Abstract
We classify several topological properties of a Tychonoff space X by means of certain locally convex topologies with a -base located between the pointwise topology and the bounded-open topology of the real-valued continuous function space .
MSC:
54C; 54D; 54E; 46A; 46B
1. Introduction
Unless otherwise stated, X stands for an infinite Tychonoff space. We denote by the linear space of real-valued continuous functions on X equipped with the pointwise topology . The topological dual of is denoted by , or by when provided with the weak* topology. We represent by the space equipped with the compact-open topology . A family of subsets of a set X is called a resolution for X if it covers X and verifies that for . Let us recall that a Hausdorff topological space X is called a -space if the closure of each functionally bounded set in X is compact. Every realcompact space is a -space, but the converse is not true. If E is a locally convex space, the bidual of E is the topological dual of the strong dual of E. By the strong dual of E, we mean the dual of E equipped with the strong topology of the uniform convergence on the bounded sets in E. We denote by and the weak locally convex topology of and the bounded-open topology of , respectively. If A is a nonempty set in a real linear space L, we represent by the (real) absolutely convex hull of A. The linear subspace of consisting of those bounded functions is denoted by .
A base of (absolutely convex) neighborhoods of the origin in a locally convex space E such that if is called a -base. Let us mention that the notion of a -base, originally introduced in the realm of locally convex spaces [1], has been extended to topological groups and general topological spaces, sometimes under the name of -base, by some authors (see [2,3]). However, in this paper we keep the original name. Trivially, if E is a metrizable locally convex space with a decreasing base of locally convex neighborhoods of the origin, the family , where for all is a -base for E. The converse is not true, i.e., a locally convex space with a -base need not be metrizable (see p. 107, [4]).
A locally convex space E with a -base has metrizable compact sets, since, in this case, the weak dual is quasi-Suslin [5], hence trans-separable [6] (properties not defined in this paper can be found in [7,8,9,10]). Therefore, if there is a locally convex topology on with a -base such that , every completely regular topology on such that is angelic and has metrizable compact sets. Research on -bases and their generalizations remain active since [1]. For recent results on this topic see [3,11,12,13,14]. Research on the existence of locally convex topologies on stronger than the pointwise topology has been studied in [15,16]. In this paper, we enlarge the classification of topological properties on X provided in (Theorem 3.1, [15]).
2. A Preliminary Result
A resolution for X is functionally bounded if it consists of functionally bounded sets in X. A space X is said to be strictly angelic [17] if it is angelic and all separable compact subsets of X are first countable. A classification of topologies on X by locally convex topologies on is provided in [15] (see also (Theorem 98, [4])).
Theorem 1
(Ferrando–Gabriyelyan–Kąkol (Theorem 3.1, [15])). If X is a Tychonoff space, the following properties hold
- 1.
- There exists a metrizable locally convex topology on such that if and only if X is a σ-compact space.
- 2.
- There exists a metrizable locally convex topology on such that if and only if there is an increasing sequence of functionally bounded subsets of X swallowing the compact sets of X.
- 3.
- There exists a metrizable locally convex topology on such that if and only if there is an increasing sequence of functionally bounded subsets of X covering X, or equivalently, if and only if is σ-compact.
- 4.
- There is a metrizable locally convex topology on such that if and only if X is countable.
- 5.
- There is a locally convex topology on with a -base such that if and only if X has a compact resolution.
- 6.
- There is a locally convex topology on with a -base such that if and only if X has a functionally bounded resolution swallowing the compact sets.
- 7.
- There exists a locally convex topology on with a -base such that if and only if X has a functionally bounded resolution. Equivalently, if and only if is K-analytic. In this case is strictly angelic.
- 8.
- There is a locally convex topology on with a -base such that if and only if X is countable.
3. A More Complete Classification
To enlarge the previous classification, we need the following result.
Lemma 1.
Let X be completely regular. If Q is a metrizable and compact subspace of X, there exists a continuous linear extender map , i.e., such that for every .
Proof.
Since Q is (homeomorphic to) a metrizable compact subspace of the Stone–Čech compactification of X, according to [18] (see also (Proposition 4.1, [19])) there exists a continuous linear map such that which embeds in (as a closed subspace). If is the restriction map , it turns out that is a continuous linear map from into such that for every .
Hence, has a closed graph when both spaces are regarded as Banach spaces. This implies in particular that is continuous, since the supremum norm topology on is stronger than the relative compact-open topology of . Thus, is a continuous linear extender map, as stated. □
Theorem 2.
If X is a Tychonoff space, the following properties hold
- 1.
- The compact-open topology on is metrizable if and only if X is a hemicompact space (Arens’ theorem (Theorem 7, [20]), see also (Theorem 2.5, [21])).
- 2.
- The bounded-open topology on is metrizable if and only if there is an increasing sequence of functionally bounded subsets of X swallowing the functionally bounded sets of X.
- 3.
- The weak topology on is metrizable if and only if X is countable and compact sets in X are finite.
- 4.
- The pointwise topology on is metrizable if and only X is countable.
- 5.
- The compact-open topology on has a -base if and only if X has a compact resolution that swallows the compact sets (Theorem 2, [22]).
- 6.
- The bounded-open topology on has a -base if and only if X has a functionally bounded resolution that swallows the functionally bounded sets.
- 7.
- The weak topology on has a -base if and only if X is countable and compact sets in X are finite.
- 8.
- The pointwise topology on has a -base if and only X is countable (Corollary 15.2, [7]).
Proof.
The proof of statements and is similar (for see (Theorem 12, [23])).
Let us prove statement . If is weakly metrizable, the weakly bounded sets are metrizable, so X is countable by (Theorem 2.3, [21]). This fact also follows from statement of Theorem 1. We claim that, in addition, the compact sets of X are finite. Otherwise, there exists an infinite compact set Q in X. However, since X is countable, Q is metrizable. By Lemma 1 there is a continuous linear extender from into , i.e., such that for every . If in , given there is with for every , so for . Thus, in , and hence in , i.e., , which means that embeds the Banach space in as a closed subspace. Since the weak topology is inherited by linear subspaces, the space is linearly homeomorphic to a subspace of . In other words, the Banach space is weakly metrizable. As the weak topology of a Banach space is metrizable if and only if it is finite-dimensional, it turns out that Q must be finite, a contradiction. Conversely, if X is both countable and has finite compact sets, the former statement guarantees that is metrizable whereas the latter implies that coincides with . Hence, the weak topology is metrizable.
The proof of statement is similar to that of statement . The only difference is that the weak topology of the Banach space now carries a -base. However, if a locally convex space in its weak topology has a -base, it is countable-dimensional (Proposition 11.2, [7]). Thus, must be countable-dimensional, so finite-dimensional by the Baire category theorem. This ensures that the compact set Q must be finite. □
Remark 1.
In (Theorem 2.3, [21]), it is shown that if X is countable, the bounded sets of are weakly metrizable. According to statement of the previous theorem, if X contains an infinite compact set, then cannot be weakly metrizable. Thus, if X is countable and contains an infinite compact set, then is not metrizable but has metrizable bounded sets.
Example 1.
As is a countable space with infinite compact sets which is not hemicompact, neither nor are metrizable, the former statement by Arens’ theorem and the latter by the previous remark, but is metrizable. Of course, has no -base. In fact, since is not a Polish space, Christensen’s theorem (Theorem 94, [4]) prevents from having a compact resolution that swallows the compact sets. This also implies that has no -base, by statement of Theorem 2.
Example 2.
Let be equipped with the discrete topology and choose . Then, provided with the relative topology of is not discrete but has finite compact sets. Thus, the weak topology on is metrizable. In fact, clearly, , although .
Example 3.
is an uncountable hemicompact space, hence is metrizable, but and are not.
Example 4.
If K is a countable infinite compact set, obviously both and are metrizable, but is not.
Example 5.
The Sorgenfrey line is a Lindelöf space which is not σ-compact, so and by statement of Theorem 1 there is no metrizable, locally convex topology on such that .
Example 6.
The space has a -base by statement of Theorem 2, but there is no metrizable topology such that , since is not σ-compact.
4. The Interval
In this section, we deal with the interval . Before stating our main result for this case, we need to establish two auxiliary results concerning the weak* dual of the space . We regard (the canonical homeomorphic copy of) X in as a Hamel basis of , and denote by the image of in . If and , we write to represent the action of the linear functional u on f, in particular .
Lemma 2.
Let E denote the dual of . If X is a μ-space, then E coincides with the bidual of .
Proof.
If X is a -space, then is the strong dual of (see (Lemma 2.2, [24])). Therefore, and hence, E coincides with the bidual of . □
Lemma 3.
Let X be a μ-space. If A is a bounded set in , there are a compact set in X and a real number with such that .
Proof.
Since X is a -space, if A is a bounded set in , then is a neighborhood of the origin in , so there is and a compact set in X such that
which implies that , where the bipolar of A is taken in as well as the polar of . The fact that yields so that . We claim that .
Assume that with . Since X is a Hamel basis of , we have that , with for . First, we show that the support of u is contained in , so that for . Indeed if for some , there is such that and for every . Note that since if with and for , then . As , this contradicts the fact that . Thus, for .
Now, we show that , so that . Since the support of u is finite and is contained in , there are pairwise disjoint open sets in with , so there are compactly supported continuous functions such that for . This means that , has compact support, , and for . For , set , where denotes the sign of the real number , and define . Clearly , , and for every , because of the supports of the functions are pairwise disjoint.
Since and there exists with and for such that . Consequently,
and due to the fact that and it follows that
as required. □
If X is a -space and denotes the family of all bounded sets in , as a consequence of Lemma 2, the space E coincides with (see (23.2.(1), [8]))
where the closure is in E under the weak* topology of E. Hence, if there is a bounded set A in with . However according to Lemma 3, given , there is a compact in X together with some with . Thus, if designates the family of all functionally bounded sets in X, then
where the closures are in . Therefore, the following property holds.
Remark 2.
If X is a μ-space and stands for the family of all compact sets in X, then
where E is the dual of .
Proof.
Because of the bipolar theorem, one has . □
This motivates the following definition.
Definition 1.
Let E be the dual of . We say that a countable family of compact sets in X is a complete sequence if there exists a sequence of positive integers such that
If X is a metrizable hemicompact space and is a family of compact sets that swallows the compact sets in X, then is a complete sequence. If is a base of neighborhoods of the origin in , there are by Lemma 3 a compact set S in X and such that . If then for every . If we define if for some and otherwise, then
If is a dual pair, a locally convex topology for F is called a polar topology if is the topology of the uniform convergence on the sets of a family of -bounded sets in E. Obviously, both the weak topology and the strong topology are polar topologies for F. We are ready to establish our last classification result, which reads as follows.
Theorem 3.
Let X be a Tychonoff space. There is a metrizable polar topology on such that if and only if X contains a complete sequence.
Proof.
Let us denote by E the dual of the locally convex space . Assume that is a metrizable polar topology on with a decreasing base of neighborhoods of the origin enjoying that . By the first statement of Theorem 1, we know that X is a -compact space, so a -space.
As is a polar topology, there exists a family of bounded sets in such that for each , there are some with . According to Lemma 3, there is a compact set in X and some such that , so that . Set . Since for each and , we get
Thus, the sequence is complete.
Assume conversely that there is a complete sequence in X. We may suppose that as well as that . If not, replace by the sequence with and the sequence by with . Since for every , if is complete so is . Then, for each , define . Clearly, and is a base of neighborhoods of the origin of a metrizable locally convex topology for . Observe that is stronger than , since
On the other hand, as , the family of all compact sets in X, we see that is weaker than . □
5. Conclusions
We enlarged the classification of some topological properties on X provided in (Theorem 3.1, [15]) by using specific locally convex topologies with a -base lying between the pointwise topology and the bounded-open topology of the real-valued continuous function space . Our main results are
Theorem 4.
Let X be a Tychonoff space.
- 1.
- The weak topology on is metrizable if and only if it has a -base, and if and only if X is countable and compact sets in X are finite.
- 2.
- There is a metrizable polar topology on such that if and only if X contains a complete sequence.
In the second statement, we say that a countable family of compact sets in X is a complete sequence if there exists a sequence of positive integers such that covers the dual of .
Author Contributions
Conceptualization, M.L.-P. and J.C.F.; formal analysis, M.L.-P. and J.C.F.; investigation, M.L.-P. and J.C.F.; writing—original draft preparation, M.L.-P. and J.C.F.; writing—review and editing, M.L.-P. and J.C.F.; visualization, M.L.-P. and J.C.F. All authors have read and agreed to the published version of the manuscript.
Funding
This research was funded in part by grant PGC2018-094431-B-I00 of Ministry of Science, Innovation and Universities of Spain.
Institutional Review Board Statement
Not applicable.
Informed Consent Statement
Not applicable.
Data Availability Statement
Not applicable.
Acknowledgments
The authors thank the Editor of the Special Issue “Mathematical Analysis and Applications III”, Hari Mohan Srivastava, for his invitation to write this feature paper. Moreover, we thank the anonymous reviewers for their comments and suggestions to improve this paper.
Conflicts of Interest
The authors declare no conflict of interest.
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