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Abstract and Applied Analysis

Volume 2014 (2014), Article ID 458098, 5 pages

http://dx.doi.org/10.1155/2014/458098
Research Article

Generalized Metric Spaces Do Not Have the Compatible Topology

1Department of Basic Sciences, Faculty of Engineering, Kyushu Institute of Technology, Tobata, Kitakyushu 804-8550, Japan

2Department of Mathematics, Faculty of Science, King Abdulaziz University, Jeddah, Saudi Arabia

Received 12 May 2014; Accepted 9 July 2014; Published 4 August 2014

Academic Editor: Wei-Shih Du

Copyright © 2014 Tomonari Suzuki. 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

We study generalized metric spaces, which were introduced by Branciari (2000). In particular, generalized metric spaces do not necessarily have the compatible topology. Also we prove a generalization of the Banach contraction principle in complete generalized metric spaces.

1. Introduction

In 2000, Branciari in [1] introduced a very interesting concept whose name is “ -generalized metric space.”

Definition 1 (see Branciari [1]). Let be a set, let be a function from into , and let . Then is said to be a -generalized metric space if the following hold: (N1) if and only if for any ;(N2) for any ;(N3) for any , such that are all different.

Example 2. Every metric space is a -generalized metric space.

A -generalized metric space is also said to be a generalized metric space.

Definition 3 (see Branciari [1]). Let be a set and let be a function from into . Then is said to be ageneralized metric space if the following hold: (G1) if and only if for any .(G2) for any .(G3) for any such that are all different.

The concept of “generalized metric space” is very similar to that of “metric space.” However, it is very difficult to treat this concept because does not necessarily have the topology which is compatible with ; see Example 7. So this concept is very interesting to researchers. See also [2, 3].

Motivated by the above, in this paper, we study generalized metric spaces. In particular, generalized metric spaces do not necessarily have the compatible topology. Also we prove a generalization of the Banach contraction principle in complete generalized metric spaces.

2. -Generalized Metric Space

Throughout this paper we denote by the set of all positive integers.

In this section, we study -generalized metric space. In particular, we give examples in order to understand this concept deeply.

Lemma 4. Let be a bounded metric space and let be a real number satisfying Let and be two subsets of with and . Define a function from into by Then is a generalized metric space.

Proof. (N1) and (N2) are obvious. Let us prove (N3). Let be all different. Put In the case where , (N3) holds because . In the other case, where , without loss of generality, we may assume . Then we have and from the definition of . Hence, Thus (N3) holds.

Definition 5. Let be a -generalized metric space. Then a net is said to converge to if and only if .

Definition 6. Let be a topological space with topology . Let be a function from into satisfying (N1)–(N3) with some . Then is compatible with if and only if the following are equivalent for any net in and : (a) .(b) converges to in .

The following is a very important example.

Example 7. Let Define a function from into by Then the following hold: (i) is not a metric space;(ii) is a generalized metric space;(iii) does not have a topology which is compatible with .

Proof. Since is not a metric space. Define a metric on by for . Put Then is equal to the defined by Lemma 4 with . Therefore, is a generalized metric space. In order to show (iii), we will show that the following does not hold. If a net converges to and for every a net converges to , then has a subnet converging to ; see [4, page 77].

We have that converges to and converges to for every . However, since for , a net does not converge to . Therefore there does not exist a topology which is compatible with .

Remark 8. For , . For , if and only if and for any .

Remark 9. Indeed, let be the topology induced by a subbase: where . Since we have Hence is an open neighborhood of . So a sequence does not converge to in . Since , is not compatible with .

We can easily make an example of a -generalized metric space which is not a -generalized metric space for .

Example 10. Put and let satisfy . Define a function from into by Then the following hold: (i) is not a -generalized metric space for with ;(ii) is a -generalized metric space for with .

Proof. (N1) and (N2) obviously hold. Let satisfy . Since (N3) does not hold. So is not a -generalized metric space. Let satisfy . Let be all different. Then we have Thus (N3) holds. Hence is a -generalized metric space.

We give some definitions. The reason of these definitions is that does not necessarily have the topology which is compatible with . So does not necessarily have the uniformity which is compatible with .

Definition 11. Let be a -generalized metric space. (a)A sequence is said to be Cauchy if and only if .(b) is said to be complete if and only if every Cauchy sequence converges to some point in .(c) is said to be Hausdorff if and only if implies .

Lemma 12. Let be a -generalized metric space and let such that are all different and are all different. Then holds.

Proof. In the case where , the conclusion obviously holds from (N1). In the other case, where , the conclusion obviously holds from (N3).

3. The CJM Fixed Point Theorem

In this section, we generalize the CJM fixed point theorem; see Ćirić [5], Jachymski [6], and Matkowski [7, 8].

Theorem 13. Let be a complete -generalized metric space and let be a CJM contraction on ; that is, the following hold: (i)for every , there exists such that implies for any ;(ii) implies for any .

Then has a unique fixed point of . Moreover, for any .

Proof. We first note that is nonexpansive by (ii); that is for any . Fix and define a sequence in by for . We next show that converges to a fixed point of , dividing the following three cases: (a)there exists such that ;(b) for all and there exist such that and ;(c) are all different.

In the first case, is a fixed point of . By (N1), converges to . In the second case, from (ii), we have is strictly decreasing. So, since , we have This is a contradiction. Thus, the second case cannot be possible. In the third case, from (ii), we have is strictly decreasing for any . So converges to some . Then we note that for every . Arguing by contradiction, we assume . From (i), there exists such that

From the definition of , there exists such that . Then we have . This is a contradiction. Therefore we obtain . That is, holds for any . Thus holds. Fix . Then, by (i), there exists such that

Let such that for all with . We will show for by induction. For , we have and, thus, (23) holds. We assume (23) holds for some with . We have, by (N3), Hence . We put We note . By (N3), we have Thus, (23) holds for . So, by induction, (23) holds for every . Therefore we have shown Since is arbitrary, we obtain that is Cauchy. Since is complete, converges to some point . We have by Lemma 12 and the nonexpansiveness of for sufficiently large . As tends to , we obtain . Thus, is a fixed point of . The uniqueness of the fixed point is obviously followed by (ii).

Remark 14. In [9], there is another fixed point theorem which is independent of Theorem 13.

By Theorem 13, we obtain a generalization of the Banach contraction principle [10, 11].

Corollary 15 (see Branciari [1]). Let be a complete -generalized metric space and let be a contraction on ; that is, there exists such that for any . Then has a unique fixed point of . Moreover, for any .

Remark 16. The authors in [12] stated the proof in [1] is incorrect and gave a proof under the assumption that is Hausdorff and . See also [13].

In order to show that Theorem 13 is a generalization of Theorem 3.1 in [14], we prove the following. See also [15]. The idea on the proof of the following proposition appears in [16, 17].

Proposition 17. Let be a -generalized metric space and let be a mapping on . Assume that there exist functions from into such that the following hold: (i) for any ;(ii) is nondecreasing;(iii) for any with .

Then is a CJM contraction.

Proof. Since for any , (ii) of the definition of CJM contraction obviously holds. We will show (i) of the definition of CJM contraction. Fix . From (iii), we can put We choose such that Let satisfy . In the case where , we have because . In the case where , we have which implies . In the other case, where , we have which implies . Hence we have in all cases. Therefore is a CJM contraction.

Conflict of Interests

The author declares that there is no conflict of interests regarding the publication of this paper.

Acknowledgments

The author wishes to express his sincere thanks to the referees for their careful reading and many suggestions. The author is supported in part by Grant-in-Aid for Scientific Research from Japan Society for the Promotion of Science.

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