# Some Topological Properties ofCbX

Authors: Juan Carlos Ferrando
Journal: Journal of Function Spaces
 (2014)

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## Abstract

IfX is a completely regular space, first we characterize those spaces CbX whose compact sets are metrizable. Then we use this result to provide a general condition for X to ensure the metrizability of compact sets in CbX. Finally, we characterize those spaces CbX that have a G-basis.


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## Body

## 1. Preliminaries

We start by recalling that a subsetA of a topological space X is called (functionally)bounded [1, Chapter 0] if fA is a bounded set in R for every real-valued continuous function f on X. In what follows, unless otherwise stated, X will be a Hausdorff completely regular space. We will denote by CpX, CcX, or CbX the ring CX of all real-valued continuous functions defined on X provided with the pointwise convergence topology, the compact-open topology, or the bounded-open topology, that is, the locally convex topology on CX of uniform convergence on the bounded sets of X, respectively. An account of the different uniform convergence topologies on CX can be found in [2]. We will denote by LX the topological dual of CpX and by LpX the linear space LX equipped with the weak* topology [1]. From now onwards a covering {Aα:α∈Σ} of X with Σ⊆NN, where NN is equipped with the product (discrete) topology and Aα⊆Aβ if α≤β, will be referred to as anorderedΣ-covering, although if Σ=NN we will speak ofordered covering rather than ordered NN-covering. Every K-analytic space X contains an ordered covering consisting of compact sets [3]. In [4, Definition 2.1] an unbounded subspace Σ of NN is called σ-complete if each bounded sequence αnn=1∞ in Σ satisfies that sup⁡n∈Nαn∈Σ. Clearly Σ=NN is σ-complete. The main result of [4] is the following.Theorem 1 (see [4, Theorem  2.8]).
If there exists a dense subspaceY of X that admits an ordered Σ-covering consisting of Y-bounded subsets indexed by a σ-complete set, then every compact set in Cb(X) is metrizable. Particularly, if Y has an ordered covering made up of Y-bounded sets, every compact set in Cb(X) is metrizable.A uniform spaceX,N is calledtransseparable if for each vicinity N of N there is a countable set Q in X such that NQ=⋃x∈Q‍UNx=X, where UNx={y∈X:x,y∈N}; see [5, Section 6.4]. Equivalently, X,N is transseparable if each uniform cover of X admits a countable subcover [6]. A family F of functions from a uniform space X,N into a uniform space Y,M is calleduniformly equicontinuous [7, X.2.1 Definition 2] if for each V∈M there is U∈N such that fx,fy∈V, whenever f∈F and x,y∈U. Here we have a characterization of transseparable spaces X,N in terms of uniformly equicontinuous sets of CX,τN, where τN stands for the uniform topology on X.Theorem 2 (see [8, Theorem  1]).
A uniform spaceX,N is transseparable if and only if every pointwise bounded uniformly equicontinuous set of functions from X,N into R is metrizable in CcX,τN or, equivalently, in CpX,τN.Particularly, by Ascoli’s theorem, if every compact set ofCcX,τN is metrizable, then X,N is transseparable. Separable uniform spaces as well as Lindelöf uniform spaces are transseparable and for uniform pseudometrizable spaces transseparability is equivalent to separability. The class of transseparable uniform spaces is hereditary, productive, and closed under uniform continuous images. Some applications of transseparable spaces can be found in [5, Section 6.4]. Regarding transseparability, the following result is useful.Theorem 3 (see [9, Theorem]).
LetX,N be a uniform space. If X has an ordered covering consisting of N-precompact sets, then X,N is transseparable.This result is easily generalized for orderedΣ-coverings indexed by σ-complete sets.Theorem 4 (see [4, Lemma  2.7]).
LetX,N be a uniform space. If X has an ordered Σ-covering consisting of N-precompact sets such that Σ is σ-complete, then X,N is transseparable.Finally, a locally convex space (lcs)E is said to have a G-base (or a G-basis) [5, Chapter 1] if there exists a basis {Uα:α∈NN} of (absolutely convex) neighborhoods of the origin in E such that Uβ⊆Uα whenever α≤β. As is well known, if a locally convex space E has a G-base, every compact set in E is metrizable.In what follows we continue the research started in [4] on CbX spaces and extend the investigation of papers [10, 11] on spaces CcX to the CbX case. We begin by characterizing the spaces CbX whose compact sets are metrizable in terms of a particular uniformity M on X (cf. Theorem 5). Then we use this characterization to prove Theorem 8 below, which extends Theorem 4. Finally, we characterize those CbX spaces having a G-base.


## 2. Metrizable Compact Sets in SpacesCbX

First we characterize thoseCbX spaces whose compact sets are metrizable in terms of a particular uniformity on X. This result is a translation of [10, Theorem 3] to the bounded-open topology framework.Theorem 5.
The compact sets ofCbX are metrizable if and only if X,M, where M is the uniformity on X generated by the pseudometrics
(1)dAx,y=sup⁡f∈Afx-fy
for each compact set A of CbX, is transseparable.Proof.
LetE be the topological dual of CbX. Let us denote by KCbX the family of all compact sets of CbX and by ρE,CbX the locally convex topology on E of uniform convergence on all compact sets of CbX. If δ stands for the canonical homeomorphic embedding of X into LpX, then δ(X)⊆LX⊆E and the topology ρX=δ-1ρE,CbX on X is stronger than the original topology of X, since ρE,CbXδX is stronger than the topology induced on δX by the weak* topology σE,CX of E. This latter fact implies that CX⊆CX,ρX algebraically.
Assuming that all compact sets ofCbX are metrizable, the dual E equipped with the locally convex topology ρE,CbX is transseparable [12, Theorem 2]. This topology ρE,CbX generates a unique admissible translation-invariant uniformity N on E, so that τN=ρE,CbX. Setting f~ instead of f∈CX when f is considered as a linear functional on E, observe that u,v∈N∈N with u,v∈E if and only if there are A∈KCbX and ϵ>0 such that sup⁡f∈A〈f~,u-v〉<ϵ. Particularly, the relative uniformity of N on δX satisfies that
(2)δx,δy∈N∩δX×δX⟺sup⁡f∈A⁡f~,δx-f~,δy<ϵ.
This defines a uniformity M on X such that x,y∈M∈M if and only if there are A∈KCbX and ϵ>0 with sup⁡f∈Afx-fy<ϵ. Given that X,M and δX equipped with the relative uniformity of N are uniformly homeomorphic and the class of transseparable spaces is hereditary and closed under uniform continuous images, it follows that the uniform space X,M is transseparable.
Assume conversely thatX,M is transseparable whenever M is the uniformity on X generated by the pseudometrics dAx,y=sup⁡f∈Afx-fy for each A∈KCbX and let A be a fixed compact subset of CbX. As observed above, A is contained in the space CX,τM=CX,ρX. Moreover, A is a (pointwise bounded) uniformly equicontinuous set of functions from X,M to R since, given ϵ>0, if dAx,y<ϵ, obviously fx-fy<ϵ whenever f∈A. Hence, according to Theorem 2, the set A is metrizable in CcX,τM. Since τM=ρX is stronger than the original topology on X, there are more compact sets in X than in X,τM, which ensures that the topology on A inherited from CcX is stronger than that inherited from CcX,τM. Since A is compact in CbX and the bounded-open topology is stronger than the compact-open, the three topologies, that is, the bounded-open, the compact-open from X, and the compact-open from X,τM, coincide on A. Thus A is metrizable.Next we extend [4, Theorem  2.8] by locating a class of completely regular spaces X, wider than those spaces having an ordered Σ-covering indexed by a σ-complete set of NN, with the property that compact sets of CbX are metrizable.Definition 6.
A subspaceΣ of NN will be called fair if every uniform space X,N that admits an ordered Σ-covering consisting of N-precompact sets is transseparable.As an obvious consequence of Theorem4, every σ-complete set of NN, particularly the whole space NN, is fair.Lemma 7.
LetX,N be a uniform space. If M stands for the uniform structure on X generated by the pseudometrics (1) when A runs over the compact sets of CbX,τN, then every τN-bounded set of X is M-precompact.Proof.
Note that the completely regular spaceX,τM is homeomorphic to δX equipped with the relative topology of ρE,CbX,τN on E of uniform convergence on all compact sets of CbX,τN, where E stands for the topological dual of CbX,τN. If B is a τN-bounded set of X, then
(3)B⋄:=f∈CX,τN:sup⁡x∈B⁡fx≤1
is a neighborhood of the origin in CbX,τN and, consequently, the absolutely convex set
(4)B⋄0:={u∈E:sup⁡f∈B⋄⁡f~,u≤1}
is a weak* compact subset of E, that is, a σE,CX,τN-compact set. Since ρE,CbX,τN coincides with σE,CbX,τN on the CbX,τN-equicontinuous subsets of E, then it follows that B⋄0 is a ρE,CbX,τN-compact set. Given that δB is contained in B⋄0∩LX, we conclude that δB¯ρE,CbX,τN is ρE,CbX,τN-compact, so that δB is ρE,CbX,τN-precompact in the locally convex space E,ρE,CbX,τN. This is tantamount to saying that B is an M-precompact subset of X.Theorem 8.
LetY be a dense subspace of a completely regular space X which admits an ordered Σ-covering consisting of Y-bounded sets. If the index set of the covering is fair, then every compact set of CbX is metrizable.Proof.
It suffices to prove the theorem assuming thatY=X, since the restriction map S:CbX→CbY given by Sf=fY is a continuous (linear) injection, so if each compact set of CbY is metrizable, every compact set of CbX is metrizable as well. So, let us assume that there exists an ordered Σ-covering Aα:α∈Σ of X consisting of bounded sets.
IfM denotes the uniformity on X generated by the pseudometrics (1) when A runs over the compact sets of CbX, Lemma 7 informs us that the family Aα:α∈Σ is an ordering Σ-covering of X consisting of M-precompact sets. Since Σ is fair, the uniform space X,M is transseparable. Consequently, an application of Theorem 5 shows that every compact set in CbX is metrizable.Lemma 9.
IfM is the uniformity on X generated by the pseudometrics (1) when A runs over the compact sets of CbX, then τM coincides with the original topology of X on each X-bounded set.Proof.
LetB be a bounded set of X. If E stands for the topological dual of CbX, we proceed as in the proof of Lemma 7 to show that ρE,CbXδB=σE,CXδB. Moreover, if F stands for the topological dual of CX,τM, then X,τM coincides with δB when this latter is provided with the relative weak* topology σF,CX,τMδB of F. So bearing in mind the definition of τM we conclude that
(5)σF,CX,τMδB=ρE,CbXδB=σE,CXδB
which implies that τM coincides on B with the original topology of X.Theorem 10.
LetX be k-space. If every admissible uniformity on X is transseparable, then compact sets of CbX are metrizable.Proof.
As a consequence of the previous lemma,X and X,τM have the same compact sets. But since X is a k-space, its topology coincides with the strongest topology on X that has the same compact sets as the original one. Consequently, τM coincides with the original topology of X, which shows that M is an admissible uniformity on X. So, if the hypothesis holds, the uniform space X,M must be transseparable. In this case Theorem 5 ensures that every compact set of CbX is metrizable.The fact thatτM coincides with the original topology of X on each X-bounded set does not mean that X and X,M have the same bounded sets. But that can be taken for granted if X is a normal space, as the following theorem shows.Theorem 11.
IfX is a normal space, then X and X,τM have the same bounded sets. Consequently CbX embeds into CbX,τM.Proof.
Assume without loss of generality thatB is a closed X-bounded subset of X. If f∈CX,τM, Lemma 9 shows that fB∈CB, where B is considered as a topological subspace of X. So there is g∈CX such that gB=fB. But since B is X-bounded, it holds that
(6)sup⁡x∈Bfx=sup⁡x∈Bgx<∞,
so that B is τM-bounded. Conversely, if B is τM-bounded set of X, the fact that τM is finer than the original topology of X implies that B is X-bounded. On the other hand, since the identity map φ:X,τM→X is continuous, the first part of the theorem ensures that the inclusion map f↦f∘φ from CbX into CbX,τM is a linear embedding.


## 3. Characterizing SpacesCbX with a G-Base

It is shown in [11, Theorem 2] that the space Cc(X) has a G-base if and only if X has an ordered covering consisting of compact sets which swallows the compact sets of X, that is, such that each compact set of X is contained in some member of the covering. A similar strategy provides the following characterization for CbX.Theorem 12.
The spaceCb(X) has a G-base if and only if X has an ordered covering consisting of bounded sets which swallows the bounded sets of X.Proof.
For each bounded setB⊆X and ϵ>0 define
(7)B,ϵ=f∈CX:sup⁡x∈Bfx≤ϵ.
If {Bα:α∈NN} is an ordered covering of X consisting of bounded sets, define
(8)Uα=Bα,α1-1
for α∈NN and put U={Uα:α∈NN}. Clearly U is a family of absolutely convex and absorbing sets in C(X) such that Uβ⊆Uα whenever α≤β, composing a filter base. The reader may easily check that U is a G-base for a locally convex topology τ on CX with τp≤τ≤τb. Now assume that {Bα:α∈NN} swallows the bounded sets and let V be a neighborhood of the origin of CbX. If B is a bounded set in X with B,ϵ⊆V for some ϵ>0, choosing γ∈NN such that B⊆Bγ and γ1-1<ϵ, then Uγ⊆B,ϵ⊆V. This shows that τ=τb, so {Uα:α∈NN} is a G-base for CbX.
Conversely, suppose thatCbX has a G-base {Uα:α∈NN}. For every set U in CX define a corresponding set U⋄ in X by writing
(9)U⋄=x∈X:fx≤1∀f∈U.
Clearly U⋄ is closed in X and U⊆V implies that U⋄⊇V⋄. If B is a closed bounded set in X and ϵ>0, then B,ϵ⋄⊆B since if x∈X∖B, there is f∈CX with fx=2 and fB=0, so that f∈B,ϵ and x∉B,ϵ⋄. If B is bounded and 0<ϵ≤1, then B⊆B,ϵ⋄; hence B,ϵ⋄=B for every closed bounded set B. In addition, if U is a neighborhood of the origin in CbX, then U⋄ is a bounded subset of X. Indeed, if B is a bounded set in X such that B,ϵ⊆U for some ϵ>0, by the previous observations U⋄⊆B,ϵ⋄⊆B and hence U⋄ is bounded because it is subset of a bounded set.
Now for anyB closed bounded in X there is some Uα⊆B,1, which means that B=B,1⋄⊆Uα⋄. This shows that the family A={Uα⋄:α∈NN} swallows all bounded sets, so particularly it covers X. As in addition Uα⋄⊆Uβ⋄ for α≤β, it follows that A is an ordered covering of X consisting of bounded sets.



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