák.
Is there an ideal
Last week we have shown (33.6) that, in particular,
The ideal
In fact, we have shown even more, see 33.8.
Today we aim to show that the ideal
The ideal
Also recall
An ideal
and
If
We will also show the following example.
The ideal
The ideal
The sets
Let
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
In fact,
Assume to the contrary that for some branch
Enumerate the tree
Recall the definition of the Katětov order for ideals:
We say that an ideal
M. Hrušák has a program of finding "Katětov critical" ideals, i.e. given a
property
If an ideal
If
Let
Further, the sets
We now turn to descriptive complexity. Recall that Farah, motivated by the GalvinPrikry theorem, proved
If a family of
For the definition of semiselective see 33.10.
If a semiselective ideal is tall then it is not analytic.
If
Given an ideal
Each tall,
We know that
We can now generalize Farah's theorem:
If a family of
Thus we have the following picture:
3511.2.2015
35.1O. Guzmán: Generalized diamonds
The notation
It is clear that
Let
Let
Given two sets
Devlin and Shelah proved that
If
Let
The following proposition disproves the conjecture 32.18 of M. Hrušák (see also definition 32.17).
It is consistent that
Let
First notice that if there is a normal AD family of size
Can the above be reversed, i.e. does
The above is not clear, since pseudointersections of
35.2J. CancinoManríquez: Ideal independent families
(joint work with O. Guzman and A. Miller)
A family
If
Any AD family as well as an independent family is ideal independent.
If
The spread generating number (
Does
Let
Assume to the contrary that
In the Miller model
It is consistent that
Shelah has a model where
35.3J. CancinoManríquez: Irresolvable spaces
(joint work with M. Hrušák and D. MezaAlcántara)
In the following all spaces are assumed to be regular and crowded.
A topological space
A space is irresolvable if all dense sets have nonempty interior.
The irresolvability number
Does
The
Let
Assume to the contrary that
Finally
3625.2.2015
36.1D. Chodounský: Mathias like reals
Given a filter
\(r\) is a pseudointersection of\(\mathcal F\) , and For each
\(\mathcal F\) universal\(H\) there is a finite\(a\in[r]^{<\omega}\) which is also in\(H\) .
Recall that a set
Every Mathiaslike real becomes a Mathias real possibly after intersecting it with a further Cohen real.
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.
Assume
Fix a
Assume a forcing
Proof (idea): Use the previous proposition, the fact that if a real is
Mathiaslike for
36.2D. Chodounský: Eventually different reals
Recall the relation
thus we can write (a part of) Cinchoń's diagram as follows:
After adding two infinitely equal reals we get a Cohen real.
The previous theorem motivates the definition:
A forcing is half Cohen iff it adds infinitely equal reals.
Suslin ccc forcings add half Cohen reals iff they add Cohen reals.
There is a forcing adding a half Cohen real which does not add a Cohen real.
An model
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.
Given a filter
where a filter is
To show that 1 implies 3. we show something stronger:
If a model with a Mathiaslike real does not contain an eventually
different real then the filter is
Let
Next 3. immediately implies 2. so it remains to show that 2. implies 1. This is
a normal forcing argument: let
.
Now use
3718.3.2015
37.1J. Grebík: Free sequences in Boolean algebras.
Let
A strictly decreasing sequence is free but not independent.
The cardinal invariant
Recall that for an ultrafilter
The following cardinal invariant is less known.
The cardinal invariant
If
Proof (Hint): Construct
If
Let
For each
Now extend
The second inequality is proved similarly: Let
If
The cardinal invariant
So we have the following picture:
Is it consistent that
Note that countable support iteration does not work, since in the required model
we need to have
The following question might be simpler:
Is it consistent that
Is
Is
Is
It is consistent that
If
3825.3.2015
38.1J. Verner: AntiPpoints
A point
Does
38.2D. Chodounský
38.2.1A Dynamical question
Given a disjoint sequence
Whenever
Fix an interval partition
Given an interval partition
If
38.2.2Indestructible gaps
It is consistent that there is an indestructible nonHausdorff gap.
An
\(L_\alpha\cap R_\alpha=\emptyset\) for each\(\alpha<\omega_1\) \(L_\alpha\subseteq^* L_\beta, R_\alpha\subseteq^* R_\beta\) for each\(\alpha<\beta\)
It is a gap if there is no separator, i.e. an
Each Hausdorff gap is oriented and each oriented gap is special.
A special gap is indestructible by forcings preserving
A Hausdorff gap exists.
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,
Is it consistent that there are special and not oriented or oriented and not Hausdorff gaps?
There is a special nonHausdorff gap.
After adding
We will show how to force an oriented nonHausdorff gap. Recall that
the definition of condition
It should be evident that if a gap satisfies
Is there a tower which generates a meager ideal?
If there is a tall meager
It is consistent that there is an oriented nonHausdorff gap.
We will add a gap which is oriented but its left side does not
satisfy
The forcing is Knaster.
Let
We now show by contradiction that the left part does not satisfy
It is not hard to see that forcing used in the proof is equivalent to
adding
3915.4.2015
39.1News from Singapore
Let
YPFA implies
39.2J. Wódka: Monotone spaces
39.2.1What I am working on...
A function
39.2.2Monotone spaces
A metric space
A metrizable space is a GOspace iff there is a matric which makes it a monotone space.
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:
 J. Mycielski: New settheoretical axioms derived from a lean metamathematics, JSL 1995
 A. Enayat: On the LeibnizMycielski axiom in set theory, Fund. Math. 181 (2004)
 A. Enayat: Leibnizian models of set theorey, JSL 69 (3), 2004
 A. Enayat: Models of set theory with definable ordinals, Arch. Math. Logic 44 (2005)
 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
Let
It is clear that if
An
If
An ultrafilter
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.
Let
An
Any structure with a definable linear order is weakly Leibnizian.
It follows that there are arbitrarily large weakly Leibnizian structures (independent of the size of the language).
An
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.
From the definitions it immediately follows that
A consistent extension
Recall that
The proof of the above theorem uses the following standard theorem of Model theory.
If
The universe is setgeneric over
Grigorieff showed that the forcing going from
In this context recall
also, I recall that about 20 years ago a master's student of Sy Friedman proved that
The universe is classgeneric over
However, it was probably not written up.
If
The following are equivalent
\(M\models V=OD\) (note this is the same as\(V=HOD\) )\(M\) has a definable (without parameters) global well ordering The theory of
\(M\) has a pointwise definable model.
A model
Every consistent extension of
If a consistent extension of
A Paris model has countably many ordinals. However, it may be arbitrarily wide.
Let
Mostowski and RyllNardzewski 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 BalcarVopěnka theorem:
If two models of
The LeibnizMycielski principle (
This is motivated by the classical Leibniz principle of the identity of indiscernibles:
If two objects are distinct they differ in some property
The following are equivalent for a complete consistent
extension
\(T\) has a model with no indiscernibles (i.e. a Leibnizian model).\(T\) proves the LeibnizMycielski principle.
The proof uses the reflection theorem and Paris models.
The KinnaWagner selection principle (
The KinnaWagner selection principle is equivalent to the statement
(
The KinnaWagner selection principle is weaker than the Axiom of Choice.
The KinnaWagner selection principle has a global version (similar to the
global version of
The following are equivalent:
 This global version of the KinnaWagner selection principle
 There is a definable injective class function
\(H\) from\(V\) into subsets of\(2^{<Ord}\)  The LeibnizMycielski principle holds.
We thus have the following picture where none of the implications can be reversed:
Is there a model of
This is connected to the following theorem of Kanovei:
There is a countable set of reals none of which are definable.
Can the above set be of the form
Yes!
4113.5.2015
41.1D. Chodounský: YPFA and the continuum.
Recall that last time we were proving that PID
(9.8) implies
A pair
\(T_\alpha\subseteq^* T_\beta\) for each\(\alpha<\beta<\lambda\) ;\(A_\alpha\cap T_\beta=^*\emptyset\) for each\(\alpha<\kappa,\beta<\lambda\)
If
\((\forall\beta<\omega_1)(\forall n<\omega) (\{\alpha<\beta:A_\alpha\cap T_\beta\subseteq n\}<\omega)\)
Given a
Assume PID. If the first alternative holds for
The strategy is as follows:
Let
\(M\) is a countable elementary submodel of\(H(\theta)\) ;\(I\) almost covers every element of\(\mathcal I\) in\(M\) ; and\(\alpha<\kappa\) is not contained in any\(K\in\mathcal K\cap M\) .
We order suitable triples as follows:
\(p\supseteq q\) ; and 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
The game
This game is closed and hence determined. We say that
a set
Given
4220.5.2015
42.1J. Grebík: Oscillations of real numbers
The motivation for the following is the Prikry problem:
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. Ycc posets, which cannot add Random reals and look for one which does not add Cohen.
If a Suslin ccc forcing adds an unbounded real then it adds a Cohen real.
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.
There is a
The above forcing is, in fact, the Laver forcing with a nowhere dense ultrafilter.
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 Ycc forcing which does not add
Cohen reals. The idea is to take an unbounded sequence of functions
If
Given two functions
The following is easy.
\(osc(f,g)<\omega\iff f\leq^* g\vee g\leq^* f\) .\(osc\) , 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.
Let
PFA implies
The proof will use the following notion of coding:
Given
A given
Otherwise let
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
Given two reals
A real
A forcing
We will now define three variants of the Hechler forcing.
The first variant of Hechler forcing,
We aim to show that
Let
Let
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.
Given a function
If
Taking
Given a strictly
Since
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
If
The forcing
Let
The family
The fact that
Now let
The family
For any
Now pick any
If
An algebra
Hence we have the following theorem:
If
It is still not clear, however, whether
45.2D. Chodounský: The new version of the Ycc paper.
A forcing
 if
\(p\leq q\) then\(w(p)\supseteq w(q)\) ,  if
\(p\) is compatible with\(q\) then there is\(r\leq p,q\) such that\(w(r)=w(p)\cup w(q)\) , and  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\(p_\beta^1\) .
A suitable forcing is Ycc.
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.
If
However we will be more interested in a different theorem from the paper:
Recall that, when
Dilip claims he can replace
In the following a suitable partition of
\[ \lim_{n\to\infty}\frac{A\cap I_n}{I_n}=0 \] \[ \lim_{n\to\infty} \max\{ab:a,b\in A\cap I_n\}=\infty \] for some suitable partition. Then\(A\in\mathcal Z_0\) .
Given an interval
If
Given two interval partitions
Given a suitable interval parition there are functions
If
For each suitable partition
It remains to prove the lemma. It is sufficient to find suitable functions
4615.7.2015
46.1J. Zapletal: Kσcliques
Define the following cardinal invariant
If
It follows that
Is it consistent that
46.2M. Hrušák: Incomparable families and trees in P(ω)/fin
Does there exists a family
\(\mathcal I\) is closed under complements; if
\(A,B\in\mathcal I\) then\(A\not\subseteq^* B\) ; and  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 selfdual matroids.
If
Recall that
Is there a family of infinite subsets of
Each maximal family of incomparable (non
Let
We will call
Disjoint nodes in
Let
Is it consistent that
It is consistent that
If there is a completely separable mad family then there is a maximal
tree of height
A base tree is a maximal tree.
Is
It is easy to see that
If
There is no maximal tree all of whose levels would be countable.
Is (it consistent that) there is a maximal
46.3Balcar's question
There is an independent splitting family.
The following can be found in J. Brendle: There is an independent splitting family.
There is an independent splitting family of size
Let
46.4Hrušák: Miscellanea (Interlude)
Consider the ideal
Is
Each
Can every
First add
Without first adding
Killing density zero always adds an unbounded real.
It is consistent that all towers generate a meager filter.
Force with, e.g., Hechler; having
It is consistent that every tower generates a nonmeager filter.
{}
Proof: Start with CH and force
Does there exist a meager tower and
Does
Recall that a tight gap is a gap consisting of functions which has no
partial separator, i.e. a system
475.8.2015
47.1M. Doucha: Continuous Scott Analysis
Joint work with A. Nies, I. Ben Yaacov and T. Tsankov. We consider the logic
Given a formula
Given a countable language
Given
Two tuples
Two tuples
The Scott rank of
If
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.)
A topological group
Recall that a topological group
A topological group
\(u_L=u_R\) ; or the uniform completion w.r.t.
\(u_L\) (or\(u_R\) ) is a group; or  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,
 there is a biinvariant compatible metric on G, i.e.
\((\forall x,y,a,b\in G)d(x,y)=d(axb,ayb)\) .
The following are straightforward examples:
 All Abelian groups are uniform.
 All compact groups are uniform.
A Heisenberg group, i.e. a group of 3by3 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 biinvariant compatible metric.
If
There is a universal separable Abelian group with a biinvariant metric. I.e. each separable Abelian group with a biinvariant metric is isometric to a subgroup of this universal group.
There is a separable group with a biinvariant metric bounded by
On the other hand there is no separable universal group for groups with a uniform unbounded biinvariant metrics.
A uniform Banach group is a Banach space
There is a uniform Banach group which has
47.3M. Doucha: Lipschitz free Banach spaces
47.4D. Chodounský: Work with S. Geschke
What is the relationship between the two right and leftshift dynamical
systems on
Consistently, e.g. under PFA, the above are different, i.e. there is
no automorphism
4812.8.2015
48.1B. Farkas: Towers in filters
The following is joint work with Joerg Brendle.
Given an ideal
If
The Random graph is a countable graph
The Solecki's ideal
The ideal
The ideal
If
The following diagram shows various
We know that
M. Hrušák and his students have been able to reduce the question whether
The ideal
The following are equivalent:
\(\mathcal{ED}_{fin}\leq_{KB}\mathcal I\) ;\(\mathcal I\) is not weak Q, (i.e. if for each partition of; and\(\omega\) into finite sets there is a\(\mathcal I\) positive selector) (if
\(\mathcal I\) is Borel)\(\mbox{non}^*(\mathcal I)\geq\mathfrak{t}\) .
If
The van der Waerden ideal
Assume, aiming towards a contradiction, that
Given
Recall Mazur's and Solecki's characterization of definable ideals:
Let
\(\mathcal I\) is\(F_\sigma\) iff there is a lscsm\(\varphi\) such that\(\mathcal I=Fin(\varphi)\) (Mazur)\(\mathcal I\) is an analytic Pideal iff there is a lscsm\(\varphi\) such that\(\mathcal I = Exh(\varphi)\) (Solecki),
where
\(\varphi(\emptyset)=\emptyset\) ;\(\varphi(A\cup B)\leq\varphi(A)+\varphi(B)\) ;\(\varphi(A)\leq\varphi(B)\) for\(A\subseteq B\) and\(\varphi(A)=\sup\{\varphi(A\cap n):n<\omega\}\)
and
Even more is true: an ideal
It is easy to see that under CH all Pideals contain a tower (they are even generated by a tower). This can be generalized:
After adding
Let
After adding a Cohen real there is
(Note that for the claim it is sufficient that
Recursively use the claim to construct a
The above proof does not work for longer extensions. So we now show how to build long towers in tall analytic Pideals.
A forcing notion
If
If a forcing dominates an ideal, a simple iteration will force a tower into the ideal.
The Localization forcing (Loc) dominates all tall analytic Pideals.
The proof of the lemma is essentially proving that
Let
If
In the above theorem it is not clear what happens outside of
Does the towerspectrum
The main question, however, is:
Assume
Assume
If
We prove only the second part. Let
Is it possible that
Assume
The problem with the above theorem is that we do not know how to preserve
towers in
4911.11.2015
49.1Victor TorresPerez: Shelah's SemiStationary Reflection Principle
Suppose we have a family of intervals of some linear order. Under what conditions can we color these intervals by two colors such that samecolored intervals are disjoint?
A family
A family
A family
Rado's conjecture is independent of ZFC modulo a supercompact cardinal.
Rado's conjecture implies the SCH
(singular cardinal hypothesis),
Recall, that
For an uncountable cardinal
Strong Chang's Conjecture says that there are arbitrarily large
49.1.1Rado's conjecture vs. Martin's Maximum
Recall that Martin's Maximum (MM) says that for every stationary setpreserving
forcing and
A
Martin's Maximum implies
1.
The following are consistent, modulo a supercompact cardinal:
1. MM and
For the following recall that CH implies
Assume that Rado's Conjecture is true then
1.
Strong Chang's Conjecture is equivalent to the SemiStationary Reflection Principle (SSR).
Note that MM implies WRP which implies SSR.
SSR implies the missing negations of
SSR implies
PID implies
Given two sets of ordinals
Given
The SemiStationary Reflection Principle (SSR) principle
says that for every semistationary
For
SSR for
Assume that there is a square sequence
If there is a stationary, "weakly full"
We use the square sequence to define the set
Take
There is a sufficiently large
Given
The fact that
5016.12.2015
50.1Vojtáš: Recommender systems
516.1.2016
51.1Š. Stejskalová: Itay Neeman: Forcing with sequences of models of two types. (I)
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
Given a condition
Last time we proved the following facts.
The residuum
If
A set
Note that
I (Jonathan) am not quite sure the following lemma was actually stated in this way...
If
Suppose that
We aim to show that the forcing we are talking about is stronglyproper w.r.t. some models. First the relevant definitions.
Given a forcing
Given
Given a forcing
The relationship between proper and strongly proper is the same relation as between ccc and semiCohen.
The following, while probably not literally true, is nevertheless true in spirit.
A forcing
Suppose
\(M[G]\preceq H(\theta)[G]\) and\(M[G]\cap V=M\) ; and given a
\(P\) name\(\dot{f}\in M\) for a function defined on some ordinal then\((\dot{f}\cap M)^G = \dot{f}^G\upharpoonright M\) .
A set
A set
If
Let
If
5320.1.2016
53.1Š. Stejskalová: Itay Neeman: Forcing with sequences of models of two types. (III)
5410.2.2016
54.1José de Jesús PelayoGómez
The cutandchoose game for an ideal
Player I has a winning strategy in
If
If
Player I has a winning strategy in
Let
If Player I does not have a winning strategy in
The corollary directly follows from the following observation:
If player I has a winning strategy in
The Solecki ideal is an ideal on the countable set
The ideal is generated by clopen sets containing a branch. Note
Does player II have a winning strategy in
(This is, of course, motivated by the question whether
If the player II has a winning strategy in
The above conjecture is false, even for
There is an
Fix
The proof is technical and will be skipped.
Similarly we define that
Note that
Is there a Borel ideal
For each analytic ideal
54.2Oswaldo Guzmán: Sierpińsky's principle and meager sets
The principle (*) is the statement that there
is a family
CH implies (*).
A family
If
(*) iff there is a Luzin set.
If there is a Luzin set then () and () implies that
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.
In fact, the proof gives a meager weakLuzin set. So one can ask:
Does
No!
We will show a sketch of the proof of the theorem. First we need a lemma.
If
 each
\(f_\alpha\) is a partial function from\(\omega\) to\(\omega\) ;  the domains of the functions form an A.D. family;
 for every
\(g:\omega\to\omega\) there is an\(\alpha<\omega_1\) such that\(g\cap f_\alpha=\omega\)
Enumerate
Let
The family
Let
The family
Clear.
We now sketch the No answer to the question 54.24. We define
a forcing
\(s_p\in\omega^{<\omega}\) ;\(M_p=\langle M^p_0,\ldots,M^p_n\rangle\) is an\(\in\) chain of countable elementary submodels of\(H((2^{\mathfrak c}))^+\) ;\(F_p:M_p\to\omega^{\omega}\) is a function such that\(F_p(M^p_i)\) is Cohen over\(M_i\) and, if\(i<n\) , then\(F_p(M^p_i)\in M^p_{i+1}\) ; and\(s_p\cap F_p(M^p_i)=\emptyset\) 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.
The forcing
 it is proper (but it cannot be strongly proper because of the next property);
 if
\(X\) is a nonmeager weakLuzin set then it forces that\(X\) is not weakLuzin; and  it does not destroy category, i.e. it forces that the old reals are not meager in the extension; in fact, even
 the countable support iteration of
\(P_{cat}\) (of any length) does not destroy category
Assuming the consistency of an inaccessible cardinal it is consistent
that
Do a countable support iteration of
We first prove properness.
Let
Let
Next we prove 3.
If
Let
The condition
Let
5517.2.2016
55.1M. Olšák: Random Reals and Suslin Trees (R. Laver, 1987)
If
In contrast with this a single Cohen real adds a Suslin tree.
Let
For every
Since
Moreover, fix a countable
Consider the forcing
\(f(\alpha,n)\wedge f(\beta,n)\Vdash \alpha\perp_{<_T}\beta\) ;\(\mu(f(\alpha,n)) = 0\) or\(\mu(f(\alpha,n)) > 1/2\) ; and\(f(\alpha,n)\in B_\alpha\) .
The order on
\(\mbox{dom}(f)\supseteq\mbox{dom}(g)\) ;\( (\forall (\alpha,n)\in\mbox{dom}(g))f(\alpha,n)\leq g(\alpha,n)\) ; and\( (\forall (\alpha,n)\in\mbox{dom}(g))g(\alpha,n) > 0 \rightarrow f(\alpha,n) > 0)\) .
Let
Suppose
The lemma gives us a generic
Given
We still need to prove Lemma 55.5. First, however, observe:
Fix
This is clear (use
Fix
Does Laver, Miller or Sacks add a Suslin tree?
Under
The random forcing adds a poset which is ccc but not powerfully ccc so
random forcing destroys
5624. 2. 2016
56.1J. Verner: Ramsey theory of metric spaces
572.3.2016
57.1J. Verner: Ramsey theory of metric spaces (ctd.)
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.:
Assume that
There is no analytic ideal with
An ideal
The ideal
The ideal
If
There is a coanalytic example
57.3J. Grebík: A Fraïssé class without a Katětov functor
This talk is motivated by the problem:
Is there a Fraïssé class without a Katětov functor?
which appeared in Kubiś, Mašulović:Katetov functors, published as preprint 1412.1850 on the ArXive.
Let
A category
 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\(Y\) ;  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.\(X_0,X_1\rightarrow Z\) ;  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  it is "essentially" countable, i.e. it has only countably many pairwise nonisomorphic objects.
Finite graphs, finite linear orders, finite groups, finite Boolean algebras and finite dimensional vector spaces all form a Fraïssé class.
Given a Fraïssé class
Given
An object
If
A Fraïssé class
Given a finite metric space
A Katětov functor on
The answer to problem 57.8 is yes.
There is a Fraïssé class
Let
589.3.2016
58.1W. Bielas: Cantorval
58.2P. Simon
There is a Cantor set
The construction giving the cantor set
58.3J. Verner: A simple lemma supposedly proved by my uncle
Given
Assume that
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
The class of all finite groups has no pushouts, but it is a Fraïssé class (this is difficult). Does it have a Katětov functor?
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.
Let
If there is a Katětov functor
Simple threedimensional arrow chasing using the defining condition of the Katětov functor (i.e. that there is a natural transformation from the identity + onepoint extensions).
Assume that
If
We now proceed to prove Theorem 57.19. We just give
the definition of
Let
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
Fix an
 There is a natural transformation
\(\eta\) from the inclusion to\(F_n\) ; and  every onepoint extension
\(\langle x,t\rangle\in\mathcal C_n\) embeds into\(F(x)\) (in\(\mathcal C_{n+1}\) ).
Then we can extend
 either there is
\(v\in x\) with\(t_\xi<v<t_\psi\) ; or \(supp(\xi)<supp(\psi)\) ; or the biggest
\(v\in supp(\xi)\Delta supp(\psi)\) is in\(\xi\) ; or \(supp(\xi)=supp(\psi)\) 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
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
If two sequences
Assume now that
Given a compact metric space we define the
center of distances of X as follows:
If two sequences
Given a sequence
The set of subsums is symmetric, i.e. if
The sumbsum set of a positive summable sequence is either:
 a finite union of closed intervals
 a Cantor set
 a symmetric Cantorval
A set
Let
If
Let
\(X\subseteq[0,5/3]\) and\([2/3,1]\subseteq X\) is a component of\(X\) ;\(x\mapsto 5/3x\) is a symmetry of\(X\) ;\([0,1/2)\cap X=1/4X\) and\([1/2,2/3]\cap X=X\cap[0,1/6] + 1/2\) ;\(\lambda(X)=1\) ; the set
\([0,2/3]\setminus X\) is closed under the functions\(1/4x, 1/2+1/4x\) and\(1/4(5/3x)\) whose ranges are disjoint;
If a sequence
6130.3.2016
61.1Jerzy Król: Connections between forcing and physics
61.1.1History
 P. Benioff: Models of Zermelo Frankel set theory as carriers for the mathematics of physics. I Journal of mathematical Physics, 17, 618 (1976)
 G. Takeuti: Two Applications of Logic to Mathematics Princeton, 1976 IS and PUP
 R. A. van Wesep: Hidden variables in quantum mechanics: Generic models, settheoretic 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
Every locally compact group carries a leftinvariant
Assume

If
\(G\) is locally compact,\(\varphi\) measurablem and\(add(\mathcal N)=\mathfrak c\) then\(\varphi\) is continuous. 
If
\(G\) is a Baire group,\(\varphi\) Bairemeasurable and\(add(\mathcal M)=\mathfrak c\) then\(varphi\) is continuous.
It is known (Banach, Otis ?) that if
If
We will first show how to prove the result due to Banach for separable
Let
A nonempty subset
Open sets are extrameasurable while singletons are not.
Does there exist an extrameasurable null set if
Assume that
 There is a noncontinuous measurable homomorphisms
\(\varphi:G\to H\) . 
There is a countable family
\(\mathcal A=\{A_n:n<\omega\}\) of extremeasurable 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
Assume now that
If
Assume
We will only show the proof for Polish groups where additivity of
null is the same. We may assume that
6320.4.2016
63.1D. Chodounský: Towers and gaps
Recall that a tower (or a chain) is a sequence
Also recall that a pregap is a pair of towers
A pregap is a gap iff there is no
Recall that a gap is called special (or indestructible) if it is a gap
in every bigger model which preserves
A gap is special iff it is indestructible by ccc forcing.
A gap is special iff there is a
Note, the PID axiom implies that every gap is special. On the other hand, PID
is consistent with CH, in particular with
Assume
 The set
\(\{\alpha:X_\alpha\leq\omega\}\) is at most countable; and  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
Given a nonspecial tower
The forcing
We now start with a model of GCH (and, e.g.,
Assume to the contrary and let
If it were homogeneous in
Finite support iteration of
Using the delta system lemma on supports.
6427.4.2016
64.1J. Verner: J. Brendle & M. Hrušák: no meager towers
64.2M. Hrušák: CDH spaces
A separable metric space
Does there exist a connected meager in itself (meager for short) CDH
A space
If a CDH
In the Cohen model every
In the Cohen model there is no connected meager CDH space.
CH implies that there is a connected meager
CDH space
Can such a space be constructed in the plane?
There is a connected
First notice that if there is a
Under CH there is a connected
64.3M. Hrušák: Gruff ultrafilters
An ultrafilter
Do gruff ultrafilters exist?
Yes, if
Yes, if
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
A MAD family
 There is
\(I\in\mathcal I(\mathcal A)\) which hits each\(a\in X\) or  There is
\(Y\in[\omega]^\omega\) such that\(\bigcup Y\in\mathcal I(\mathcal A)\) .
It is consistent that no such families exist.
Given an ideal
A MAD family
The property (+) is very strong. Moreover (+) MAD families cannot be destroyed by definable forcings. Also property (+) implies the ShelahSteprans property.
Is it consistent that there are raving MAD families with of
size
A Borel ideal
An nonmeager ideal is ShelahSteprans (even raving).
64.5M. Hrušák: Sequential order of compact spaces
Does there exist a compact sequential space
There is one of order
If
Under PFA if the sequential order coincides with the
CantorBendixon rank then the order is at most
Under
654.5.2016
65.1D. Chodounský: Independent families
We will show Shelah's result that the following is consistent:
The independence number
For a set
In this notation a system
The ultrafilter number
Note the consistency of
An independent system
A dense independent system is a maximal independent system. Moreover the modfinite equality and inclusion in the above definition can be replaced by strict equality and inclusion without changing the notion.
Given an independent family
An independent family
A forcing is Cohen preserving if each open dense subset of
A forcing is Cohen preserving if it is Cohen preserving for a
single countable
The definition of
An independent family is selective if it is dense and the filter
A filter
A filter
A filter
A filter
We now construct one step of our iteration. Given an ultrafilter
Sacks property probably implies Cohenpreserving.
For our purposes we shall call as sequence
Given a tree
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
Given a uniformly branching tree
If
Given
If
The forcing
6611.5.2016
66.1D. Chodounský: Independent families II.
Recall that last time we defined, given an ideal
An
Given an ideal
If
If I had a winning strategy then he would have a winning strategy such
that each
If
Enumerate
The forcing has the Sacks property.
Let
\(e_0(T^\prime_0)\cup\Delta_0\subseteq e_0(T_0)\) ; and\(e_j(T_0)=e^\prime_j\) .
Continue by induction. Since this is not a winning strategy for player I
we can assume that he lost. Then
6710.5.2016
67.1J. Blobner
67.2D. Chodounský
Recall Definition (12.16) of the Katětov order. Also
recall that there is, up to graph isomorphism, a unique graph on
The random ideal
Note that the random ideal is an
An ideal
The Solecki ideal is the ideal on clopen subsets of
The Solecki ideal is an
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
For each
For each
Given
Under MA the algebras
(Vojtáš, P.:
Boolean isomorphism between partial orderings of convergent and divergent series and infinite subsets of
It is consistent that the above algebras are not isomorphic.
The proof works by showing that the
Fuchino S., Mildenberger H., Shelah S. and Vojtáš P.: On absolutely divergent series, Fund. Math. 160 (1999)
Given
688.6.2016
68.1P. Vojtáš: Divergent series
We consider the structure
Let
For each
68.2M. Hrušák: Strong measure zero
If
For topological groups we can equivalently define that
A set
The following is known:
 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 GMSproperty)  Under the BorelConjecture (SMZ consists precisely of countable sets) every group has the GMS property.
Under CH the group
Under CH (or some other reasonable condition for counteraxamples
to GMS) if
A group
6929.6.2016
69.1M. Doležal: The ideal of σporous sets
This is joint work with Preiss and Zelený.
A set
 If
\(B\subseteq A\) and\(P(x,A)\) then\(P(x,B)\) ; \(P(x,A)\iff (\exists r>0)(P(x,B_r(x)\cap A))\) ; and If
\(P(x,A)\) then\(P(x,\overline{A})\) .
The intended meaning of
The standard porosity relation
The strong porosity relation
Let
A porosity relation
\(\psi(0)=0\) ;\(0<\psi(r)\leq r\) for\(r>0\) ; and if
\(P(x,C)\) then\(P(x,\{y:d(y,C)\leq\psi(d(x,y))\})\) .
It satisfies condition
Let
(idea) Characterize
Fix two sequences
\(x_{n+1}\in B(x_n,(R_nr_n))\) \(S_{n+1}^i\subseteq B(x_{n+1},R_{n+1})\)
Player two wins if either the result of the game
If
Write
If
(hint) There is a
 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\(s\subseteq t\) ;  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  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, closuredescending 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
If There is
If Case 1 fails and
If both previous cases fail then
Finally we modify the game
703.8.2016
70.1Samuel Goméz da Silva: Star covering properties
A space
The motivation for the above definition is the following result of Flesichman:
A Hausdorff space
Proof (Idea): Every open cover has a discrete kernel, i.e. a maximal
set
Paracompact
The
By maximality closed discrete subsets of
If
(by contradiction) Assume
Since, if
Does
If
Together with the following
If
this shows that
Let
If
Since normal
An AD family
The space
It follows that
Does
The diamond principle
Recall that the diamond principle gives for all Borel
70.1.1The selective version of (a)
The following was introduced by Casarta, di Maio, Kočinac in 2011:
A space
If
Note that under CH, (a) and selectively (a) are equivalent for
Let
 If
\(\mathcal A<\mathfrak p\) then\(X\) is (a)  If
\(\mathcal A<\mathfrak d\) then\(X\) is selectively (a)  If
\(\mathcal A\) is MAD then it is selectively (a) iff\(\mathcal A<\mathfrak d\) .
Proof (idea): 2. Let
 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).
It is consistent that there is a selectively (a) non (a)
Separable,
Proof (by contradiction): Fix a countable dense
{}
7113.10.2016
71.1E. Thuemmel: Extremally disconnected groups
We present the following result of Reznichenko and Sipacheva:
It is consistent that there are no countable,
nontrivial (i.e.
The uncountable case is still open.
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.
There is an extremally disconnected group under
We show that if there is no no rapid filter then there is no nontrivial countable extremally disconnected group.
A filter
The proof strategy will be to construct two disjoint sets
We also use the following theorem of Malychin.
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.
Call
If
We show that
If
We show that
Call
If
The thick subsets of
Let
From now on we assume there is no rapid filter.
If
It is sufficient to assume, in the above lemma, that
First we may assume that the
We now apply the lemma to get
 for each
\(g\not\in U_n\) we have\(\xi^\prime_1\cap gU_{n+1}<\omega\) ; and \(\lim_{g\in\xi^\prime_1}(\min g)=\infty\) . for every
\(F\) neighbourhood of\(e\) there are distinct\(a,b\in F\) with\(ab\in\xi^\prime_1\)
If
Let
Now partition
It is clear that the
723.11.2016
72.1M. Pawliuk: Fraïssé classes & the Hrushovski property
We start with a motivating question.
Let
If
The result intuitively says that topological dynamics of
72.1.1Three examples
Consider
The characterization of
Consider the Random graph, the unique countable graph such that
for any disjoint finite sets of vertices
The characterization is again equivalent to saying that every type over a finite set of parameters is realized.
Consider the rational Urysohn space
Again, the characterization is equivalent to realizing types over finite sets of parameters.
We now consider directed graphs.
A directed graph is a set
If
\((\mathbb Q,<)\)  The random complete digraph (tournament)
 The random complete
\(n\) partite digraph  ... 15.
What are the dynamical properties of
The answer for the above questions is known for all
Let
This theorem is interesting even for
The following is a big open question:
Does the theorem still hold if we replace graph by tournament,
i.e. a complete digraph? (Yes for
The following is a little less known but still quite big.
Is there any assymetric class (e.g. digraphs) with the Hrushovski property?
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)
Suppose
 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. \(LO(\mathcal K)\) , the is a Ramsey class
Suppose
\(M(G)=X_{\mathcal K^*}\) \(\mathcal K^*\) is a Ramsey class.
We say that
If
\(\mu_A(A^*)\in[0,1]\) for each\(A^*\in\mathcal K^*\) expansion of\(A\) .\(\mu_A\) is a probability measure. 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
7315.2.2017
73.1J. Verner: Chains of Ppoints
744.1.2017
74.1D. Chodounský: Preserving ultrafilters
Whenever we add an unbounded real all non Ppoint 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
For the second part first construct a tree
 At every level all nodes either split or don't split
 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\(A=\{s:s(n)=1\}\) .  For every level
\(n\) there are\(s,t\) on the level such that\(s(n)\neq t(n)\)
Given a perfect tree
A similar argument shows that after adding a real at least one MAD family is destroyed.
A definable (Suslin) ccc forcing either contains a Cohen real or a Random real (JudahShelah) which are splitting, so no definable ccc forcing preserves ultrafilters.
After any iteration of definable forcings of length
A typical argument showing that Ppoints exist either uses some cardinal invariants
(e.g.
After adding a Silver real the filter generated by a groundmodel Ppoint
is nolonger
Assume
If
Assume CH. Then for all definable proper
\(P\) preserves\(P\) points.\(P\) does not add splitting reals and has the weak Laver property
Definability means that, either:
\(P\) is of the form\(Borel(X)\setminus \mathcal I\) for some\(\Pi^1_1\) on\(\Sigma^1_1\) \(\sigma\) ideal\(\mathcal I\) on\(X\) . 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\(X\) .
A forcing
If
The combinatorial core of Zapletal's theorem is the essence of the following proposition:
A forcing
Suppose
Now consider the set
On the other hand assume that
7525.1.2017
75.1E. Thuemmel: The Ppoint game for coideals
A Boolean algebra
An ideal
Given an ideal
Player I has a winning strategy iff either
The next proposition says that a strenghtening of nondistributivity (*) is in fact equivalent to the condition in the above theorem. It is rather easy to see and we omit the proof.
An ideal is either not distributive or not
An ideal has (*) iff there is an
We use ideas of Galving and Shelah. Proposition
75.5 immediately gives a winning strategy for Player I
in case
If
Easy.
If
An ideal
Is there an Analytic Ramsey ideal?
If
We use a trick due to S. Sierpińsky. Let
If
If the above conjecture is true then there is no analytic tall Ramsey ideal.
75.2Saeed Ghasemi: Scattered Cstar algebras
Let
An concrete Cstar algebra is a normclosed subspace of
An abstract Cstar algebra is a
Every abstract Cstar algebra is a concrete Cstar algebra.
The finitedimensional Cstar algebras are exactly the matrix algebras
over
For the rest of this talk assume
Let
For the following fix a basis
An element
The Boolean algebra
A family
MA implies that
It is not known whether
Recall that given an AD family
Note that if
Every commutative Cstar algebra is of the form
This allows us to generalize topological notions to Cstar algebras, e.g.
 A Cstar algebra is connected iff it has no nontrivial projections
 A compactification of space corresponds to the unitization of a Cstar algebra
 An algebra is "compact" if it is unital
 There is a CzechStone compactification (universal unitization, given by the multiplier algebra of A), onepointcompactification (a minimal unitization)
 etc.
To define the multiplier of
Note that whenever
Unfortunately, although the most interesting part of the talk followed, I was too confused with Cstar algebras to continue recording it.
768.1.2017
76.1S. NavarroFlores  Ramsey spaces
Recall that we write
A family
A family
Let
Every analytic ideal contains a cofinal
(w.r.t. inclusion)
Does every coanalytic coideal contain a
7715.2.2017
77.1O. Guzmán: Desctructibility of MAD families
Let
Kunen showed that there is a Cohen indestructible family under CH
(to show that
Does there always exist a Cohen indestructible MAD family?
There is a MAD family
Let
If
Define
Let
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,
Adding a real destroys
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.
It is known that
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.
Is there an ultrafilter which survives one Sacks but not two?
The usual proof of
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
real. In particular adding a dominating real kills all MAD families.
A MAD family
If a
If
Forcing with countable AD families is
It is interesting that the above generic MAD family can be destroyed without adding dominating or splitting reals.
Given an ideal
If
Let
Under suitable LC assumptions this is true for all definable ideals.
A MAD family is ShelahSteprans if
If
If
MathiasPrikry forcing with
Is it consistent that
An ideal
A MAD family
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
It is consistent that there are no ShelahSteprans MAD families.
If an ideal
We now aim to prove 77.21. So let
An ideal
An ideal is not a weak Pideal iff it is Katětov above
The first player has a winning strategy iff
If
If
This is easy for
II has a winning strategy iff it is not ShelahSteprans.
II has a winning strategy iff he has a winning tactic.
7822.2.2017
78.1Bohuslav Balcar: 19432017
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.
Given an ideal
II has a centered winning strategy iff
Assume first that
Assume, on the other hand, that
If
We apply this theorem to the ideal
78.3O. Guzmán
Given an ideal
A coideal
If
Let
An ideal
An ultrafilter
There is a MAD family
Is there a +Ramsey MAD family?
Yes, assuming either
In fact, the following is true!
There is a +Ramsey MAD family (in ZFC).
The proof breaks down into two cases, depending on whether
The cardinal
This is similar to the characterization of
Let
On the other hand we show how to construct a non +Ramsey ideal of cofinality
Let
Let
If
Assume
Now for every
A Borel subset of
However, since
If
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
80.1Š. Stejskalová: Rado conjecture
A cardinal
If GCH holds below a measurable it holds at the measurable. If it holds below a supercompact then it holds everywhere.
A tree
Rado Conjecture on
In the above the width of the tree is not limited!
Note that in ZFC there is a special Aronszajn tree on
The following theorem summarizes some facts about RC, proved by several authors including Todorcević.
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:
If there is a supercompact cardinal then GCH and
Assume
The Lévy collapse
Assume that
Assume
 There is an elementary embedding
\(j^*:V[G]\to M[H]\) with\(j\upharpoonright V=j\) \(j[G]\subseteq H\)
By the above notes and elementarity
If
By the above lemma, since
To get the second part we use Mitchell forcing
Mitchell forcing is
The rest of the proof is similar to the above using the fact that