A Characterization of the Existence of a Fundamental Bounded Resolution for the Space in Terms of

Abstract

We characterize in terms of the topology of a Tychonoff space the existence of a bounded resolution for that swallows the bounded sets, where is the space of real-valued continuous functions on equipped with the compact-open topology.

1. Preliminaries

In the sequel, unless otherwise stated, is a nonempty completely regular Hausdorff space. We represent by the ring of real-valued continuous functions defined on equipped with the pointwise topology . As usual, we denote by the wea dual of . When is equipped with the compact-open topology we write . As in [1], we denote by the linear space of real-valued continuous and bounded functions defined on . If is regarded as a subspace of , we denote this space by . Since is dense in , both and have the same dual. The Banach space equipped with the supremum norm has recently been studied in [2]. Let us recall that a family of subsets of a set is called a resolution for if it covers and whenever ; i.e., for every (see [3, Chapter 3]). If is a topological vector space, a resolution consisting of bounded sets (see [4, Definition 1.4.5]) is referred to as a bounded resolution. A bounded resolution on that swallows the bounded sets, i.e., such that for each bounded set in there is with , is referred to as a fundamental bounded resolution. The existence of such a resolution in a locally convex space has shown to be equivalent to the existence of a so-called -base of absolutely convex neighborhoods of the origin in the strong dual of . Besides, fundamental bounded resolutions are essential in order to get a proper extension of the class of -spaces (see [5] for details). Fundamental compact resolutions, i.e., resolutions consisting of compact sets which swallow the compact sets, have been widely used, even in Banach space theory [6], since they were introduced in [7]. A well-known result of Christensen [8, Theorem 3.3] asserts that a metrizable space is Polish if and only if has a fundamental compact resolution. Moreover, it has been shown in [9, Theorem 2] that has a -base of neighborhoods of the origin if and only if has a fundamental compact resolution. In this note we provide two characterizations, in terms of the domain space , of the existence of a fundamental bounded resolution for , one by means of certain uniformity for and the other purely topological. Our main motivation is the two following -theoretic results.

Theorem 1 ([7, Theorem 3.7] or [10, Problem 216]). The space has a fundamental compact resolution if and only if is countable and discrete.

Theorem 2 ([11, Theorem 3.3]). The space has a fundamental bounded resolution if and only if is countable.

A space is called -analytic if there is an upper semicontinuous compact-valued map from the product space , where is equipped with the discrete topology, into such that . A family of functions from a uniform space into a uniform space is called uniformly equicontinuous [12, Chapter 7, Problem G] if for each there is such that whenever and . Let be a uniformity for a (nonempty) set and denote by the uniform topology defined by . A base of the uniformity is called a -base if whenever . There is no loss of generality by assuming that each is a symmetric vicinity.

2. Fundamental Bounded Resolutions for

Although it can be easily seen that each metrizable locally convex space has a fundamental bounded resolution, if the locally convex space has a fundamental bounded resolution, unlike what happens with , the space needs not be countable and moreover needs not be metrizable.

Proposition 3. Let be a metrizable space. Then has a fundamental bounded resolution if and only if is -compact.

Proof. If is -compact then is weakly -analytic by [13, Proposition 2.2], or it has a compact resolution that swallows the compact sets by [14, Corollary 2.10]. In any case has a bounded resolution. Since is a -space then is complete, hence locally complete. So, it follows from Valdivia’s [3, Theorem 3.5] that there exists a resolution of consisting of Banach disks that swallows the bounded sets of , which shows that has a fundamental bounded resolution. Conversely, if has a fundamental bounded resolution then has a bounded resolution, so [3, Corollary 9.2] shows that is -compact.

Example 4. According to the previous proposition has a fundamental bounded resolution, but is neither countable nor discrete. On the other hand, has also a fundamental bounded resolution, but , although is countable, is not discrete. Observe that is metrizable, but is not. Of course, if is hemicompact, or even compact, then is metrizable, or even a Banach space, and in this case has obviously a fundamental bounded resolution.

Theorem 5. The space has a fundamental bounded resolution if and only if , where is the uniformity for generated by the pseudometrics for each bounded set of , has a -base.

Proof. Let be the topological dual of and denote by the family of all bounded sets of and by the strong topology on . By identifying with its canonical homeomorphic embedding in , note that . The strong topology generates a unique admissible translation-invariant uniformity on , so that . By considering also as a linear functional on , observe that for each there is such that Hence belongs to the relative uniformity of to if and only if there exists such that If is a fundamental bounded resolution for , by setting we obtain a -base of . If then , and if there is such that whenever , so if is such that , clearly .
Conversely, if the uniform structure for generated by the family of pseudometrics , where has a -base , this entails that for each there is such that for every . Setting for each , clearly if and . Consequently is a fundamental bounded resolution for .

Let be a nonempty completely regular Hausdorff topological space and let denote the family of all compact sets of .

Lemma 6. A subset of is bounded for the compact-open topology if and only if there exists a sequence of subsets of such that(1)for each , if then (2) for each (3)(4)If a set is such that for each with for all , then

Proof. If is -bounded and , define Clearly for each and if there is with so that , which shows that . In addition, if the relation holds by construction. Finally, if we set then . Therefore, if verifies that then so that . Hence satisfies the required conditions.
Conversely, if there exists a sequence of satisfying the four conditions of the statement of the lemma (actually, only the first and the third conditions are required) then clearly is -bounded on , for if there is with such that .

In what follows the fourth condition above, which is independent of , will be referred to as the closure condition of , and we shall say that the family is closed.

Definition 7. A collection of closed subsets of will be called a covering net of if is an increasing covering of for each such that whenever for all .

Theorem 8. The space has a fundamental bounded resolution if and only if there is a covering net such that if is an increasing covering of by closed sets, there exists such that for all .

Proof. If there exists a covering net for which satisfies the property of the statement of the theorem, the sets compose a fundamental bounded resolution for . Indeed, each set is -bounded by virtue of the previous lemma, since is an increasing covering of consisting of closed sets such that for all . Besides whenever since , and if is a -bounded subset of , according to the previous lemma there exists an increasing covering of consisting of closed sets. Therefore, by the condition in the statement of the theorem, there exists such that for all . Now, if then for all and , in particular for each and all , which shows that . Hence , which proves that is a fundamental bounded resolution for .
Conversely, if has a fundamental bounded resolution and we set then is an increasing covering of such that whenever for all . In addition for and by the definition of . Moreover, satisfies the closure condition, for if then , so if then , which means that .
Now, if is any increasing covering of consisting of closed sets, define and observe that if then , which according to the preceding lemma ensures that is a -bounded set. Since is a fundamental bounded resolution for there exists such that . Now if then so that, in particular, . We claim that . Indeed, since for each , we have that holds for each such that for every by virtue of the definition of . Therefore, the closure property of yields . This shows that for every , which, bearing in mind the properties of the family established before, guarantees that is a covering net for consisting of closed sets that satisfies the required property.

In what follows we shall refer to a Tychonoff space satisfying the conditions of the statement of Theorem 8 as a -space. It is shown in [5, Proposition 3.2] that if is a -space or, which is equivalent, if has a fundamental bounded resolution, then has a countable -network at the origin [15, 16]. The next theorem shows that in order for to be a -space it suffices that have a fundamental bounded resolution.

Theorem 9. Let be completely regular. The space has a fundamental bounded resolution if and only if is a -space.

Proof. If is a -space, Theorem 8 ensures that has a fundamental bounded resolution, which implies that , as a subspace of , also has a fundamental bounded resolution. Conversely, if has a closed fundamental bounded resolution , for each let denote the closure of in . Let us show that is a fundamental bounded resolution for . Indeed, let us denote by the family of all compact sets of and pick an arbitrary bounded subset of . For each and , set and define if and if . Clearly , besides for and for all . Define and note that is an adherent point of in . Therefore, is contained in the closure of in . If is any compact subset of , the fact that is -bounded guarantees that which shows that is a -bounded subset of . Hence, there is such that , so that . Consequently is a fundamental bounded resolution for , as stated. Another application of Theorem 8 shows that is a -space.

The existence of a bounded resolution on a locally convex space does not imply the existence of a fundamental bounded resolution for , as the following example shows.

Example 10. If is an infinite Talagrand compact set, the wea dual of has a bounded resolution but it has no fundamental bounded resolution.

Proof. Since is -analytic, is also -analytic by [17, Proposition 0.5.13]. So has even a compact resolution by [3, Theorem 3.2]. Suppose by contradiction that has a fundamental bounded resolution . Identifying with its homeomorphic copy in , for each set and consider the family . We claim that is a functionally bounded subset of . Indeed, if according to [17, Proposition 0.5.11] there exists a (unique) continuous functional of such that . Since is bounded, there is such that for all . In particular, for every . So is a functionally bounded resolution in . If is a functionally bounded subset of , then , considered as a subset of , is bounded. Therefore, there is with . Hence swallows the functionally bounded subsets of . Since is Lindelöf and hence a -space, the family consists of compact subsets and swallows the compact sets of . So has a fundamental compact resolution. But according to Theorem 1, the space should be countable and discrete. Therefore, being compact is finite, a contradiction.

Data Availability

The data used to support the findings of this study are available from the corresponding author upon request.

Conflicts of Interest

The author declares that there are no conflicts of interest regarding the publication of this paper.