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Cerpa-Torres, Manuel Felipe; Rincón-Villamizar, Michael A.

doi:https://doi.org/10.1155/2019/7146073

International Journal of Mathematics and Mathematical Sciences

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Research Article | Open Access

Volume 2019 | Article ID 7146073 | https://doi.org/10.1155/2019/7146073

Isomorphisms from Extremely Regular Subspaces of into Spaces

Manuel Felipe Cerpa-Torres¹and Michael A. Rincón-Villamizar¹

Academic Editor: Luca Vitagliano

Received30 Oct 2019

Accepted03 Dec 2019

Published27 Dec 2019

Abstract

For a locally compact Hausdorff space K and a Banach space X, let be the Banach space of all X-valued continuous functions defined on K, which vanish at infinite provided with the sup norm. If X is , we denote as . If be an extremely regular subspace of and is an into isomorphism, what can be said about the set-theoretical or topological properties of K and S? Answering the question, we will prove that if X contains no copy of , then the cardinality of K is less than that of S. Moreover, if and is also a subalgebra of , the cardinality of the αth derivative of K is less than that of the αth derivative of S, for each ordinal Finally, if and , then K is a continuous image of a subspace of S. Here, is the geometrical parameter introduced by Jarosz in 1989: As a consequence, we improve classical results about into isomorphisms from extremely regular subspaces already obtained by Cengiz.

1. Introduction

All Banach spaces here are assumed to be real. For a locally compact Hausdorff space K and Banach space X, let be the Banach space of all X-valued continuous functions defined in K which vanish at infinite, endowed with the supremum norm. If K is compact, we write instead of . When X is , these spaces are denoted by and , respectively. For a set A, we denote its cardinality by . We denote if X and Y are isomorphic Banach spaces and X ↪ Y if Y contains a subspace isomorphic to X. If and is an isomorphism, the number is called distortion of T; if , , we write . Other notations and definitions used here can be found in [1].

The classical Banach–Stone theorem established that if and are isometric, then K and S are homeomorphic [2–5]. Amir and Cambern, independently, showed that the conclusion of Banach–Stone theorem holds if we assume that [6, 7]. Moreover, it is well-known that 2 is the best number in this result [8]. On the other hand, the number 2 can be replaced by 3 in the case of countable compact metric spaces [9].

Many authors have obtained generalizations of the Banach–Stone theorem, both in the scalar and vector-valued case (see [10–14]). Among the many generalizations of this theorem, we would like to highlight the following one. First, we recall a definition introduced by Cengiz [15].

Definition 1. A closed subspace of is called extremely regular if for each , open with and , there is such that for all . An extremely regular subspace of is called extremely regular subalgebra if whenever .
For examples of extremely regular subspaces (subalgebras), see [15]. The next result was obtained by Cengiz [16].

Theorem 1. Let and be extremely regular subspaces of and , respectively. If , then K and S are homeomorphic.

Despite this, the conclusion of the Banach–Stone is far from to be valid for arbitrary isomorphisms. Thus, it is natural asking.

Problem 1. Which set-theoretical and topological properties are preserved by isomorphisms of extremely regular subspaces?
So, the aim of this paper is to give answers to the problem posed above. In this direction, Cengiz showed that, under additional topological hypothesis on K and S, existence of an isomorphism between extremely regular subspaces of and , respectively, implies that [17, 18]. Our first result shows that Cengiz’s result holds for arbitrary locally compact Hausdorff spaces. So, we prove the following.

Theorem 2. Let and be extremely regular subspaces of and , respectively. If , then .

In order to prove Theorem 2, we establish the following vector-valued result, which generalizes ([19], Theorem 2).

Theorem 3. Let K be infinite. For an extremely regular subspace of and a Banach space X, we have

Recall that if S is a topological space, the derivative of S, denoted by , is obtained by deleting from S its isolated points. If α is an ordinal, we define the αth derivative of S as follows: , , and , in the case where α is a limit ordinal.

The following result is an extension of ([19], Theorem 5) for extremely regular subspaces.

Theorem 4. Let X be a Banach space containing no copy of . Let be an extremely regular subalgebra of and an into isomorphism. For each ordinal α such that is infinite, we have

Moreover, if is finite, then so is .

On the other hand, a well-known generalization of the Banach–Stone theorem was given by Holsztyński who showed that if is isometrically isomorphic to a subspace of , then K is a continuous image of a subset of S [20]. Jarosz extended this result for extremely regular subspaces of and into isomorphisms with distortion less than 2 [21]. Moreover, in the setting of metric spaces, Jarosz obtained a vector-valued extension of Holsztyński’s theorem for isomorphisms from extremely regular subspaces of into spaces [14]. Our last theorem is a vector-valued version of the previously mentioned Jarosz’s result which is valid for all locally compact Hausdorff spaces. Before stating it, we recall a parameter introduced by Jarosz [14] for all Banach spaces X:

Theorem 5. Let be an extremely regular subspace of and X a Banach space satisfying . If is an into isomorphism satisfying , then K is a continuous image of a subset of S.

Paper is organized as follows: in the first section, we will establish some properties for extremely regular subspaces of . In the remaining sections, we will prove Theorems 2, 3, 4, and 5. The idea of the proof of these theorems is inducing a set-valued map from isomorphisms defined on extremely regular subspaces and then improving the arguments given in [19] and [22] in order to obtain the wanted conclusions.

2. On Extremely Regular Subspaces

In this section, we prove some properties about extremely regular spaces. We start by stating the notation used throughout the paper.

The unit ball of a Banach space X is denoted by . We identify dual space with the space of regular countably additive bounded measures and denote it by ([23], p. 222). The space will be equipped with the topology inherited from . The total variation of a measure on a Borel set E is denoted by and its norm by . As usual, denotes the Dirac measure at .

Let be given and is a fundamental system of open neighborhoods of k. Consider the set and define a partial order in as follows: if and only if and . Note that is a directed set. Let be an extremely regular subspace of . It is easy to see that there exists a net in satisfying(1)(2) for all

We write to indicate that above conditions are satisfied.

The next result is proved in ([24], Lemma 1) for compact spaces. The proof also works in the locally compact case, so we restate it as follows.

Lemma 1. Let be an extremely regular subspace of and given. Suppose that . If , then

It is well known that if K is infinite, then contains a subspace isometric to . We will prove an analogous result for extremely regular subspaces of .

Proposition 1. Let K be infinite and an extremely regular subspace of . Then contains a subspace isomorphic to .

Proof. Let be a sequence of distinct elements of K and a sequence of open subsets of K satisfying if and for all For each , let be such that(1)(2), if We claim that is a weakly unconditionally Cauchy sequence. Let be a bounded sequence and fixed. ThenOn the other hand, if is given, we haveFrom inequalities (5) and (6), it follows thatSince is arbitrary, we haveSo, we prove the claim. Now, observe that . By the well-known Bessaga–Pełczyński theorem ([1], Theorem 2.4.11), we conclude that embeds in .

3. On Isomorphisms from Extremely Regular Spaces into Spaces

The aim of this section is proving Theorem 3, and as a consequence, we give a proof of Theorem 2 which was proved by Cengiz under additional topological hypothesis. First, we state two lemmas. A proof of the following one can be found in ([25], Proposition 5.1).

Lemma 2. Let X be a Banach space. If satisfy for some , then there are with such that .

The following result was proved for compact spaces in ([24], Theorem 2.4). By modifying slightly the proof showed in [24], we see that result also holds in the locally compact case.

Theorem 6. Let be an extremely regular subspace of and be an into isomorphism. For each , we have

To go on, we introduce a set-valued map which plays a fundamental role in our proofs.

Notation 1. Let be an extremely regular space and an into isomorphism. For any , we define a set-valued map from S to K as follows:where is defined by for all
Given a set-valued map from K to S, for any , we defineIn the case where is a singleton, we write instead of .

Lemma 3. Let X be a Banach space containing no copy of and an extremely regular space of . Let be an into isomorphism. For any and , is finite.

Proof. Suppose that there is such that is infinite and take a sequence of distinct elements of . For each , there is such thatFix with By Lemma 1, if we haveSo, there exists such that if and thenSet and choose satisfying Thus, andBy setting , we have(1)(2) for all (3)Now by induction, assume that there are disjoint open subsets of K, , and with satisfying(1)(2) for all (3)for all . Hence, by Lemma 1, if , we haveThus, there exists such that if and , thenLet be open with and for all and set . If satisfies and , then andFinally, we set . Thus(1)(2) for all (3)Hence, there exists a sequence of mutually disjoint open subsets of K, a sequence in , and a strictly increasing sequence of natural numbers such that(1)(2) for all (3)for all . By arguing as in Proposition 1, it is easy to check that is a weakly unconditional Cauchy sequence. Sincewe conclude that embeds in X by ([1], Theorem 2.4.11), contradicting the hypothesis. Thus, must be finite for each

Lemma 4. If , then for all , we have .

Proof. Notice that embeds isometrically in via the isomorphism , where for all . Let be the into isomorphism given by for . By Theorem 6, we haveSince and , it follows thatThus, there is , satisfying .
Now, if , from Lemma 1 we obtainWhence, there are and such that ; that is, .

Proof of Theorem 1. Let be an into isomorphism. We claim that S is infinite. Indeed, if S were finite, say , then . Since K is infinite, by Proposition 1. Whence, embeds in and by ([26], Theorem 1) we conclude that which is impossible. So, S must be infinite. Now, let be fixed. If is given, by Lemma 4, we have . By taking , we obtain that . Thus,On the other hand, by Lemma 3, is finite for all Since S is infinite, the equation above implies thatThus,

Proof of Theorem 2. Suppose that , then and If K were finite, then S also must be finite by Theorem 3. By ([15], Main theorem), we have and , where and . So, If K and S are infinite, then by Theorem 2, we have and . Thus, by the Cantor–Schröder–Bernstein theorem, we conclude that .

4. On Isomorphisms from Extremely Regular Spaces into Spaces with Distortion Less than 3

In this section, we prove Theorem 4. We need the following lemma for isomorphisms from extremely regular subalgebras of into spaces. The proof follows by adapting the argument used in ([19], Proposition 3.1) for extremely regular subalgebras. First, we state a notation.

Notation 2. For and , we denote

Lemma 5. Let be an extremely regular subalgebra of . Let be an into isomorphism with and fix such that If satisfy and , thenwhere

Proof of Theorem 3. Let α be an ordinal and satisfying Also let η be as in Lemma 5. We shall use the fact that closed subalgebras of are lattices ([23], Lemma 4.48). Consider the collectionfor each and defineObserve thatis family of compact subsets of S. So, if we prove that this family has the finite intersection property, we will immediately conclude that for .
Let then and are in . So, by Lemma 5, we haveThus by compactness, the family has nonempty intersection. It follows thatNow, consider the set-valued map from K to S given by . We will prove that is finite for all If not, there is such that is infinite. Since K is regular, there is a sequence of open disjoint subsets of K such that . For each , pick . Also, for each , let be such that , , and for all . Note that and in view of , we have for each . Whence,It is not difficult to show that is weakly unconditionally convergent. By (33) and ([1], Theorem 2.4.11), we conclude that embeds in X which is impossible. Henceforth, is finite for all .
Now, if and , we have which implies that where Thus,Consequently, if is finite, then so is . Now if is infinite, from the equation above, we havethat is , and we are done.

5. On Isomorphisms from Extremely Regular Spaces into Spaces with Distortion Less than

Finally, we prove Theorem 5. We remark that Jarosz obtained the statement of theorem in the metric setting ([14], Theorem 1). By following a new approach, we show that conclusion of theorem holds in the general topological case.

Proof of Theorem 4. Let be a into isomorphism with . By considering the isomorphism , we have and . So, we may assume that and . Let such thatThe following claim is a consequence of Lemma 4.

Claim 1. For each , we have .

Claim 2. For with , we have .
Suppose that and let . There are satisfyingLet be given, thenIf and , it follows from Lemma 1 thatThus, there is such that if and , thenIn a similar way, there exists such that if and , we haveChoose and open subsets of K with , , and , and let be such that . Then, and . In view of inequalities (40) and (41), we obtainHence,By Lemma 2, there are with satisfyingSince and , we infer thatArbitrariness of implies that which contradicts (36). So, we are done.
Now, let and define by if and only if . Claims 1 and 2 show that ϕ is a surjective function.

Claim 3. The function ϕ is continuous.
Let be a net in converging to and assume that . Let be given. Since , there is such thatNow, if , we have by Lemma 1Since , there is a compact neighborhood with such that for any , there exists satisfying . In view of (47), there is open with and such thatwhereas , there is such that andBecause of , there is such thatSo, if , by Lemma 1 we getFrom (51), there is open with , , and satisfyingIt follows from (49), (52), and Lemma 2 that there exist satisfying andNotice that , so, from the inequality above, we obtainSince was chosen arbitrarily, we have which is impossible by (36). This completes the proof of Theorem 4.

Data Availability

No data were used to support this study.

Conflicts of Interest

The authors declare that they have no conflicts of interest.

Acknowledgments

The second author thanks Programa de Movilidad de la Universidad Industrial de Santander for supporting this work (request 2958).

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Copyright

Copyright © 2019 Manuel Felipe Cerpa-Torres and Michael A. Rincón-Villamizar. 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.

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