The topology of ultrafilters as subspaces of 2 ω
aa r X i v : . [ m a t h . GN ] N ov THE TOPOLOGY OF ULTRAFILTERS AS SUBSPACES OF ω ANDREA MEDINI AND DAVID MILOVICH
Abstract.
Using the property of being completely Baire, countable densehomogeneity and the perfect set property we will be able, under Martin’s Ax-iom for countable posets, to distinguish non-principal ultrafilters on ω up tohomeomorphism. Here, we identify ultrafilters with subpaces of 2 ω in the ob-vious way. Using the same methods, still under Martin’s Axiom for countableposets, we will construct a non-principal ultrafilter U ⊆ ω such that U ω iscountable dense homogeneous. This consistently answers a question of Hruˇs´akand Zamora Avil´es. Finally, we will give some partial results about the rela-tion of such topological properties with the combinatorial property of being aP-point. By identifying a subset of ω with an element of the Cantor set 2 ω in the obviousway (which we will freely do throughout the paper), it is possible to study thetopological properties of any X ⊆ P ( ω ). We will focus on the case X = U , where U is an ultrafilter on ω . The case X = F , where F is simply a filter on ω , has beenstudied extensively (see Chapter 4 in [3]). From now on, all filters and ideals areimplicitly assumed to be on ω .First, we will observe that there are many (actually, as many as possible) non-homeomorphic ultrafilters. However, the proof is based on a cardinality argument,hence it is not ‘honest’ in the sense of Van Douwen: it would be desirable to find‘quotable’ topological properties that distinguish ultrafilters up to homeomorphism.This is consistently achieved in Section 3 using the property of being completelyBaire (see Corollary 9 and Theorem 11), in Section 4 using countable dense homo-geneity (see Theorem 15 and Theorem 21) and in Section 6 using the perfect setproperty (see Theorem 28 and Corollary 31).In Section 5, we will adapt the proof of Theorem 21 to obtain the countabledense homogeneity of the ω -power, consistently answering a question of Hruˇs´akand Zamora Avil´es from [10] (see Corollary 26).In Section 7, using a modest large cardinal assumption, we will obtain a stronggeneralization of the main result of Section 6 (see Theorem 35).Finally, in Section 8, we will investigate the relationship between the property ofbeing a P-point and the above topological properties; many questions on this frontremain open. Proposition 1.
Let U , V ⊆ ω be non-principal ultrafilters. Define U ∼ = V if thetopological spaces U and V are homeomorphic. Then the equivalence classes of ∼ = have size c .Proof. To show that each equivalence class has size at least c , simply use home-omorphisms of 2 ω induced by permutations of ω and an almost disjoint family ofsubsets of ω of size c (see, for example, Lemma 9.21 in [12]). Date : November 27, 2011.
By Lavrentiev’s lemma (see Theorem 3.9 in [13]), if g : U −→ V is a homeo-morphism, then there exists a homeomorphism f : G −→ H that extends g , where G and H are G δ subsets of 2 ω . Since there are only c such homeomorphisms, itfollows that an equivalence class of ∼ = has size at most c . (cid:3) Corollary 2.
There are c pairwise non-homeomorphic non-principal ultrafilters. Notation and Terminology
Our main reference for descriptive set theory is [13]. For other set-theoreticnotions, see [3] or [12]. For notions that are related to large cardinals, see [14]. Forall undefined topological notions, see [6].By space we mean separable metrizable topological space, with a unique excep-tion in Section 6. For every s ∈ <ω
2, we will denote by [ s ] the basic clopen set { x ∈ ω : s ⊆ x } . Given a tree T ⊆ <ω
2, we will denote by [ T ] the set of branchesof T , that is [ T ] = { x ∈ ω : x ↾ n ∈ T for all n ∈ ω } .Given a function f and A ⊆ dom( f ), we will denote by f [ A ] the image of A under f , that is f [ A ] = { f ( x ) : x ∈ A } .A space X is homogeneous if whenever x, y ∈ X there exists a homeomorphism f : X −→ X such that f ( x ) = y .Define the homeomorphism c : 2 ω −→ ω by setting c ( x )( n ) = 1 − x ( n ) for every x ∈ ω and n ∈ ω . Using c , one sees that every ultrafilter U ⊆ ω is homeomorphicto its dual maximal ideal J = 2 ω \ U = c [ U ].A perfect set in a space X is a non-empty closed subset P of X with no isolatedpoints. Recall that P is a perfect set in 2 ω if and only if it is homeomorphic to 2 ω .A Bernstein set is a subset B of X = 2 ω such that B and X \ B both intersectevery perfect set in X . Given such a set B , since 2 ω is homeomorphic to 2 ω × ω ,one actually has | P ∩ B | = c and | P ∩ ( X \ B ) | = c for every perfect set P in X .For every x ⊆ ω , define x = ω \ x and x = x . Given a family A ⊆ P ( ω ), a word in A is an intersection of the form \ x ∈ τ x w ( x ) for some τ ∈ [ A ] <ω and w : τ −→
2. Recall that A is an independent family ifevery word in A is infinite.A family F ⊆ P ( ω ) has the finite intersection property if T σ is infinite for all σ ∈ [ F ] <ω . Given such a family, we will denote by hFi the filter generated by F .Let Cof be the collection of all cofinite subsets of ω . Recall that an ultrafilter U isnon-principal if and only if Cof ⊆ U . In particular, every non-principal ultrafilteris dense in 2 ω . For any fixed x ∈ ω , define x ↑ = { y ∈ ω : x ⊆ y } .Whenever x, y ∈ P ( ω ), define x ⊆ ∗ y if x \ y is finite. Given C ⊆ P ( ω ), a pseudointersection of C is a subset x of ω such that x ⊆ ∗ y for all y ∈ C . Givena cardinal κ , a non-principal ultrafilter U is a P κ -point if every C ∈ [ U ] <κ has apseudointersection in U . A P -point is simply a P ω -point.A family I ⊆ P ( ω ) has the finite union property if S σ is coinfinite for all σ ∈ [ I ] <ω . Given such a family, we will denote by hIi the ideal generated by I .Let Fin be the collection of all finite subsets of ω . For any fixed x ∈ ω , define x ↓ = { y ∈ ω : y ⊆ x } .Given C ⊆ P ( ω ), a pseudounion of C is a subset x of ω such that y ⊆ ∗ x for all y ∈ C . A maximal ideal J is a P -ideal if c [ J ] is a P-point. HE TOPOLOGY OF ULTRAFILTERS AS SUBSPACES OF 2 ω Basic properties
In this section, we will notice that some topological properties are shared by allnon-principal ultrafilters. It is easy to realize that every principal ultrafilter
U ⊆ ω is homeomorphic to 2 ω .Since any maximal ideal J (actually, any ideal) is a topological subgroup of 2 ω under the operation of symmetric difference (or equivalently, sum modulo 2), everyultrafilter U = c [ J ] is also a topological group. In particular, every ultrafilter U isa homogeneous topological space.The following proposition is Lemma 3.1 in [8]. Proposition 3 (Fitzpatrick, Zhou) . Let X be a homogeneous topological space.Then X is a Baire space if and only if X is not meager in itself.Proof. One implication is trivial. Now assume that X is not a Baire space. Since X is homogeneous, it follows easily that B = { U : U is a non-empty meager open set in X } is a base for X . So X = S B is the union of a collection of meager open sets. Hence X is meager by Banach’s category theorem (see Theorem 16.1 in [20]).For the convenience of the reader, we sketch the proof in our particular case.Fix a maximal C ⊆ B consisting of pairwise disjoint sets. Observe that X \ S C isclosed nowhere dense. For every U ∈ C , fix nowhere dense sets N n ( U ) such that U = S n ∈ ω N n ( U ). It is easy to check that S U ∈C N n ( U ) is nowhere dense in X forevery n ∈ ω . (cid:3) Given any ultrafilter
U ⊆ ω , notice that c is a homeomorphism of 2 ω such that2 ω is the disjoint union of U and c [ U ]. In particular, U must be non-meager andnon-comeager in 2 ω by Baire’s category theorem. Actually, it follows easily fromthe 0-1 Law that no non-principal ultrafilter U can have the property of Baire (seeTheorem 8.47 in [13]). In particular, no non-principal ultrafilter U can be analytic(see Theorem 21.6 in [13]) or co-analytic. Corollary 4.
Let
U ⊆ ω be an ultrafilter. Then U is a Baire space.Proof. If U were meager in itself, then it would be meager in 2 ω , which is a contra-diction. (cid:3) On the other hand, by Theorem 8.17 in [13], no non-principal ultrafilter can bea Choquet space (see Section 8.C in [13]).3.
Completely Baire ultrafilters
Definition 5.
A space X is completely Baire if every closed subspace of X is aBaire space.For example, every Polish space is completely Baire. For co-analytic spaces, theconverse is also true (see Corollary 21.21 in [13]).In the proof of Theorem 11, we will need the following characterization (seeCorollary 1.9.13 in [16]). Observe that one implication is trivial. Lemma 6 (Hurewicz) . A space is completely Baire if and only if it does not containany closed homeomorphic copy of Q . ANDREA MEDINI AND DAVID MILOVICH
The following (well-known) lemma is the first step in constructing an ultrafilterthat is not completely Baire.
Lemma 7.
There exists a perfect subset P of ω such that P is an independentfamily.Proof. We will give three proofs. The first proof simply shows that the classicalconstruction of an independent family of size c (see for example Lemma 7.7 in [12])actually gives a perfect independent family. Define I = { ( ℓ, F ) : ℓ ∈ ω, F ⊆ ℓ } . Since I is a countably infinite set, we can identify 2 I and 2 ω . The desired indepen-dent family will be a collection of subsets of I . Consider the function f : 2 ω −→ I defined by f ( x ) = { ( ℓ, F ) : x ↾ ℓ ∈ F } . It is easy to check that f is a continuous injection, hence a homeomorphic embed-ding by compactness. It follows that P = ran( f ) is a perfect set. To check that P is an independent family, fix τ ∈ [ P ] <ω and w : τ −→
2. Suppose that τ = f [ σ ],where σ = { x , . . . x k } and x , . . . , x k are distinct. Choose ℓ large enough so that x ↾ ℓ, . . . , x k ↾ ℓ are distinct. It follows that( ℓ ′ , { x ↾ ℓ ′ : x ∈ σ and w ( f ( x )) = 1 } ) ∈ \ y ∈ τ y w ( y ) for every ℓ ′ ≥ ℓ , which concludes the proof.The second proof is also combinatorial. We will inductively construct k n ∈ ω and a finite tree T n ⊆ <ω n ∈ ω so that the following conditions aresatisfied.(1) k m < k n whenever m < n < ω .(2) T m ⊆ T n whenever m ≤ n < ω .(3) All maximal elements of T n have length k n . We will use the notation M n = { t ∈ T n : dom( t ) = k n } .(4) For every t ∈ T n there exist two distinct elements of T n +1 whose restrictionto k n is t .(5) Given any v : M n −→
2, there exists i ∈ k n +1 \ k n such that t ( i ) = v ( t ↾ k n )for every t ∈ M n +1 .In the end, set T = S n<ω T n and P = [ T ]. Condition (4) guarantees that P isperfect. Next, we will verify that condition (5) guarantees that P is an independentfamily. Fix τ ∈ [ P ] <ω and w : τ −→
2. For all sufficiently large n ∈ ω , some v ∈ M n v ( x ↾ k n ) = w ( x ) for all x ∈ τ . By condition (5), there exists i ∈ k n +1 \ k n such that x ( i ) = ( x ↾ k n +1 )( i ) = v ( x ↾ k n ) = w ( x )for all x ∈ τ .Start with k = 0 and T = { ∅ } . Given k n and T n , define k n +1 = k n + 2 | M n | + 1.Fix an enumeration { v j : j ∈ | M n | } of all functions v : M n −→
2. Let T n +1 consist of all initial segments of functions t : k n +1 −→ t ↾ k n ∈ M n and t ( k n + j ) = v j ( t ↾ k n ) for all j < | M n | . Then, condition (5) is clearly satisfied.Since there is no restriction on t ( k n + 2 | M n | ), condition (4) is also satisfied. HE TOPOLOGY OF ULTRAFILTERS AS SUBSPACES OF 2 ω The third proof is topological. Fix an enumeration { ( n i , w i ) : i ∈ ω } of all pairs( n, w ) such that n ∈ ω and w : n −→
2. Define R i = (cid:26) x ∈ (2 ω ) n i : \ j ∈ n i x w i ( j ) j is infinite (cid:27) for every i ∈ ω and observe that each R i is comeager. By Exercise 8.8 and Theorem19.1 in [13], there exists a comeager subset of the Vietoris hyperspace K (2 ω ) con-sisting of perfect sets P ⊆ ω such that { x ∈ P n i : x j = x k whenever j = k } ⊆ R i for every i ∈ ω . It is trivial to check that any such P is an independent family. (cid:3) We remark that, in some sense, the last two proofs that we have given of theabove lemma are the same. The Vietoris hyperspace K (2 ω ) is naturally homeo-morphic to the space X of pruned subtrees of <ω { T ∈ X : T ∩
2. Moreover, the set { T ∈ X : [ T ] is an independent family } is comeager in X because the combinato-rial proof’s rule for constructing T n +1 from T n only needs to be followed infinitelyoften.The authors propose to call the following Kunen’s closed embedding trick . Theorem 8 (Kunen) . Fix a zero-dimensional space C . There exists a non-principalultrafilter U ⊆ ω that contains a homeomorphic copy of C as a closed subset.Proof. Fix P as in Lemma 7. Since P is homeomorphic to 2 ω , we can assume that C is a subspace of P . Observe that the family G = C ∪ { ω \ x : x ∈ P \ C } has the finite intersection property because P is an independent family. Any non-principal ultrafilter U ⊇ G will contain C as a closed subset. (cid:3) Corollary 9.
There exists an ultrafilter
U ⊆ ω that is not completely Baire.Proof. Simply choose C = Q . (cid:3) Since 2 ω is homeomorphic to 2 ω × ω , one can easily obtain the following strenght-ening of Theorem 8. Observe that, since any space has at most c closed subsets,the result cannot be improved. Theorem 10.
Fix a collection C of zero-dimensional spaces such that |C| ≤ c .There exists a non-principal ultrafilter U ⊆ ω that contains a homeomorphic copyof C as a closed subset for every C ∈ C . The next theorem, together with Corollary 9, shows that under MA(countable)the property of being completely Baire is enough to distinguish ultrafilters up tohomeomorphism.
Theorem 11.
Assume that
MA(countable) holds. Then there exists a non-principalultrafilter
U ⊆ ω that is completely Baire.Proof. Enumerate as { Q η : η ∈ c } all subsets of 2 ω that are homeomorphic to Q .By Lemma 6, it will be sufficient to construct a non-principal ultrafilter U suchthat no Q η is a closed subset of U .We will construct F ξ for every ξ ∈ c by transfinite recursion. In the end, let U be any ultrafilter extending S ξ ∈ c F ξ . By induction, we will make sure that thefollowing requirements are satisfied. ANDREA MEDINI AND DAVID MILOVICH (1) F µ ⊆ F η whenever µ ≤ η < c .(2) F ξ has the finite intersection property for every ξ ∈ c .(3) |F ξ | < c for every ξ ∈ c .(4) The potential closed copy of the rationals Q η is dealt with at stage ξ = η +1:that is, either ω \ x ∈ F ξ for some x ∈ Q η or there exists x ∈ F ξ such that x ∈ cl( Q η ) \ Q η .Start by letting F = Cof. Take unions at limit stages. At a successor stage ξ = η + 1, assume that F η is given. First assume that there exists x ∈ Q η suchthat F η ∪ { ω \ x } has the finite intersection property. In this case, simply set F ξ = F η ∪ { ω \ x } .Now assume that F η ∪ { ω \ x } does not have the finite intersection property forany x ∈ Q η . It is easy to check that this implies Q η ⊆ hF η i . Apply Lemma 12with F = F η and Q = Q η to get x ∈ cl( Q η ) \ Q η such that F η ∪ { x } has the finiteintersection property. Finally, set F ξ = F η ∪ { x } . (cid:3) Lemma 12.
Assume that
MA(countable) holds. Let F be a collection of subsetsof ω with the finite intersection property such that |F| < c . Let Q be a non-emptysubset of ω with no isolated points such that Q ⊆ hFi and | Q | < c . Then thereexists x ∈ cl( Q ) \ Q such that F ∪ { x } has the finite intersection property.Proof. Consider the countable poset P = { s ∈ <ω q ∈ Q and n ∈ ω such that s = q ↾ n } , with the natural order given by reverse inclusion.For every σ = { x , . . . , x k } ∈ [ F ] <ω and ℓ ∈ ω , define D σ,ℓ = { s ∈ P : there exists i ∈ dom( s ) \ ℓ such that s ( i ) = x ( i ) = · · · = x k ( i ) = 1 } . Using the fact that Q ⊆ hFi , it is easy to see that each D σ,ℓ is dense in P .For every q ∈ Q , define D q = { s ∈ P : there exists i ∈ dom( s ) such that s ( i ) = q ( i ) } . Since Q has no isolated points, each D q is dense in P .Since |F| < c and | Q | < c , the collection of dense sets D = { D σ,ℓ : σ ∈ [ F ] <ω , ℓ ∈ ω } ∪ { D q : q ∈ Q } has also size less than c . Therefore, by MA(countable), there exists a D -genericfilter G ⊆ P . Let x = S G ∈ ω . The dense sets of the form D σ,ℓ ensure that F ∪ { x } has the finite intersection property. The definition of P guarantees that x ∈ cl( Q ). Finally, the dense sets of the form D q guarantee that x / ∈ Q . (cid:3) Question 1.
Can the assumption that MA(countable) holds be dropped in Theo-rem 11? 4.
Countable dense homogeneity
Definition 13.
A space X is countable dense homogeneous if for every pair ( D, E )of countable dense subsets of X there exists a homeomorphism f : X −→ X suchthat f [ D ] = E . HE TOPOLOGY OF ULTRAFILTERS AS SUBSPACES OF 2 ω We will start this section by consistently constructing an ultrafilter that is notcountable dense homogeneous. We will use Sierpi´nski’s technique for killing home-omorphisms (see [19] or Appendix 2 of [5] for a nice introduction). The key lemmais the following.
Lemma 14.
Assume that
MA(countable) holds. Let D be a countable independentfamily that is dense in ω . Fix D and D disjoint countable dense subsets of D .Then there exists A ⊆ ω satisfying the following requirements. • A is an independent family. • D ⊆ A . • If G ⊇ D is a G δ subset of ω and f : G −→ G is a homeomorphism suchthat f [ D ] = D , then there exists x ∈ G such that { x, ω \ f ( x ) } ⊆ A .Proof. Enumerate as { f η : η ∈ c } all homeomorphisms f η : G η −→ G η such that f η [ D ] = D , where G η ⊇ D is a G δ subset of 2 ω .We will construct A ξ for every ξ ∈ c by transfinite recursion. In the end, set A = S ξ ∈ c A ξ . By induction, we will make sure that the following requirements aresatisfied.(1) A µ ⊆ A η whenever µ ≤ η < c .(2) A ξ is an independent family for every ξ ∈ c .(3) |A ξ | < c for every ξ ∈ c .(4) The homeomorphism f η is dealt with at stage ξ = η + 1: that is, thereexists x ∈ G η such that { x, ω \ f η ( x ) } ⊆ A ξ .Start by letting A = D . Take unions at limit stages. At a successor stage ξ = η + 1, assume that A η is given.List as { w α : α ∈ κ } all the words in A η , where κ = |A η | < c by (3). It is easyto check that, for any fixed n ∈ ω , α ∈ κ and ε , ε ∈
2, the set W α,n,ε ,ε = { x ∈ G η : | w α ∩ x ε ∩ f η ( x ) ε | ≥ n } is open in G η . It is also dense, because D \ ( F ∪ f − η [ F ]) ⊆ W α,n,ε ,ε , where F consists of the finitely many elements of A η that appear in w α . Therefore,each W α,n,ε ,ε is comeager in 2 ω . Recall that MA(countable) is equivalent tocov( M ) = c (see Theorem 7.13 in [4] or Theorem 2.4.5 in [3]). It follows that theintersection W = \ { W α,n,ε ,ε : n ∈ ω , α ∈ κ and ε , ε ∈ } is non-empty. Now simply pick x ∈ W and set A ξ = A η ∪ { x, ω \ f η ( x ) } . (cid:3) Theorem 15.
Assume that
MA(countable) holds. Then there exists an ultrafilter
U ⊆ ω that is not countable dense homogeneous.Proof. Fix D , D and A as in Lemma 14. Let U ⊇ A be any ultrafilter. Assume,in order to get a contradiction, that U is countable dense homogeneous. Let g : U −→ U be a homeomorphism such that g [ D ] = D . By Lavrentiev’s lemma, itis possible to extend g to a homeomorphism f : G −→ G , where G is a G δ subsetof 2 ω (see Exercise 3.10 in [13]). By Lemma 14, there exists x ∈ G such that { x, ω \ f ( x ) } ⊆ A ⊆ U , contradicting the fact that f ( x ) = g ( x ) ∈ U . (cid:3) Question 2.
Can the assumption that MA(countable) holds be dropped in Theo-rem 15?
ANDREA MEDINI AND DAVID MILOVICH
When first trying to prove Theorem 15, we attempted to construct a non-principal ultrafilter U such that no homeomorphism g : U −→ U would be suchthat g [Cof] ∩ Cof = ∅ . This is easily seen to be impossible by choosing g to be themultiplication by any coinfinite x ∈ U . Actually, something much stronger holdsby the following result of Van Mill (see Proposition 3.4 in [17]). Definition 16 (Van Mill) . A space X has the separation property if for every count-able subset A of X and every meager subset B of X there exists a homeomorphism f : X −→ X such that f [ A ] ∩ B = ∅ . Proposition 17 (Van Mill) . Let G be a Baire topological group acting on space X that is not meager in itself. Then, for all subsets A and B of X with A countableand B meager, the set of elements g ∈ G such that gA ∩ B = ∅ is dense in G . Corollary 18.
Every Baire topological group has the separation property.
Corollary 19.
Every ultrafilter
U ⊆ ω has the separation property. It is easy to see that, for Baire spaces, being countable dense homogeneous isstronger than having the separation property. On the other hand, the product of2 ω and the one-dimensional sphere S is a compact topological group that has theseparation property but is not countable dense homogeneous (see Corollary 3.6 andRemark 3.7 in [17]). Theorem 15 consistently gives a zero-dimensional topologicalgroup with the same feature. Notice that such an example cannot be compact (oreven Polish) by the following paragraph.Recall that a space X is strongly locally homogeneous if it admits an open base B such that whenever U ∈ B and x, y ∈ U there exists a homeomorphism f : X −→ X such that f ( x ) = y and f ↾ X \ U is the identity. For example, any homogeneouszero-dimensional space is strongly locally homogeneous. For Polish spaces, stronglocal homogeneity implies countable dense homogeneity (see Theorem 5.2 in [1]).In [18], Van Mill constructed a homogeneous Baire space that is strongly locallyhomogeneous but not countable dense homogeneous. Actually, his example doesnot even have the separation property (see Theorem 3.5 in [18]), so it cannot be atopological group by Corollary 18. In this sense, our example from Theorem 15 isbetter than his. On the other hand, his example is constructed in ZFC, while oursneeds MA(countable). Furthermore, his example can be easily modified to haveany given dimension (see Remark 4.1 in [18]).Next, we will construct (still under MA(countable)) a non-principal ultrafil-ter that is countable dense homogeneous. In [2], Baldwin and Beaudoin usedMA(countable) to construct a homogeneous Bernstein subset of 2 ω that is count-able dense homogeneous. Both examples give a consistent answer to Question 389in [9], which asks whether there exists a countable dense homogeneous space that isnot completely metrizable. In [7], using metamathematical methods, Farah, Hruˇs´akand Mart´ınez Ranero showed that the answer to such question is ‘yes’ in ZFC.The following lemma will be one of the key ingredients. The other key ingredientis the poset used in the proof of Lemma 22, which was inspired by the poset usedin the proof of Lemma 3.1 in [2]. Lemma 20.
Let f : 2 ω −→ ω be a homeomorphism. Fix a non-principal max-imal ideal J ⊆ ω and a countable dense subset D of J . Then f restricts to ahomeomorphism of J if and only if cl( { d + f ( d ) : d ∈ D } ) ⊆ J . HE TOPOLOGY OF ULTRAFILTERS AS SUBSPACES OF 2 ω Proof.
Assume that f restricts to a homeomorphism of J . It is easy to check thatthe function g : 2 ω −→ ω defined by g ( x ) = x + f ( x ) has range contained in J .Since g is continuous, its range must be compact, hence closed in 2 ω .Now assume that cl( { d + f ( d ) : d ∈ D } ) ⊆ J . Let x ∈ ω . Fix d n ∈ D for n ∈ ω so that lim n →∞ d n = x . By continuity, x + f ( x ) = lim n →∞ ( d n + f ( d n )) ∈ J . The proof is concluded by observing that if a, b ∈ ω are such that a + b ∈ J , theneither { a, b } ⊆ J or { a, b } ⊆ ω \ J . (cid:3) Theorem 21.
Assume that
MA(countable) holds. Then there exists a non-principalultrafilter
U ⊆ ω that is countable dense homogeneous.Proof. For notational convenience, we will construct a maximal ideal
J ⊆ ω containing all finite sets that is countable dense homogeneous. Enumerate as { ( D η , E η ) : η ∈ c } all pairs of countable dense subsets of 2 ω .We will construct I ξ for every ξ ∈ c by transfinite recursion. In the end, let J be any maximal ideal extending S ξ ∈ c I ξ . By induction, we will make sure that thefollowing requirements are satisfied.(1) I µ ⊆ I η whenever µ ≤ η < c .(2) I ξ has the finite union property for every ξ ∈ c .(3) |I ξ | < c for every ξ ∈ c .(4) The pair ( D η , E η ) is dealt with at stage ξ = η + 1: that is, either ω \ x ∈ I ξ for some x ∈ D η ∪ E η or there exists x ∈ I ξ and a homeomorphism f η :2 ω −→ ω such that f η [ D η ] = E η and { d + f η ( d ) : d ∈ D η } ⊆ x ↓ .Observe that, by Lemma 20, the second part of condition (4) guarantees that anymaximal ideal J extending I ξ will be such that f η : 2 ω −→ ω restricts to ahomeomorphism of J .Start by letting I = Fin. Take unions at limit stages. At a successor stage ξ = η + 1, assume that I η is given. First assume that there exists x ∈ D η ∪ E η such that I η ∪ { ω \ x } has the finite union property. In this case, we can just set I ξ = I η ∪ { ω \ x } .Now assume that I η ∪ { ω \ x } does not have the finite union property for any x ∈ D η ∪ E η . It is easy to check that this implies D η ∪ E η ⊆ hI η i . Let x and f be given by applying Lemma 22 with I = I η , D = D η and E = E η . Finally, set I ξ = I η ∪ { x } and f η = f . (cid:3) Lemma 22.
Assume that
MA(countable) holds. Let
I ⊆ ω be a collection ofsubsets of ω with the finite union property and assume that |I| < c . Fix twocountable dense subsets D and E of ω such that D ∪ E ⊆ hIi . Then there existsa homeomorphism f : 2 ω −→ ω and x ∈ ω such that f [ D ] = E , I ∪ { x } still hasthe finite union property and { d + f ( d ) : d ∈ D } ⊆ x ↓ .Proof. Consider the countable poset P consisting of all triples of the form p =( s, g, π ) = ( s p , g p , π p ) such that, for some n = n p ∈ ω , the following requirementsare satisfied. • s : n −→ • g is a bijection between a finite subset of D and a finite subset of E . • π is a permutation of n Furthermore, we require the following compatibility conditions to be satisfied. Con-dition (1) will actually ensure that { d + f ( d ) : d ∈ ω } ⊆ x ↓ . Notice that this isequivalent to ( d + f ( d ))( i ) ≤ x ( i ) for all d ∈ ω and i ∈ ω .(1) ( t + π ( t ))( i ) = 1 implies s ( i ) = 1 for every t ∈ n i ∈ n .(2) π ( d ↾ n ) = g ( d ) ↾ n for every d ∈ dom( g ).Order P by declaring q ≤ p if the following conditions are satisfied. • s q ⊇ s p . • g q ⊇ g p . • π q ( t ) ↾ n p = π p ( t ↾ n p ) for all t ∈ n q d ∈ D , define D dom d = { p ∈ P : d ∈ dom( g p ) } . Given p ∈ P and d ∈ D \ dom( g p ), one can simply choose e ∈ E \ ran( g p ) such that e ↾ n p = π p ( d ↾ n p ). This choice will make sure that q = ( s p , g p ∪ { ( d, e ) } , π p ) ∈ P .Furthermore it is clear that q ≤ p . So each D dom d is dense in P .For each e ∈ E , define D ran e = { p ∈ P : e ∈ ran( g p ) } . As above, one can easily show that each D ran e is dense in P .For every σ = { x , . . . , x k } ∈ [ I ] <ω and ℓ ∈ ω , define D σ,ℓ = { p ∈ P : there exists i ∈ n p \ ℓ such that s p ( i ) = x ( i ) = · · · = x k ( i ) = 0 } . Next, we will prove that each D σ,ℓ is dense in P . So fix σ and ℓ as above. Let p = ( s, g, π ) ∈ P with n p = n . Find n ′ ≥ ℓ, n such that the following conditionshold. • All d ↾ n ′ for d ∈ dom( g ) are distinct. • All e ↾ n ′ for e ∈ ran( g ) are distinct. • x ( n ′ ) = · · · = x k ( n ′ ) = d ( n ′ ) = e ( n ′ ) = 0 for all d ∈ dom( g ), e ∈ ran( g ).This is possible because I has the finite union property and σ ∪ dom( g ) ∪ ran( g ) ⊆ hIi . We can choose a permutation π ′ of n ′ π ′ ( d ↾ n ′ ) = g ( d ) ↾ n ′ for every d ∈ dom( g ) and π ′ ( t ) ↾ n = π ( t ↾ n ) for all t ∈ n ′
2. Extend s to s ′ : n ′ −→ s ′ ( i ) = 1 for every i ∈ [ n, n ′ ). It is clear that p ′ = ( s ′ , g, π ′ ) ∈ P and p ′ ≤ p .Now let π ′′ be the permutation of n ′ +1 π ′′ ( t ) = π ′ ( t ↾ n ′ ) ⌢ t ( n ′ )for all t ∈ n ′ +1
2. Extend s ′ to s ′′ : n ′ + 1 −→ s ′′ ( n ′ ) = 0. It is easy tocheck that p ′′ = ( s ′′ , g, π ′′ ) ∈ D σ,ℓ and p ′′ ≤ p ′ .Since |I| < c , the collection of dense sets D = { D σ,ℓ : σ ∈ [ I ] <ω , ℓ ∈ ω } ∪ { D dom d : d ∈ D } ∪ { D ran e : e ∈ E } has also size less than c . Therefore, by MA(countable), there exists a D -genericfilter G ⊆ P . Define x = S { s p : p ∈ G } . To define f ( y )( i ), for a given y ∈ ω and i ∈ ω , choose any p ∈ G such that i ∈ n p and set f ( y )( i ) = π p ( y ↾ n p )( i ). (cid:3) Question 3.
Can the assumption that MA(countable) holds be dropped in Theo-rem 21?
HE TOPOLOGY OF ULTRAFILTERS AS SUBSPACES OF 2 ω By Theorem 2.3 in [10], every analytic countable dense homogeneous space mustbe completely Baire. So the following question seems natural. See also Theorem2.6 in [10].
Question 4.
Is a countable dense homogeneous ultrafilter
U ⊆ ω necessarilycompletely Baire?5. A question of Hruˇs´ak and Zamora Avil´es
The main result of [10] states that, given a Borel subset X of 2 ω , the followingstatements are equivalent. • X ω is countable dense homogeneous. • X is a G δ .Question 3.2 in the same paper asks whether there exists a non- G δ subset X of2 ω such that X ω is countable dense homogeneous. By a rather straightforwardmodification of the proof of Theorem 21, we will give a consistent answer to suchquestion (see Corollary 26).Our example is also relevant to the second half of Question 387 in [9], which asksto characterize the zero-dimensional spaces X such that X ω is countable densehomogeneous.Observe that, given any ideal I ⊆ ω , the infinite product I ω inherits the struc-ture of topological group using coordinate-wise addition. The following lemma isproved exactly like the corresponding half of Lemma 20. Lemma 23.
Let f : (2 ω ) ω −→ (2 ω ) ω be a homeomorphism. Fix a non-principalmaximal ideal J ⊆ ω and a countable dense subset D of J ω . If cl( { d + f ( d ) : d ∈ D } ) ⊆ J ω then f restricts to a homeomorphism of J ω . Theorem 24.
Assume that
MA(countable) holds. Then there exists a non-principalultrafilter
U ⊆ ω such that U ω is countable dense homogeneous.Proof. For notational convenience, we will construct a maximal ideal
J ⊆ ω con-taining all finite sets such that J ω is countable dense homogeneous. Enumerate as { ( D η , E η ) : η ∈ c } all pairs of countable dense subsets of (2 ω ) ω .We will construct I ξ for every ξ ∈ c by transfinite recursion. In the end, let J be any maximal ideal extending S ξ ∈ c I ξ . By induction, we will make sure thatthe following requirements are satisfied. Let P η = S i ∈ ω π i [ D η ∪ E η ], where π i :(2 ω ) ω −→ ω is the natural projection.(1) I µ ⊆ I η whenever µ ≤ η < c .(2) I ξ has the finite union property for every ξ ∈ c .(3) |I ξ | < c for every ξ ∈ c .(4) The pair ( D η , E η ) is dealt with at stage ξ = η +1: that is, either ω \ x ∈ I ξ forsome x ∈ P η or there exists x i ∈ I ξ for every i ∈ ω and a homeomorphism f η : (2 ω ) ω −→ (2 ω ) ω such that f η [ D η ] = E η and { d + f η ( d ) : d ∈ D η } ⊆ Q i ∈ ω ( x i ↓ ).Observe that, by Lemma 23, the second part of condition (4) guarantees that anymaximal ideal J extending I ξ will be such that f η : (2 ω ) ω −→ (2 ω ) ω restricts to ahomeomorphism of J ω .Start by letting I = Fin. Take unions at limit stages. At a successor stage ξ = η + 1, assume that I η is given. First assume that there exists x ∈ P η such that I η ∪ { ω \ x } has the finite union property. In this case, we can just set I ξ = I η ∪ { ω \ x } .Now assume that I η ∪ { ω \ x } does not have the finite intersection property forany x ∈ P η . It is easy to check that this implies P η ⊆ hI η i , hence D η ∪ E η ⊆ hI η i ω .Let x i for i ∈ ω and f be given by applying Lemma 25 with I = I η , D = D η and E = E η . Finally, set I ξ = I η ∪ { x i : i ∈ ω } and f η = f . (cid:3) Lemma 25.
Assume that
MA(countable) holds. Let
I ⊆ ω be a collection ofsubsets of ω with the finite union property and assume that |I| < c . Fix twocountable dense subsets D and E of (2 ω ) ω such that D ∪ E ⊆ hIi ω . Then thereexists a homeomorphism f : (2 ω ) ω −→ (2 ω ) ω and x i ∈ ω for i ∈ ω such that f [ D ] = E , I ∪ { x i : i ∈ ω } still has the finite union property and { d + f ( d ) : d ∈ D } ⊆ Q i ∈ ω ( x i ↓ ) .Proof. We will make a natural identification of (2 ω ) ω with 2 ω × ω . Namely, we willidentify a sequence ( x i ) i ∈ ω with the function x given by x ( i, j ) = x i ( j ).Consider the countable poset P consisting of all triples of the form p = ( s, g, π ) =( s p , g p , π p ) such that, for some m = m p ∈ ω and n = n p ∈ ω , the followingrequirements are satisfied. • s : m × n −→ • g is a bijection between a finite subset of D and a finite subset of E . • π is a permutation of m × n { d + f ( d ) : d ∈ (2 ω ) ω } ⊆ Q i ∈ ω ( x i ↓ ). Noticethat this is equivalent to ( d + f ( d ))( i, j ) ≤ x ( i, j ) = x i ( j ) for all d ∈ ω × ω and( i, j ) ∈ ω × ω .(1) ( t + π ( t ))( i, j ) = 1 implies s ( i, j ) = 1 for every t ∈ m × n i, j ) ∈ m × n .(2) π ( d ↾ ( m × n )) = g ( d ) ↾ ( m × n ) for every d ∈ dom( g ).Order P by declaring q ≤ p if the following conditions are satisfied. • s q ⊇ s p . • g q ⊇ g p . • π q ( t ) ↾ ( m p × n p ) = π p ( t ↾ ( m p × n p )) for all t ∈ m q × n q d ∈ D , define D dom d = { p ∈ P : d ∈ dom( g p ) } . Given p ∈ P and d ∈ D \ dom( g p ), one can simply choose e ∈ E \ ran( g p ) suchthat e ↾ ( m p × n p ) = π p ( d ↾ ( m p × n p )). This choice will make sure that q =( s p , g p ∪ { ( d, e ) } , π p ) ∈ P . Furthermore it is clear that q ≤ p . So each D dom d is densein P .For each e ∈ E , define D ran e = { p ∈ P : e ∈ ran( g p ) } . As above, one can easily show that each D ran e is dense in P .For each σ = { x , . . . , x k } ∈ [ I ] <ω and ℓ ∈ ω , define D σ,ℓ = { p ∈ P : there exists j ∈ n p \ ℓ such that s p (0 , j ) = · · · = s p ( m p − , j ) = x ( j ) = · · · = x k ( j ) = 0 } . HE TOPOLOGY OF ULTRAFILTERS AS SUBSPACES OF 2 ω Next, we will prove that each D σ,ℓ is dense in P . So fix σ and ℓ as above. Let p = ( s, g, π ) ∈ P with m p = m and n p = n . Find m ′ ≥ m and n ′ ≥ ℓ, n such thatthe following conditions hold. • All d ↾ ( m ′ × n ′ ) for d ∈ dom( g ) are distinct. • All e ↾ ( m ′ × n ′ ) for e ∈ ran( g ) are distinct. • x ( n ′ ) = · · · = x k ( n ′ ) = d i ( n ′ ) = e i ( n ′ ) = 0 for all d ∈ dom( g ), e ∈ ran( g )and i ∈ m ′ .This is possible because I has the finite union property and σ ∪ { d i : d ∈ dom( g ) , i ∈ ω } ∪ { e i : e ∈ ran( g ) , i ∈ ω } ⊆ hIi . We can choose a permutation π ′ of m ′ × n ′ π ′ ( d ↾ ( m ′ × n ′ )) = g ( d ) ↾ ( m ′ × n ′ ) for every d ∈ dom( g ) and π ′ ( t ) ↾ ( m × n ) = π ( t ↾ ( m × n )) for all t ∈ m ′ × n ′
2. Extend s to s ′ : m ′ × n ′ −→ s ′ ( i, j ) = 1 for every( i, j ) ∈ ( m ′ × n ′ ) \ ( m × n ). It is clear that p ′ = ( s ′ , g, π ′ ) ∈ P and p ′ ≤ p .Now let π ′′ be the permutation of m ′ × ( n ′ +1) π ′′ ( t )( i, j ) = ( π ′ ( t ↾ ( m ′ × n ′ ))( i, j ) if ( i, j ) ∈ m ′ × n ′ t ( i, j ) if ( i, j ) ∈ m ′ × { n ′ } for all t ∈ m ′ × ( n ′ +1)
2. Extend s ′ to s ′′ : m ′ × ( n ′ + 1) −→ s ′′ ( i, j ) = 0for all ( i, j ) ∈ m ′ ×{ n ′ } . It is easy to check that p ′′ = ( s ′′ , g, π ′′ ) ∈ D σ,ℓ and p ′′ ≤ p ′ .We will need one last class of dense sets. For any given ℓ ∈ ω , define D ℓ = { p ∈ P : m p ≥ ℓ } . An easier version of the above argument shows that each D ℓ is in fact dense.Since |I| < c , the collection of dense sets D = { D σ,ℓ : σ ∈ [ I ] <ω , ℓ ∈ ω } ∪ { D dom d : d ∈ D } ∪ { D ran e : e ∈ E } ∪ { D ℓ : ℓ ∈ ω } has also size less than c . Therefore, by MA(countable), there exists a D -genericfilter G ⊆ P . Define x i = S { s p ( i, − ) : p ∈ G } for every i ∈ ω . To define f ( y )( i, j ),for a given y ∈ ω × ω and ( i, j ) ∈ ω × ω , choose any p ∈ G such that ( i, j ) ∈ m p × n p and set f ( y )( i, j ) = π p ( y ↾ ( m p × n p ))( i, j ). (cid:3) Corollary 26.
Assume that
MA(countable) holds. Then there exists a non- G δ subset X of ω such that X ω is countable dense homogeneous. Question 5.
Can the assumption that MA(countable) holds be dropped in Theo-rem 24?
Question 6.
Is there an analytic non- G δ subset X of 2 ω such that X ω is countabledense homogeneous? Co-analytic?6. The perfect set property
Definition 27.
Let X be a space. We will say that A ⊆ X has the perfect setproperty if A is either countable or it contains a perfect set.It is a classical result of descriptive set theory, due to Souslin, that every analyticsubset of a Polish space has the perfect set property (see, for example, Theorem29.1 in [13]).The following is an easy application of Kunen’s closed embedding trick. Theorem 28.
There exists an ultrafilter
U ⊆ ω with a closed subset of cardinality c that does not have the perfect set property.Proof. Fix a Bernstein set B in 2 ω , then apply Theorem 8 with C = B . (cid:3) Next, we will consistently construct a non-principal ultrafilter U such that everyclosed subset of U has the perfect set property. Actually, we will get a much strongerresult (see Theorem 29).Recall that a play of the strong Choquet game on a topological space ( X, T ) isof the form I ( q , U ) ( q , U ) · · · II V V · · · ,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 T n ∈ ω U n = ∅ . The topological space ( X, T ) is strong Choquet ifII has a winning strategy in the above game. See Section 8.D in [13].Define an A -triple to be a triple of the form ( T , A, Q ) such that the followingconditions are satisfied. • T is a strong Choquet, second-countable topology on 2 ω that is finer thanthe standard topology. • A ∈ T . • Q is a non-empty countable subset of A with no isolated points in thesubspace topology it inherits from T .By Theorem 25.18 in [13], for every analytic A there exists a topology T as above.Also, by Exercise 25.19 in [13], such a topology T necessarily consists only ofanalytic sets. In particular, all A-triples can be enumerated in type c . Theorem 29.
Assume that
MA(countable) holds. Then there exists a non-principalultrafilter
U ⊆ ω such that A ∩ U has the perfect set property for every analytic A ⊆ ω .Proof. Enumerate as { ( T η , A η , Q η ) : η ∈ c } all A-triples, making sure that eachtriple appears cofinally often. Also, enumerate as { z η : η ∈ c } all subsets of ω .We will construct F ξ for every ξ ∈ c by transfinite recursion. By induction, wewill make sure that the following requirements are satisfied.(1) F µ ⊆ F η whenever µ ≤ η < c .(2) F ξ has the finite intersection property for every ξ ∈ c .(3) |F ξ | < c for every ξ ∈ c .(4) By stage ξ = η + 1, we must have decided whether z η ∈ U : that is, z εη ∈ F ξ for some ε ∈ Q η ⊆ F η then, at stage ξ = η + 1, we will deal with A η : that is, thereexists x ∈ F ξ such that x ↑ ∩ A η contains a perfect subset.In the end, let U = S ξ ∈ c F ξ . Notice that U will be an ultrafilter by (4).Start by letting F = Cof. Take unions at limit stages. At a successor stage ξ = η + 1, assume that F η is given. First assume that Q η * F η . In this case, simplyset F ξ = F η ∪ { z εη } for a choice of ε ∈ Q η ⊆ F η . Apply Lemma 30 with F = F η , A = A η , Q = Q η and T = T η to get a perfect set P ⊆ A such that F η ∪ { T P } has the finite intersectionproperty. Let x = T P . Set F ξ = F η ∪ { x, z εη } , for some ε ∈ HE TOPOLOGY OF ULTRAFILTERS AS SUBSPACES OF 2 ω Finally, we will check that U has the required property. Assume that A is ananalytic subset of 2 ω such that A ∩ U is uncountable. By Theorem 25.18 in [13],there exists a second-countable, strong Choquet topology T on 2 ω that is finer thanthe standard topology and contains A . Since every second countable, uncountableHausdorff space contains a non-empty countable subspace with no isolated points,we can find such a subspace Q ⊆ A ∩ U . Since cf( c ) > ω , there exists µ ∈ c suchthat Q ⊆ F µ . Since we listed each A-triple cofinally often, there exists η ≥ µ suchthat ( T , A, Q ) = ( T η , A η , Q η ). Condition (5) guarantees that U ∩ A will contain aperfect subset. (cid:3) Lemma 30.
Assume that
MA(countable) holds. Let F be a collection of subsetsof ω with the finite intersection property such that |F| < c . Suppose that ( T , A, Q ) is an A -triple with Q ⊆ F . Then there exists a perfect subset P of A such that F ∪ { T P } has the finite intersection property.Proof. Fix a winning strategy Σ for player II in the strong Choquet game in (2 ω , T ).Also, fix a countable base B for (2 ω , T ). Let P be the countable poset consisting ofall functions p such that, for some n = n p ∈ ω , the following conditions hold.(1) p : ≤ n −→ Q × B . We will use the notation p ( s ) = ( q ps , U ps ).(2) U p ∅ = A .(3) For every s, t ∈ ≤ n
2, if s and t are incompatible (that is, s * t and t * s )then U ps ∩ U pt = ∅ .(4) For every s ∈ n q ps ↾ , U ps ↾ ) ( q ps ↾ , U ps ↾ ) · · · ( q ps ↾ n , U ps ↾ n )II V ps ↾ V ps ↾ · · · V ps ↾ n is a partial play of the strong Choquet game in (2 ω , T ), where the opensets V ps ↾ i played by II are the ones dictated by the strategy Σ.Order P by setting p ≤ p ′ whenever p ⊇ p ′ .For every ℓ ∈ ω , define D ℓ = { p ∈ P : n p ≥ ℓ } . Since Q has no isolated points and T is Hausdorff, it is easy to see that each D ℓ isdense.For any fixed ℓ ∈ ω , consider the partition of 2 ω in clopen sets P ℓ = { [ s ] : s ∈ ℓ } ,then define D ref ℓ = { p ∈ P : { U ps : s ∈ n p } refines P ℓ } . Let us check that each D ref ℓ is dense. Given p ∈ P and ℓ ∈ ω , let n = n p and q s = q ps for every s ∈ n
2. Since Q has no isolated points, it is possible, for every s ∈ n
2, to choose q s = q s such that q s ∈ V ps ∩ Q . Then choose j ≥ ℓ big enough sothat [ q s ↾ j ] ∩ [ q s ↾ j ] = ∅ for every s ∈ n
2. Now simply extend p to a condition p ′ : ≤ n +1 −→ Q × B by defining p ′ ( s ⌢ ε ) = ( q εs , U εs ) for every s ∈ n ε ∈ U εs ∈ B is such that q εs ∈ U εs ⊆ V ps ∩ [ q εs ↾ j ]. It is easy to realize that p ′ ∈ D ref ℓ .For any fixed σ = { x , . . . , x k } ∈ [ F ] <ω and ℓ ∈ ω , define D σ,ℓ = { p ∈ P : there exists i ∈ ω \ ℓ such that x ( i ) = x ( i ) = · · · = x k ( i ) = 1 for all x ∈ U ps for all s ∈ n p } . Let us check that each D σ,ℓ is dense. Given p ∈ P , σ and ℓ as above, let n = n p and q s = q ps for every s ∈ n
2. Notice that \ s ∈ n q ps ∩ \ σ is an infinite subset of ω , because Q ⊆ F by assumption. So there exists i ∈ ω with i ≥ ℓ such that q ps ( i ) = x ( i ) = · · · = x k ( i ) = 1for every s ∈ n
2. Since Q has no isolated points, it is possible, for every s ∈ n
2, tochoose q s = q s such that q s ∈ V ps ∩ [ q ps ↾ ( i + 1)] ∩ Q . Then choose j ≥ i + 1 bigenough so that [ q s ↾ j ] ∩ [ q s ↾ j ] = ∅ for every s ∈ n
2. Now simply extend p to acondition p ′ : ≤ n +1 −→ Q × B by defining p ′ ( s ⌢ ε ) = ( q εs , U εs ) for every s ∈ n ε ∈
2, where each U εs ∈ B is such that q εs ∈ U εs ⊆ V ps ∩ [ q εs ↾ j ]. It is easy to realizethat p ′ ∈ D σ,ℓ .Since |F| < c , the collection of dense sets D = { D ℓ : ℓ ∈ ω } ∪ { D ref ℓ : ℓ ∈ ω } ∪ { D σ,ℓ : σ ∈ [ F ] <ω , ℓ ∈ ω } has also size less than c . Therefore, by MA(countable), there exists a D -genericfilter G ⊆ P . Let g = S G : <ω −→ Q × B . Given s ∈ <ω
2, pick any p ∈ G suchthat s ∈ dom( p ) and set U s = U ps . For any x ∈ ω , since Σ is a winning strategy forII, we must have T n ∈ ω U x ↾ n = ∅ . Using the dense sets D ref ℓ , one can easily showthat such intersection is actually a singleton. Therefore, letting f ( x ) be the uniqueelement of T n ∈ ω U x ↾ n yields a well-defined function f : 2 ω −→ A . Using condition(3) in the definition of P , one sees that f is injective.Next, we will show that f is continuous in the standard topology, hence a home-morphic embedding by compactness. Fix x ∈ ω and let y = f ( x ). Fix ℓ ∈ ω . Since G is a D -generic filter, there must be p ∈ D ref ℓ ∩ G . Let n = n p . Notice that thisimplies U x ↾ n = U px ↾ n ⊆ [ y ↾ ℓ ], hence f ( x ′ ) ∈ [ y ↾ ℓ ] whenever x ′ ∈ [ x ↾ n ].Therefore P = ran( f ) is a perfect subset of A . Finally, using the dense sets D σ,ℓ one can show that F ∪ { T P } has the finite intersection property. (cid:3) Corollary 31.
Assume that
MA(countable) holds. Then there exists a non-principalultrafilter
U ⊆ ω such that every closed subset of U has the perfect set property. Question 7.
Can the assumption that MA(countable) holds be dropped in Theo-rem 29?Observe that if Q ⊆ ω is homeomorphic to Q in the standard topology, A =cl( Q ) and T A is the topology obtained by declaring A open, then ( T A , Q, A ) is anA-triple because T A is Polish (see Lemma 13.2 in [13]). It follows easily that theultrafilter constructed in Theorem 29 cannot contain closed copies of the rationals,hence it is completely Baire by Lemma 6. Question 8.
Is an ultrafilter
U ⊆ ω such that A ∩ U has the perfect set propertywhenever A is an analytic subset of 2 ω necessarily completely Baire?We also remark that if Γ ⊆ P (cid:0) ω (cid:1) is closed under c and U is such that A ∩ U has the perfect set property for all A ∈ Γ, then A \ U has the perfect set propertyfor all A ∈ Γ. HE TOPOLOGY OF ULTRAFILTERS AS SUBSPACES OF 2 ω Extending the perfect set property
Assuming V = L , there exists an uncountable co-analytic set A that does notcontain any perfect set (see Theorem 25.37 in [12]). It follows that MA(countable)is not enough to extend Theorem 29 to all co-analytic sets. This section is devotedto attaining a positive result for the co-analytic case. Actually, we will obtaina much stronger result (see Theorem 35). We will need a modest large cardinalassumption, a larger fragment of MA, and the negation of CH. Lemma 32.
Assume that
U ⊆ ω is a P ω -point. If A ⊆ ω is such that everyclosed subspace of A has the perfect set property, then A ∩ U has the perfect setproperty.Proof. Let A be as above, and assume that A ∩U is uncountable. Choose B ⊆ A ∩U such that | B | = ω . Since U is a P ω -point, there is a pseudointersection x of B in U .For some n ∈ ω , uncountably many elements of B are in the closed set C = ( x \ n ) ↑ .By hypothesis, A ∩ C contains a perfect set P . We now have A ∩ U ⊇ P as desired.Thus, A ∩ U has the perfect set property. (cid:3) It is not hard to verify that the hypothesis on A in the above lemma is optimal.Let x and x be complementary infinite subsets of ω . Identify each P ( x i ) withthe perfect set { x ∈ ω : x ( n ) = 0 for all n ∈ x − i } . Fix a Bernstein subset B i of P ( x i ) and set A i = B i ∪ P ( x − i ) for each i ∈
2. Each A i has the perfect setproperty. However, if U ⊆ ω is an ultrafilter, then some A i ∩ U lacks the perfectset property. Indeed, if x i ∈ U , then y ∈ U for some subset y ⊆ x i such that x i \ y is infinite. The perfect set y ↑ ∩P ( x i ) contains c many elements of B i , so A i ∩ U has size c as well. However, A i ∩ U ⊆ B i , so A i ∩ U does not contain a perfect set.The following lemma is essentially due to Ihoda (Judah) and Shelah (see Theorem3.1 in [11]). Given a class Γ, we define PSP(Γ) to mean that every X ∈ Γ ∩ P (2 ω )has the perfect set property. Lemma 33.
The existence of a Mahlo cardinal is equiconsistent with
MA( σ - centered) + ¬ CH + PSP( L ( R )) . Proof.
Any generic extension by the Levy collapse Col( ω, κ ) of an inaccessible car-dinal κ to ω satisfies PSP( L ( R )) (see the proof of Theorem 11.1 in [14]). By theproof of Lemma 1.1 in [11], if κ is inaccessible and P is a forcing poset that satisfiesthe following conditions, then every generic extension V [ G ] of V by P is such that L ( R ) V [ G ] = L ( R ) V [ H ] for some V -generic filter H ⊆ Col( ω, κ ).(1) P has the κ -cc.(2) P forces κ = ω .(3) For every R ⊆ P of size less than κ , there exists Q ⊆ P such that | Q | < κ , R ⊆ Q , and Q is completely embedded in P by the inclusion map.Assuming that there exists a Mahlo cardinal κ , the proof of Theorem 3.1 in [11]constructs a generic extension V [ G ] of V by a forcing P such that V [ G ] satisfiesMA( σ -centered) + ¬ CH, using a forcing poset P that satisfies conditions (1), (2)and (3). Therefore, PSP( L ( R )) also holds in V [ G ].Conversely, PSP( L ( R )) implies that all injections of ω into 2 ω are outside of L ( R ), which in turn implies ω L [ r ]1 < ω for all reals r . The proof of Theorem 3.1in [11] shows that if MA( σ -centered) + ¬ CH holds and ω L [ r ]1 < ω for all reals r ,then ω is Mahlo in L . (cid:3) For the convenience of the reader, we include the proof of the following standardlemma.
Lemma 34.
Assume that
MA( σ - centered) holds. Then there exists a P c -point U .Proof. Enumerate all subsets of ω as { z η : η ∈ c } . We will construct F ξ for every ξ ∈ c by transfinite recursion. By induction, we will make sure that the followingrequirements are satisfied.(1) F µ ⊆ F η whenever µ ≤ η < c .(2) F ξ has the finite intersection property for every ξ ∈ c .(3) |F ξ | < c for every ξ ∈ c .(4) By stage ξ = η + 1, we must have decided whether z η ∈ U : that is, z εη ∈ F ξ for some ε ∈ ξ = η + 1, we will make sure that F ξ contains a pseudointersectionof F η .Start by letting F = Cof. Take unions at limit stages. At a successor stage ξ = η + 1, assume that F η is given.Since MA( σ -centered) implies p = c (see Theorem 7.12 in [4]), there exists aninfinite pseudointersection x of F η . Now simply set F ξ = F η ∪ { x, z εη } for a choiceof ε ∈ U = S ξ ∈ c F ξ . Notice that U will be an ultrafilter by (4). Since p = c is regular (see Theorem 7.15 in [4]), condition (5) implies that U is a P c -point. (cid:3) It is well-known that MA(countable) is not a sufficient hypothesis for the abovelemma. Consider the Cohen model W = V [( c α : α < ω )], where V (cid:15) CH and each c α is an element of 2 ω that avoids all meager Borel sets with Borel codes in V [( c β : β < ω , β = α )]. Observe that every x ∈ ω is in V [( c α : α ∈ I )] for some countableset I ⊆ ω . In this model, cov( M ) = c = ω , so MA(countable) + ¬ CH holds(see Theorem 7.13 in [4]). However, if
U ∈ W is a non-principal ultrafilter, then U ∩ V [( c α : α < ω )] is a subset of U of size ω with no infinite pseudointersection. Theorem 35.
It is consistent, relative to a Mahlo cardinal, that there exists anon-principal ultrafilter
U ⊆ ω such that A ∩ U has the perfect set property for all A ∈ P (cid:0) ω (cid:1) ∩ L ( R ) . On the other hand, if there exists such an ultrafilter U , then ω is inaccessible in L .Proof. Assume that MA( σ -centered)+ ¬ CH+ PSP( L ( R )) holds, which is consistentrelative to a Mahlo cardinal by Lemma 33. By Lemma 34, there exists a P c -point U .Since ¬ CH holds, U is a P ω -point. Fix A ∈ P (cid:0) ω (cid:1) ∩ L ( R ). Every closed subspace C of A is also in L ( R ) because C = A ∩ [ T ] for some tree T ⊆ <ω . By PSP( L ( R )),all such C have the perfect set property. So A ∩ U has the perfect set property byLemma 32.For the second half of the theorem, assume that U ⊆ ω is a non-principalultrafilter such that A ∩ U has the perfect set property for all A ∈ P (cid:0) ω (cid:1) ∩ L ( R ).First, observe that given A as above, c [ A ] is in L ( R ) too, so A ∩ U and c [ A ] ∩ U have the perfect set property. Since A = ( A ∩ U ) ∪ ( A ∩ c [ U ]) = ( A ∩ U ) ∪ c [ c [ A ] ∩ U ] , it follows that A itself has the perfect set property. So PSP( L ( R )) holds, whichimplies ω L [ r ]1 < ω for all reals r . Therefore ω is inaccessible in L . (cid:3) HE TOPOLOGY OF ULTRAFILTERS AS SUBSPACES OF 2 ω Question 9.
What is the exact consistency strenght of a non-principal ultrafilter
U ⊆ ω such that A ∩ U has the perfect set property for all A ∈ P (cid:0) ω (cid:1) ∩ L ( R )?In particular, does the Levy collapse Col( ω, κ ) of an inaccessible cardinal κ to ω force such an ultrafilter? 8. P -points Given a non-principal ultrafilter
U ⊆ ω , it seems natural to investigate whetherthere is any relation between the topological properties of U that we studied so farand combinatorial properties of U . In order to construct several kinds of non-P-points, we will essentially use an idea from [15]. Definition 36. A mixed independent family is a pair ( F , A ) of collections of subsetsof ω such that \ σ ∩ \ x ∈ τ x w ( x ) is infinite whenever σ ∈ [ F ] <ω , τ ∈ [ A ] <ω and w : τ −→
2. A dual mixed indepen-dent family is a pair ( I , B ) of collections of subsets of ω such that ( c [ I ] , c [ B ]) is amixed independent family. Lemma 37.
Let ( F , A ) be a mixed independent family such that A is infinite.Then there exists a non- P -point U extending F ∪ A .Proof.
Fix a countably infinite subset B of A . It is easy to check that G = F ∪ A ∪ { ω \ x : x ⊆ ∗ y for every y ∈ B} has the finite intersection property. Let U be any ultrafilter extending G . It is clearthat B has no pseudointersection in U . (cid:3) Similarly, one can prove the following.
Lemma 38.
Let ( I , B ) be a dual mixed independent family such that B is infinite.Then there exists a maximal ideal J extending I ∪ B that is not a P -ideal. We will begin by studying the relation between P-points and completely Baireultrafilters.
Theorem 39.
There exists a non- P -point U ⊆ ω that is not completely Baire.Proof. We will use the same notation as in the proof of Theorem 8. Choose C = Q ,so that any ultrafilter extending G will contain a closed copy of Q . Now simplyapply Lemma 37 to ( ∅ , G ). (cid:3) Theorem 40.
Assume that
MA(countable) holds. Then there exists a P -point U ⊆ ω that is completely Baire.Proof. Enumerate all countable collections of subsets of ω as {C η : η ∈ c } . Thesetup of the construction will be as in the proof of Theorem 11, but we will dodifferent things at even and odd successor stages.Start by letting F = Cof. Take unions at limit stages. At a successor stage ξ = 2 η + 1, assume that F η is given, then take care of Q η as in the proof ofTheorem 11. At a successor stage ξ = 2 η + 2, assume that F η +1 is given, thentake care of C η as follows.First assume that there exists x ∈ C η such that F η +1 ∪ { ω \ x } has the finiteintersection property. In this case, we can just set F ξ = F η +1 ∪{ ω \ x } . Now assume that F η +1 ∪{ ω \ x } does not have the finite intersection property for any x ∈ C η . Itis easy to check that this implies C η ⊆ hF η +1 i . Recall that MA(countable) implies d = c (see, for example, Proposition 5.5 and Theorem 7.13 in [4]). So, by Proposition6.24 in [4], there exists a pseudointersection x of C η such that F η +1 ∪ { x } has thefinite intersection property. Finally, set F ξ = F η +1 ∪ { x } . (cid:3) Question 10.
For a non-principal ultrafilter
U ⊆ ω , is being a P-point equivalentto being completely Baire?Now we turn to the relation between P-points and countable dense homogeneousultrafilters. Theorem 41.
Assume that
MA(countable) holds. Then there exists a non-principalultrafilter
U ⊆ ω that is countable dense homogeneous but not a P -point.Proof. For notational convenience, we will actually construct a maximal ideal
J ⊆ ω that is countable dense homogeneous but not a P-ideal.The setup of the construction will be as in the proof of Theorem 21, but wewill simultaneously construct B ξ for ξ ∈ c so that the following conditions will besatisfied. In the end, set B = S ξ ∈ c B ξ and apply Lemma 38 to ( I , B ).(1) B µ ( B η whenever µ < η < c .(2) ( I ξ , B ξ ) is a dual mixed independent family for every ξ ∈ c .(3) |B ξ | < c for every ξ ∈ c .Start by letting ( I , B ) = (Fin , ∅ ). Take unions at limit stages. At a successorstage ξ = η + 1, assume that ( I η , B η ) is given. First get x by applying Lemma 42with I = I η , B = B η and ( D, E ) = ( D η , E η ). Then, as in the proof of Lemma 14,use MA(countable) to get y / ∈ B η such that ( I η ∪ { x } , B η ∪ { y } ) is still a dual mixedindependent family. Finally, set ( I ξ , B ξ ) = ( I η ∪ { x } , B η ∪ { y } ). (cid:3) The following lemma is easily proved by modifying the proof of Lemma 22 (sub-stitute the dense sets D σ,ℓ with the obviously defined dense sets D σ,τ,w,ℓ ). Lemma 42.
Assume that
MA(countable) holds. Let ( I , B ) be a dual mixed inde-pendent family such that |I| < c and |B| < c . Fix two countable dense subsets D and E of ω such that D ∪ E ⊆ hIi . Then there exists a homeomorphism f : 2 ω −→ ω and x ∈ ω such that f [ D ] = E , ( I ∪{ x } , B ) is still a dual mixed independent familyand { d + f ( d ) : d ∈ D } ⊆ x ↓ . Theorem 43.
Assume that
MA(countable) holds. Then there exists a non- P -point U ⊆ ω that is not countable dense homogeneous.Proof. Let A be as in Lemma 14. By the proof of Theorem 15, no ultrafilterextending A is countable dense homogeneous. Now simply apply Lemma 37 to( ∅ , A ). (cid:3) Theorem 44.
Assume that
MA(countable) holds. Then there exists a P -point U ⊆ ω that is countable dense homogeneous.Proof. For notational convenience, we will actually construct a maximal ideal
J ⊆ ω that is countable dense homogeneous and a P-ideal. Enumerate all countablecollections of subsets of ω as {C η : η ∈ c } . The setup of the construction will beas in the proof of Theorem 21, but we will do different things at even and oddsuccessor stages. HE TOPOLOGY OF ULTRAFILTERS AS SUBSPACES OF 2 ω Start by letting I = Fin. Take unions at limit stages. At a successor stage ξ = 2 η + 1, assume that I η is given, then take care of ( D η , E η ) as in the proof ofTheorem 21. At a successor stage ξ = 2 η + 2, assume that I η +1 is given, then takecare of C η as follows.First assume that there exists x ∈ C η such that I η +1 ∪ { ω \ x } has the finiteunion property. In this case, we can just set I ξ = I η +1 ∪ { ω \ x } . Now assumethat I η +1 ∪ { ω \ x } does not have the finite union property for any x ∈ C η . It iseasy to check that this implies C η ⊆ hI η +1 i . As in the proof of Theorem 40, it ispossible to get a pseudounion x of C η such that I η +1 ∪ { x } has the finite unionproperty. Finally, set I ξ = I η +1 ∪ { x } . (cid:3) Question 11.
Is a P-point
U ⊆ ω necessarily countable dense homogeneous?Finally, we will investigate the relation between P-points and the perfect setproperty. Theorem 45.
There exists a non- P -point U ⊆ ω with a closed subset of cardinality c that does not have the perfect set property.Proof. We will use the same notation as in the proof of Theorem 8. Choose C to bea Bernstein set in 2 ω , so that any ultrafilter extending G will have a closed subsetwithout the perfect property. Now simply apply Lemma 37 to ( ∅ , G ). (cid:3) Theorem 46.
Assume that
MA(countable) holds. Then there exists a P -point U ⊆ ω such that A ∩ U has the perfect set property whenever A is an analyticsubset of ω .Proof. Enumerate all countable collections of subsets of ω as {C η : η ∈ c } . Thesetup of the construction will be as in the proof of Theorem 29, but we will dodifferent things at even and odd successor stages.Start by letting F = Cof. Take unions at limit stages. At a successor stage ξ = 2 η + 1, assume that F η is given, then take care of ( T η , A η , Q η ) and z η as inthe proof of Theorem 29. At a successor stage ξ = 2 η + 2, assume that F η +1 isgiven, then take care of C η as in the proof of Theorem 40. (cid:3) Question 12.
For a non-principal ultrafilter
U ⊆ ω , is being a P-point equivalentto A ∩ U having the perfect set property whenever A is an analytic subset of 2 ω ?Observe that Lemma 32 might be viewed as a partial answer to Question 12. Acknowledgements.
The first author thanks Jan van Mill for making him noticeProposition 1 and the fact that the topology of ultrafilters as subspaces of 2 ω was‘unexplored territory’. Together with Question 3.2 from [10], that was the mainmotivation for this article. The first author also thanks Kostas Beros for suggestingthe second proof of Lemma 7 and helping to shape the present proof of Lemma 22.Both authors thank Ken Kunen for kindly ‘donating’ Theorem 8 to them, and theanonymous referee for valuable comments on an earlier version of this paper. References [1]
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