The Scientific World Journal

Volume 2014, Article ID 640323, 5 pages

http://dx.doi.org/10.1155/2014/640323

## Topologies on Superspaces of TVS-Cone Metric Spaces

^{1}School of Mathematical Sciences, Soochow University, Suzhou 215006, China^{2}Department of Mathematics, Ningde Normal University, Fujian 352100, China^{3}Department of Mathematics, Zhangzhou Normal University, Zhangzhou 363000, China

Received 6 August 2013; Accepted 31 October 2013; Published 22 January 2014

Academic Editors: R. Abu-Saris, P. Bracken, S. Jafari, and B. Nagy

Copyright © 2014 Xun Ge and Shou Lin. 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

This paper investigates superspaces and of a tvs-cone metric space , where and are the space consisting of nonempty subsets of and the space consisting of nonempty compact subsets of , respectively. The purpose of this paper is to establish some relationships between the lower topology and the lower tvs-cone hemimetric topology (resp., the upper topology and the upper tvs-cone hemimetric topology to the Vietoris topology and the Hausdorff tvs-cone hemimetric topology) on and , which makes it possible to generalize some results of superspaces from metric spaces to tvs-cone metric spaces.

#### 1. Introduction

Ordered normed spaces and cones have many applications in applied mathematics, for instance, in using Newton’s approximation method [1–4] and in optimization theory [5]. By using an ordered Banach space instead of the set of real numbers as the codomain for a metric, -metric and -normed spaces were introduced in the mid-20th century ([2], see also [3, 4, 6]). Huang and Zhang [7] reintroduced such spaces under the name of cone metric spaces. In some results about metric spaces, can metric spaces be relaxed to cone metric spaces? This is an interesting question and many relevant results have been obtained (see [7–11], e.g.). Recently, Khani and Pourmahdian [9] proved that each cone metric space is metrizable, which shows that some improvements by relaxing metric spaces to cone metric spaces are trivial. This leads us to discuss more general cone metric spaces, which further addresses the relationship between metric topology and geometry. In our discussion, it is interesting to consider certain topological groups in place of Banach spaces in the definition of cone metric spaces, which can serve as a topic for further studies [9]. In fact, Du [12] introduced and investigated tvs-cone metric spaces by replacing Banach spaces with topological vector spaces in the definition of cone metric spaces. In the past years, tvs-cone metric spaces have aroused many mathematical scholars’ interests and some interesting results have been obtained (see [12–15], e.g.). As an important result for tvs-cone metric spaces, it is proved that each tvs-cone metric space is metrizable ([13, 14], e.g.), which makes it meaningless to research topological properties of tvs-cone metric spaces. However, we notice that some results related to nontopological properties, for example, metric properties (including hemimetric properties), are not direct consequences of known theorems. In particular, we are interested in tvs-cone hemimetric properties on superspaces of tvs-cone metric spaces.

In fact, superspace is an important concept in topological spaces theory. For superspaces of metric spaces, we often deal with six topologies: the lower topology, the lower hemimetric topology, the upper topology, the upper hemimetric topology, the Vietoris topology, and the Hausdorff hemimetric topology. What relationships are there among these topologies? It is an interesting question. Let and be the space consisting of nonempty subsets of and the space consisting of nonempty compact subsets of , respectively. And then the following two theorems are well known (see [16], e.g.).

Theorem 1. *Let be a metric space and let be a subset of . Then the following hold.*(1)*If is open in the lower topology on , then is open in the lower hemimetric topology on .*(2)*If is open in the upper hemimetric topology on , then is open in the upper topology on .*

*Theorem 2. Let be a metric space. Then the following hold.(1)The lower topology and the lower hemimetric topology coincide on .(2)The upper topology and the upper hemimetric topology coincide on .(3)The Vietoris topology and the Hausdorff hemimetric topology coincide on .*

*As a concrete exploration for tvs-cone metric properties, the following question arises from Theorems 1 and 2 naturally.*

*Question 1. *Can Theorems 1 and 2 be generalized from metric space to tvs-cone metric space?

*This paper investigates superspaces and of a tvs-cone metric space . The purpose of this paper is to establish some relationships between the lower topology and the lower tvs-cone hemimetric topology (resp., the upper topology and the upper tvs-cone hemimetric topology and the Vietoris topology and the Hausdorff tvs-cone hemimetric topology) on and , respectively. These results answer Question 1 affirmatively and make it possible to generalize the discussions for superspaces from metric spaces to tvs-cone metric spaces.*

*Throughout this paper, , , and denote the set of all natural numbers, the set of all positive real numbers, and the set of all nonnegative real numbers, respectively.*

*2. TVS-Cone Metric Spaces*

*2. TVS-Cone Metric Spaces*

*Definition 3 (see [12, 14]). *Let be a topological vector space with its zero vector . A subset of is called a tvs-cone in if the following are satisfied.(1) is a closed in with a nonempty interior .(2) and .(3).

*Remark 4. *Let be a topological vector space with a tvs-cone . It is clear that from Definition 3(2). In addition, it is easy to see that . In fact, pick . Then and when . If , then there is such that . By Definition 3(3), . This contradicts that . So .

*Definition 5 (see [12, 14]). *Let be a topological vector space with a tvs-cone . Some partial orderings , , and on with respect to are defined as follows, respectively. Let .(1) if .(2) if and .(3) if .

*Remark 6. *For the sake of convenience, we also use notations “”, “,” and “” on with respect to . The meanings of these notations are clear and the following hold:(1),(2),(3),(4).

*Definition 7 (see [10]). *A tvs-cone in a topological vector space is called strongly minihedral if each subset of bounded above has a supremum, equivalently, if each subset of bounded below has an infimum.

*In this paper, we always suppose that a tvs-cone in a topological vector space is strongly minihedral.*

*Lemma 8. Let be a topological vector space with a tvs-cone . Then the following hold.(1)If , then for each .(2)If and , then .(3)If and , then there is such that and .*

*Proof. * Let ; that is, . Then there is an open neighborhood of in such that . If , then from Definition 3, where . Note that and is an open subset of . So ; that is .

Let and . Then and ; that is, and . So there is an open neighborhood of in such that . Write . Note that is an open subset of , and from Definition 3. So ; that is, ; hence, . It follows that .

Let and ; that is, . Then there is such that for all and for all . Put , where . Then from the above and . It is clear that and ; that is, and . So and .

*We give the definition of tvs-cone metric, which is very similar to the well-known definition of metric.*

*Definition 9 (see [14]). *Let be a nonempty set and let be a topological vector space with a tvs-cone . A mapping is called a tvs-cone metric on , and is called a tvs-cone metric space if the following are satisfied.(1) for all and if and only if .(2) for all .(3) for all .

*Note that hemimetric takes values in the extended nonnegative real numbers ([16]). We let as a possible value of the mapping in the following definition, where and the following hold.(a) for each .(b) for each .*

*Definition 10. *Let be a nonempty set and let be a topological vector space with a tvs-cone . A mapping is called a tvs-cone hemimetric on , and is called a tvs-cone hemimetric space if the following and are satisfied.(1) for all .(2) for all .

*Proposition 11. Let be a tvs-cone hemimetric space. Put for and , and put . Then is a base for some topology on .*

*Proof. *It is clear that . Let and . Since , . Put ; then . We claim that . In fact, if , then ; hence, , and so . Using the same way, we can obtain that there is such that . By Lemma 8, there is such that and . Let ; then and , so and , and hence . This has proved that . Note that . Consequently, is a base for some topology on . In fact, put such that ; then is a topology on and is a base for .

*3. Superspaces of TVS-Cone Metric Spaces*

*3. Superspaces of TVS-Cone Metric Spaces*

*Definition 12. *Let be a tvs-cone metric space and let denote the topology on induced by the tvs-cone metric described in Proposition 11. For an arbitrary nonempty subset of , and denote the subfamilies and of , respectively.(1) is called the lower topology on , where is generated by the subbase .(2) is called the upper topology on , where is generated by the base .(3) is called the Vietoris topology on , where is generated by and together.

*Definition 13. *Let be a tvs-cone metric space. For , put , , and and , where . Then , , and are tvs-cone hemimetrics on . Let , , and denote the topologies on induced by , , and described in Proposition 11*,* respectively.(1)The topology is called the lower tvs-cone hemimetric topology.(2)The topology is called the upper tvs-cone hemimetric topology.(3)The topology is called the Hausdorff tvs-cone hemimetric topology.

*Remark 14. *It is often preferable to restrict the six topologies , , , , and to , and the relative topologies on will still be denoted by , and so forth. Also, if (resp., , ) is restricted to , then it does not take .

*In this section, we need to use the following notation.*

*Notation.* Let be a tvs-cone metric space, , , , and . Consider the following:(1),(2),(3),(4),(5),(6),(7).

*Theorem 15. Let be a tvs-cone metric space and let be a subset of . Then the following hold.(1)If is open in the lower topology on , then is open in the lower tvs-cone hemimetric topology on .(2)If is open in the upper tvs-cone hemimetric topology on , then is open in the upper topology on .*

*Proof. * Let be open in the lower topology on . Without loss of generality, we can assume that is an element in the subbase for the lower topology ; that is, for some . Let , then . Pick ; then there is such that since is open in . Let ; then ; that is, . Since , , hence for some . Thus, , which means that . It follows that . This proves that . So is an interior point of in the lower tvs-cone hemimetric topology and the proof is completed.

Let be open in the upper tvs-cone hemimetric topology on . Let ; then there is such that . Note that is open in . So is open in the upper topology on . Clearly, . On the other hand, if , that is, , then , and hence . This has proved that . Consequently, is an open neighborhood of for the upper topology on and the proof is completed.

*Remark 16. * The converses of both and in Theorem 15 are not true (even if is a metric space). Moreover, there is no simple relationship between the Vietoris topology and the Hausdorff tvs-cone hemimetric topology on , which is similar to or in Theorem 15 (see [16], e.g.).

It is clear that “” in Theorem 15 can be replaced by . Furthermore, we have the better results for the topologies on superspaces (see the following).

*Lemma 17. Let be a tvs-cone metric space. If is a compact subset of , then, for any , there is a finite subset of such that .*

*Proof. *Let be a compact subset of and let . Then is an open cover of ; there is a finite subset of such that covers . It follows that .

*Lemma 18. Let be a tvs-cone metric space. If with compact in and open in , then there is such that .*

*Proof. *Let with compact in and open in . By Proposition 11, for each , there is such that . Put ; then from Lemma 8. Since is an open cover of and is compact, there is a finite subset of such that covers . By Lemma 8, there is such that for each . We claim that . In fact, let ; then there is such that , that is, . Furthermore, there is such that ; that is, . By Lemma 8, . It follows that . This has proved that .

*Theorem 19. Let be a tvs-cone metric space. Then the following hold.(1)The lower topology and the lower tvs-cone hemimetric topology coincide on .(2)The upper topology and the upper tvs-cone hemimetric topology coincide on .(3)The Vietoris topology and the Hausdorff tvs-cone hemimetric topology coincide on .*

*Proof. * Assume that is open in the lower hemimetric topology on . Let ; then there is such that . Since is compact, by Lemma 17, there is a finite subset of such that . We write for each and put . Note that for each . It is clear that and is an element of the base for the lower topology on . Let . For each , we claim that . In fact, let ; then . Since , . Pick ; then ; hence, ; that is, . This proves that . Furthermore, . Thus, ; that is, . This proves that . It follows that is an interior point of for the lower topology on . Consequently, is open in the lower topology on . Combining Remark 16(2), the proof is completed.

Let be open in the upper topology on . Without loss of generality, we can assume that is an element of the base for the upper topology on ; that is, for some . Let ; then and is compact. By Lemma 18, there exists such that . If ; then ; hence, . It follows that . Consequently, . This has proved that is an open neighborhood of for the upper tvs-cone hemimetric topology on . Combining Remark 16, the proof is completed.

*We need to prove that and . Let be open in the Vietoris topology on . Without loss of generality, we only need to assume that is an element of the subbase for the lower topology or an element of the base for the upper topology . If , then by . So, for each , there is such that . It is clear that . In fact, if , then . Note that . So . It follows that . So is an interior point of for the Hausdorff tvs-cone hemimetric topology . Consequently, is open in the Hausdorff tvs-cone hemimetric topology on . By a similar way, if , then is open in the Hausdorff tvs-cone hemimetric topology on . This has proved that . Conversely, let be open in the Hausdorff tvs-cone hemimetric topology on . Then, for each , there is such that . We claim that . In fact, if , then and . So ; hence, . This proves that . By and , and are open in and , respectively. It follows that is open in . So is an interior point of for the Vietoris topology on . Consequently, is open in the Vietoris topology on . This has proved that .*

*Conflict of Interests*

*Conflict of Interests**The authors declare that there is no conflict of interests regarding the publication of this paper.*

*Acknowledgments*

*Acknowledgments**The authors would like to thank the referees for reviewing this paper and offering their valuable comments. This work is supported by the National Natural Science Foundation of China (no. 11201414, 11171162, 61070245, 11226085, and 11301367), Doctoral Fund of Ministry of Education of China (no. 20123201120001), China Postdoctoral Science Foundation (no. 2013M541710) and Jiangsu Province Postdoctoral Science Foundation (no. 1302156C).*

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