Abstract
This paper provides two new Velichko-like theorems for the weak counterpart of the locally convex space of all real-valued functions defined on a Tychonoff space X equipped with the compact-open topology .
Keywords:
Velichko’s theorem; K-analytic framed; angelic space; σ-compact space; (relatively) sequentially complete set MSC:
54C35; 46A03; 54H05; 46A50
1. Preliminaries
Henceforth, unless otherwise stated, X will be a nonempty completely regular Hausdorff space. We represent by the linear space of real-valued continuous functions defined on X equipped with the pointwise topology . The topological dual of is denoted by , or by when provided with the weak* topology , so that . The linear space equipped with the compact-open topology is represented by . In what follows, we shall denote by the weak counterpart of the locally convex space , i.e., the space equipped with the weak topology , where E stands for the dual of .
The current work supplements the research carried out in [1,2] and is related to [3], Chapter 9, and [4,5,6]. It must be regarded as a part of the important growth of -theory and -theory that is taking place nowadays (see, for example, [7,8,9,10]).
2. Introduction
A set M of is called (relatively) sequentially complete if each Cauchy sequence of contained in M converges in to a function (respectively, to some ). The classical theorem of N. V. Velichko ([11], 1.2.1 Theorem) together with two different generalizations reads as follows:
Theorem 1.
Each of the following statements implies that X is finite:
- 1.
- The space is σ-compact (Velichko).
- 2.
- The space is σ-countably compact (Tkachuk and Shakhmatov, [12]).
- 3.
- The space is σ-bounded relatively sequentially complete (Ferrando, Ka̧kol and Saxon, [2], Corollary 3.2).
Here, is said to have a -topological property if there is a sequence of subsets of , such that each set enjoys this -topological property. Also, the term bounded is meant in the locally convex sense ([13], 1.4.5 Definition). The first and second statements of the previous theorem are equivalent by virtue of Theorem 2 below, which is a part of [14], Proposition 1 (see also [3], Proposition 9.6).
In the next example, denotes the linear subspace of consisting of bounded functions equipped with the relative pointwise topology. This example shows that the third statement of Theorem 1 does not work for instead of .
Example 1.
If is covered by a sequence of pointwise-bounded sequentially complete bounded sets, X need not be finite.
Proof.
If B stands for the closed unit ball of the Banach space , then . So, is a sequence of pointwise-bounded sets. Since is sequentially complete, if is a Cauchy sequence in contained in B, there exists , such that in . But, as for all , we obtain for all . Thus, , which shows that B is sequentially complete in . Consequently, each set is sequentially complete. But is infinite. □
Let us recall that a resolution for a topological space X is a covering of X, such that whenever coordinate-wise, i.e., if for every . If E is a locally convex space, a resolution for E is called bounded if each is a bounded set in E. An increasing covering of a locally convex space E consisting of absolutely convex bounded sets is a trivial example of a bounded resolution for E, by setting for each . The next result was obtained in [14], Proposition 1.
Theorem 2
(Ferrando–Ka̧kol). The space has a bounded resolution if and only if is both K-analytic-framed in and angelic.
The last statement of Theorem 1 is actually a consequence of [3], Lemma 9.5 together with the following characterization for X to be a P-space, in which the requirement of pointwise boundedness for the covering sequence, which is required in the third statement of Theorem 1, has been dropped ([2], Theorem 3.1).
Theorem 3
(Ferrando–Ka̧kol–Saxon). The space is σ-relatively sequentially complete if and only if X is a P-space.
In this paper, we are going to prove the following two Velichko-type theorems for the weak counterpart of .
Theorem 4.
The space is covered by a sequence of relatively sequentially complete sets if and only if X is a P-space.
Theorem 5.
The space is covered by a sequence of pointwise-bounded relatively sequentially complete sets if and only if X is finite.
Theorems 4 and 5 are, respectively, the -version of Theorem 3 and of the third statement of Theorem 1.
3. An Auxiliary Result
Recall that a sequence in is called pointwise eventually constant (cf. [2]) if for each there exists , such that for all but finitely many . So, if is a pointwise eventually constant sequence in , there always exists some , such that pointwise on X. We shall require the following theorem, which is contained (but not explicitly stated) in [2], it being a consequence of [2], Theorem 1.1, and of the equivalence of statements (7) and (7’) of paper [2] (see [2], p. 910).
Theorem 6.
The following statements are equivalent.
- 1.
- Every uniformly bounded pointwise eventually constant sequence converges in .
- 2.
- X is a P-space.
If X is a compact space and is a regular countably additive real-valued measure defined on the Borel algebra of X, we shall denote by the linear space of all (classes of) real-valued -measurable functions defined on X, and we shall denote by the linear space of regular countably additive Borel real measures on . As for the pointwise topology, a subset M of is called relatively sequentially complete if each Cauchy sequence of contained in M converges in to some .
Theorem 7.
Let X be a compact set. The space is σ-relatively sequentially complete if and only if X is finite.
Proof.
First, let us show that every uniformly bounded sequence in that is pointwise convergent in is a Cauchy sequence in . So, let be a uniformly bounded sequence in pointwise convergent in . If B denotes the closed unit ball of the Banach space , there is , such that for every , so that for all . As pointwise on X, then and
for every . Thus, using the fact that , we establish that is a Cauchy sequence in , as stated.
The following argument is based on (but not identical to) the proof of [2], Theorem 3.1. Let be a sequence of weakly relatively sequentially complete subsets of . As is a Banach space, the Baire category theorem provides , such that has an interior point where the closure is in the norm-topology of . So, if B stands again for the closed unit ball of , there is , such that
If is a uniformly bounded pointwise eventually constant sequence in , there is , such that in and there exists , such that for every . So, is a uniformly bounded pointwise eventually constant sequence in that converges to in . Clearly, for each there is , such that
where denotes the norm of .
All this implies that the sequence is uniformly bounded and converges pointwise to in , so that, according to the first part of the proof, is a Cauchy sequence in . But, as and is weakly relatively sequentially complete, it follows that in . Particularly, one has . As , this means that . Therefore, converges in .
Hence, according to Theorem 6, X must be a P-space. But every compact P-space is finite (see [15], Problem 4K). □
Remark 1.
Theorem 7 clearly implies the well-known fact that if X is a compact space the Banach space is weakly sequentially complete if and only if X is finite (note that if the compact set X is infinite, then contains an isomorphic copy of , which is not weakly sequentially complete).
4. Proofs of Theorems 4 and 5
In the present section, we prove Theorems 4 and 5, which are concerned with the weak topology of rather than the pointwise topology . We shall need the following result, which is an extension of a classic result of -theory (see [16], Proposition 4.1), which we have borrowed from [17].
Lemma 1
([17], 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 .
4.1. Proof of Theorem 4
Proof.
Suppose that is covered by a sequence consisting of weakly relatively sequentially complete sets. We claim that all compact sets of X are finite.
Assume for the sake of contradiction that there is an infinite compact set Q in X that is, hence, metrizable by the Urysohn metrizability theorem. By Lemma 1, there is a linear-continuous map , such that , i.e., a continuous linear extender when is regarded as a Banach space. Consequently, the linear map is weak-to-weak continuous and is, hence, uniformly continuous for the weak topologies. Let us observe that is a countable covering of consisting of weakly relatively sequentially complete sets. In fact, if is a weakly Cauchy sequence in then is a weakly Cauchy sequence in , due to being uniformly continuous, so that there is , such that in . Now the restriction map defined by is continuous, which implies that T is also weak-to-weak continuous. Therefore, in . Hence, setting and employing , it follows that in . Thus, the set is relatively sequentially complete in , which shows, as stated, that the sequence is a covering of consisting of weakly relatively sequentially complete sets. Now the application of Theorem 7 guarantees that Q is finite, as desired.
As every compact set of X is finite, we obtain , which means that the space is -relatively sequentially complete. Hence, Theorem 3 yields that X is a P-space.
Conversely, if X is a P-space, the compact sets of X are finite and, consequently, we obtain . So, it follows from Theorem 3 that is -relatively sequentially complete. □
4.2. Proof of Theorem 5
Proof.
Finite X ensures a suitable sequence, with each , where B is the closed unit ball in the finite-dimensional Banach space . To prove the converse, recall the well-known property that a Tychonoff space X is pseudocompact if and only if is -bounded (in the locally convex sense). Hence, if we suppose that space is covered by a sequence of pointwise-bounded weakly relatively sequentially complete sets, then X is pseudocompact, and by Theorem 4 we obtain that X is a P-space. Therefore, X must be finite. □
We thank our anonymous reviewer for the simplification of this proof by employing the mentioned characterization of pseudocompactness of a Tychonoff space X.
Remark 2.
It is shown in [3], Proposition 9.18, that if has a fundamental sequence consisting of bounded sets (i.e., a sequence of bounded sets that swallows the bounded sets) then X is finite. Of course, this property does not hold for , because if X is any compact set and B stands for the closed unit ball of then is a fundamental sequence of bounded sets for .
5. Conclusions
In this paper, we have provided a complement to the research of [1,2] with Theorems 4 and 5, stating that if X is a Tychonoff space, then is covered by a sequence of
- Weakly relatively sequentially complete sets if and only if X is a P-space.
- Pointwise-bounded weakly relatively sequentially complete sets if and only if X is finite.
Author Contributions
The authors S.L.-A., M.L.-P. and S.M.-L. contributed equally to this work. 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 the Ministry of Science, Innovation and Universities of Spain.
Institutional Review Board Statement
Not applicable.
Informed Consent Statement
Not applicable.
Data Availability Statement
Data are contained within the article.
Acknowledgments
The authors wish to thank the referees for their careful reading as well as for valuable comments and suggestions that have improved both the organization and readability of the paper.
Conflicts of Interest
The authors declare no conflict of interest.
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