the sense of Van Douwen. It would be desirable to get quotable - PowerPoint PPT Presentation

Preliminaries Many non-homeomorphic ultrafilters Main results Basic properties Bonus materials Overview of the results The topology of ultrafilters as subspaces of 2 ω Andrea Medini 1 David Milovich 2 1 Department of Mathematics University of Wisconsin - Madison 2 Department of Engineering, Mathematics, and Physics Texas A&M International University April 2, 2012 The topology of ultrafilters as subspaces of 2 ω Andrea Medini, David Milovich

Preliminaries Many non-homeomorphic ultrafilters Main results Basic properties Bonus materials Overview of the results All ultrafilters are non-principal and on ω . By identifying a subset of ω with an element of 2 ω in the obvious way, we can view any ultrafilter U as a subspace of 2 ω . Proposition (folklore) There are 2 c non-homeomorphic ultrafilters. Proof. Using Lavrentiev’s lemma, one sees that the homeomorphism classes have size c . So there must be 2 c of them. The above proof is a cardinality argument: it is not ‘honest’ in the sense of Van Douwen. � It would be desirable to get ‘quotable’ topological properties that distinguish ultrafilters up to homeomorphism. The topology of ultrafilters as subspaces of 2 ω Andrea Medini, David Milovich

Preliminaries Many non-homeomorphic ultrafilters Main results Basic properties Bonus materials Overview of the results Similar investigations have been carried out for filters: a delicate interplay emerged between Baire property and Lebesgue measurability. However, these matters are trivial for ultrafilters. Notice that that 2 ω = U ⊔ c [ U ] , where c : 2 ω − → 2 ω is the complement homeomorphism. (So J = c [ U ] is the dual ideal.) Proposition (folklore) Every ultrafilter U ⊆ 2 ω has the following properties. U is non-meager and non-comeager. U does not have the Baire property. U is not Lebesgue measurable. U is not analytic and not co-analytic. U is a Baire space. U is a topological group (hence a homogeneous space). The topology of ultrafilters as subspaces of 2 ω Andrea Medini, David Milovich

Preliminaries Many non-homeomorphic ultrafilters Main results Basic properties Bonus materials Overview of the results The distinguishing properties From now on, all spaces are separable and metrizable. Recall the following definitions. Definition A space X is completely Baire if every closed subspace of X is a Baire space. A space X is countable dense homogeneous if for every pair ( D , E ) of countable dense subsets of X there exists a homeomorphism h : X − → X such that h [ D ] = E . Given a space X , a subset A of X has the perfect set property if A is countable or A contains a homeomorphic copy of 2 ω . The topology of ultrafilters as subspaces of 2 ω Andrea Medini, David Milovich

Preliminaries Many non-homeomorphic ultrafilters Main results Basic properties Bonus materials Overview of the results Main results Theorem Assume MA(countable). Let P be one of the following topological properties. P = being completely Baire. j P = countable dense homogeneity. P = every closed subset has the perfect set property. Then there exist ultrafilters U , V ⊆ 2 ω such that U has property . � P and V does not have property P Question Can the assumption of MA(countable) be dropped? The topology of ultrafilters as subspaces of 2 ω Andrea Medini, David Milovich

Preliminaries The negative results Main results The positive results Bonus materials Kunen’s closed embedding trick Theorem (Kunen, private communication) Let C be a zero-dimensional space. Then there exists an ultrafilter U ⊆ 2 ω with a closed subspace homeomorphic to C. By choosing C = Q or C = a Bernstein set one obtains the following corollaries. Corollary There exists an ultrafilter V ⊆ 2 ω that is not completely Baire. Corollary There exists an ultrafilter V ⊆ 2 ω with a closed subset that does not have the perfect set property. The topology of ultrafilters as subspaces of 2 ω Andrea Medini, David Milovich

Preliminaries The negative results Main results The positive results Bonus materials Proof of Kunen’s trick Lemma (folklore) There exists a perfect set P ⊆ 2 ω such that P is an independent family: that is, every word x 1 ∩ · · · ∩ x m ∩ ω \ y 1 ∩ · · · ∩ ω \ y n is infinite, where x 1 , . . . , x m , y 1 , . . . , y n ∈ P are distinct. Let C be the space you want to embed in V as a closed subset. Since P ∼ = 2 ω , assume C ⊆ P . Now simply define G = C ∪ { ω \ x : x ∈ P \ C } . Notice that G has the finite intersection property because P is independent. Any ultrafilter V ⊇ G will intersect P exactly on C . The topology of ultrafilters as subspaces of 2 ω Andrea Medini, David Milovich

Preliminaries The negative results Main results The positive results Bonus materials An ultrafilter that is not countable dense homogeneous We will use Sierpi´ nski’s technique for killing homeomorphisms. Lemma Assume MA(countable). Fix D 1 and D 2 disjoint countable dense subsets of 2 ω such that D = D 1 ∪ D 2 is an independent family. Then there exists A ⊇ D satisfying the following conditions. A is an independent family. If G ⊇ D is a G δ subset of 2 ω and f : G − → G is a homeomorphism such that f [ D 1 ] = D 2 , then there exists x ∈ G such that { x , ω \ f ( x ) } ⊆ A . In the end, let V be any ultrafilter extending A . The topology of ultrafilters as subspaces of 2 ω Andrea Medini, David Milovich

Preliminaries The negative results Main results The positive results Bonus materials Enumerate as { f η : η ∈ c } all such homeomorphisms. We will construct an increasing sequence of independent families A ξ for ξ ∈ c . Set A 0 = D and take unions at limit stages. We will take care of f η at stage ξ = η + 1, using cov ( M ) = c . List as { w α : α ∈ κ } all the words in A η . It is easy to check that, for any fixed n ∈ ω , α ∈ κ and ε 1 , ε 2 ∈ 2, W α, n ,ε 1 ,ε 2 = { x ∈ G η : | w α ∩ x ε 1 ∩ f η ( x ) ε 2 | ≥ n } is open dense in G η , so comeager in 2 ω . So pick x in the intersection of every W α, n ,ε 1 ,ε 2 . The topology of ultrafilters as subspaces of 2 ω Andrea Medini, David Milovich

Preliminaries The negative results Main results The positive results Bonus materials An aside: the separation property The following property, among Baire spaces, is a weakening of countable dense homogeneity. Definition (Van Mill, 2009) A space X has the separation property if, given any A , B ⊆ X such that A is meager and B is countable, there exists an → X such that h [ A ] ∩ B = ∅ . homeomorphism h : X − Theorem (Van Mill, 2009) Every Baire topological group has the separation property. Corollary Every ultrafilter U ⊆ 2 ω has the separation property. The topology of ultrafilters as subspaces of 2 ω Andrea Medini, David Milovich

Preliminaries The negative results Main results The positive results Bonus materials A countable dense homogeneous ultrafilter Any ultrafilter U is homeomorphic to its dual maximal ideal J . So, for notational convenience, we will construct an increasing sequence of ideals I ξ , for ξ ∈ c . In the end, let J be any maximal ideal extending � ξ ∈ c I ξ . The idea is to use the following lemma. Lemma Let f : 2 ω − → 2 ω be a homeomorphism. Fix a maximal ideal J ⊆ 2 ω and a countable dense subset D of J . Then f restricts to a homeomorphism of J iff cl ( { d + f ( d ) : d ∈ D } ) ⊆ J . Enumerate as { ( D η , E η ) : η ∈ c } all pairs of countable dense subsets of 2 ω . At stage ξ = η + 1, make sure that either ω \ x ∈ I ξ for some x ∈ D η ∪ E η , or there exists an homeomorphism f : 2 ω − → 2 ω and x ∈ I ξ such that f [ D η ] = E η and { d + f ( d ) : d ∈ D η } ⊆ x ↓ . The topology of ultrafilters as subspaces of 2 ω Andrea Medini, David Milovich

Preliminaries The negative results Main results The positive results Bonus materials To construct f : 2 ω − → 2 ω and x , use MA(countable) on the poset P consisting of all triples p = ( s , g , π ) = ( s p , g p , π p ) such that, for some n = n p ∈ ω , the following conditions hold. s : n − → 2. g is a bijection between a finite subset of D and a finite subset of E . π is a permutation of n 2. ( t + π ( t ))( i ) = 1 implies s ( i ) = 1 for every t ∈ n 2 and i ∈ n . π ( d ↾ n ) = g ( d ) ↾ n for every d ∈ dom ( g ) . Order P by declaring q ≤ p if the following conditions hold. s q ⊇ s p . g q ⊇ g p . π q ( t ) ↾ n p = π p ( t ↾ n p ) for all t ∈ n q 2. The topology of ultrafilters as subspaces of 2 ω Andrea Medini, David Milovich

Preliminaries The negative results Main results The positive results Bonus materials An ultrafilter U such that A ∩ U has the perfect set property whenever A is analytic Recall that a play of the strong Choquet game on a topological space ( X , T ) is of the form I ( q 0 , U 0 ) ( q 1 , U 1 ) · · · II V 0 V 1 · · · , where U n , V n ∈ T are such that q n ∈ V n ⊆ U n and U n + 1 ⊆ V n for every n ∈ ω . Player II wins if � n ∈ ω U n � = ∅ . The topological space ( X , T ) is strong Choquet if II has a winning strategy in the above game. The topology of ultrafilters as subspaces of 2 ω Andrea Medini, David Milovich