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I. Ramabhadra Sarma 1 and J. Madhusudana Rao 2 and P. Sumati Kumari 3 and D. Panthi 4
Academic Editor:Kyung Soo Kim
1, Department of Mathematics, Acharya Nagarjuna University, Guntur 522004, India
2, Department of Mathematics, Vijaya College of Engineering, India
3, Department of Mathematics, K L University, India
4, Department of Mathematics, Nepal Sanskrit University, Nepal
Received 22 July 2014; Accepted 18 August 2014; 14 October 2014
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.
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 Arandelovic and Keckic [6] have made important contributions in this area. The basic definition needed by most of these studies is that of a symmetric space. If X is a nonempty set, a function : X×X[arrow right]R+ is called a dislocated symmetric on X if d(x,y)=0 implies that x=y and d(x,y)=d(y,x) for all x,y∈X . A dislocated symmetric (simply d -symmetric) on X is called symmetric on X if d(x,x)=0 for all x in X . 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 "d " denotes a dislocated distance on a nonempty set X . xn , yn , x , y , and so forth are elements of X and Ci for 1...4;i...4;5 and Wi for 1...4;i...4;3 indicate properties of sequences in (X,d) . Consider
: C1 : lim d(xn ,yn )=0=lim...d(xn ,x)[implies]lim...d(yn ,x)=0 ,
: C2 : lim d(xn ,x)=0=lim...d(yn ,x)[implies]lim...d(xn ,yn )=0 ,
: C3 : lim d(xn ,yn )=0=lim...d(yn ,zn )[implies]lim...d(xn ,zn )=0 .
A space in which C1 is satisfied is called coherent by Pitcher and Chittenden [7]. Niemytzki [5] proved that a coherent symmetric space (X,d) is metrizable, and in fact there is a metric ρ on X such that (X,d) and (X,ρ) have identical topologies and also that lim...d(xn ,x)=0 if and only if lim...ρ(xn ,x)=0 .
Cho et al. [8] have introduced
: C4 : lim...d(xn ,x)=0[implies]lim...d(xn ,y)=d(x,y) for all y in X ,
: C5 : lim...d(xn ,x)=lim...d(xn ,y)=0[implies]x=y .
The following properties were introduced by Wilson [4]:
: W1 : for each pair of distinct points a,b in X there corresponds a positive number r=r(a,b) such that r<inf...c∈X d(a,c)+d(b,c) ,
: W2 : for each a∈X , for each k>0 , there corresponds a positive number r=r(a,k) such that if b is a point of X such that d(a,b)...5;k and c is any point of X then d(a,c)+d(c,b)...5;r ,
: W3 : for each positive number k there is a positive number r=r(k) such that d(a,c)+d(c,b)...5;r for all c in X and all a,b in X with d(a,b)...5;k .
2. Implications among the Axioms
Proposition 1.
In a d- symmetric space (X,d) , C3 [implies]C1 [implies]C5 , C3 [implies]C2 , and C4 [implies]C5 .
Proof.
Assume that C3 holds in (X,d) and let lim...d(xn ,yn )=0 and lim...d(xn ,x)=0 . Put zn =x ∀n so that [figure omitted; refer to PDF] By C3 , lim...d(yn ,zn )=0 ; that is, lim...d(yn ,x)=0 .
Hence [figure omitted; refer to PDF] Assume that C1 holds in (X,d) and let lim...d(xn ,x)=0 and lim...d(xn ,y)=0 . Put yn =y ∀n ; then [figure omitted; refer to PDF] By C1 , lim...d(yn ,x)=0 ; that is, lim...d(y,x)=0 .
Consider lim...d(x,y)=0 ; this implies that x=y . Hence C5 holds. Thus [figure omitted; refer to PDF] Assume that C3 holds and let lim...d(xn ,x)...=...0 and lim...d(yn ,x)=0 .
Put zn =x ∀n ; then lim...d(xn ,zn )=lim...d(zn ,yn )=0 .
By C3 , lim...d(xn ,yn )=0 . Hence [figure omitted; refer to PDF] Assume that C4 holds and let lim...d(xn ,x)=0 and lim...d(xn ,y)=0 .
By C4 , lim...d(xn ,y)=d(x,y) . Hence d(x,y)=0 . Hence x=y .
The following proposition explains the relationship between Wilson's axioms [4] W1 , W2 , and W3 and the Ci 's.
Proposition 2.
Let (X,d) be a d- symmetric space; then
: (i) W1 ...C5 , (ii) W2 ...C1 , and (iii) W3 ...C3 .
Proof.
( i ) Assume W1 . Suppose lim...d(a,xn )=lim...d(b,xn )=0 but a...0;b .
Then [figure omitted; refer to PDF] By [figure omitted; refer to PDF] equations (6) and (7) are contradictory. Hence a=b . Thus W1 [implies]C5 .
Suppose that W1 fails. Then there exist a...0;b in X such that for every n there corresponds xn in X such that d(a,xn )+d(b,xn )<1/n : [figure omitted; refer to PDF] Thus if W1 fails then C5 fails. That is, C5 [implies]W1 . Hence W1 ...C5 .
( ii ) Assume W2 . Then for each a∈X and each k>0 there corresponds r>0 such that, for all b∈X with d(a,b)...5;k and ∀x∈X , d(a,x)+d(b,x)...5;r .
Suppose that C1 fails. There exist a∈X , {bn } , and {cn } in X such that lim...d(a,bn )=lim...d(bn ,cn )=0 but lim...d(a,cn )...0;0.
Since lim...d(a,cn )......0;...0 there exists k...>...0 and a subsequence (cnk ) such that [figure omitted; refer to PDF] Since [figure omitted; refer to PDF] this implies that lim...{d(a,bn )+d(bn ,cn )}...0;0 , a contradiction.
Conversely assume that W2 fails. Then there exist a∈X and k>0 such that ∀n>0 ∃ bn ∈X and cn ∈X such that [figure omitted; refer to PDF] This implies that lim...d(a,cn )=lim...d(bn ,cn )=0 but lim...d(a,bn )...0;0 .
Hence C1 fails.
(iii) Assume W3 . Suppose that C3 fails. Then there exist sequences {an } , {bn } , and {cn } in X such that lim...d(an ,bn )=lim...d(bn ,cn )=0 but lim...d(an ,cn )...0;0 .
Since W3 holds, ∀k>0 there corresponds r>0 such that for all a , b with [figure omitted; refer to PDF]
Since lim...d(an ,cn )...0;0 there exists a positive number ∈ and a subsequence of positive integers {nk } such that d(ank ,cnk )> ∈ . Choose r1 corresponding to ∈ so that [figure omitted; refer to PDF]
Thus [figure omitted; refer to PDF]
This contradicts the assumption that lim...d(an ,bn )=lim...d(bn ,cn )...=...0 .
Hence [figure omitted; refer to PDF] Assume that W3 fails.
Then there exists k>0 such that, ∀ positive integer n , there exist an , bn , and cn with [figure omitted; refer to PDF]
Hence [figure omitted; refer to PDF]
Hence C3 fails.
Hence [figure omitted; refer to PDF]
This completes the proof of the proposition.
We introduce the following.
Axiom C . Every convergent sequence satisfies Cauchy criterion. That is, if (xn ) is a sequence in X , x∈X and lim...d(xn ,x)=0; then given ∈ >0 ∃ N(∈)∈N such that d(xn ,xm )< ∈ whenever m , n...5;N(∈) we have the following.
Proposition 3.
In a d- symmetric space (X,d) , C1 [implies]C[implies]C2 .
Proof.
For C1 [implies]C , suppose that a sequence (xn ) in (X,d) is convergent to x but does not satisfy Cauchy criterion. Then ∃r>0 such that for every positive integer k there correspond integers mk , nk such that [figure omitted; refer to PDF] Let [figure omitted; refer to PDF] Then [figure omitted; refer to PDF] But lim...d(yk ,zk )...0;0 ; this contradicts C1 .
Proof.
For C[implies]C2 , suppose that lim...d(xn ,x)=lim...d(yn ,x)=0 .
Let (zn ) be the sequence defined by z2n-1 =xn and z2n =yn . Then lim...d(zn ,x)=0 . Hence (zn ) satisfies Cauchy criterion.
Given ∈>0 ∃N(∈)∈N such that d(zn ,zm )< ∈ for m,n...5;N(∈) :
: [implies]d(z2n-1 ,z2n )< ∈ for n...5;N(∈) ,
: [implies]lim...d(xn ,yn )< ∈ for n...5;N(∈) ,
: [implies]lim...d(xn ,yn )=0 .
3. Examples for Nonimplications
Example 4.
A d- symmetric space in which the triangular inequality fails and C1 through C5 hold.
Let X=[0,1] . Define d on X×X as follows: [figure omitted; refer to PDF] Clearly d is a d- symmetric space. d does not satisfy the triangular inequality since d(0.1,0.2)+d(0.2,0.1)=0.6<1=d(0.1,0.1) .
We show that C1 through C5 holds. We first show that lim...d(xn ,x)=0 iff x=0 and lim...xn =0 in R .
If x......0;...0 then lim...d(xn ,x)...=... xn +x...5;x>0 . Hence lim...d(xn ,x)...5;x>0 .
If x...=...0 then lim...d(xn ,0)...=...0 or xn . Hence lim...d(xn ,x)=0...lim...xn =0 in R .
Now we show that lim...d(xn ,yn )...=...0 if and only if lim...xn =lim...yn =0 in R .
Consider lim...d(xn ,yn )=0[implies]d(xn ,yn )<1/2 for large n :
: [implies]d(xn ,yn )=xn +yn or 0 for large n ,
: [implies] either xn =yn =0 or d(xn ,yn )=xn +yn for large n ,
: [implies]lim...xn =lim...yn =0 in R .
Conversely if lim...xn ...=...lim...yn ...=...0 in R then lim...d(xn ,yn )=0 or xn +yn for large n .
Hence lim...d(xn ,yn )=0 .
Verification of validity of C1 through C5 is done as follows.
C 1 : let lim...d(xn ,yn )=0 and lim...d(xn ,x)=0; then lim...xn =lim...yn =0 in R and x=0 .
Hence d(yn ,x)=d(yn ,0)=yn or 0 . This implies that lim...d(yn ,x)=0 .
C 2 : let d(xn ,x)=d(yn ,x)=0 . Then x=0 and lim...xn =lim...yn =0 in R .
Hence lim...d(yn ,xn )=0 .
C 3 : let d(xn ,yn )=d(yn ,zn )=0 ; then lim...xn =lim...yn =lim...zn =0 in R .
Hence lim...d(xn ,zn )=0 .
C 4 : let lim...d(xn ,x)=0 . Then x=0 and lim...xn =0 .
If y=0 , 0...4;d(xn ,y)...4;xn . Hence lim...d(xn ,y)=0=d(x,y) .
If y...0;0 , d(xn ,y)=xn +y . Hence lim...d(xn ,y)=y=0+y=d(x,y) .
C 5 : let lim...d(xn ,x)=0 and lim...d(xn ,y)=0 .
Then x=0,y=0 and lim...xn =0 . Hence x=y .
Example 5.
A d- symmetric space (X,d) in which C1 [hence C5 ] holds while Cj does not hold for j=2,3,4 .
Let X=[0,∞) . Define d on X×X as follows: [figure omitted; refer to PDF] Clearly (X,d) is a d- symmetric space. We show that C1 , C5 hold.
Let lim...d(xn ,x)=0=lim...d(xn ,yn ) .
If x...0;0, d(xn ,x)>x if xn ...0;0 . [figure omitted; refer to PDF] This implies that [figure omitted; refer to PDF] Thus lim...d(xn ,x)=0[implies]x=0 and (xn ) can be split into two subsequences (xn (1) ),(xn (2) ) , where (xn (1) )=0 ∀n , (xn (2) )...0;0 for every n and if (xn (2) ) is infinite subsequence lim...(xn (2) )=∞ . We consider the case where both (xn (1) ) and (xn (2) ) are infinite sequences as when one is a finite sequence the same proof works with minor modifications. Consider [figure omitted; refer to PDF] If we show that yn (2) cannot be positive for infinitely many n , it will follow that lim...d(xn (2) ,yn (2) )=lim...d(xn (2) ,0)=0 so that lim...d(0,yn )=0 . Hence C1 holds.
If yn (2) ......0;...0 for infinitely many n , say {ynk (2) } is the infinite subsequence of {yn (2) } with ynk (2) ...0;0 ∀nk , then d(xnk (2) ,ynk (2) )=xnk (2) +ynk (2) >xnk (2) so that lim... d(xnk (2) ,ynk (2) )...5;lim...xnk (2) ...5;∞ contradicting the assumption that lim...d(xn ,yn )=0 . Thus C1 holds. Since C1 [implies]C5 , C5 holds.
C 2 does not hold since d(n,0)=1/n while d(n,n)=2n ∀n so that lim...d(n,n)...0;0 .
C 3 does not hold since lim...d(n,0)=lim...d(0,n) while lim...d(n,n)=∞ .
C 4 does not hold since lim...d(n,0)=0 but lim...d(n,2)=∞ while d(0,2)=1/2 .
Example 6.
A d- symmetric space (X,d) in which C2 holds but C1 , C3 , C4 , and C5 fail.
Let X=[0,1]∪{2} . Define d on X×X as follows: [figure omitted; refer to PDF] Clearly (X,d) is a d- symmetric space.
We first show that if {xn } in X converges to x in (X,d) then x∈{0,2} .
Suppose that x...0;0 and x...0;2 ; then x∈(0,1] :
: [implies]lim...d(xn ,x)=0=xn +x or x ,
: [implies]lim...d(xn ,x)...5;x>0 ,
: [implies]lim...d(xn ,x)...0;0 .
Hence if lim...d(xn ,x)=0 then x∈{0,2} .
: C1 fails: xn =1/n,yn =2 , and x=0 ; [figure omitted; refer to PDF]
: C2 holds: suppose that lim...d(xn ,x)=lim...d(yn ,x)=0 ; then x∈{0,2} .
Case 1.
If x=2 , lim...d(xn ,x) [arrow right]0[implies]d(xn ,x) =xn eventually and lim...xn =0 in R . Hence ∃N∈N∋xn <1 and yn <1 for n...5;N .
Here d(xn ,yn )=xn +yn . This implies that lim...d(xn ,yn )=0 .
Case 2.
If x=0 , [figure omitted; refer to PDF]
If lim...d(xn ,0)=0 , d(xn ,0)= xn eventually and lim...xn =0 in R .
Similarly d(yn ,0)= yn eventually and lim...yn =0 in R .
As in Case 1 it follows that [figure omitted; refer to PDF]
Thus C2 holds.
: C3 fails since C3 [implies]C1 .
: C5 fails: let xn =1/n , x=0,and y=2 [figure omitted; refer to PDF]
: C4 fails since C4 [implies]C5 .
Example 7.
A d- symmetric space (X,d) in which C4 holds but C1 fails.
Let X=N∪{0} . Define d on X×X as follows: [figure omitted; refer to PDF] If {xn } in X and lim...d(xn ,0)=0 then xn is eventually odd.
If x...0;0,d(xn ,x) cannot be 1 so xn +x is even or odd and |xn -x|=1 .
But in this case d(xn ,x)=|1/xn -1/x| so that d(xn ,x)...0;0 .
Thus d(xn ,x)=0...x=0 and xn is eventually odd.
If m is a fixed even integer and xn is odd, xn +m is odd and eventually >2.
So [figure omitted; refer to PDF] If m is a fixed odd integer and xn is odd, xn +m is even.
So d(xn ,m)=|1/m-1/xn | so that lim...d(xn ,0)=0[implies]lim...d(xn ,m)=d(0,m) .
If m...=...0 and xn is odd eventually [figure omitted; refer to PDF] If m=0 and xn =0 eventually [figure omitted; refer to PDF] Hence C4 holds in (X,d) .
C 1 does not hold: let xn =2n-1 and yn =2n : [figure omitted; refer to PDF] Hence d(xn ,0)=d(xn ,yn )=0 and d(yn ,0)...0;0 .
Example 8.
A d- symmetric space (X,d) in which C3 holds but C4 does not hold.
Let X=[0,1]∪{2} . Define d on X×X as follows: [figure omitted; refer to PDF] Clearly (X,d) is a d- symmetric space which is not a symmetric space.
We first show that if {xn } converges to x in (X,d) then x∈{0,2} .
Suppose that 0...0;x...0;2 ; then x∈(0,1] : [figure omitted; refer to PDF] Since lim...d(x,xn )=0 ∃N∋d(x,xn )<1 for n...5;N
: [implies]d(x,xn )=x+xn ...5;x for n...5;N ,
: [implies]lim...d(x,xn )...0;0 , a contradiction.
We now show that lim...d(xn ,yn )=0 if and only if lim...xn =lim...yn =0 . Consider [figure omitted; refer to PDF]
Conversely if lim...xn =lim...yn =0 then ∃N∈N∋xn <1,yn <1 for n...5;N
[implies] lim ... d ( x n , y n ) = 0 or xn +yn for large n .
Hence d(xn ,yn )=0 .
Thus d(xn ,yn )=0 if and only if lim...xn =lim...yn =0 .
As a consequence we have [figure omitted; refer to PDF] Hence C3 holds in (X,d) .
C 4 fails: xn =1/(n+1) for n...5;1 : [figure omitted; refer to PDF]
Example 9.
A d- symmetric space (X,d) in which C4 holds but C2 ,C3 fail to hold.
Let X=N∪{0,∞} . Define d on X×X as follows: [figure omitted; refer to PDF] If m,n∈N , [figure omitted; refer to PDF] Clearly (X,d) is a d- symmetric space which is not a symmetric space.
We show that if lim...d(xn ,x)=0 then x=0 and {xn } consists of two subsequences {yn } and {zn } , one of which may possibly be finite, where yn =0 ∀n and 0...0;zn ∈N ∀n and lim...(1/zn )=0 (in case {zn } is an infinite sequence).
To prove this we first note that lim...d(xn ,x)=0[implies]x...0;∞ and xn ...0;∞ eventually.
If x∈N , d(xn ,x)=1/x or 1 or |1/xn -1/x| .
Hence lim...d(xn ,x)=0[implies]x∉N ; hence x=0 .
Further d(xn ,0)=0 or 1/xn . Consequently {xn } may be split into two sequences {yn } and {zn } as described above.
We show that C4 holds. Assume that lim...d(xn ,x)=0 . Then x=0 .
Let m∈N and yn =0 ∀n . Then d(yn ,m)=d(o,m)=1/m .
So lim...d(yn ,m)=d(o,m) .
If zn ...0;0 ∀n,and lim...(1/zn )=0 the d(zn ,m)=|1/zn -1/m| for n>m so that lim...d(zn ,m)=1/m=d(0,m) .
Thus if m∈N and lim...d(xn ,x)=0 then lim...d(xn ,m)=d(x,m) .
Clearly this holds when m=∞ or m=0 as well.
Hence C4 holds.
C 2 does not hold: let xn =x,yn =n+1,and x=0 : [figure omitted; refer to PDF] C3 does not hold: [figure omitted; refer to PDF] C5 holds since C4 [implies]C5 .
Remarks. From this example we can conclude that
(1) C5 does not imply C2 as otherwise, since C4 [implies]C5 it would follow that C4 [implies]C2 which does not hold as is evident from the above example,
(2) in a d- symmetric space, convergent sequences are necessarily Cauchy sequences.
Example 10.
A d- symmetric space (X,d) in which C4 holds but C2 , C3 fail to hold.
Let X=N∪{0} . Define d on X×X as follows: [figure omitted; refer to PDF] Clearly (X,d) is a d -symmetric space.
We first characterize all convergent sequences in (X,d) .
Suppose that lim...d(xn ,x)=0 . We show that x=0 .
If x is odd and xn is even d(xn ,x)=1 if xn >x+2 .
So lim...d(xn ,x)...0;0 . Thus xn is even for at most finitely many n .
We may thus assume that xn is odd ∀n .
The d(xn ,x)=1/xn +1/x so that d(xn ,x)...5;1/x>0 .
Hence x cannot be odd. Now suppose that x>0 and x is even.
Then d(xn ,x)=1 if xn =0 if xn is odd and |xn -x|>2 while d(xn ,x)=1/xn +1/x if xn +x is even or |xn -x|=1 . In all cases lim...d(xn ,x)...0;0 .
Hence the only possibility is x=0 .
We now show that the following are equivalent.
(a) lim...d(xn ,x)=0 in R ,
(b) there exists a positive integer N such that xn is positive and even, only if n<N .
Assumption (b): xn is odd or zero if n...5;N so that lim...d(xn ,0)=lim...(1/xn )=0 .
Hence (b)[implies] (a).
Assumption (a): since d(2m,0)=1 for m∈N , it follows that at most finitely many terms of {xn } can be even. This proves (b). Thus lim...d(xn ,x)=0...x=0 and ∃N∈N∋xn is "0" or odd for n...5;N .
Consequently C5 holds.
C 1 does not hold: let xn =2n+1,yn =2n and x=0 ; [figure omitted; refer to PDF] But lim...d(yn ,x)=1 since lim...d(2n,0)=1 ∀n .
C 2 holds: assume that lim...d(xn ,x)=0=lim...d(yn ,x) .
Then x=0 and then there exists N such that xn is "0" or odd and yn =0 or odd for n...5;N and lim...(1/xn )=lim...(1/yn )=0 .
If xn =yn =0 , d(xn ,yn )=0 .
If xn =0 , yn is odd, d(xn ,yn )=1/yn .
If yn =0 , xn is odd, d(xn ,yn )=1/xn .
If xn is odd and yn is odd, d(xn ,yn )=1/xn +1/yn .
Consequently lim...d(xn ,yn )=0 .
C 3 does not hold: let xn =0 , yn =2n+1 , and zn =2n : [figure omitted; refer to PDF] so that lim...d(xn ,yn )=lim...d(yn ,zn )=0 but lim...d(xn ,zn )=1 .
C 4 does not hold: let xn =2n+1 , x=0 , and y=3: [figure omitted; refer to PDF]
Example 11.
The following example shows that there exist symmetric spaces in which C does not hold.
Let X={0,1/2,1/3,1/4,...} .
Define d(x,x)=0 , d(x,y)=d(y,x) [figure omitted; refer to PDF] Then (X,d) is a symmetric space; {1/n} converges to 0 but is not a Cauchy sequence.
Acknowledgments
The authors thank the editor and anonymous referees for their constructive comments, suggestions, and ideas which helped them to improve the paper.
Conflict of Interests
The authors declare that there is no conflict of interests regarding the publication of this paper.
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Abstract
Dislocated symmetric spaces are introduced, and implications and nonimplications among various kinds of convergence axioms are derived.
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