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Topology and its Applications 104 (2000) 27–38 Filters, consonance and hereditary Baireness A. Bouziad 1 Université de Rouen, Département de Mathématiques, CNRS UPRES-A 6085, 76821 Mont Saint Aignan, France Received 8 December 1997; received in revised form 29 September 1998 Abstract A topological space is called consonant if, on the set of all closed subsets of X, the co-compact topology coincides with the upper Kuratowski topology. For a filter F on the set of natural numbers ω, let X F = ω ∪ {∞} be the space for which all points in ω are isolated and the neighborhood system of is {A ∪ {∞}: A F }. We give a combinatorial characterization of the class Φ of all filters F such that the space X F is consonant and all its compact subsets are finite. It is also shown that a filter F belongs to Φ if and only if the space C p (X F ) of real-valued continuous functions on X F with the pointwise topology is hereditarily Baire. 2000 Elsevier Science B.V. All rights reserved. Keywords: Consonance; Upper Kuratowski–Painlevé convergence; Co-compact topology; P -set; Filter; Hereditarily Baire space AMS classification: 54B20; 54C35 1. Introduction and preliminaries For a topological space X, we consider two topologies on the hyperspace F (X) of all closed subsets of X. The co-compact topology τ co , generated by the family of all sets of the form {F F (X): F K = ∅}, where K is an arbitrary compact subset of X, and the upper Kuratowski topology τ uk associated with the upper Kuratowski–Painlevé convergence. Recall that a net (F γ ) γ Γ F (X) upper Kuratowski–Painlevé converges to F F (X) means that \ n [ {F γ : γ > α}: α Γ o F. The space X is said to be consonant if the equality τ uk = τ co holds. The class of consonant spaces was introduced and studied first by Dolecki, Greco and Lechicki [7]. Consonant 1 Email: [email protected]. 0166-8641/00/$ – see front matter 2000 Elsevier Science B.V. All rights reserved. PII:S0166-8641(99)00014-0

Filters, consonance and hereditary Baireness

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Page 1: Filters, consonance and hereditary Baireness

Topology and its Applications 104 (2000) 27–38

Filters, consonance and hereditary Baireness

A. Bouziad1

Université de Rouen, Département de Mathématiques, CNRS UPRES-A 6085,76821 Mont Saint Aignan, France

Received 8 December 1997; received in revised form 29 September 1998

Abstract

A topological space is called consonant if, on the set of all closed subsets ofX, the co-compacttopology coincides with the upper Kuratowski topology. For a filterF on the set of natural numbersω, letXF = ω∪{∞} be the space for which all points inω are isolated and the neighborhood systemof∞ is {A ∪ {∞}: A ∈ F}. We give a combinatorial characterization of the classΦ of all filtersFsuch that the spaceXF is consonant and all its compact subsets are finite. It is also shown that a filterF belongs toΦ if and only if the spaceCp(XF ) of real-valued continuous functions onXF withthe pointwise topology is hereditarily Baire. 2000 Elsevier Science B.V. All rights reserved.

Keywords:Consonance; Upper Kuratowski–Painlevé convergence; Co-compact topology;P -set;Filter; Hereditarily Baire space

AMS classification: 54B20; 54C35

1. Introduction and preliminaries

For a topological spaceX, we consider two topologies on the hyperspaceF(X) of allclosed subsets ofX. The co-compact topologyτco, generated by the family of all setsof the form {F ∈ F(X): F ∩ K = ∅}, whereK is an arbitrary compact subset ofX,and the upper Kuratowski topologyτuk associated with the upper Kuratowski–Painlevéconvergence. Recall that a net(Fγ )γ∈Γ ⊂ F(X) upper Kuratowski–Painlevé converges toF ∈ F(X) means that⋂{⋃

{Fγ : γ > α}: α ∈ Γ}⊂ F.

The spaceX is said to beconsonantif the equalityτuk = τco holds. The class of consonantspaces was introduced and studied first by Dolecki, Greco and Lechicki [7]. Consonant

1 Email: [email protected].

0166-8641/00/$ – see front matter 2000 Elsevier Science B.V. All rights reserved.PII: S0166-8641(99)00014-0

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28 A. Bouziad / Topology and its Applications 104 (2000) 27–38

spaces have, in a way, a “nice” structure of compact subsets; this is illustrated, forexample, by locallyCech-complete spaces and Hausdorffkω-spaces, which are knownto be consonant [14]. Here, we are interested in consonance of countable Hausdorff spaceswhich have exactly one non-isolated point.

For a filter F on the setω of natural numbers, letXF = ω ∪ {∞} be the spacetopologized by isolating all points ofω and by using the family{A ∪ {∞}: A ∈ F} asa neighborhood system of∞. All filters considered in this note are supposed to be non-principal. Notice that every countably infinite space with only one non-isolated point ishomeomorphic to a spaceXF for some filterF on ω. In the existing literature, oneencounters two types of filtersF such thatXF is consonant: countably generated ones(in this case the spaceXF is completely metrizable) and the filters corresponding to thecountable Fréchet–Urysohn fanSω. Let us note that the consonance ofSω is a useful toolin consonance theory; see [14].

In a recent paper Arab and Calbrix in [1, Theorem 4.3] proved that, ifF is P -pointof ω∗, thenXF is consonant. We shall show that, conversely, ifF is an ultrafilter andXF is consonant, thenF is aP -point ofω∗ (Proposition 2.3). By a well-known result ofRudin, the Continuum Hypothesis implies the existence ofP -points inω∗, and by a resultof Shelah the existence ofP -points inω∗ in ZFC only is impossible. (For more details,consult [9,13].) Consequently, it follows from Proposition 2.3 that, not only there existultrafiltersF onω such thatXF is not consonant, but it is usually always the case.

Recall that a subsetA of a spaceX is said to be aP -set of X, if the intersection ofcountably many neighborhoodsofA inX is again a neighborhoodofA inX. A pointx ∈Xis aP -point of X if {x} is aP -set ofX. Let βω denote theCech–Stone compactificationof the discrete spaceω andω∗ = βω \ ω. For any filterF onω, letA(F) be the (closed)subset ofω∗ of all ultrafilters onω finer thanF . It is easy to check thatA(F) is aP -set ofω∗ if, and only if, for every sequence(An)n∈ω ⊆F there isA ∈F such thatA⊆∗ An (thatisA\An is finite) for eachn ∈ ω. Let us say thatF is aP -filter if A(F) is aP -subset ofω∗.

In Section 2 a classΦ of filters, defined by a simple combinatorial condition, isintroduced and it is shown thatΦ coincides with the class of all filtersF on ω, forwhich the spaceXF is consonant and all its compact subsets are finite. By using thevery definition ofΦ, we show thatΦ includes allP -points ofω∗ and is stable undercountable intersections (see Proposition 2.4, where a bit more is proved), which givesanother proof and an extension of the result of Arab and Calbrix cited above. Some otherstability theorems, which are not valid in consonance theory, are shown to be true in theclass of all spacesXF for F ∈Φ (see Proposition 2.5 and Corollary 2.2).

For a spaceY , let Cp(Y ) denote the space of continuous real functions onY with thepointwise convergence topology. Tkachuk and Pytkeev have given characterizations ofspacesY for which Cp(Y ) is a Baire space (see [15,17,18]). Nothing seems to have beknown about hereditary Baire property ofCp(Y ) until Gul’ko and Sokolov [10] provedthat, for an ultrafilterF , Cp(XF ) is hereditarily Baire if and only ifF is a P -filter. InSection 3, a connection is established between consonance ofXF and the hereditary Baireproperty ofCp(XF ). We prove thatCp(XF ) is hereditarily Baire if and only if the spaceXF is consonant and all its compact subsets are finite, if and only ifF is aP -filter and

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A. Bouziad / Topology and its Applications 104 (2000) 27–38 29

Cp(XF ) is a Baire space (see Theorem 3.1). SinceCp(XF ) is a Baire space for any ultrafil-terF (a result of Lutzer and McCoy [11, Theorem 5.1]), Theorem 3.1 completes the aboveresult of Gul’ko and Sokolov [10]. At the end, in Corollary 4.1, Theorem 3.1 is appliedto show that every filterF such thatA(F) is P -set of cellularity less than the continuumbelongs toΦ.

Referring the reader to the papers cited in the references for more information onconsonance theory, in our note we rely exclusively on the following characterizations ofconsonant spaces established respectively in [7,3]. Let us only mention that in Lemmas 1.1and 1.2 below, compact spaces are not assumed to be Hausdorff.

Following [7], a non-empty collectionH of open sets in a spaceX is calledcompactifthe following condition holds:

(i) if U ∈H andV ⊆X is an open set such thatU ⊆ V , thenV ∈H;(ii) for every collectionU of open sets inX such that

⋃U ∈H, there is a finite sub-

collectionV ⊆ U such that⋃V ∈H.

Lemma 1.1. LetX be a regular space. Then, the following conditions are equivalent:(1) The spaceX is consonant.(2) Every compact collectionH of open sets inX is trivial in the sens there is a compact

setK ⊆X so thatV ∈H, for any open setV ⊆X satisfyingK ⊆ V .

For a spaceX, let 2X denote the set of all non-empty closed subsets ofX. Recall thata set-valued mapϕ :Y → 2X , whereY is a topological space, is said to belower semi-continuous(briefly l.s.c.), if for every open setU ⊆X the set

ϕ−1(U)= {y ∈ Y : ϕ(y)∩U 6= ∅}is open inY .

Lemma 1.2. A regular spaceX is consonant if, and only if, for every l.s.c. set-valued mapϕ :Y → 2X , whereY is compact, there is a compact setK ⊆X such thatϕ−1(K)= Y .

2. A class of filters

Let [ω]ω denote the set of all infinite subsets ofω. We denote byΦ the class of all filtersF onω for which the following property holds:

For every sequence(An)n∈ω ⊆ F such that{0, . . . , n} ⊆An for eachn ∈ ω, there isI ∈ [ω]ω such that

⋂k∈I Ak ∈F .

For every (countable) collectionU ⊆ [ω]ω, write limU to denote the set ofn ∈ ω suchthatn /∈ U only for finitely manyU ∈ U . For a filterF onω, the following conditions areequivalent:

(i) F ∈Φ;(ii) for every countable infinite collectionU ⊆F such that limU ∈F , there is an infiniteV ⊆ U such that

⋂V ∈F .

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30 A. Bouziad / Topology and its Applications 104 (2000) 27–38

Proposition 2.1. Compact subsets ofXF are finite for everyF ∈Φ.

Proof. Let A⊆ ω be an infinite set and let us show that there isB ∈ F such thatA \ Bis infinite. For eachn ∈ ω let in =min{k ∈ A: k > n} andAn = {in}c. ThenAn ∈ F and{0, . . . , n} ⊆ An, for everyn ∈ ω. Hence, there isI ⊆ ω infinite such that the setB =⋂k∈I An belongs toF . The subset{in: n ∈ I } of A is infinite andB ∩{in: n ∈ I } = ∅. 2

Proposition 2.2. Let F be a filter onω. ThenF ∈ Φ if, and only if, the spaceXF isconsonant and all compacts subsets ofXF are finite.

Proof. If F ∈Φ then, by Proposition 2.1, all compacts subsets ofXF are finite. Supposethat XF is not consonant. Then, by Lemma 1.1, there is a compact nontrivial familyH of open sets inXF . For eachn ∈ ω there isAn ∈ F such that{0, . . . , n} ⊆ An andAn ∪ {∞} /∈ H. Let I ⊆ ω be an infinite set so that the setB =⋂k∈I Ak belongs toF .SinceH is a compact family,{0, . . . , l} ∪ B ∪ {∞} ∈H fore somel ∈ ω. Choosek ∈ Iso that{0, . . . , l} ⊆ Ak . Then, sinceB ⊆ Ak, compactness ofH implies the contradictionAk ∪ {∞} ∈H.

Conversely, suppose thatF /∈ Φ and let us show thatXF is not consonant. Let(An)n∈ω ⊆ F be a sequence satisfying{0, . . . , n} ⊆ An for everyn ∈ ω, and suppose thatthere is noI ∈ [ω]ω for which

⋂k∈I Ak ∈ F . We assume without loss of generality that

ω \ An is non-empty for eachn ∈ ω, and so the set-valued mapϕ :XF0 → 2XF definedby ϕ(n) = ω \An andϕ(∞) = {∞}, is well defined. Here, the symbolF0 denotes theFréchet filter onω, i.e., the filter of all cofinite subsets ofω. Since, for everyA ∈F the set{k ∈ ω: A⊆ Ak} is finite, ϕ is l.s.c. (at the point∞∈ XF0). The spaceXF0 is compact,but there is no compactK ⊆ XF such thatϕ−1(K) = XF0. Indeed, letK be a compactsubset ofXF ; sinceK is finite, there isl ∈ ω so thatK ⊆ {0, . . . , l} ∪ {∞}, in particularϕ(l)∩K = ∅. It follows from Lemma 1.2 thatXF is not consonant.2Proposition 2.3. Let F be an ultrafilter onω. Then the following conditions areequivalent:

(1) F is aP -point ofω∗;(2) F belongs toΦ;(3) XF is consonant.

Proof. We have only to show that(1) ⇔ (2). To show that(1) ⇒ (2), let (An)n∈ω ⊆Fbe so that{0, . . . , n} ⊆An for everyn ∈ ω. SinceF is aP -point ofω∗ there isA ∈F suchthatA⊆∗ An for everyn ∈ ω. PutB0=A0, k0= 0 and suppose thatB0, . . . ,Bn andk0<

k1< · · ·< kn has been constructed. Then, choose an integerkn+1> kn satisfyingn⋃i=0

A \Bi ⊆ {0, . . . , kn+1}

and putBn+1 = Akn+1. Let I1, I2 be two infinite disjoint subsets ofω. Notice thatA ⊆Bn∪Bm for eachn 6=m, henceA⊆ (⋂k∈I1Bk)∪ (

⋂k∈I2Bk). SinceF is an ultrafilter, for

somei ∈ {1,2} we have⋂k∈Ii Bk ∈F .

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A. Bouziad / Topology and its Applications 104 (2000) 27–38 31

To show that(2) ⇒ (1), let (An)n∈ω ⊆ F be a decreasing sequence. PickI ∈ [ω]ω sothat the setB =⋂k∈I {0, . . . , k} ∪Ak belongs toF . Then, for eachn ∈ ω, if k ∈ I is suchthatk > n, we haveB \An ⊆ B \Ak ⊆ {0, . . . , k}. 2

To state the next result we borrow a small cardinal number from [2]. A familyD ⊆ [ω]ωis called agroupwise densefamily if it is open(that is, ifI ∈ [ω]ω andJ ∈D are such thatI ⊆∗ J thenI ∈ D), and if for every familyC of infinitely many pairwise disjoint finite(non-empty) subsets ofω, the union of some subfamily ofC is inD. Then

g=min{|D|: D is a set of groupwise dense families in[ω]ω with

⋂D= ∅}.

It is known thatℵ16 g (see [2] for more details concerning the cardinalg).

Proposition 2.4. Let κ < g. If (Fη)η<κ ⊆ Φ, then the filterF =⋂η<κ Fn belongs toΦ.In particular,Φ is stable with respect to taking countable intersections.

Proof. Let (An)n∈ω ⊆F be a sequence so that{0, . . . , n} ⊆ An for eachn ∈ ω. For everyη < κ letDη be the set of allI ∈ [ω]ω such that

⋂n∈I An ∈ Fη. Let us verify thatDη is a

groupwise dense family; sinceκ < g, this will imply that⋂η<κ Dη 6= ∅. It is clear thatDη

is open. LetC = (Fn)n∈ω be a family of pairwise disjoint family of finite sets inω. Picka subsequence(kn)n∈ω of the integers so that{0, . . . , n} ⊂⋂i∈Fkn Ai , and letJ ∈ [ω]ω besuch that

⋂n∈J Bn ∈Fη, whereBn =⋂i∈Fkn Ai . Then

⋃n∈J Fkn ∈Dη. 2

It is proved in [1, Theorem 4.3], thatXF is consonant for everyP -pointF of ω∗. Thefollowing gives an extension of this result.

Corollary 2.1. LetF =⋂η<κ Fη, whereκ < g andFη is aP -point ofω∗ for everyη < κ .ThenXF is consonant.

Nogura and Shakhmatov constructed in [14] two filtersF andG on ω such that thespacesXF andXG are consonant, but the topological sumXF ⊕XG is not consonant. Thefollowing result shows that the situation is rather different for filters fromΦ.

Proposition 2.5. Let (Fλ)λ∈Λ be a family of filters inΦ. Then, the topological sum⊕λ∈ΛXFλ is a consonant space.

Proof. By [14, Theorem 8.3], it suffices to show that, for every finite setI ⊆ Λ, thetopological sum

⊕λ∈I XFλ is consonant. We prove only the case|I | = 2, the general

case is similar. So, letF1 andF2 be two filters inΦ and suppose thatX =XF1 ⊕XF2 isnot consonant. LetK denote the set of (the two) non-isolated points ofX. Then, by Lem-ma 1.1, there is a nontrivial compact collectionH of open sets inX. For eachn ∈ ω, chooseAn ∈F1 andBn ∈F2 so that{0, . . . , n} ⊆ An ∩ Bn and(An ⊕Bn) ∪K /∈H, and letI bean infinite subset ofω so that

A=⋂i∈IAn ∈F1 and B =

⋂n∈I

An ∈F2.

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32 A. Bouziad / Topology and its Applications 104 (2000) 27–38

SinceH is a compact family, there isl ∈ ω such that[{0, . . . , l} ∪A] ⊕ [{0, . . . , l} ∪ B] ∪K ∈H. Fix n ∈ I so that{0, . . . , l} ⊆An∩Bn. Then, sinceA⊆An andB ⊆ Bn, we obtainthe contradiction(An ⊕Bn)∪K ∈H. 2

It is proved in [4, Corollary 8] that ifX1, . . . ,Xn is a finite sequence of regular spaces,then the consonance ofX1⊕ · · · ⊕Xn implies the consonance ofX1 × · · · ×Xn. Hencethe following statement is a consequence of Proposition 2.5.

Corollary 2.2. Let F1, . . . ,Fn be a finite sequence inΦ. Then, the product spaceXF1 × · · · ×XFn is a consonant space.

Remark 2.1. The proof of the implication (2)⇒ (1) in Proposition 2.3 shows that, ifF ∈Φ, thenF is aP -filter. The converse is not true. LetF1 be the filter of all subset ofdensity one inω. It easy to see thatA(F1) is P -set ofω∗ (see [19, Example 7.15]) andall compact subsets ofXF1 are finite. Moreover,F1 is a Borel subset of the Cantor space{0,1}ω; hence, it follows from [1, Theorem 4.1] (see also Lemma 3.2 below) thatXF1 isnot consonant.

3. Cp(XF ) and consonance ofXF

A space is a Baire space provided that every countable intersection of dense open setsis dense, and it is hereditarily Baire if all its closed subspaces are Baire. The main resultof this section shows that the spaceCp(XF ) of real-valued continuous functions onXFwith the pointwise topology is hereditarily Baire if, and only if,F ∈ Φ. In the proof weuse Debs’ characterization of hereditarily Baire metrizable space in terms of the (strong)Choquet gameΓ . Recall that the gameΓ in a spaceX is a two players (I and II) game oflengthω and is defined as follows: Player I begins and chooses a couple(V0, x0), whereV0 is an open subset ofX andx0 ∈ V0; next, player II takes an open setU0 of X such thatx0 ∈ U0⊆ V0. At thenth step player I takes(Vn, xn), whereVn ⊆Un−1 is an open set ofXandxn ∈ Vn, then II takes an open setUn of X such thatxn ∈ Un ⊆ Vn. The player I winsthe play if

⋂n∈ω Un = ∅. By Debs result [5], a metrizable spaceX is hereditarily Baire

if, and only if, player I has no winning strategy in the gameΓ in X. Besides the Choquetgame, we shall use the Banach–Mazur game which is similar to the Choquet game in everythings except that player I has only to take a non-empty open setVn of X and player IIreplies by choosing a non-empty open setUn ⊆ Vn ⊂ Un−1. It is well known that a spaceX is a Baire space if, and only if, player I has no winning strategy in the Banach–Mazurgame (see [5]). It is easily seen that in the Choquet game and in the Banach–Mazur game,if one of the players is restricted to choose his moves in a given base ofX, then the newgame is equivalent to the corresponding old one. This fact will be used without furthercomment.

In what follows, when a filterF onω is considered as a topological space, the topologyin question is the one induced by the Cantor space{0,1}ω. We will use the standardfact that the subspaceF of {0,1}ω is homeomorphic to a closed subspace ofCp(XF ).

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A. Bouziad / Topology and its Applications 104 (2000) 27–38 33

Consequently, ifCp(XF ) is hereditarily Baire, then so isF . Let us also point out that thecombinatorial criterion of Baireness ofCp(XF ) given in [11, Theorem 5.1] is preciselythe one given in [6, Lemma 2.3] to characterize Baireness ofF ; thusCp(XF ) is a Bairespace if, and only if,F is a Baire space.

For a finite subsetF ofω andA⊆ ω we use the symbol〈F,A〉 to denote the set{B ⊆ ω:F ∩A= F ∩B}. Recall that the collection of all sets of the form〈F,A〉, F finite, is baseof the Cantor space{0,1}ω.

Proposition 3.1. LetF be a filter onω and suppose thatF (as a subspace of{0,1}ω) ishereditarily Baire. ThenF ∈Φ.

Proof. Let (An)n∈ω ⊆ F be such that{0, . . . , n} ⊆ An for eachn ∈ ω. We are going toconstruct a subsequence(kn)n∈ω of the integers and a strategyσ for player I in the gameΓ in the spaceF , so that thenth move of I is of the form(Vn,Akn ∩ Akn−1 ∩ · · · ∩ Ak0).Put k0 = 0, G0 = {0} andσ(∅) = (〈G0,A0〉,A0). Suppose that the strategyσ has beendefined up to thenth move andk0, . . . , kn has been constructed. LetUn be the last moveof player II. SinceAkn ∩Akn−1 ∩ · · · ∩Ak0 ∈Un, pick a finite setFn ⊆ ω so that

〈Fn,Akn ∩Akn−1 ∩ · · · ∩Ak0〉 ⊆Un,and a natural numberkn+1> kn such thatFn ⊆Akn+1; and let

Gn+1= Fn ∪ {0, . . . , n+ 1}and

Bn+1=Akn+1 ∩Akn ∩ · · · ∩Ak0.

Define σ(U0, . . . ,Un) = (〈Gn+1,Bn+1〉,Bn+1). SinceFn ⊆ Akn+1 and Fn ⊆ Gn+1, itfollows thatσ(Un)⊆Un. This finishes the inductive construction ofσ and(Akn)n∈ω.

SinceF is hereditarily Baire, player II has a winning game againstσ , say(Un)n∈ω. LetA ∈ F be such thatA ∈⋂n∈ω Un. We claim thatA ⊆⋂n∈ω Akn . Indeed, letx ∈ A andsuppose thatx ∈ Ackl for somel ∈ ω. Let n >max{x, kl}; thenx ∈Gn andx /∈ Bn. SinceA ∈ 〈Gn,Bn〉 we obtain the contradictionx /∈A. 2

Gul’ko and Sokolov [10] have proved that an ultrafilterF is aP -point ofω∗ if and onlyif Cp(XF ) is hereditarily Baire. Therefore, by Proposition 2.3, the converse of Proposi-tion 3.1 holds ifF is an ultrafilter. Following some ideas from [10], we shall show that thisconverse is true for any filter.

Lemma 3.1. Let F ∈ Φ. Given infinite collectionsU0,U1, . . . of sets inF such that,for someB ∈ F , B ⊆ lim Un for everyn ∈ ω, it is possible to chooseAn ∈ Un so that⋂n∈ω An ∈F .

Proof. WriteUn = {Akn: k ∈ ω} and letx0< x1< · · · be an enumeration ofB. We supposewithout loss of generality that{x0, x1, . . . , xn} ⊆ Akn for every k,n ∈ ω. (If necessary,

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34 A. Bouziad / Topology and its Applications 104 (2000) 27–38

for eachn ∈ ω fix kn so that{x0, . . . , xn} ⊆ Akn for everyn ∈ ω and k > kn, and take{Akn: k > kn} in place ofUn.) For everyn ∈ ω, let

Bn =n⋂i=0

An−ii .

ThenB ⊆ limn Bn. Indeed, letk ∈ ω; for eachn < k choose someik so thatxk ∈ Ain foreveryi > ik and letl =max{k, in0 + n0}, wherein0 =max{i0, . . . , ik−1}; thenxk ∈ Bm foreverym> l.

Now, sinceF ∈Φ, there is an increasing sequence(kn)n∈ω ⊆ ω such that⋂n∈ω Bkn ∈

F , which implies that⋂n∈ω A

kn−nn ∈F . 2

The following is needed in the proof of the more general Proposition 3.2.

Lemma 3.2. If F ∈Φ, thenCp(XF ) is a Baire space.

Proof. Suppose thatCp(XF ) is not a Baire space. According to [11, Theorem 5.1]there is a sequence(Vn)n∈ω of infinite collection of finite disjoint sets inω, such thatω \⋃n∈ω Fn /∈ F for any choice ofFn ∈ Vn. Write Un = {Fc: F ∈ Vn}; thenω = lim Unfor everyn ∈ ω. By Lemma 3.1, for eachn ∈ ω there isFn ∈ Un such thatω\⋃n∈ω Fn ∈F ,which contradicts the choice of(Vn)n∈ω. 2

In their paper [10], Gul’ko and Sokolov introduce a game∆ in the spaceXF which canbe defined as follows: Player I begins and chooses a setA0 ∈F , then player II takes a finitesetF0 ⊆ A0; at thenth steep, player I chooses a setAn ∈ F disjoint fromFn−1 such thatAn ⊆An−1 and then player II replies by taking a finite setFn ⊆An. Player II wins the playif⋃n∈ω Fn ∈ F . In fact, this is a slightly modified presentation of Gul’ko and Solokov’s

game: The original definition of∆ in [10] asks thatAcn /∈F ; but, since only ultrafilters wasconsidered there, this means thatAn ∈F . Since the filters considered here are not assumedto be ultrafilters, we adopt the above description of∆. The following fact is proved in [10](Lemma 1 and the implication (5)⇒ (6) in the main result), whenF is an ultrafilter; butthe proof given there works for any filter without any changes in the arguments.

Lemma 3.3. LetF be a filter onω. If player I has a winning strategy in the gameΓ inCp(XF ), then player I has winning strategy in the game∆ in XF .

Now, we are ready to state the converse of Proposition 3.1.

Proposition 3.2. LetF be a filter onω. If F ∈Φ, thenCp(XF ) is hereditarily Baire(thusF is hereditarily Baire).

Proof. By Debs result mentioned above and Lemma 3.3, it suffices to prove that player Ihas no winning strategy for the game∆. Suppose that this is not true, and letσ be sucha strategy. We are going to define a winning strategyτ for player I in the Banach–Mazur

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game on the subspaceF of {0,1}ω. This will imply thatF is not a Baire space and thenCp(XF ) is not a Baire space (see the comment preceding Proposition 3.1). Next, in orderto finish the proof, it suffices to apply Lemma 3.2 to get a contradiction.

Every move (exceptσ(∅)) of player I, following the strategyσ , depends on some finitesubset ofω previously chosen by player II, so the familyA of all these moves is countable.SinceF is aP -filter (see Remark 2.1), there is a setB ∈ F such thatB ⊆∗ A for eachA ∈ A. For sake of simplicity, without loss of generality, we suppose thatσ(∅) = ω andB = ω and so all moves of player I in the game∆ following σ are cofinite. Every playcompatible with the strategyτ (that we are going to define) will produce another playcompatible withσ .

Putm0= 0 and letn0 ∈ ω be such that{n0, . . .} ⊆ σ(∅) and define

τ (∅)= ⟨{0, . . . , n0}, {n0, . . .}⟩

(where{n0, . . .} denotes the set{n ∈ ω: n > n0}). Suppose that (in the Banach–Mazurgame) player II takesU0 = 〈F0,B0〉 ⊆ τ (∅), where B0 ∈ F . Fix an integerm1 >max{1, n0} such thatF0⊆ {0, . . . ,m1}, an integern1 such that{n1, . . .} ⊆ σ({n0, . . . ,m1},and define

τ (U0)=⟨{0, . . . , n1}, {n1, . . .} ∪B0 ∩ F0

⟩.

Notice thatm0 < n0 6m1 < n1, thusτ (U0) is a legal move of player I, i.e.,τ (U0)⊂ U0.Now, supposeτ has been defined up to thekth steep andm0 < n0 6 m1 < n1 6 · · · <nk−1 6 mk−1 has been constructed. LetUk = 〈Fk,Bk〉 be the least move of player II inthe Banach–Mazur game, whereBk ∈ F . Choose an integermk > max{k,nk−1} so thatFk ⊆ {0, . . . ,mk}, an integernk satisfying{nk, . . .} ⊆ σ({n0, . . . ,m1}, . . . , {nk−1, . . . ,mk})and define

τ (U1,U2, . . . ,Uk)=⟨{0, . . . , nk}, {nk, . . .} ∪Bk ∩Fk ⟩.

The inductive construction ofτ is finished.Suppose thatτ is not a winning strategy for player I, and let(Un)n∈ω be a winning

play for player II againstτ . Let C ∈ F be such thatC ∈⋂n∈ω〈Fn,Bn〉. We claim thatC ⊆⋃k∈ω{nk, . . . ,mk+1}, where(nk)k∈ω and (mk)k∈ω are the sequences produced bythe play(Un)n∈ω; this will imply that player I lost the corresponding play in the game∆

following his (supposed winning) strategyσ , because({nk, . . . ,mk+1})k∈ω is a play forplayer II compatible withσ .

Let x ∈C and fixk ∈ ω such thatnk−16 x < nk (note thatn06 x sinceC ∈ τ (∅)); then

x ∈C ∩ {0, . . . , nk} =({nk, . . .} ∪Bk ∩Fk)∩ {0, . . . , nk} ⊆ {nk} ∪ {0, . . . ,mk},

which implies thatx ∈ {nk−1, . . . ,mk} or x ∈ {nk, . . . ,mk+1}. 2The following statement gives an overall picture of the main results of this note, except

the well-known equivalence between (2) and (3).

Theorem 3.1. LetF be a filter onω endowed with the subspace topology of the Cantorset{0,1}ω. Then the following are equivalent:

(1) XF is consonant and all its compact subsets are finite;

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36 A. Bouziad / Topology and its Applications 104 (2000) 27–38

(2) F is aP -filter andCp(XF ) is a Baire space;(3) F is aP -filter andF is a Baire space;(4) Cp(XF ) is hereditarily Baire;(5) F is hereditarily Baire;(6) F ∈Φ.

Note. The proof of Proposition 3.2 presented here is a slight modification of the reasoningfrom [10] (the proof of the implication (6)⇒ (1) in the main result). Since the filterFconsidered here is not assumed to be an ultrafilter, we had to replace this assumption byBaireness ofCp(XF ) given by Lemma 3.2.

4. An attempt to get a ZFC example

It would have been nice, if this note had ended by a ZFC example of a filter inΦ. Tomake obtaining such filters possibly easier in the future, we finish this note by a discussion.

For a subsetY of a spaceX, let rc(Y,X) denote therelative cellularityof Y in X; thatis the smallest cardinal numberm> ℵ0 such that any familyO of pairwise disjoint non-empty open subsets ofX, such thatO ∩ Y 6= ∅ for everyO ∈ O, has cardinality6 m.Notice thatrc(Y,X) 6 c(Z), for every subspaceZ of X such thatY ⊆ Z, wherec(Z) isthe (usual) cellularity ofZ.

If I ∈ [ω]ω, let I∗ be the set of all ultrafiltersF onω such thatI ∈ F . Finally, if Y isa subset ofω∗, we use the symbolF(Y ) to denote the filter onω obtained by taking theintersection of all ultrafilters inY .

Proposition 4.1. Let Y be a non-empty subset ofω∗ such that rc(Y,ω∗) < c (thecardinality of the continuum). ThenCp(XF(Y )) is a Baire space.

Proof. It suffices to show that the subspaceF(Y ) of {0,1}ω is a Baire space. Let(Fn)n∈ωbe a partition ofω by finite sets and let us show that there isI ∈ [ω]ω such that

ω \⋃n∈I

Fn ∈F(Y );

this will imply, by a theorem of Talagrand [16, Theorem 21], thatF(Y ) is Baire. LetI ⊆[ω]ω be an almost disjoint family of cardinalityc and suppose thatω \⋃n∈I Fn /∈ F(Y )for everyI ∈ I. Then, the intersection ofY with any set in

O ={(⋃

n∈IFn

)∗: I ∈ I

}is non-empty; butO is a collection of disjoint non-empty open subsets ofω∗, of cardinalityc, contradicting the fact thatrc(Y,ω∗) < c. 2

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A. Bouziad / Topology and its Applications 104 (2000) 27–38 37

The next result is a consequence of Theorem 3.1 and Proposition 4.1.

Corollary 4.1. For every closedP -setY of ω∗ such that rc(Y,ω∗) < c, the filterF(Y )belongs toΦ.

In particular, for every closed non-empty cccP -setY of ω∗, the filterF(Y ) is in Φ.Is it enough to conclude thatΦ is honestly non-empty? Unfortunately, likeP -points, it isknown that there is a model of set theory where there is no closed cccP -set ofω∗ (see [9]).

Question 4.1. Is there (in ZFC) a closedP -set ofω∗ of relative cellularity inω∗ lessthanc?

Acknowledgement

The author would like to thank S.P. Gul’ko and G.A. Sokolov for procuring themanuscript of their paper [10] to him.

While a final version of this article was being submitted, the author received a preprintby W. Marciszewski [12] containing some of the results of this paper; in particular [12]contains the equivalence of (3) and (4) in our Theorem 3.1.

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