A countable dense homogeneous topological vector space is a Baire space
aa r X i v : . [ m a t h . GN ] A p r A COUNTABLE DENSE HOMOGENEOUS TOPOLOGICALVECTOR SPACE IS A BAIRE SPACE
TADEUSZ DOBROWOLSKI, MIKO LAJ KRUPSKI, AND WITOLD MARCISZEWSKI
Abstract.
We prove that every homogeneous countable dense homogeneoustopological space containing a copy of the Cantor set is a Baire space. In par-ticular, every countable dense homogeneous topological vector space is a Bairespace. It follows that, for any nondiscrete metrizable space X , the functionspace C p ( X ) is not countable dense homogeneous. This answers a questionposed recently by R. Hern´andez-Guti´errez. We also conclude that, for anyinfinite dimensional Banach space E (dual Banach space E ∗ ), the space E equipped with the weak topology ( E ∗ with the weak ∗ topology) is not count-able dense homogeneous. We generalize some results of Hruˇs´ak, Zamora Avil´es,and Hern´andez-Guti´errez concerning countable dense homogeneous products. Introduction
In this note we only consider Tikhonov spaces.A space X is countable dense homogeneous ( CDH ) if X is separable and givencountable dense subsets D, D ′ ⊆ X , there is a homeomorphism h : X → X suchthat h [ D ] = D ′ . This is a classical notion going back to works of Cantor, Fr´echetand Brouwer. Among Polish spaces, canonical examples of CDH spaces include theCantor set, Hilbert cube, space of irrationals, and all separable complete metriclinear spaces (m.l.s.) and manifolds modeled on them. In contrast, every Borel butnot closed vector subspace of a complete m.l.s. is not
CDH . In recent years manyefforts have been put in constructing examples of
CDH non-Polish spaces (see theexcellent survey articles [AvM2, Section 14] and [HvM]). With no extra algebraicstructure, there even exist
CDH metrizable spaces that are meager (this notion isrecalled below), e.g., see [FZ] and [FHR]. An easy example of a non-metrizable
CDH space is the Sorgenfrey line.The main purpose of this work is examining the link between (
CDH ) and Baire-ness (and also hereditary Baireness) of a space, mainly, of a topological vector space.Recall that a space X is a Baire space if the Baire Category Theorem holds for X ;the latter means that every sequence ( U n ) of dense open subsets of X has a denseintersection in X . If every closed subset of X is a Baire space then we call X a hereditarily Baire space . A space is meager if it can be written as a countableunion of closed sets with empty interior. Clearly, if X is a Baire space, then X isnot meager. The reverse implication is in general not true. However, it holds forevery homogeneous space X (see [LM, Theorem 2.3]). The idea of studying therelationship between CDH and (hereditarily) Baireness is not new and goes backto Fitzpatrick and Zhou’s paper [FZ]. They showed, among other things, that if ahomogeneous
CDH metric space X contains a countable set which is not G δ , then X must be a Baire space. Later the topic was studied by Hruˇs´ak and Zamora Avil´es[HZA] who proved that that every analytic metric CDH space must be a hereditarilyBaire space. Our paper deals mainly with topological vector spaces.
Date : April 7, 2020.2010
Mathematics Subject Classification.
Primary: 54C35, 54E52, 46A03, Secondary: 22A05.
Key words and phrases.
Function space; pointwise convergence topology; C p ( X ) space; count-able dense homogeneity; Baire space; topological vector space; the property (B). Function spaces C p ( X ) provide a wide class of topological vector spaces to beinvestigated for CDH . By C p ( X ) we denote the space of all continuous real-valuedfunctions on a space X , endowed with the pointwise topology. V. Tkachuk hasasked if there exists a nondiscrete space X such that C p ( X ) is CDH . (For the caseof discrete X see below). Last year, R. Hern´andez-Guti´errez [HG2] gave the firstconsistent example of such a space X . He has asked whether a metrizable space X must be discrete, provided C p ( X ) is CDH [HG2, Question 2.6]. Our work wasinspired by his paper; in particular, we answer his question in the affirmative.Actually, combining our Theorem 5.2 with earlier results, we prove that, for ametrizable space X , the space C p ( X ) is CDH if and only if X is discrete of cardinalityless than the pseudointersection number p .In order to describe the example of Hern´andez-Guti´errez we need to recall somedefinitions and older results. Given a filter F on an infinite countable set T , weregard F as a subspace of 2 T , a topological copy of the Cantor set. We consideronly free filters on T , i.e., filters containing all cofinite subsets of T . By N F wedenote the space T ∪ {∞} , where ∞ 6∈ T , equipped with the following topology:All points of T are isolated and the family { A ∪ {∞} : A ∈ F } is a neighborhoodbase at ∞ . Recall that F is a P -filter if for every sequence ( U n ) of elements of F there exists an A ∈ F which is almost contained in every U n , i.e. A \ U n is finite forevery n . An ultrafilter (that is, a maximal filter) that is a P -filter is also called a P -point.Hereditarily Baire spaces C p ( N F ) have been characterized in terms of F by thethird named author (see, [Ma, Thm. 1.2]). Theorem 1.1.
Let F be a filter on ω . The following are equivalent:(a) F is a nonmeager P -filter;(b) F ⊂ ω is a hereditarily Baire space;(c) C p ( N F ) is a hereditary Baire space. Hern´andez-Guti´errez, Hruˇs´ak [HGH]; and Kunen, Medini, Zdomskyy [KMZ]added the following equivalent condition to this list (d) F ⊂ ω is CDH . Hern´andez-Guti´errez [HG2] proved that the above condition is equivalent to (e) C p ( N F ) is CDH . Recall that it is unknown whether nonmeager P -filters exist in ZFC. In particular,the following question remains open. Question 1.2.
Does there exist in ZFC a
CDH topological vector space which isnot Polish?
The equivalence of conditions (a)–(e) above inspired us to investigate the rela-tionship between the
CDH and (hereditarily) Baire properties of topological vectorspaces E . We prove that, if such a space E is CDH then it is a Baire space, cf.Corollary 3.4. In the last section, we provide examples of hereditarily Baire separa-ble pre-Hilbert spaces which are not
CDH . Consistently, there exist
CDH topologicalvector spaces which are not hereditarily Baire, cf. the comment after Theorem 5.5.Our techniques apply to Cartesian products. We show that for any separablespace X and any infinite cardinal number κ , if the product X κ is CDH , then X must be a Baire space, cf. Theorem 4.1. This generalizes a theorem of Hruˇs´ak andZamora Avil´es [HZA, Theorem 3.1], who proved the same result for the countablepower of a separable metric space X . We also show that if the product X × Y oftwo crowded spaces X and Y is CDH , then X contains a copy of the Cantor set ifand only if Y contains a copy of the Cantor set, cf. Theorem 4.5. This generalizes atheorem of Hern´andez-Guti´errez [HG1, Theorem 2.3], who obtained the same resultunder additional assumption that both spaces X and Y have countable π -weight. CDH
TOPOLOGICAL VECTOR SPACE IS A BAIRE SPACE 3 CDH and property (B) of topological vector spaces
Some spaces that we discuss are not
CDH because they contain nonhomeomorphiccountable dense subspaces. Therefore, it is reasonable to present the following term:A space X is uniquely separable ( US ) if X is separable and any two countable densesubspaces of X are homeomorphic. Obviously, for any X , CDH implies US .Recall that a space Y is referred to as crowded if every neighborhood of any pointcontains at least two distinct points. We will use the following stronger notion: Thespace Y is sequentially crowded if every point y ∈ Y is the limit of a sequence ofpoints of Y distinct from y .The classic characterization of the space of rationals Q yields Remark . Every crowded separable metrizable space is US .It turns out that for certain topological vector spaces the lack of having CDH ismanifested drastically by not even being US . That happens for spaces with property(B). Definition . A space X has the property (B) provided X can be covered bycountably many closed nowhere-dense sets { A n : n ∈ ω } such that for any compactset K ⊆ X there exists n ∈ ω with K ⊆ A n .This notion was introduced in [KM] and later studied by Tkachuk [Tk2] and[Tk3] under the name of Banakh property . Here is the main result of this section. Itstates, in particular, that no topological vector space can simultaneously have theproperty (B) and be US . Theorem 2.3.
If a topological group G has the property (B) and contains a non-trivial convergent sequence, then G is not US (hence, is not CDH ). In particular, ifa topological vector space E has the property (B), then E is not US . This theorem follows immediately from the next two lemmas and the trivialobservation that a topological vector space E with the property (B) must containa nonzero vector x , hence contains a nontrivial convergent sequence ( x/n ) n ∈ ω . Lemma 2.4.
Any separable homogeneous space X with a nontrivial convergentsequence contains a sequentially crowded countable dense subset. Furthermore, if,in addition X is a topological group, such a set can be taken as a subgroup of thegroup.Proof. First, we will present a proof for the case of a topological group. Let ( x n ) bea nontrivial convergent sequence in X . We may assume that lim x n is the neutralelement of X . Take any countable dense subset Y of X . Without loss of generality,we may further assume that all x n ’s are in Y . Then D , the smallest subgroup of X that contains Y , has the required properties.Now, assume that X is a separable homogeneous space with a nontrivial con-vergent sequence. By homogeneity, each point x ∈ X is the limit of a sequence ofpoints of X distinct from x . Therefore, for each countable subset A of X , there is acountable set B containing A , such that each point x ∈ A is the limit of a sequenceof points of B distinct from x . Take any countable dense subset D of X . Induc-tively, choose countable sets D ⊂ D ⊂ D ⊂ . . . , such that, for any n ∈ ω and x ∈ D n , there is a sequence of points of D n +1 , distinct from x , and, converging to x . One can easily verify that the union S n ∈ ω D n is the required dense subset. (cid:3) Lemma 2.5.
Let X be a space with a sequentially crowded countable dense subset.If X has the property (B), then it is not US . This property was considered by Arhangel’skii and van Mill in [AvM1] under the name of c -space. TADEUSZ DOBROWOLSKI, MIKO LAJ KRUPSKI, AND WITOLD MARCISZEWSKI
Proof.
Striving for a contradiction suppose that X is US . Then, every countabledense subset of X is sequentially crowded. Let { A n : n = 1 , . . . } be a countablefamily consisting of closed nowhere dense subsets of X witnessing the property(B). We will show that there is a convergent sequence C ⊆ X such that no A n contains C , yielding a contradiction. Without loss of generality, we can assumethat for every n we have A n ⊆ A n +1 . Fix a countable dense set D ⊆ X and let D = { d , d , d , . . . } be a faithful enumeration of D . Since the set A n is closedand has empty interior, for each n ≥ D n ⊆ X \ A n densein X . The set D n ∪ { d n } is sequentially crowded being countable dense in X , hencethere is a sequence ( x kn ) k ∈ ω of elements of D n converging to d n . Define A = { d } ∪ ∞ [ n =1 { x kn : k ∈ ω } . The countable set A is dense in X , because its closure contains D , and hence it issequentially crowded. Pick a sequence ( a m ) m ∈ ω of elements of A , convergent to d .The set C = { a m : m ∈ ω } ∪ { d } is as desired. Indeed, note that for each n ≥ C ∩ { x kn : k ∈ ω } is finite,for otherwise d n would be an accumulation point of C distinct from d . Therefore, C ∩ { x kn : k ∈ ω } 6 = ∅ , for infinitely many n ’s which means that for infinitely many n ’s C meets the set X \ A n . Since the family { A n : n = 1 , , . . . } is increasing, no A n contains C . (cid:3) It is easy to see, cf. [KM, page 653] that any infinite-dimensional Banach spacesendowed with the weak topology enjoys the property (B). More generally, we havethe following:
Proposition 2.6. [KM, Proposition 5.2]
Let ( E, k · k ) be a normed space and let τ be a linear topology on E , strictly weaker than the norm topology. If norm closedballs in E are τ -closed and τ -compact sets are norm bounded, then ( E, τ ) has theproperty (B).In particular, for an infinite-dimensional Banach space E , both spaces ( E, w ) and ( E ∗ , w ∗ ) possess the property (B). Corollary 2.7.
For an infinite-dimensional Banach space E , both spaces ( E, w ) and ( E ∗ , w ∗ ) are not US (hence, are not CDH ). All CDH topological vector spaces are Baire spaces
The following theorem and its consequences are our main results.
Theorem 3.1.
Every homogeneous
CDH space X containing a copy of the Cantorset is a Baire space. Recall that each homogeneous space is either meager or a Baire space. Therefore,the above theorem follows immediately from Lemmas 2.4 and 3.6 (below).Since a connected
CDH space is homogeneous (see [FL]), we have the following
Corollary 3.2.
Every connected
CDH space X with a copy of the Cantor set is aBaire space. The next corollary is an instant consequence of homogeneity of topological groups
Corollary 3.3.
Every
CDH topological group G containing a copy of the Cantorset is a Baire space. For the third corollary we need the trivial observation, that, for a topologicalvector space E , either E = { } (so it is a Baire space) or E contains a copy of thereal line, hence also of the Cantor set. CDH
TOPOLOGICAL VECTOR SPACE IS A BAIRE SPACE 5
Corollary 3.4. every
CDH topological vector space E is a Baire space. Now, let us present two lemmas needed to conclude Theorem 3.1.
Lemma 3.5.
Let C be a copy of the Cantor set in a separable space X . Thenthe space X contains a countable dense subset D such that D ∩ C is dense in C .Therefore, D is not a G δ -subset of X .Proof. Let A be any countable dense subset of X , and B any countable dense subsetof C . Put D = ( A \ C ) ∪ B . Clearly, D is as required. (cid:3) Lemma 3.6.
Let X be a meager space containing a copy of the Cantor set and asequentially crowded countable dense subset. Then the space X is not CDH .Proof.
The first part of the proof will follow closely the argument from the proofof Lemma 2.5. Striving for a contradiction, suppose that X is CDH . Let { A n : n = 1 , . . . } be a countable family of closed nowhere dense subsets of X such that X = S n ≥ A n . Without loss of generality, we can assume that for every n we have A n ⊆ A n +1 .Since X is CDH , every countable dense subset of X is sequentially crowded. Fixa countable dense set D ⊆ X and let D = { d , d , . . . } be a faithful enumerationof D . Since the set A n is closed and has empty interior, for each n ≥
1, there is acountable set D n ⊆ X \ A n dense in X . The set D n ∪ { d n } is sequentially crowdedbeing countable dense in X , hence there is a sequence ( x kn ) k ∈ ω of elements of D n converging to d n . Define D = ∞ [ n =1 { x kn : k ∈ ω } . The countable set D is dense in X , because its closure contains D .Now, let C ⊆ X be a copy of a Cantor set. By Lemma 3.5, X contains acountable dense subset B , such that B ∩ C is dense in C . Since X is CDH , thereexists a homeomorphism h of X with h ( B ) = D . Then, h ( B ∩ C ) = h ( C ) ∩ D ; hence, h ( C ) ∩ D is dense in C ′ = h ( C ), yet another copy of the Cantor set. To obtain acontradiction, we will show that for each C ′ , a copy of the Cantor set, C ′ ∩ D is notdense in C ′ . By the Baire Category Theorem, there is n such that C ′ ∩ A n hasa nonempty interior in C ′ . The set D ∩ A n , being a subset of S n − n =1 { x kn : k ∈ ω } ,has only finitely many accumulation points; hence, C ′ ∩ D is not dense in C ′ . (cid:3) Let us note that, for metrizable spaces, the assertion of Theorem 3.1 is known(see [Me, Corollary 3] or cf. remarks preceding Proposition 2.2 in [HG2]). For thesake of self-completeness we include the sketch of proof. If a separable metrizablehomogeneous space were not a Baire space then, as pointed out below Theorem 3.1,it would be meager. This would allow to construct a G δ countable dense subsetwhich, in turn, by CDH , yields that all countable subsets would be G δ (see [FZ,Theorem 3.4]). The latter contradicts our Lemma 3.5 (see also [FZ, Theorem 3.5]).Recall the following result, proved by Fitzpatrick and Zhou in [FZ]: Theorem 3.7. [FZ, Theorem 2.5]
Any
CDH space X can be represented as a count-able union of a discrete family of clopen homogeneous CDH subspaces U n . Combining the above theorem with our Theorem 3.1, we get the following:
Theorem 3.8.
Every
CDH space X is a countable union of a discrete family ofhomogeneous clopen CDH subspaces U n , such that each U n is either a Baire spaceor does not contain a copy of the Cantor set. In particular, X is a union of disjointclopen CDH subspaces U and V such that U is a Baire space, V is meager and doesnot contain a copy of the Cantor set. TADEUSZ DOBROWOLSKI, MIKO LAJ KRUPSKI, AND WITOLD MARCISZEWSKI
Proof.
The first assertion is a direct consequence of Theorems 3.7 and 3.1. To getthe second assertion, we take U = [ { U n : U n is Baire } and V = [ { U n : U n is meager } (cid:3) The following fact is probably known.
Proposition 3.9.
Let E be an infinte-dimensional normed space and F be a vectorsubspace of the dual E ∗ separating points of E . Then the space E equipped with theweak topology generated by F is not a Baire space.Proof. Let A = T { ϕ − ([ −k ϕ k , k ϕ k ]) : ϕ ∈ F } . Clearly A is closed in the weaktopology generated by F . Using the facts that F separates the points of E andeach weak neighborhood of 0 contains a nontrivial vector subspace, one can easilyverify that A has empty (weak) interior. Since the unit ball of E is contained in A , we have E = S n ∈ ω nA . This shows that E with the weak topology is meager,hence not a Baire space. (cid:3) From this fact and Corollary 3.4 we immediately obtain
Corollary 3.10.
Let E be an infinite-dimensional normed space and F be a linearsubspace of the dual E ∗ separating points of E . Then the space E equipped with theweak topology generated by F is not CDH . Corollary 3.10 does not extend to non-normed vector spaces. For the productspace R ω , which is CDH , its weak topology coincides with the product topology.4.
Cartesian products with CDH
In the paper [HZA, Theorem 3.1] it was proved that, for a separable metrizablespace X , if the product X ω is CDH , then X is a Baire space. Using our Lemma 3.6we can generalize this result as follows Theorem 4.1.
For a separable space X (not necessarily metrizable) and an infinitecardinal κ , if X κ is CDH , then X and X κ are Baire spaces. In the proof we will use two additional easy lemmas, the first one is probablyknown.
Lemma 4.2.
For any space X and an infinite cardinal κ , the product X κ is eithermeager or a Baire space.Proof. Suppose that X κ is not meager. Then the Banach Category Theorem (see[HM, Theorem 1.6]) gives us a nonempty open subspace U ⊂ X κ which, in itsrelative topology, is a Baire space. Since an open subset of Baire space is again aBaire space, without loss of generality we can assume that U is a basic open set,i.e., U = π − A ( V ), where A is a finite subset of κ , V a nonempty open subset of X A ,and π A : X κ → X A is the projection. Then, for the open continuous projection π κ \ A : X κ → X κ \ A we have π κ \ A ( U ) = X κ \ A , hence X κ \ A is a Baire space, cf.[HM, Corollary 4.2]. Obviously, X κ \ A is homeomorphic to X κ . (cid:3) Lemma 4.3.
Let X be a space of cardinality | X | > , and κ be an infinite cardinal.Then each point x ∈ X κ is contained in a copy of the Cantor set C ⊂ X κ .Proof. Let x = ( x α ) α<κ . For each n ∈ ω , pick a subset C n ⊂ X of size 2, such that x n ∈ C n . For ω ≤ α < κ put C α = { x α } . It is clear that the set C = Q α<κ C α isas required. (cid:3) CDH
TOPOLOGICAL VECTOR SPACE IS A BAIRE SPACE 7
Proof of Theorem 4.1.
Assume that the product X κ is CDH . Clearly, we can assumethat | X | >
1. We will show that X κ contains a sequentially crowded countable densesubset D . Let A be any countable dense subset of X κ . For each a ∈ A , use Lemma4.3 to find a copy C a ⊂ X κ of the Cantor set, containing a . Let D a be a countabledense subset of C a . One can easily verify that the set D = S { D a : a ∈ A } has therequired properties.Now, by Lemmas 3.6 and 4.3, X κ is not meager. Therefore, Lemma 4.2 impliesthat X κ is a Baire space. Since X is an image of X κ under an open continuousprojection, it is also a Baire space. (cid:3) R. Hern´andez-Guti´errez established in [HG1] the following result concerningCartesian products with
CDH (we are grateful to him for turning our attentionto this result):
Theorem 4.4. [HG1, Theorem 2.3]
Let X and Y be two crowded spaces of countable π -weight. If X × Y is CDH , then X contains a copy of the Cantor set if and onlyif Y contains a copy of the Cantor set. Using methods developed above, we can generalize this theorem by eliminatingthe assumption that spaces under consideration have countable π -weight (see [Tk1,1.4] for the definition of π -weight), i.e. we have: Theorem 4.5.
Let X and Y be two crowded spaces. If X × Y is CDH , then X contains a copy of the Cantor set if and only if Y contains a copy of the Cantorset. We will need the following lemma.
Lemma 4.6.
Let U be an open homogeneous CDH subset of the product X × Y of two spaces X and Y . If Y is crowded and U contains a nontrivial convergentsequence, then U contains a countable dense subset E such that for every y ∈ Y theintersection E ∩ ( X × { y } ) has at most one element.Proof. Let π Y : X × Y → Y be the projection map. By Lemma 2.4, the set U contains a countable dense subset D = { d , d , . . . } which is sequentially crowded.Since U is CDH , every countable dense subset of U is sequentially crowded. Weshall construct E inductively. The fiber X = ( X × { π Y ( d ) } ) ∩ U is a closednowhere-dense subset of U (because U is open in X × Y and Y has no isolatedpoints). Thus, the set ( D \ X ) ∪ { d } is dense in U , hence, sequentially crowded.We infer that there is a sequence ( e k ) k ∈ ω of elements of D \ X convergent to d .Passing to a subsequence we can ensure that all points in the sequence ( π Y ( e k )) k ∈ ω are distinct.Now, suppose that for n ≥ m < n , we have constructed sequences( e km ) k ∈ ω such that:(1) π Y ↾ S m Let π Y : X × Y → Y be the projection map. By symmetryit is enough to show that if X contains a copy of the Cantor set, then Y containsa copy of the Cantor set too. Suppose that X contains a copy of the Cantorset. Then, clearly, the product X × Y contains a copy of the Cantor set C ′ . ByTheorem 3.7, there is a clopen homogeneous CDH space U which meets C ′ . Since U is clopen, the intersection C = U ∩ C ′ is again a copy of the Cantor set. ByLemma 3.5, there is a countable dense subset D of U such that D ∩ C is dense in C . Applying Lemma 4.6 to U , we infer that there is a countable dense subset E of U such that the map π Y ↾ E is one-to-one. Now, since U is CDH , there existsa homeomorphism h : U → U such that h ( D ) = E . The set D ∩ C is dense in C ,so the set F = E ∩ h ( C ) is dense in h ( C ), yet another copy of the Cantor set. Itfollows that F is crowded and π Y ↾ F is one-to-one. Obviously, π Y ( F ) is crowdedand it is dense in π Y ( h ( C )). Therefore, π Y ( h ( C )) is a crowded compact metrizablesubspace of Y , so it must contain a copy of the Cantor set. (cid:3) One can easily observe that the above proof actually gives the following. Theorem 4.7. Let Q t ∈ T X t be a CDH product of crowded spaces X t . If Q t ∈ T X t contains a copy of the Cantor set, then each X t does so. Since every infinite product of crowded spaces contains a copy of the Cantor set,we immediately obtain the next corollary which generalizes [HG1, Corollary 2.5]. Corollary 4.8. Let T be an infinite set, and X t be a crowded space for t ∈ T . Ifthe product Q t ∈ T X t is CDH , then each X t contains a copy of the Cantor set. CDH of C p ( X )By Corollary 3.4, we know that if a function space C p ( X ) is CDH , then it isnecessarily a Baire space. Interestingly, there is a topological characterization ofthe space X for which the space C p ( X ) is a Baire space (see [vM, Theorem 6.4.3]).The proof of the following simple fact (cf. [Tk1, S.284]) does not require a use ofthis characterization result, however. First, recall that a subset A of X is bounded if for every f ∈ C p ( X ), the set f ( A ) is bounded in R . Note, that every finite set A is bounded, so this notion is meaningful if A is infinite. Proposition 5.1. Let X be a space containing an infinite bounded subset A . Then C p ( X ) is meager.Proof. For n ∈ ω , we set F n = { f ∈ C p ( X ) : f [ A ] ⊆ [ − n, n ] } . Clearly, the sets F n are closed and have empty interiors in C p ( X ) because A isinfinite. Moreover, since A is a bouded subset of X , we have S n ∈ ω F n = C p ( X ). (cid:3) CDH TOPOLOGICAL VECTOR SPACE IS A BAIRE SPACE 9 Combining the proposition above with Corollary 3.4 we immediately obtain: Theorem 5.2. Let X be a space containing an infinite bounded subset A . Then C p ( X ) is not CDH . Obviously, a nontrivial convergent sequence in a space X forms an infinite boundedsubset. Therefore, we obtain the next corollary which answers Question 2.6 from[HG2]. Corollary 5.3. Let X be a metrizable space. If C p ( X ) is CDH then X is discrete. In the context of the CDH property of C p ( X ) spaces, it is worth recalling that C p ( X ) is separable if and only if X admits a weaker separable metrizable topology(see [Tk1, S.174]). If such X contains additionally an infinite bounded subset, thenit actually contains a nontrivial convergent sequence. Indeed, if A is bounded so isits closure ¯ A . By [AT, Corollary 6.10.9], ¯ A is compact. Having a weaker separablemetrizable topology ¯ A is a metrizable compactum; therefore, A contains a nontrivialconvergent sequence. We are indebted to Alexander Osipov for informing us aboutthis observation. Therefore, by the above fact, in Theorem 5.2 (and also in Propo-sition 5.8 below) instead of assuming that X contains an infinite bounded subset,one can equivalently assume that X contains a nontrivial convergent sequence.For a discrete space X , we have C p ( X ) = R X . The following characterization of CDH products of the real line incorporates assertions that can be found in [SZ] and[HZA], cf. [HG2, Section 6]. Theorem 5.4. The product R X ( [0 , X , { , } X ) is CDH if and only if X is ofcardinality less than the pseudointersection number p . The above result together with our Corollary 5.3 yields: Theorem 5.5. Let X be a metrizable space. Then C p ( X ) is CDH if and only if X is discrete of cardinality less than p . Very recently G. Plebanek has shown that, for X of cardinality ω , the productspace R X is not a hereditarily Baire space . Hence, if X is a discrete space ofcardinality ω then the space C p ( X ) is not a hereditarily Baire space.This implies that, under an additional set-theoretic assumption that ω < p , theproduct R ω is an example of a topological vector space ( C p ( X ) space) which is CDH but is not a hereditarily Baire space. Thus, it is not possible to replace Baire by hereditarily Baire in Corollary 3.4, even for spaces of the form C p ( X ). However,the following question remains open. Question 5.6. Suppose that X is countable and C p ( X ) is CDH . Is C p ( X ) a hered-itarily Baire space? Note that by the results of [Ma] the existence of a countable nondiscrete X suchthat C p ( X ) is a hereditarily Baire space is equivalent to the existence of a nonmeager P -filter on ω .An obvious consequence of Corollary 5.3 is the following Corollary 5.7. If X is an uncountable separable metrizable space, then C p ( X ) isnot CDH . The last corollary can be also proved in a different, more direct way. Let us showthe following Proposition 5.8. Let X be a separable space which contains an infinite boundedsubset A . If C p ( X ) is separable, then it contains a countable dense subset E whichis a G δ -set in C p ( X ) . Jan van Mill has found a short proof of this result and has kindly allowed us to include it inour paper. His argument can be found in Appendix. Proof. By a standard inductive argument we can pick a sequence ( a n ), a n ∈ A , anda sequence ( U n ) of pairwise disjoint open sets such that a n ∈ U n , for n ∈ ω . For each n , take a continuous function g n : X → [0 , 1] such that g n ( a n ) = 1 and g n ( x ) = 0for x ∈ X \ U n . Let { h n : n ∈ ω } be dense in C p ( X ), and let r n = | h n ( a n ) | + n . Put f n = h n + r n g n and define E = { f n : n ∈ ω } . We will show that E has the requiredproperties.Take any finite set F = { x , . . . , x k } ⊆ X , and a sequence V , . . . , V k of nonemptyopen subsets of R , and consider the basic open set W = { f ∈ C p ( X ) : f ( x i ) ∈ V i , i ≤ k } in C p ( X ) given by these sequences. Observe that W contains infinitely many h n ,since C p ( X ) is crowded. On the other hand F intersects only finitely many U n .Therefore we can find n such that h n ∈ W and F ∩ U n = ∅ . Then f n | F = h n | F ,hence f n ∈ W .Let B k = { f ∈ C p ( X ) : ∀ a ∈ A | f ( a ) | ≤ k } . Clearly, all sets B k are closedand C p ( X ) = S k ∈ ω B k . From the definition of r n and f n immediately follows that | f n ( a n ) | ≥ n , therefore f n / ∈ B k for n > k .Since X is separable, each finite subset of C p ( X ) is a G δ -set (see [Tk1, S.173]).Therefore C p ( X ) \ E = [ k ∈ ω ( B k \ { f n : n ≤ k } )is an F σ -set and E is a a G δ -subset of C p ( X ). (cid:3) From Lemma 3.5 and Proposition 5.8 we immediately obtain version of Theorem5.2 for separable X .For a space X and a subset Y of R we denote by C p ( X, Y ) the subspace of C p ( X )consisting of functions with values in Y , i.e., C p ( X, Y ) = C p ( X ) ∩ Y X . Let I denotethe unit interval [0 , C p ( X, I ) we have the following counterpart ofTheorem 5.5 Theorem 5.9. Let X be a metrizable space. Then C p ( X, I ) is CDH if and only if X is discrete of cardinality less than p . For the proof we need a weaker version of Proposition 5.1 for C p ( X, I ) Proposition 5.10. Let X be a space containing a nontrivial convergent sequence.Then the space C p ( X, I ) is meager.Proof. Let ( x n ) be a sequence of distinct points of X converging to a point x ∈ X .For n ∈ ω , we set F n = { f ∈ C p ( X, I ) : ∀ ( k ≥ n ) | f ( x k ) − f ( x ) | ≤ / } . Clearly, the sets F n are closed and C p ( X, I ) = S n F n , because, for each f ∈ C p ( X, I ), f ( x n ) → f ( x ). Given n , the set F n has empty interior in C p ( X, I ), since, for any f ∈ C p ( X, I ) and any finite A ⊂ X , we can find k ≥ n such that x k / ∈ A , and afunction g ∈ C p ( X, I ) such that g | A = f | A and | g ( x k ) − g ( x ) | ≥ / (cid:3) Lemma 5.11. For a space X , if C p ( X, I ) is separable, so is the space C p ( X, (0 , .Proof. Let A be a countable dense subset of C p ( X, I ). One can easily verify that D = { (1 − /n ) f + 1 / (2 n ) : f ∈ A, n = 1 , , . . . } , is a dense subset of C p ( X, (0 , (cid:3) Proof of Theorem 5.9. For a discrete space X we have C p ( X, I ) = I X , hence the“if” part of the theorem follows directly from Theorem 5.4.To show the “only if” part, assume, towards a contradiction, that C p ( X, I ) is CDH and X is not discrete. Since X contains a nontrivial convergent sequence, by CDH TOPOLOGICAL VECTOR SPACE IS A BAIRE SPACE 11 Proposition 5.10, the space C p ( X, I ) is meager. Obviously, C p ( X, I ) contains a copyof I (constant functions), hence also a copy of the Cantor set. By Lemma 5.11,the space C p ( X, (0 , C p ( X ), byLemma 2.4, it contains a sequentially crowded countable dense subset D . Moreover,using the fact that the space C p ( X, (0 , C p ( X, I ), the set D is alsodense in C p ( X, I ). Now, the desired contradiction follows from Lemma 3.6. (cid:3) Remark . Using a similar method as above, one can also prove that, for a zero-dimensional metrizable space X , the space C p ( X, { , } ) is CDH if and only if X isdiscrete of cardinality less than p .Recall that C ∗ p ( X ) is the subspace of C p ( X ) consisting of bounded functions. It iswell known that for an infinite space X , C ∗ p ( X ) is not a Baire space (cf. Proposition3.9). Corollary 5.13. For no infinite space X , the space C ∗ p ( X ) is CDH . Our next theorem and corollary slightly strengthen the results of Osipov, cf.[HG2, Sec. 8] and [Os]. Their statements use the notion of γ -set; here is one way ofphrasing this notion: A space X is a γ -set if for any open ω -cover U of X , there is asequence ( U n ) n ∈ ω , U n ∈ U , such that every point x ∈ X belongs to all but finitelymany U n ’s. Recall that a cover U of a space X is an ω -cover if, for any finite subset F of X , there exists U ∈ U such that F ⊆ U . Theorem 5.14. Let X be a space such that X n is hereditary Lindel¨of for every n ∈ ω . If X is not a γ -set, then C p ( X ) is not US . Corollary 5.15. If X is a separable metrizable space which is not a γ -set, then C p ( X ) is not US . Theorem 5.14 is an immediate consequence of Lemma 2.4 and the next lemma.The proof of that lemma follows closely the argument from the proof of the Gerlits-Nagy theorem stating that a space X is a γ -set if and only if C p ( X ) is a Fr´echet-Urysohn space, cf. [Ar, Thm. II.3.2]. Lemma 5.16. Let X be a space such that X is not a γ -set and X n is hereditaryLindel¨of for every n ∈ ω . Then C p ( X ) contains a countable dense subset which isnot sequentially crowded.Proof. Recall that the assumption that all finite powers of X are hereditary Lindel¨ofimplies that C p ( X ) is hereditary separable (see [Ar, II.5.10]).Let U be an open ω -cover of X witnessing the fact that X is not a γ -set. Define G = { f ∈ C p ( X ) : ∃ ( U ∈ U ) f − ( R \ { } ) ⊆ U } . Using the fact that U is an ω -cover one can easily verify that the set G is densein C p ( X ). Take a countable dense subset H of G and put D = H ∪ { c } . Wewill prove that D is not sequentially crowded. To this end, we will show that nosequence ( f n ) of functions from D , distinct from c , converges to c ; here, c is aconstant function with value of 1 . Suppose the contrary. For each n find U n ∈ U with f − n ( R \ { } ) ⊆ U n . Given x ∈ X , we have f n ( x ) → 1, hence f n ( x ) = 0 for n > n . It follows that x ∈ U n for n > n , a contradiction with our assumption on U . (cid:3) Corollary 5.15 suggests the following: Question 5.17. Let X be an uncountable separable metrizable space. Is it true that C p ( X ) is not US ? Let us also comment on Question 2.5 from [HG2]. Hern´andez-Guti´errez asked if,for an uncountable Polish space X , the space C p ( X ) has a countable dense subsetwithout nontrivial convergent sequences. An affirmative answer to this questionfollows immediately from [Tk2, Cor. 3.20] and the fact that, for such X , the space C p ( X ) is hereditary separable (it has a countable network).We conclude this section with two observations related to the characterization ofHern´andez-Guti´errez [HG2, Theorem 1.2] of the CDH property of C p ( N F ) spacesthat we quoted in the introduction. Proposition 5.18. It is consistent (relative to ZFC ) that there exists ω many CDH spaces C p ( N F ) which are pairwise non-homeomorphic.Proof. It is well known that the continuum hypothesis or Martin’s Axiom impliesthe existence of 2 ω many P -ultrafilters F on ω (cf. [Bl]), and each ultrafilter F is nonmeager. By [HG2, Theorem 1.2] all corresponding function spaces C p ( N F )are CDH . One can easily observe that different filters F, F ′ generate distinct spaces C p ( N F ) , C p ( N F ′ ). Indeed, the intersection { f ∈ C p ( N F ) : f ( ∞ ) = 0 } ∩ N F is acanonical copy of F in C p ( N F ). Now, we can repeat the standard counting argument(cf. [Ma3]). We have 2 ω different CDH spaces C p ( N F ). For a given filter F , thespace C p ( N F ), being a separable metrizable space, can be homeomorphic to nomore than 2 ω other spaces C p ( N F ′ ). (cid:3) The proof of the next fact follows very closely the argument from the proof ofLemma 3.2 in [Ma], for the reader’s convenience we repeat it here. Proposition 5.19. Let ( F n ) be a sequence of nonmeager P -filters on ω . Then theproduct Q n ∈ ω C p ( N F n ) is CDH .Proof. First, recall that for every filter F the space C p ( N F ) is homeomorphic tothe space c F = { f ∈ C p ( N F ) : f ( ∞ ) = 0 } , see [Ma1, Lemma 2.1]. Hence the space Q n ∈ ω C p ( N F n ) is homeomorphic to the product Q n ∈ ω c F n . We can consider theproduct Q n ∈ ω F n as a filter F on ω × ω . To do this, we identify ( A n ) ∈ Q n ∈ ω F n with S { A n × { n } : n ∈ ω } ⊂ ω × ω . By [Ma, Corollary 2.4] (cf. [Sh, p. 327,Fact 4.3]), F is a hereditary Baire space, hence a nonmeager P -filter. Therefore,by [HG2, Theorem 1.2], C p ( N F ) and c F are CDH . One can easily verify that c F ishomeomorphic to Q n ∈ ω c F n . (cid:3) Bernstein-like direct sum decompositions of Roman Pol In this section we show that there are examples of hereditarily Baire m.l.s. whichare not CDH . Let us recall the notion introduced by R. Pol in [Po]. Let E be aninfinite-dimensional separable complete m.l.s.. A direct sum decomposition E = V ⊕ V is called a Bernstein-like direct sum decomposition of E , if every linearlyindependent Cantor set C (i.e., a topological copy of the Cantor set) in E intersectsboth subspaces V i . R. Pol showed that every infinite-dimensional separable completem.l.s. admits a Bernstein-like direct sum decomposition. We will also use Lemma3.1 from [Po]: Lemma 6.1. Let A be an analytic set in a separable complete m.l.s. E . If A contains an uncountable linearly independent set, then A contains a linearly inde-pendent Cantor set. Lemma 6.2. Let Q be a closed copy of the rationals Q in a topological vector space V . Then, for any nonempty relatively open subset U of Q , the linear subspace span U (that is, a vector subspace of V spanned by U ) is infinite-dimensional.Proof. Suppose that, for some nonempty relatively open subset U of Q , span U isfinite-dimensional. Pick a nonempty clopen subset V of Q contained in U . Then CDH TOPOLOGICAL VECTOR SPACE IS A BAIRE SPACE 13 the space F = span V is finite dimensional, hence homeomorphic to some Euclideanspace R n . Since V is a topological copy of Q which is closed in F , we arrive at acontradiction. (cid:3) Lemma 6.3. Let A be a nonempty closed subset of a complete m.l.s. E , such that,for any nonempty relatively open subset U of A , span U is infinite-dimensional.Then A contains an uncountable linearly independent set.Proof. Suppose the contrary, then the subspace F = span A is spanned by a count-able infinite set { x i : i ∈ ω } . Put F n = span { x i : i ≤ n } , n ∈ ω . Each F n is closedin F , and F = S n ∈ ω F n . By the Baire Category Theorem there is n such that U = Int A ( F n ∩ A ) = ∅ . By our assumption on A , span U is infinite-dimensional,but it is contained in F n , a contradiction. (cid:3) Proposition 6.4. Let E = V ⊕ V be a Bernstein-like direct sum decompositionof a separable infinite-dimensional complete m.l.s. E . Then each summand V i is ahereditarily Baire space.Proof. Suppose the contrary, then by the Hurewicz theorem the space V i contains aclosed copy Q of the rationals Q . Let A be a closure of Q in E . For any nonemptyrelatively open subset U of A , span( U ∩ Q ) is infinite-dimensional by Lemma 6.2.Therefore, Lemma 6.3 implies that A contains an uncountable linearly independentset. Then the set A \ Q is analytic and contains an uncountable linearly independentset. From Lemma 6.1, we infer that A \ Q contains a linearly independent Cantorset C . The set C is disjoint from V i , a contradiction. (cid:3) Proposition 6.5. Let E = V ⊕ V be a Bernstein-like direct sum decompositionof a separable infinite-dimensional complete m.l.s. E . Then each subspace V i is not CDH .Proof. Using a countable base of V i and fact that each nonempty open subset U of V i has infinite-dimensional span U , one can easily inductively construct a countabledense subset D of V i which is linearly independent. By Lemma 3.5 it is enough toshow that for any copy K of the Cantor set in V i , K ∩ D is not dense in K . Supposethat K ∩ D is dense in K . Then, for any nonempty relatively open subset U of K , U ∩ D is infinite, so span U is infinite-dimensional. Hence, by Lemma 6.3, K contains an uncountable linearly independent set. Now, by Lemma 6.1, the lattercontains a linearly independent Cantor set C . Then C is disjoint from V j , for j = i ,a contradiction. (cid:3) Question 6.6. Does there exist (in ZFC) a noncomplete CDH pre-Hilbert space? Appendix A. R ω is not a hereditarily Baire space The aim of this appendix is to give a proof of a recent (unpublished) result byPlebanek that we quoted in Section 5. This short and elegant proof is due to Janvan Mill, whom we thank for allowing us to include it in this paper. Theorem A.1 (G. Plebanek) . The product R ω is not a hereditarily Baire space.Proof (van Mill). Since the space of natural numbers ω is a closed subspace of thereal line R it is enough to prove that he product ω ω contains a closed subset A which is not a Baire space. Let A = { f : ω → ω : f is not decreasing } . Since f ∈ A if and only if, for all α < β < ω we have f ( α ) ≤ f ( β ), the subspace A isclearly closed. If f ∈ ω ω is unbounded, then it is unbounded on some countableinterval [0 , α ), so by looking at the value f ( α ) we can examine that f / ∈ A . Hencewe can write A as S n A n , where A n = { f ∈ A : f ≤ n } . 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Department of Mathematics, Pittsburg State University, Pittsburg, KS 66762, USA E-mail address : [email protected] Institute of Mathematics, University of Warsaw, Banacha 202–097 Warszawa, Poland E-mail address : [email protected] Institute of Mathematics, University of Warsaw, Banacha 202–097 Warszawa, Poland E-mail address ::