Convergence Axioms on Dislocated Symmetric Spaces

Abstract

Dislocated symmetric spaces are introduced, and implications and nonimplications among various kinds of convergence axioms are derived.

1. Introduction

A metric space is a special kind of topological space. In a metric space, topological properties are characterized by means of sequences. Sequences are not sufficient in topological spaces for such purposes. It is natural to try to find classes intermediate between those of topological spaces and those of metric spaces in which members sequences play a predominant part in deciding their topological properties. A galaxy of mathematicians consisting of such luminaries as Frechet [1], Chittenden [2], Frink [3], Wilson [4], Niemytzki [5], and Aranđelović and Kečkić [6] have made important contributions in this area. The basic definition needed by most of these studies is that of a symmetric space. If is a nonempty set, a function is called a dislocated symmetric on if implies that and for all . A dislocated symmetric (simply -symmetric) on is called symmetric on if for all in . The names dislocated symmetric space and symmetric space have expected meanings. Obviously, a symmetric space that satisfies the triangle inequality is a metric space. Since the aim of our study is to find how sequential properties and topological properties influence each other, we collect various properties of sequences that have been shown in the literature to have a bearing on the problem under study. In what follows “” denotes a dislocated distance on a nonempty set . , , , , and so forth are elements of and for and for indicate properties of sequences in . Consider , , .

A space in which is satisfied is called coherent by Pitcher and Chittenden [7]. Niemytzki [5] proved that a coherent symmetric space is metrizable, and in fact there is a metric on such that and have identical topologies and also that if and only if .

Cho et al. [8] have introduced : for all in , : .

The following properties were introduced by Wilson [4]: : for each pair of distinct points in there corresponds a positive number such that , : for each , for each , there corresponds a positive number such that if is a point of such that and is any point of then , : for each positive number there is a positive number such that for all in and all in with .

2. Implications among the Axioms

Proposition 1. In a symmetric space , , , and .

Proof. Assume that holds in and let and . Put so that By , ; that is, .
Hence Assume that holds in and let and . Put ; then By , ; that is, .
Consider ; this implies that . Hence holds. Thus Assume that holds and let and .
Put ; then .
By , . Hence Assume that holds and let and .
By , . Hence . Hence .

The following proposition explains the relationship between Wilson’s axioms [4] , , and and the ’s.

Proposition 2. Let be a symmetric space; then    ,      , and   .

Proof. Assume . Suppose but .
Then By equations (6) and (7) are contradictory. Hence . Thus .
Suppose that fails. Then there exist in such that for every there corresponds in such that : Thus if fails then fails. That is, . Hence .
Assume . Then for each and each there corresponds such that, for all with and , .
Suppose that fails. There exist , , and in such that but
Since there exists and a subsequence such that Since this implies that , a contradiction.
Conversely assume that fails. Then there exist and such that and such that This implies that but .
Hence fails.
(iii) Assume . Suppose that fails. Then there exist sequences , , and in such that but .
Since holds, there corresponds such that for all , with
Since there exists a positive number and a subsequence of positive integers such that . Choose corresponding to so that
Thus
This contradicts the assumption that .
Hence Assume that fails.
Then there exists such that, positive integer , there exist , , and with
Hence
Hence fails.
Hence
This completes the proof of the proposition.

We introduce the following.

Axiom . Every convergent sequence satisfies Cauchy criterion. That is, if is a sequence in , and then given such that whenever , we have the following.

Proposition 3. In a symmetric space , .

Proof. For , suppose that a sequence in is convergent to but does not satisfy Cauchy criterion. Then such that for every positive integer there correspond integers , such that Let Then But ; this contradicts .

Proof. For , suppose that .
Let be the sequence defined by and . Then . Hence satisfies Cauchy criterion.
Given such that for : , , .

3. Examples for Nonimplications

Example 4. A symmetric space in which the triangular inequality fails and through hold.
Let . Define on as follows: Clearly is a symmetric space. does not satisfy the triangular inequality since .

We show that through holds. We first show that and in .

If then . Hence .

If then . Hence in .

Now we show that if and only if in .

Consider for large :  or for large ,  either or for large ,  in .

Conversely if in then or for large .

Hence .

Verification of validity of through is done as follows.

 let and then in and .

Hence or . This implies that .

let . Then and in .

Hence .

let ; then in .

Hence .

let . Then and .

If , . Hence .

If , . Hence .

let and .

Then and . Hence .

Example 5. A symmetric space in which [hence ] holds while does not hold for .

Let . Define on as follows: Clearly is a symmetric space. We show that , hold.

Let .

If if . This implies that Thus and can be split into two subsequences , where , for every and if is infinite subsequence . We consider the case where both and are infinite sequences as when one is a finite sequence the same proof works with minor modifications. Consider If we show that cannot be positive for infinitely many , it will follow that so that . Hence holds.

If for infinitely many , say is the infinite subsequence of with , then so that contradicting the assumption that . Thus holds. Since , holds.

does not hold since while so that .

does not hold since while .

does not hold since but while .

Example 6. A symmetric space in which holds but , , , and fail.

Let . Define on as follows: Clearly is a symmetric space.

We first show that if in converges to in then .

Suppose that and ; then : or, , .

Hence if then .  fails: , and ;   holds: suppose that ; then .

Case 1. If , eventually and . Hence and for .

Here . This implies that .

Case 2. If ,

If , eventually and in .

Similarly eventually and .

As in Case 1 it follows that

Thus holds.  fails since .  fails: let ,   fails since .

Example 7. A symmetric space in which holds but fails.

Let . Define on as follows: If in and then is eventually odd.

If cannot be 1 so is even or odd and .

But in this case so that .

Thus and is eventually odd.

If is a fixed even integer and is odd, is odd and eventually >2.

So If is a fixed odd integer and is odd, is even.

So so that .

If and is odd eventually If and eventually Hence holds in .

does not hold: let and : Hence and .

Example 8. A symmetric space in which holds but does not hold.

Let . Define on as follows: Clearly is a symmetric space which is not a symmetric space.

We first show that if converges to in then .

Suppose that ; then : Since for   for , , a contradiction.

We now show that if and only if . Consider

Conversely if then

or for large .

Hence .

Thus if and only if .

As a consequence we have Hence holds in .

fails: :

Example 9. A symmetric space in which holds but fail to hold.

Let . Define on as follows: If , Clearly is a symmetric space which is not a symmetric space.

We show that if then and consists of two subsequences and , one of which may possibly be finite, where and and (in case is an infinite sequence).

To prove this we first note that and eventually.

If , .

Hence ; hence .

Further . Consequently may be split into two sequences and as described above.

We show that holds. Assume that . Then .

Let and . Then .

So .

If the for so that .

Thus if and then .

Clearly this holds when or as well.

Hence holds.

does not hold: let : does not hold: holds since .

Remarks. From this example we can conclude that(1) does not imply as otherwise, since it would follow that which does not hold as is evident from the above example,(2)in a symmetric space, convergent sequences are necessarily Cauchy sequences.

Example 10. A symmetric space in which holds but , fail to hold.

Let . Define on as follows: Clearly is a -symmetric space.

We first characterize all convergent sequences in .

Suppose that . We show that .

If is odd and is even .

So . Thus is even for at most finitely many .

We may thus assume that is odd .

The so that .

Hence cannot be odd. Now suppose that and is even.

Then if is odd and while   is  even  or  . In all cases .

Hence the only possibility is .

We now show that the following are equivalent.(a)in ,(b)there exists a positive integer such that is positive and even, only if .

Assumption (b): is odd or zero if so that .

Hence (b)(a).

Assumption (a): since for , it follows that at most finitely many terms of can be even. This proves (b). Thus and is “0” or odd for .

Consequently holds.

does not hold: let and ; But since .

holds: assume that .

Then and then there exists such that is “0” or odd and or odd for and .

If , .

If , is odd, .

If , is odd, .

If is odd and is odd, .

Consequently .

does not hold: let , , and : so that but .

does not hold: let , , and

Example 11. The following example shows that there exist symmetric spaces in which does not hold.

Let .

Define , Then is a symmetric space; converges to 0 but is not a Cauchy sequence.

Disclosure

Professor I. Ramabhadra Sarma is a retired professor from Acharya Nagarjuna University.

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, suggestions, and ideas which helped them to improve the paper.