Abstract and Applied Analysis
Volume 2015 (2015), Article ID 460893, 5 pages
http://dx.doi.org/10.1155/2015/460893

Research Article

Completion of a Dislocated Metric Space

P. Sumati Kumari,¹ I. Ramabhadra Sarma,² J. Madhusudana Rao,³ and D. Panthi⁴

¹Department of Mathematics, K L University, Vaddeswaram, Andhra Pradesh 522502, India
²Department of Mathematics, Acharya Nagarjuna University, Andhra Pradesh 522 510, India
³Department of Mathematics, Vijaya College of Engineering, Khammam, Telangana 507 305, India
⁴Department of Mathematics, Nepal Sanskrit University, Valmeeki Campus, Exhibition Road, Kathmandu 44500, Nepal

Received 26 August 2014; Accepted 4 November 2014

Academic Editor: Kyung Soo Kim

Copyright © 2015 P. Sumati Kumari et al. 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 provide a construction for the completion of a dislocated metric space (abbreviated -metric space); we also prove that the completion of the metric associated with a -metric coincides with the metric associated with the completion of the -metric.

1. Introduction

Completion of a metric space via Cauchy sequences can be achieved because of certain convergence properties enjoyed by the metric and the property that convergent sequences are Cauchy sequences. Lack of some of these properties in weaker forms of metric spaces comes in the way of completion process in the above lines. In semimetric spaces several new ways of completeness were invented, for example, Cauchy completeness, McAuley notions of strong and weak completeness [1], Moore completeness [2], and so on. Moshokoa [3] introduced the notion of convergence completeness for semimetric spaces and discussed completion on these lines.

For -metric spaces adoption of Van der-Waerdens completion process through Cauchy sequences is possible but is not routine, the difficulty being the mischief created by the isolated points. Here we show how to overcome this problem.

In his study of programming languages, Hitzler [4] associated a metric with every -metric by defining

We establish that the metric associated with the completion of a -metric is the completion of the metric associated with .

We recall [4] where a distance function on a set is said to be a -metric on if

(i);(ii);(iii) for all in .

If is a -metric on then is called a -metric space. Many authors (see, e.g. [5–9]) have studied fixed point theorems in -metric spaces but topology and topological aspects on this space are discussed by Sarma and Kumari [10].

The class and is an open base for the topology induced by , where . In what follows whenever we talk about topological properties of a -metric space, we refer to the topology .

In [11], the authors highlighted some convergence properties and covers a huge range of implications and nonimplications among them. By using these convergence axioms many authors (see, e.g. [12–15]) have proved fixed point theorems in certain spaces.

The presence of the triangle inequality lends the Haussdorff property for and some nice properties to . In particular satisfies properties through :

:;:;:;:;:; for all

Above mentioned convergence axioms can be found in [11]. If the triangular inequality is deleted from the axioms on then it is difficult to define the concept of completion of the resulting distance space. In such an amorphous space, even constant sequences may fail to converge. This and related difficulties compel us to retain the triangle inequality in the discussion of completeness.

Definition 1. Let and be distance spaces. A map is called an isodistance if for any one has

2. Completion

In what follows, is a -metric on a nonempty set . A complete -metric space is a -metric space in which every Cauchy sequence converges. “Cauchy sequences” in -metric spaces are defined exactly as in metric spaces.

Lemma 2. is an isolated point of if and only if or .

Proof. Suppose is an isolated point of . Then there exists such that or Conversely suppose or . If , then clearly is an isolated point of . If , then which implies that for all . Hence is an isolated point of .

Corollary 3. If , then is an isolated point of .

Proof. If , then and so for all in So or

Theorem 4. Let be a -metric space. Then there exists a complete -metric space and an isodistance such that is dense in .

Proof. Let be the collection of isolated points of and let . Let be the collection of sequences in which are ultimately a constant element lying in and denote the class of Cauchy sequences in . We define relations and , respectively, on and as follows.
If are sequences in then iff the ultimately constant value of coincides with that of .
If are sequences in then iff . Clearly is an equivalence relation. We verify that is an equivalence relation. Suppose and . Since is a Cauchy sequence in , and hence is reflexive.
Suppose for . Then . Hence is symmetric.
If and . If , then there exists an integer such that and , if . Consider

This proves that is transitive and hence an equivalence relation. Let . Then is an equivalence relation on .
Let denote /. If denotes the equivalence class in containing the sequence .
If let be the constant sequence where ,   and , the equivalence class containing .
If , , it follows from the triangle inequality that . Since are Cauchy sequences, given that , there exists a positive integer such that and for all .
This implies that proving that is a Cauchy sequence of real numbers. By the completeness of this sequence converges. The definition of makes it obvious that is independent of the choice of the representative sequences , respectively, from the classes .
We can prove similarly if and , , , exists and or equal. Provided and belong to the same equivalence class.
We define as follows:

  if and are, respectively, the ultimately constant terms of .  if , and eventually. If , , then define . If , , then define .

Verification That Is a -Metric on Clearly and for .
Suppose . Let and . We first see that either are both in or are both in .
Suppose, on the contrary, and . Let be the ultimately constant value of . Consider

Hence , contrary to the fact that .
Suppose , , and with the ultimately constant values of and , respectively.
Then .
Suppose , and . Consider

Verification of the triangular inequality is routine.
Embedding of in . Define by . It is clear that is an isodistance. We now verify that is dense in . Let and .
Case (i) ( ). In this case let “” be the ultimately constant value of .
Then by the definition of .
Then . Thus in this case.
Case (ii) ( ). There exists a positive integer such that if . Let . Then since ,  ,

Hence is dense in .
Is Complete. Let be a Cauchy sequence in , and . There exists such that implies .
There is no harm in assuming that . Since is dense in , for each positive integer , there exists in such that .
Hence

Hence is a Cauchy sequence in . Since is an isodistance, is a Cauchy sequence in .
Moreover, , if
Let denote , by the triangle inequality:

  for ; ;  proving that is complete.

Definition 5. Let and be -metric spaces. is said to be a completion of if

(i) is complete;(ii)there is an isodistance such that is dense in .

Note. If is a complete metric space then its completion is itself.

Lemma 6. Let be a -metric space and let be a completion of . Let be isodistance embedding in with dense in . Then a point of is an isolated point if and only if for some isolated point of .

Proof. Suppose is an isolated point of . If is not in , then since is dense in , there exists a sequence in such that .
By Lemma 2, it follows that is not an isolated point of , a contradiction so that for some . Hence is an isolated point of and hence that of . Since and are isometric, is an isolated point of .
Conversely, suppose is an isolated point of . If is not an isolated point of , then for each positive integer , there exists in such that . Since , either or there exists in such that .
Now

Also since .
Hence which, by Lemma 2, contradicts the fact that is an isolated point of .

Theorem 7. Let be a -metric space, and completion of , and isometrics such that is dense in . Then there exists an isodistance such that following diagram is commutative.

Proof. Consider the following:

Definition of . If and is an isolated point of , then is an isolated point of ; hence is an isolated point of .
Define . If and is not an isolated point, there exists a sequence in such that converges to in .
Since is an isodistance and is convergent and hence a Cauchy sequence, it follows that is a Cauchy sequence in . Since is an isodistance and is a Cauchy sequence, it follows that is a Cauchy sequence in . Since is complete, there exists such that . Clearly this is independent of the choice of the sequence in .
Define . Clearly and bijection.
Is an Isodistance. If , and .
So
If and , where , then

The arguments for the cases when and or and are similar. Hence is an isodistance. Interchanging the places of and , we get in a similar way an isodistance such that .
Since and , we have and .
Since is dense in and in , we get = identity on and is identity on .
Hence and are bijections.

3. Completion of the Metric Associated with a -Metric

If is a -metric on then is a metric on if is defined by when and for all in .

Suppose is the completion of ; then gives rise to a metric defined by for all and for all .

Also, the metric space has a metric space as its completion. In this section, we prove that the metric spaces and are isometric.

Definition 8. Let be a -metric space. Define on by

is a metric on and is called the metric associated with .
Clearly and whenever . If , and .
Write . Then and .
The collection and generate the same topology on . However, convergent sequences in are not necessarily the same since constant sequences are convergent sequences with respect to , while this holds with respect to for with only.
Existence of points with positive self-distance leads to unpleasantness in the extension of the concept of continuity in metric spaces as well. This is evident from the following.

Example 9. Let be a -metric on a set which is not a metric. So that the set is nonempty. If is a metric associated with then the identity map is continuous in the usual sense. But if , the constant sequence converges in while it does not converge in .
If we call sequentially -continuous if .
If and is a sequence in , we say that is -Cauchy sequence or simply -Cauchy if is a Cauchy sequence in .

Proposition 10. either

(i) eventually or(ii) can be split into subsequences and where for every , for any and .

Proof. Routine.

Proposition 11. If a sequence in is -Cauchy then is -Cauchy. Conversely if is -Cauchy and is not eventually constant, then is -Cauchy.

Proof. Since , -Cauchy -Cauchy.
Conversely suppose that is -Cauchy, given such that if and . So if , , and , then .
Since is not eventually constant and , there exists such that . Then

Thus if is not eventually constant then for all and , . Hence is -Cauchy.

Example 12. Let and ; then

If is any eventually nonconstant sequence in , then is -Cauchy if and only if there exists such that for . This implies that .
However, if , then , such that for , .
Hence for , . However, constant sequences are not -Cauchy but -Cauchy.

Theorem 13. Let be a metric space, the metric associated with on , the completion of , and the metric associated with on . Then is the completion of . In particular if is a complete metric space then is a complete metric space. We prove that

(i) is dense in ;(ii)every -Cauchy sequence in is -convergent.

Proof of (i). Let . Then there exists a sequence in such that since ,  . So that .
This implies that is dense in .

Proof of (ii). Let be -Cauchy in . If is eventually constant, then there exist and such that for .
In this case for ; hence is -convergent.
Suppose is not eventually constant. Then is a -Cauchy sequence. Since is complete, there exists such that . Since , .
Hence is -convergent to . This completes the proof of (ii).

Disclosure

I. Ramabhadra Sarma is a retired professor from Acharya Nagarjuna University, Andhra Pradesh, India.

Conflict of Interests

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

Acknowledgments

The authors thank the editor and anonymous referees for their constructive comments and ideas which helped us to improve the paper.

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