Journal of Function Spaces
Volume 2014 (2014), Article ID 195262, 4 pages
http://dx.doi.org/10.1155/2014/195262

Research Article

Some Topological Properties of

Juan Carlos Ferrando

Centro de Investigación Operativa, Universidad Miguel Hernández, 03202 Elche, Spain

Received 14 October 2014; Revised 8 December 2014; Accepted 9 December 2014; Published 22 December 2014

Academic Editor: Ismat Beg

Copyright © 2014 Juan Carlos Ferrando. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract.

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

1. Preliminaries

We start by recalling that a subset of a topological space is called (functionally) bounded [1, Chapter 0] if is a bounded set in for every real-valued continuous function on . In what follows, unless otherwise stated, will be a Hausdorff completely regular space. We will denote by , , or the ring of all real-valued continuous functions defined on provided with the pointwise convergence topology, the compact-open topology, or the bounded-open topology, that is, the locally convex topology on of uniform convergence on the bounded sets of , respectively. An account of the different uniform convergence topologies on can be found in [2]. We will denote by the topological dual of and by the linear space equipped with the weak topology [1]. From now onwards a covering of with , where is equipped with the product (discrete) topology and if , will be referred to as an ordered -covering, although if we will speak of ordered covering rather than ordered -covering. Every -analytic space contains an ordered covering consisting of compact sets [3]. In [4, Definition 2.1] an unbounded subspace of is called -complete if each bounded sequence in satisfies that . Clearly is -complete. The main result of [4] is the following.

Theorem 1 (see [4, Theorem  2.8]). If there exists a dense subspace of that admits an ordered -covering consisting of -bounded subsets indexed by a -complete set, then every compact set in is metrizable. Particularly, if has an ordered covering made up of -bounded sets, every compact set in is metrizable.

A uniform space is called transseparable if for each vicinity of there is a countable set in such that , where ; see [5, Section 6.4]. Equivalently, is transseparable if each uniform cover of admits a countable subcover [6]. A family of functions from a uniform space into a uniform space is called uniformly equicontinuous [7, X.2.1 Definition 2] if for each there is such that , whenever and . Here we have a characterization of transseparable spaces in terms of uniformly equicontinuous sets of , where stands for the uniform topology on .

Theorem 2 (see [8, Theorem  1]). A uniform space is transseparable if and only if every pointwise bounded uniformly equicontinuous set of functions from into is metrizable in or, equivalently, in .

Particularly, by Ascoli’s theorem, if every compact set of is metrizable, then 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]). Let be a uniform space. If has an ordered covering consisting of -precompact sets, then is transseparable.

This result is easily generalized for ordered -coverings indexed by -complete sets.

Theorem 4 (see [4, Lemma  2.7]). Let be a uniform space. If has an ordered -covering consisting of -precompact sets such that is -complete, then is transseparable.

Finally, a locally convex space (lcs) is said to have a -base (or a -basis) [5, Chapter 1] if there exists a basis of (absolutely convex) neighborhoods of the origin in such that whenever . As is well known, if a locally convex space has a -base, every compact set in is metrizable.

In what follows we continue the research started in [4] on spaces and extend the investigation of papers [10, 11] on spaces to the case. We begin by characterizing the spaces whose compact sets are metrizable in terms of a particular uniformity on (cf. Theorem 5). Then we use this characterization to prove Theorem 8 below, which extends Theorem 4. Finally, we characterize those spaces having a -base.

2. Metrizable Compact Sets in Spaces

First we characterize those spaces whose compact sets are metrizable in terms of a particular uniformity on . This result is a translation of [10, Theorem 3] to the bounded-open topology framework.

Theorem 5. The compact sets of are metrizable if and only if , where is the uniformity on generated by the pseudometrics

for each compact set of , is transseparable.

Proof. Let be the topological dual of . Let us denote by the family of all compact sets of and by the locally convex topology on of uniform convergence on all compact sets of . If stands for the canonical homeomorphic embedding of into , then and the topology on is stronger than the original topology of , since is stronger than the topology induced on by the weak topology of . This latter fact implies that algebraically.
Assuming that all compact sets of are metrizable, the dual equipped with the locally convex topology is transseparable [12, Theorem 2]. This topology generates a unique admissible translation-invariant uniformity on , so that . Setting instead of when is considered as a linear functional on , observe that with if and only if there are and such that . Particularly, the relative uniformity of on satisfies that

This defines a uniformity on such that if and only if there are and with . Given that and equipped with the relative uniformity of are uniformly homeomorphic and the class of transseparable spaces is hereditary and closed under uniform continuous images, it follows that the uniform space is transseparable.
Assume conversely that is transseparable whenever is the uniformity on generated by the pseudometrics for each and let be a fixed compact subset of . As observed above, is contained in the space . Moreover, is a (pointwise bounded) uniformly equicontinuous set of functions from to since, given , if , obviously whenever . Hence, according to Theorem 2, the set is metrizable in . Since is stronger than the original topology on , there are more compact sets in than in , which ensures that the topology on inherited from is stronger than that inherited from . Since is compact in and the bounded-open topology is stronger than the compact-open, the three topologies, that is, the bounded-open, the compact-open from , and the compact-open from , coincide on . Thus is metrizable.

Next we extend [4, Theorem  2.8] by locating a class of completely regular spaces , wider than those spaces having an ordered -covering indexed by a -complete set of , with the property that compact sets of are metrizable.

Definition 6. A subspace of will be called fair if every uniform space that admits an ordered -covering consisting of -precompact sets is transseparable.

As an obvious consequence of Theorem 4, every -complete set of , particularly the whole space , is fair.

Lemma 7. Let be a uniform space. If stands for the uniform structure on generated by the pseudometrics (1) when runs over the compact sets of , then every -bounded set of is -precompact.

Proof. Note that the completely regular space is homeomorphic to equipped with the relative topology of on of uniform convergence on all compact sets of , where stands for the topological dual of . If is a -bounded set of , then

is a neighborhood of the origin in and, consequently, the absolutely convex set

is a weak compact subset of , that is, a -compact set. Since coincides with on the -equicontinuous subsets of , then it follows that is a -compact set. Given that is contained in , we conclude that is -compact, so that is -precompact in the locally convex space . This is tantamount to saying that is an -precompact subset of .

Theorem 8. Let be a dense subspace of a completely regular space which admits an ordered -covering consisting of -bounded sets. If the index set of the covering is fair, then every compact set of is metrizable.

Proof. It suffices to prove the theorem assuming that , since the restriction map given by is a continuous (linear) injection, so if each compact set of is metrizable, every compact set of is metrizable as well. So, let us assume that there exists an ordered -covering of consisting of bounded sets.
If denotes the uniformity on generated by the pseudometrics (1) when runs over the compact sets of , Lemma 7 informs us that the family is an ordering -covering of consisting of -precompact sets. Since is fair, the uniform space is transseparable. Consequently, an application of Theorem 5 shows that every compact set in is metrizable.

Lemma 9. If is the uniformity on generated by the pseudometrics (1) when runs over the compact sets of , then coincides with the original topology of on each -bounded set.

Proof. Let be a bounded set of . If stands for the topological dual of , we proceed as in the proof of Lemma 7 to show that . Moreover, if stands for the topological dual of , then coincides with when this latter is provided with the relative weak topology of . So bearing in mind the definition of we conclude that

which implies that coincides on with the original topology of .

Theorem 10. Let be -space. If every admissible uniformity on is transseparable, then compact sets of are metrizable.

Proof. As a consequence of the previous lemma, and have the same compact sets. But since is a -space, its topology coincides with the strongest topology on that has the same compact sets as the original one. Consequently, coincides with the original topology of , which shows that is an admissible uniformity on . So, if the hypothesis holds, the uniform space must be transseparable. In this case Theorem 5 ensures that every compact set of is metrizable.

The fact that coincides with the original topology of on each -bounded set does not mean that and have the same bounded sets. But that can be taken for granted if is a normal space, as the following theorem shows.

Theorem 11. If is a normal space, then and have the same bounded sets. Consequently embeds into .

Proof. Assume without loss of generality that is a closed -bounded subset of . If , Lemma 9 shows that , where is considered as a topological subspace of . So there is such that . But since is -bounded, it holds that

so that is -bounded. Conversely, if is -bounded set of , the fact that is finer than the original topology of implies that is -bounded. On the other hand, since the identity map is continuous, the first part of the theorem ensures that the inclusion map from into is a linear embedding.

3. Characterizing Spaces with a -Base

It is shown in [11, Theorem 2] that the space has a -base if and only if has an ordered covering consisting of compact sets which swallows the compact sets of , that is, such that each compact set of is contained in some member of the covering. A similar strategy provides the following characterization for .

Theorem 12. The space has a -base if and only if has an ordered covering consisting of bounded sets which swallows the bounded sets of .

Proof. For each bounded set and define

If is an ordered covering of consisting of bounded sets, define

for and put . Clearly is a family of absolutely convex and absorbing sets in such that whenever , composing a filter base. The reader may easily check that is a -base for a locally convex topology on with . Now assume that swallows the bounded sets and let be a neighborhood of the origin of . If is a bounded set in with for some , choosing such that and , then . This shows that , so is a -base for .
Conversely, suppose that has a -base . For every set in define a corresponding set in by writing

Clearly is closed in and implies that . If is a closed bounded set in and , then since if , there is with and , so that and . If is bounded and , then ; hence for every closed bounded set . In addition, if is a neighborhood of the origin in , then is a bounded subset of . Indeed, if is a bounded set in such that for some , by the previous observations and hence is bounded because it is subset of a bounded set.
Now for any closed bounded in there is some , which means that . This shows that the family swallows all bounded sets, so particularly it covers . As in addition for , it follows that is an ordered covering of consisting of bounded sets.

Conflict of Interests

The author declares that there is no conflict of interests regarding the publication of this paper.

Acknowledgment

The author is grateful to the referees for useful comments and suggestions. This paper is partially supported by Grant PROMETEO/2013/058 of the Conserjería de Educación, Cultura y Deportes of Generalidad Valenciana.

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