Note on countable unions of Corson countably compact spaces

Note on countable unions of Corson countably compact spaces

Ondˇrej F.K. Kalenda

Abstract. We show that a compact spaceK has a dense set ofGδ points if it can be covered by countably many Corson countably compact spaces. If these Corson countably compact spaces may be chosen to be dense inK, thenKis even Corson.

Keywords: Corson countably compact space,Gδpoint, Corson compact space, Valdivia compact space

Classification: 54D30

Introduction

Corson compact and countably compact spaces, i.e. compact and countably compact subsets of the space

Σ(Γ) ={x∈R^Γ: suppx={γ∈Γ :x(γ)6= 0} is countable}

equipped with the pointwise topology play an important role in theory of nonsep- arable Banach spaces. Functional analytic properties of Corson compacta were studied for example in [AM], [AMN], [O] or [V1]. In the present paper we study certain topological properties of these classes of spaces.

It is well known that any Corson compact space has a dense set ofG_δ points (an easy proof is given in [K4, Theorem 3.3]). As Corson compact spaces are stable to continuous images (by [MR] or [G]), the same obviously holds for con- tinuous images of Corson compact spaces. On the other hand, there are Corson countably compact spaces with noG_δ points (take for example [0,1]^Γ∩Σ(Γ) for Γ uncountable). Corson countably compact spaces are stable to quotient images by [G] but not to general continuous images (an easy example is the space [0, ω₁] which is a continuous image of the Corson countably compact space [0, ω₁) but is not itself Corson — for a more general formulation see [K5, Theorem 2.5]). Hence the following problem [K5, Question 1] seems to be quite natural.

Supported by Research grants GAUK 277/2001, GA ˇCR 201/03/0933 and MSM 113200007.

Problem 1. LetK be a compact space which is a continuous image of a Corson countably compact space. DoesK have a dense set of G_δ points?

We do not know the general answer to this problem but our Theorem 1 gives a partial positive answer for compact spaces which are finite unions of Corson countably compact spaces. The stated question is natural in itself but moreover it is related to problems concerning the structure of open continuous images of Valdivia compacta. Let us recall the definition of Valdivia compacta.

LetK be a compact space and A⊂K. We say thatA is a Σ-subset ofK if there is a homeomorphic injection h:K→R^Γ for a set Γ with A=h⁻¹(Σ(Γ)).

A compactK is Valdivia if it has a dense Σ-subset. Recall also that a dense set A⊂Kis a Σ-subset if and only ifAis a Corson countably compact space andK is the ˇCech-Stone compactification ofA(i.e.,K=βA), see [K4, Proposition 1.9].

Valdivia compact spaces are not stable to continuous images (see [V2], [K1] or [K4, Section 3.3]). However, the following question remained open until recently.

Problem 2. Letϕ:K→Lbe an open continuous surjection between compact spaces. SupposeK is Valdivia. IsL Valdivia as well?

A counterexample was recently found in [KU]. A partial positive answer is given in [K2, Theorem 4.5] (see also [K4, Theorem 3.24]). It is proved there that the answer is positive provided L has a dense set of G_δ points. In fact, it is proved slightly more — ifL has a dense set ofG_δ points andA⊂K is a dense Σ-subset ofK, thenϕ(A) is a dense Σ-subset ofL. This more general result does not hold without the assumption onL(an easy example is given in Remark after Theorem 4.5 in [K2], see also [K4, Remark 3.25]). It follows that the positive answer to Problem 1 would give positive answer to the following question.

Problem 3. Letϕ:K→Lbe an open continuous surjection between compact spaces. SupposeAis a dense Σ-subset of Ksuch that ϕ(A) =L. IsLCorson?

This problem may look a bit artificially in itself but the answer to it may help to better understand Problem 2. (Note that the counterexample of [KU] uses cohomology theory.)

Main results

Our main results are the following two theorems.

Theorem 1. Let K be a compact space such thatK =S∞

i=1Bi where Bi is a Corson countably compact space for eachi∈N. ThenK has a dense set of G_δ points.

Theorem 2. Let K be a compact space such thatK =S∞

i=1Bi where Bi is a Corson countably compact space dense inKfor eachi∈N. ThenK is a Corson compact space.

Note that Theorem 1 yields a partial positive answer to Problem 1. Compact spaces which are finite unions of Corson countably compact spaces form a subclass of continuous images of Corson countably compact spaces (as finite union is a continuous image of a finite topological sum; cf. [K5, Lemma 2.2]). It follows from [S] that a compact space from Theorem 1 need not be a continuous image of Corson countably compact space. (In fact, the quoted paper contains an example of a compact space which is a countable union of Corson (even Eberlein) compact subspaces without being Corson. It is easy to check that this space is not even a continuous image of a Valdivia compact space.)

The author knows no example of a concrete compact space for which Theorem 1 would yield a non-trivial result. The assumptions are satisfied for example by Corson compact spaces, by the spaces [0, α] forα < ω1·ω1 or by the space from [S]. In all these examples it is easy to check they have a dense set ofGδ points.

However, the real meaning of Theorem 1 is different. Its purpose is to show that there are no non-trivial spaces satisfying its assumptions.

Remark also that Theorem 2 gives a positive answer to Problem 3 for the case K =L×C whereC is a countable compact space andϕ being the natural projection of K ontoL. (The answer to Problem 2 in this case is trivial asLis homemomorphic to a clopen subset ofK.)

Another consequence of Theorem 2 is the following. IfK is a super-Valdivia compact space (i.e. the family of dense Σ-subsets ofK coversK), then eitherK is Corson orK cannot be covered by countably many Corson countably compact subspaces.

Theorems 1 and 2 are easily seen to be equivalent. Indeed, suppose Theorem 1 holds and letKandBi,i∈N, be as in Theorem 2. ThenKhas, by Theorem 1, a dense set ofGδpoints. EachBicontains allGδpoints ofK (by Lemma 3 below) and so for any pairi, j∈Nthe setB_i∩B_j is dense inK, and henceB_i=B_j (by Lemma 2). HenceK=B₁=B₂=. . . and thusK is Corson.

Conversely, suppose that Theorem 2 holds and letK andB_i,i ∈N, be as in Theorem 1. LetU ⊂Kbe a nonempty open set. We can construct by induction nonempty open setsV_i,i∈Nsuch that

U ⊃V₁ ⊃V₁⊃V₂⊃. . .

and that for eachi∈Nthe set Vi∩Bi is either empty or dense inVi. Further, put H =T

i∈NVi. This is clearly a nonempty closedG_δ subset ofK. Moreover, H∩Bi is dense in H wheneverH∩Bi 6=∅ (by Lemma 3). Hence H is Corson by Theorem 2. It follows thatH has aG_δ point. Such a point is also a G_δ point ofK. This completes the proof.

Auxiliary results

In this section we collect some auxiliary results needed to prove Theorems 1 and 2. Most of them are known but we recall their formulation.

Lemma 1. Let X be a Corson countably compact space. Then the following holds.

(a) X is Fr´echet-Urysohn, i.e. whenever x ∈ X and A ⊂ X are such that x∈Athere is a sequencexn∈Awithxn→x.

(b) Cis compact for anyC⊂X countable. In particular, if Y is a space con- tainingX, thenC⊂X wheneverC⊂X is countable, i.e.X is countably closed inY.

The point (a) follows from [N, Theorem 2.1], see also [K4, Lemma 1.6]. The point (b) is obvious.

Lemma 2. LetKbe a space,AandBtwo Corson countably compact subsets of KandM be any subset of K. If M∩A∩B is dense inM, thenM∩A=M∩B.

This is an easy consequence of Lemma 1. It was observed in [K2, Lemma 2.15].

Lemma 3. Let K be a regular space and A ⊂K a dense countably compact subset. ThenG∩Ais dense inGfor eachG_δ setG⊂K.

This well-known result is proved for example in [K4, Lemma 1.11].

Lemma 4. Let K be a regular space, G a G_δ subset of K and x∈ G. Then there is a closedG_δ subsetH ⊂Ksuch thatx∈H ⊂G.

This lemma is an easy consequence of the definition of regular spaces.

Lemma 5. LetK be a compact space containing no copy of [0, ω2]andB ⊂K a dense subset which is a continuous image of a Corson countably compact space.

Then for eachx∈Kthere isC⊂Bwith cardinality at mostℵ₁ such thatx∈C.

Proof: We will use ideas of the proof of [K3, Theorem 1]. Let f:A →B be a continuous surjection whereA⊂Σ(Γ) is a Corson countably compact space. As, due to [K4, Lemma 1.8 and Proposition 1.9],A=βA(the closure is taken inR^Γ), there is a continuous extensiong : A →K of f. For any uncountable cardinal numberκput

Aκ ={x∈A: card suppx < κ}.

ThenAℵ₁ =A. We will prove thatg(Aℵ₂) =K.

Suppose on the contrary thatg(Aℵ₂)$K. Then we can put

˜

τ= min{κ:g(Aℵ₂)$g(Aκ)}.

Obviously such a ˜τ exists, ˜τ >ℵ₂ and ˜τ is not a limit cardinal. Hence ˜τ=τ⁺for some cardinalτ ≥ ℵ₂. Letx∈g(A_τ⁺)\g(Aℵ₂) andy ∈A_τ⁺ satisfyg(y) =x.

Then clearlyy∈A_τ⁺\Aτ, i.e. card suppy=τ. Letϕ: [0, τ]→Abe a continuous injection satisfying conditions

(i) card suppϕ(α)≤max{cardα,ℵ₀}for allα≤τ;

(ii) ϕ(τ) =y.

Such a mapping exists due to [K3, Theorem 2].

Ifτ is singular, then there is an infinite cardinalλ < τ and cardinals (θγ)_γ<λ withθγ< τ forγ < λandτ = sup

γ<λ

θγ. By the definition ofτ we haveg(ϕ(θγ))∈ g(Aℵ₂). Hence there arezγ ∈Aℵ₂,γ < λwithg(zγ) =g(ϕ(θγ)). Then clearly

{zγ:γ < λ} ⊂A_λ⁺.

The image byg of the set on the left-hand side is compact and hence x∈ {g(ϕ(θγ)) : γ < λ} ⊂g

{zγ: γ < λ}

⊂g(A_λ⁺).

Moreover,g(A_λ⁺) =g(Aℵ₂), hencex∈g(Aℵ₂), a contradiction.

Henceτis a regular cardinal. Puth=g◦ϕ. At first let us note thatx=h(τ)∈/ h([0, τ)). Henceh⁻¹(k) is bounded in [0, τ) for eachk∈h([0, τ)). Therefore, by regularity ofτ, the seth⁻¹(h([0, η])) is bounded in [0, τ) for eachη < τ. So we can choose by transfinite induction ordinalsηα< τ forα < τ such that

(a) η_α+1>suph⁻¹(h([0, ηα]));

(b) ηα= sup

β<α

η_β forα < τ limit.

Now it is clear thatK contains a homeomorphic copy of [0, τ] and thus also that of [0, ω2]. But this contradicts our assumptions.

Therefore g(Aℵ2) = K. Choose x ∈ K arbitrarily and find y ∈ Aℵ2 with g(y) = x. If suppy is countable then y ∈ A, thus x ∈ B and we can take C ={x}. If card suppy =ℵ₁ then there is (by [K1, Proposition 2.7] or also by [K3, Theorem 2]) a continuous injectionφ: [0, ω1]→Awithφ([0, ω1))⊂Aand φ(ω1) =y. Therefore we can takeC=g(φ([0, ω1))).

Proof of Theorem 2

LetK be a compact space such thatK=S∞

i=1B_i whereB_i is, for eachi∈N, a dense subset ofK which is a Corson countably compact space. If all the B_i’s are equal,Kis Corson by definition. Hence suppose that at least two of them are different, sayB₁6=B₂. The proof will be done is several steps.

Step 1. K contains no copy of [0, ω₂].

Suppose, on the contrary, that there is a homeomorphic copy L of [0, ω₂] such that L ⊂ K. As K = S∞

i=1B_i, there is some i with the cardinality of B_i ∩L being ℵ₂. But B_i ∩L is homeomorphic to a Corson countably com- pact subspace of [0, ω₂] and hence, by [K5, Theorem 2.5], it has cardinality at mostℵ₁, a contradiction. (Recall the basic idea of the proof of the quoted theo- rem: Suppose thatM ⊂[0, ω₂] is a Corson countably compact subset of cardina- lityℵ₂. Then it is easy to show thatM contains a closed subset homeomorphic

to M^′ = {α < ω₂; αis either isolated or of countable cofinality}. HenceM^′ is Corson. But this can be easily led to a contradiction.)

Step 2. We can suppose without loss of generality thatK has weightℵ₁. PutM₁={x}, wherex∈B₂\B₁. AsK contains no copy of [0, ω₂], we can, due to Lemma 5, construct by induction a sequence of setsM_k⊂Ksuch that each M_k has cardinality at mostℵ₁ and M_k ⊂B_i∩M_k+1 for alli, k ∈N. Put H = S∞

k=1M_k. ThenH∩B_iis dense inH for alli∈NandH∩B₁6=H∩B₂. Moreover, H has weightℵ₁. Indeed,H∩B1 is a Corson countably compact space of density at mostℵ₁. Hence it is homeomorphic to someC⊂Σ([0, ω₁))∩[0,1]^[0,ω1). Then clearly C has weight at most ℵ₁. Further C = βC by [K4, Proposition 1.9].

ThereforeH is a continuous image ofC and thus the weight ofH is at mostℵ₁. On the other hand, the weight ofH cannot be countable, otherwiseH would be metrizable and henceH ∩B₁=H∩B₂=H.

Step 3. We can suppose without loss of generality that for any pair i, j∈N eitherBi=Bj orBi∩Bj=∅.

Letη = (η₁, η₂) :N→N² be a bijection. Due to Lemma 2 we can construct by induction nonempty open setsV_k, k∈Nsuch that

• V_k+1⊂V_k for allk∈N;

• B₁∩B₂∩V₁=∅;

• B_η1(k)∩V_k=B_η2(k)∩V_k orB_η1(k)∩B_η2(k)∩V_k=∅for allk∈N. AsB16=B2, it follows from Lemma 2 thatB1∩B2 is not dense inK. Hence we can chooseV₁ satisfying the appropriate condition. We continue by the obvious induction using Lemma 2.

PutH =T

k∈NVk. ThenH is clearly a nonempty closedGδsubset ofK. Then Bi∩H is dense inH for everyi∈N(by Lemma 3). Moreover,H is the union of these Corson countably compact spaces, any two of them are either identical or disjoint and at least two of them are different.

AsH ⊂K, the weight ofH is at mostℵ₁. AsB₁∩B₂∩H =∅,H cannot be metrizable and hence the weight ofH is equal toℵ₁.

Step 4. Assume that K has weight ℵ₁ and B₁, B₂, . . . are pairwise disjoint dense Corson countably compact spaces. ThenK\S∞

i=1Bi 6=∅.

For i∈ N letAi ⊂ Σ([0, ω1)) be homeomorphic to Bi and letLi denote the closure of A_i in R^Γ. Then L_i is compact and L_i = βA_i (see [K4, Lemma 1.8 and Proposition 1.9]). So there is a continuous surjectiong_i :L_i→K such that g_i↾A_iis a homeomorphism ofA_i ontoB_i. Then it clearly holdsg_i(A_i) =B_i and g_i(L_i\A_i) =K\B_i.

We putF₀ =K. Further we will construct pointsxⁱ_γ ∈Ai, nonempty closed G_δ sets Gⁱ_γ ⊂Li, H_γⁱ ⊂K and Fγ ⊂K for 1 ≤γ < ω₁ and i ∈N and points

y_γⁱ ∈L_i for 1≤γ < ω₁ isolated andi∈Nin the following way.

(i) y¹_γ+1∈g₁⁻¹(Fγ)\A₁ for allγ < ω₁.

(ii) G¹_γ+1={x∈g⁻₁¹(Fγ) :x↾[0, γ] =y_γ+1¹ ↾[0, γ]

&∀δ∈(0, γ] :x↾suppx¹_δ=y_γ+1¹ ↾suppx¹_δ} for allγ < ω1. (iii) xⁱ_γ+1∈Gⁱ_γ+1∩A_i\ {xⁱ_δ: 0< δ ≤γ}for allγ < ω₁ andi∈N.

(iv) g_i(xⁱ_γ+1)∈H_γ+1ⁱ ⊂K\g_i(L_i\Gⁱ_γ+1) for allγ < ω₁ andi∈N. (v) yⁱ⁺¹_γ+1∈g_i+1⁻¹(gi(xⁱ_γ+1)) for allγ < ω₁ andi∈N.

(vi) Gⁱ⁺¹_γ+1={x∈g⁻_i+1¹(H_γ+1ⁱ ) :x↾[0, γ] =y_γ+1ⁱ⁺¹ ↾[0, γ] &∀δ∈(0, γ] : x↾suppxⁱ⁺¹_δ =yⁱ⁺¹_γ+1↾suppxⁱ⁺¹_δ }for allγ < ω₁ and i∈N. (vii) F_γ+1=T∞

i=1H_γ+1ⁱ for allγ < ω₁.

(viii) xⁱ_λ= lim_γ<λxⁱ_γ for allλ < ω₁ limit andi∈N.

(ix) Gⁱ_λ=T

γ<λGⁱ_γ for allλ < ω₁ limit andi∈N.

(x) F_λ=H_λⁱ =T

γ<λH_γⁱ =T

γ<λFγ for allλ < ω₁ limit andi∈N.

Let us show that the construction may be done. We can surely putF0 =K.

This is a nonempty closedG_δ subset ofK.

Suppose thatγ < ω₁ and we have already constructed xⁱ_δ, Gⁱ_δ, H_δⁱ andF_δ for δ∈(0, γ] andi∈N;yⁱ_δforδ∈(0, γ] isolated andi∈N.

Choosey_γ+1¹ as in (i). This is possible, asFγ is a nonemptyGδset and hence (by Lemma 3) we haveFγ∩B₂6=∅, soFγ\B₁6=∅ and thusg⁻₁¹(Fγ)\A₁6=∅.

DefineG¹_γ+1 as in (ii). Then G¹_γ+1 is closed andG_δ in L1 as it is of the form {x∈ g₁⁻¹(Fγ) :x↾C =y_γ+1¹ ↾ C} for a countable setC and g⁻₁¹(Fγ) is closed andG_δ.

Further suppose that k ∈ N and we have constructed xⁱ_γ+1 and H_γ+1ⁱ for 1≤i < k andy_γ+1ⁱ andGⁱ_γ+1 for 1≤i≤k.

By Lemma 3 the set G^k_γ+1∩A_k is dense in G^k_γ+1. If this intersection were countable, we would haveG^k_γ+1 ⊂A_k (by Lemma 1(b)). But y^k_γ+1 ∈G^k_γ+1\A_k (this follows from (i) in case k = 1 and from (v) in case k > 1). Therefore G^k_γ+1∩A_kis uncountable and we can choosex^k_γ+1 as in (iii).

As G^k_γ+1 is G_δ and L_k compact, the set K\g_k(L_k\G^k_γ+1) is a G_δ subset ofK. Moreover, this set containsg(x^k_γ+1) as x^k_γ+1 ∈G^k_γ+1∩A_k. Hence, due to Lemma 4, we can choose a closedG_δ set H_γ+1^k satisfying (iv). Further, choose y_γ+1^k+1as in (v) and defineG^k+1_γ+1by (vi). AgainG^k+1_γ+1 is a closedG_δsubset ofL_k+1 containingy^k+1_γ+1.

We have already constructed x^k_γ+1, y^k_γ+1, G^k_γ+1 and H_γ+1^k for all k ∈ N.

Remark that we have

(*) Fγ⊃H_γ+1¹ ⊃H_γ+1² ⊃. . . .

Next defineF_γ+1 according to (vii). This is clearly a nonempty closedG_δsubset ofK.

Next suppose thatλ < ω1 is a limit ordinal and we have constructedxⁱ_γ,Gⁱ_γ, H_γⁱ andFγ for allγ∈(0, λ) andi∈nandyⁱ_γ forγ∈(0, λ) isolated andi∈N.

Fixi ∈ N. Let us show that the net{xⁱ_γ:γ < λ} converges. Let δ < ω₁ be arbitrary. If δ /∈ S

0<γ<λsuppxⁱ_γ, then xⁱ_γ(δ) = 0 forγ ∈(0, λ) and hence the netxⁱ_γ(δ) converges to 0. Next suppose δ∈suppxⁱ_γ0 for someγ₀∈(0, λ). Then for each γ ∈ (γ₀, λ) we have xⁱ_γ ∈ Gⁱ_γ ⊂ Gⁱ_γ0+1, and hence xⁱ_γ(δ) = y_γ^k₀₊₁(δ).

Thus the net xⁱ_γ(δ), γ ∈ (γ₀, λ) is constant and so convergent. This completes the proof that (viii) can be fulfilled. Moreover,xⁱ_λ∈A_i asA_iis countably closed in L_i (Lemma 1(b)). Further, define Gⁱ_λ as in (ix). It is obviously a closed G_δ set. Further, asxⁱ_γ ∈ Gⁱ_γ for all γ ∈(0, λ) and the family (Gⁱ_γ:γ ∈ (0, λ)) is a decreasing family of closed sets, we getxⁱ_λ∈Gⁱ_λ.

Finally defineH_λⁱ andFλ as in (x). The definition is correct due to (*). More- over, we have

g⁻_i ¹(F_λ) = \

γ<λ

g⁻_i ¹(H_γⁱ)⊂ \

γ<λ

Gⁱ_γ=Gⁱ_λ for alli∈N.

This completes the construction.

Fixi ∈N. In the same way as we proved the existence ofxⁱ_λ for λlimit, we can prove that the net xⁱ_γ, γ < ω1 converges to some xⁱ ∈ Li. This time we havex_i ∈/ A_i. Indeed, the mapping φ : [1, ω₁] → L_i defined by φ(α) = xⁱ_α for α ∈ [1, ω₁) and φ(ω₁) = xⁱ is continuous and φ ↾ [1, ω₁) is one-to-one due to condition (iii). Thus, there isα < ω₁ such that φis one-to-one on [α, ω₁]. Then φ([α, ω₁]) is homeomorphic to [0, ω₁] and therefore it cannot be contained in Ai. Asφ([α, ω₁))⊂A_i, we getxⁱ=φ(ω₁)∈/A_i.

Let us show that the net yⁱ_γ, γ ∈ (0, ω₁) isolated, converges also to xⁱ. Fix arbitrary δ < ω₁. For any γ ≥ δ we have xⁱ_γ+1 ∈ Gⁱ_γ+1 and hence xⁱ_γ+1(δ) = y_γ+1ⁱ (δ). Thereforexⁱ(δ) = limγxⁱ_γ+1(δ) = limγyⁱ_γ+1(δ).

Further, we claim thatg₁(x¹) =g₂(x²) =. . .. Indeed, leti∈Nbe arbitrary.

Then

g_i+1(xⁱ⁺¹) = lim

γ g_i+1(y_γ+1ⁱ⁺¹) = lim

γ g_i(xⁱ_γ) =g_i(xⁱ).

We conclude by noting thatgi(xⁱ)∈/Bi (asxⁱ ∈/ Ai) for alli∈N.

Step 5. Assume that K has weight ℵ₁, n >1 and B₁, . . . , Bn are pairwise disjoint dense Corson countably compact spaces. Then K\(B₁∪ · · · ∪Bn)6=∅.

We can perform the same construction as in Step 4 with the obvious changes

— we replace the setNby{1, . . . , n} (in (iii), (iv), (vii)–(x)) or by{1, . . . , n−1}

(in (v) and (vi)). In this way we obtain the same result.

The proof is now completed. By Step 2 and Step 3 we can suppose thatKhas weightℵ₁,B₁ 6=B₂ and that for anyi, j ∈Nwe haveBi∩Bj =∅ orBi =Bj. Step 5 shows thatKcannot be covered by finitely many ofBi’s. Therefore we can suppose that all theBi’s are pairwise disjoint. Step 4 then yields a contradiction.

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Department of Mathematical Analysis, Faculty of Mathematics and Physics, Charles University, Sokolovsk´a 83, 186 75 Praha 8, Czech Republic

E-mail: [email protected]

(Received May 15, 2003,revised March 25, 2004)