# Convergence Axioms on Dislocated Symmetric Spaces

Authors: I. Ramabhadra Sarma; J. Madhusudana Rao; P. Sumati Kumari; D. Panthi
Journal: Abstract and Applied Analysis
 (2014)

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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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## Body

## 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 X is a nonempty set, a function :X×X→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 Cifor 1≤i≤5 and Wifor 1≤i≤3 indicate properties of sequences in (X,d). ConsiderC1:  limd(xn,yn)=0=lim⁡d(xn,x)⇒lim⁡d(yn,x)=0,C2:  limd(xn,x)=0=lim⁡d(yn,x)⇒lim⁡d(xn,yn)=0,C3:  limd(xn,yn)=0=lim⁡d(yn,zn)⇒lim⁡d(xn,zn)=0.A space in whichC1 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 introducedC4: lim⁡d(xn,x)=0⇒lim⁡d(xn,y)=d(x,y) for all y in X,C5: lim⁡d(xn,x)=lim⁡d(xn,y)=0⇒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∈Xd(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)≥k and c is any point of X then d(a,c)+d(c,b)≥r,W3: for each positive number k there is a positive number r=r(k) such that d(a,c)+d(c,b)≥r for allc in X and all a,b in X with d(a,b)≥k.


## 2. Implications among the Axioms

Proposition 1.
In ad-symmetric space (X,d), C3⇒C1⇒C5, C3⇒C2, and C4⇒C5.Proof.
Assume thatC3 holds in (X,d) and let lim⁡d(xn,yn)=0 and lim⁡d(xn,x)=0. Put zn=x∀n so that
(1)lim⁡d(xn,zn)=lim⁡d(xn,x)=0=lim⁡d(xn,yn)=lim⁡d(yn,xn).
By C3, lim⁡d(yn,zn)=0; that is, lim⁡d(yn,x)=0.
Hence(2)C3⇒C1.
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
(3)lim⁡d(xn,yn)=lim⁡d(xn,x)=0.
By C1, lim⁡d(yn,x)=0; that is, lim⁡d(y,x)=0.
Considerlim⁡d(x,y)=0; this implies that x=y. Hence C5 holds. Thus
(4)C1⟹C5.
Assume that C3 holds and let lim⁡d(xn,x)=0 and lim⁡d(yn,x)=0.
Putzn=x∀n; then lim⁡d(xn,zn)=lim⁡d(zn,yn)=0.
ByC3, lim⁡d(xn,yn)=0. Hence
(5)C3⟹C2.
Assume that C4 holds and let lim⁡d(xn,x)=0 and lim⁡d(xn,y)=0.
ByC4, 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≠b.
Then(6)lim⁡{d(a,xn)+d(b,xn)}=0buta≠b.
By
(7)W1∃r>0∋∀x,d(a,x)+d(b,x)≥r,
equations (6) and (7) are contradictory. Hence a=b. Thus W1⇒C5.
Suppose thatW1 fails. Then there exist a≠b in X such that for every n there corresponds xn in X such that d(a,xn)+d(b,xn)<1/n:(8)⟹lim⁡d(a,xn)=lim⁡d(b,xn)=0buta≠b.
Thus if W1 fails then C5 fails. That is, C5⇒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)≥kand ∀x∈X, d(a,x)+d(b,x)≥r.
Suppose thatC1 fails. There exist a∈X, {bn}, and {cn} in X such that lim⁡d(a,bn)=lim⁡d(bn,cn)=0but lim⁡d(a,cn)≠0.
Sincelim⁡d(a,cn)≠0there exists k>0 and a subsequence (cnk) such that
(9)d(a,cnk)>k∀nk.
Since
(10)d(a,cnk)>k,d(a,bnk)+d(bnk,cnk)≥r,
this implies that lim⁡{d(a,bn)+d(bn,cn)}≠0, a contradiction.
Conversely assume thatW2 fails. Then there exist a∈X and k>0 such that ∀n>0∃bn∈X and cn∈X such that
(11)d(a,bn)≥kbutd(a,cn)+d(bn,cn)<1n.
This implies that lim⁡d(a,cn)=lim⁡d(bn,cn)=0but lim⁡d(a,bn)≠0.
HenceC1 fails.
(iii) AssumeW3. Suppose that C3 fails. Then there exist sequences {an}, {bn}, and {cn} in X such that lim⁡d(an,bn)=lim⁡d(bn,cn)=0but lim⁡d(an,cn)≠0.
SinceW3 holds, ∀k>0 there corresponds r>0 such that for all a, b with
(12)d(a,b)≥k,d(a,c)+d(b,c)≥r∀c.
Sincelim⁡d(an,cn)≠0there exists a positive number ∈ and a subsequence of positive integers {nk} such that d(ank,cnk)>∈. Choose r1 corresponding to ∈ so that
(13)d(ank,bnk)+d(bnk,cnk)≥r1.
Thus(14)lim⁡{d(ank,bnk)+d(bnk,cnk)}≠0.
This contradicts the assumption thatlim⁡d(an,bn)=lim⁡d(bn,cn)=0.
Hence(15)W3⟹C3.
Assume that W3 fails.
Then there existsk>0 such that, ∀ positive integer n, there exist an, bn, and cn with
(16)d(an,bn)≥kbutd(an,cn)+d(bn,cn)<1n.
Hence(17)lim⁡d(an,bn)≠0butlim⁡d(an,cn)=lim⁡d(cn,bn)=0.
HenceC3 fails.
Hence(18)C3⟹W3.
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≥N(∈) we have the following.Proposition 3.
In ad-symmetric space (X,d), C1⇒C⇒C2.Proof.
ForC1⇒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
(19)mk+1>nk+1>mk>nk,d(xmk,xnk)>γ∀k.
Let
(20)yk=xmk,zk=xnk∀k.
Then
(21)lim⁡d(yk,x)=0,lim⁡d(zk,x)=0.
But lim⁡d(yk,zk)≠0; this contradicts C1.Proof.
ForC⇒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≥N(∈):⇒d(z2n-1,z2n)<∈forn≥N(∈),⇒lim⁡d(xn,yn)<∈forn≥N(∈),⇒lim⁡d(xn,yn)=0.


## 3. Examples for Nonimplications

Example 4.
Ad-symmetric space in which the triangular inequality fails and C1 through C5 hold.
LetX=[0,1]. Define d on X×X as follows:
(22)d(x,y)={x+yifx≠y,1ifx=y≠0,0ifx=y=0.
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 thatC1 through C5 holds. We first show that lim⁡d(xn,x)=0iffx=0and lim⁡xn=0 in R.Ifx≠0 then lim⁡d(xn,x)=xn+x≥x>0. Hence lim⁡d(xn,x)≥x>0.Ifx=0 then lim⁡d(xn,0)=0orxn. Hence lim⁡d(xn,x)=0⇔lim⁡xn=0 in R.Now we show thatlim⁡d(xn,yn)=0 if and only if lim⁡xn=lim⁡yn=0 in R.Considerlim⁡d(xn,yn)=0⇒d(xn,yn)<1/2 for large n:⇒d(xn,yn)=xn+yn or 0 for large n,⇒ either xn=yn=0 or d(xn,yn)=xn+yn for large n,⇒lim⁡xn=lim⁡yn=0 in R.Conversely iflim⁡xn=lim⁡yn=0 in R then lim⁡d(xn,yn)=0 or xn+yn for large n.Hencelim⁡d(xn,yn)=0.Verification of validity ofC1 through C5 is done as follows.C
1
:  let lim⁡d(xn,yn)=0and lim⁡d(xn,x)=0;then lim⁡xn=lim⁡yn=0 in R and x=0.Henced(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.Hencelim⁡d(yn,xn)=0.C
3
: let d(xn,yn)=d(yn,zn)=0; then lim⁡xn=lim⁡yn=lim⁡zn=0 in R.Hencelim⁡d(xn,zn)=0.C
4
: let lim⁡d(xn,x)=0. Then x=0 and lim⁡xn=0.Ify=0, 0≤d(xn,y)≤xn. Hence lim⁡d(xn,y)=0=d(x,y).Ify≠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.Thenx=0,y=0 and lim⁡xn=0. Hence x=y.Example 5.
Ad-symmetric space (X,d) in which C1 [hence C5] holds while Cj does not hold for j=2,3,4.LetX=[0,∞). Define d on X×X as follows:
(23)d(x,y)={x+yifx≠0≠y,1xifx≠0=y,1yifx=0≠y,0ifx=0=y.
Clearly (X,d) is a d-symmetric space. We show that C1, C5 hold.Letlim⁡d(xn,x)=0=lim⁡d(xn,yn).Ifx≠0,d(xn,x)>x if xn≠0. 						(24)=1xifxn=0.
This implies that
(25)lim⁡d(xn,x)≥min⁡{x,1x}>0.
Thus lim⁡d(xn,x)=0⇒x=0 and (xn) can be split into two subsequences (xn(1)),(xn(2)), where (xn(1))=0∀n, (xn(2))≠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
(26)lim⁡d(xn,yn)=0⟹lim⁡d(xn(1),yn(1))=lim⁡d(xn(2),yn(2))=0.
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.Ifyn(2)≠0 for infinitely many n, say {ynk(2)} is the infinite subsequence of {yn(2)} with ynk(2)≠0∀nk, then d(xnk(2),ynk(2))=xnk(2)+ynk(2)>xnk(2) so that lim⁡d(xnk(2),ynk(2))≥lim⁡xnk(2)≥∞ contradicting the assumption that lim⁡d(xn,yn)=0. Thus C1 holds. Since C1⇒C5, C5 holds.C
2 does not hold since d(n,0)=1/n while d(n,n)=2n∀nso that lim⁡d(n,n)≠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.
Ad-symmetric space (X,d) in which C2 holds but C1, C3, C4, and C5 fail.LetX=[0,1]∪{2}. Define d on X×X as follows:
(27)d(x,y)={x+yif0≤x≤1,0≤y≤1xif0≤x≤1,y=2yifx=2,0≤y≤11if{x=2,y∈{0,2}orx∈{0,2},y=2.
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 thatx≠0 and x≠2; then x∈(0,1]:⇒lim⁡d(xn,x)=0=xn+xorx,⇒lim⁡d(xn,x)≥x>0,⇒lim⁡d(xn,x)≠0.Hence iflim⁡d(xn,x)=0 then x∈{0,2}.C1 fails: xn=1/n,yn=2, and x=0;
(28)d(xn,yn)=1n,d(xn,x)=1n,d(yn,x)=1⟹lim⁡d(xn,yn)=0=d(xn,x)butlim⁡d(yn,x)≠0.C2 holds: suppose that lim⁡d(xn,x)=lim⁡d(yn,x)=0; then x∈{0,2}.Case 1.
Ifx=2, lim⁡d(xn,x)→0⇒d(xn,x)=xn eventually and lim⁡xn=0inR. Hence ∃N∈N∋xn<1 and yn<1 for n≥N.Hered(xn,yn)=xn+yn. This implies that lim⁡d(xn,yn)=0.Case 2.
Ifx=0,
(29)d(xn,0)={1ifxn=2or0,xnif0≤xn≤1.Iflim⁡d(xn,0)=0, d(xn,0)=xn eventually and lim⁡xn=0 in R.Similarlyd(yn,0)=yn eventually and lim⁡yn=0inR.As in Case1 it follows that
(30)lim⁡d(xn,yn)=lim⁡(xn+yn)=0.ThusC2 holds.C3 fails since C3⇒C1.C5 fails: let xn=1/n, x=0,andy=2(31)lim⁡d(xn,0)=lim⁡(1n)=0=lim⁡d(xn,2)C4 fails since C4⇒C5.Example 7.
Ad-symmetric space (X,d) in which C4 holds but C1 fails.LetX=N∪{0}. Define d on X×X as follows:
(32)d(m,n)=d(n,m)∀m,n∈X,d(0,n)={1nifnisodd,1ifniseven,d(0,0)=0,d(m,n)={|1m-1n|ifm+nisevenorm+nisoddand|m-n|=1,1ifm+nisoddand|m-n|>2.
If {xn} in X and lim⁡d(xn,0)=0 then xn is eventually odd.Ifx≠0,d(xn,x) cannot be 1 so xn+x is even or odd and |xn-x|=1.But in this cased(xn,x)=|1/xn-1/x|so that d(xn,x)≠0.Thusd(xn,x)=0⇔x=0 and xnis eventually odd.Ifm is a fixed even integer and xnis odd, xn+m is odd and eventually >2.So(33)lim⁡d(xn,m)=1=d(0,m).
If m is a fixed odd integer and xnis odd, xn+m is even.Sod(xn,m)=|1/m-1/xn|so that lim⁡d(xn,0)=0⇒lim⁡d(xn,m)=d(0,m).Ifm=0 and xn is odd eventually
(34)d(xn,0)=1nsolim⁡d(xn,m)=lim⁡1n=0=d(0,m).
If m=0 and xn=0 eventually
(35)d(xn,0)=1nsolim⁡d(xn,m)=lim⁡1n=0=d(0,m).
Hence C4 holds in (X,d).C
1 does not hold: let xn=2n-1 and yn=2n:
(36)d(xn,0)=12n-1,d(xn,yn)=12n-1-12n,d(yn,0)=1.
Hence d(xn,0)=d(xn,yn)=0 and d(yn,0)≠0.Example 8.
Ad-symmetric space (X,d) in which C3 holds but C4 does not hold.LetX=[0,1]∪{2}. Define d on X×X as follows:
(37)d(x,y)={x+yif0≤x≠y≤1,1ifx=y≠0orx=y=2orx∈(0,1]andy=2,2ifx=0&y=2orx=2andy=0,0ifx=y=0.
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 that0≠x≠2; then x∈(0,1]:
(38)⟹d(x,xn)={x+xnif0<x≠xn≤1,1ifx=xn≠0orxn=2orx∈(0,1]andxn=2.
Since lim⁡d(x,xn)=0∃N∋d(x,xn)<1 for n≥N⇒d(x,xn)=x+xn≥x for n≥N,⇒lim⁡d(x,xn)≠0, a contradiction.We now show thatlim⁡d(xn,yn)=0 if and only if lim⁡xn=lim⁡yn=0. Consider
(39)lim⁡d(xn,yn)=0⟹∃N∈N∋d(xn,yn)<1forn≥N⟹lim⁡d(xn,yn)=xn+ynor0forn≥N⟹eitherxn=yn=0ord(xn,yn)=xn+ynfffffffffffffffffffffffffffffffffffffffffffffffffffffffforn≥N⟹lim⁡xn=lim⁡yn=0.Conversely iflim⁡xn=lim⁡yn=0 then ∃N∈N∋xn<1,yn<1forn≥N⇒
lim
⁡
d
(
x
n
,
y
n
)
=
0 or xn+yn for large n.Henced(xn,yn)=0.Thusd(xn,yn)=0 if and only if lim⁡xn=lim⁡yn=0.As a consequence we have(40)lim⁡d(xn,yn)=0=lim⁡d(yn,zn)⟹lim⁡d(xn,zn)=0.
Hence C3 holds in (X,d).C
4 fails: xn=1/(n+1)forn≥1:
(41)d(xn,0)=1n+1⟹lim⁡d(xn,0)=0,d(xn,2)=1∀n⟹lim⁡d(xn,2)=1butd(0,2)=2.Example 9.
Ad-symmetric space (X,d) in which C4 holds but C2,C3 fail to hold.LetX=N∪{0,∞}. Define d on X×X as follows:
(42)d(m,∞)=d(∞,m)=1ifm∈X,d(m,0)=d(0,m)=1mifm∈N,d(0,0)=0.
If m,n∈N,
(43)d(m,n)={|1m-1n|if|m-n|≥2,1if|m-n|≤1.
Clearly (X,d) is a d-symmetric space which is not a symmetric space.We show that iflim⁡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≠zn∈N∀n and lim⁡(1/zn)=0 (in case {zn} is an infinite sequence).To prove this we first note thatlim⁡d(xn,x)=0⇒x≠∞ and xn≠∞eventually.Ifx∈N, d(xn,x)=1/xor1or|1/xn-1/x|.Hencelim⁡d(xn,x)=0⇒x∉N; hence x=0.Furtherd(xn,0)=0or1/xn. Consequently {xn} may be split into two sequences {yn} and {zn} as described above.We show thatC4 holds. Assume that lim⁡d(xn,x)=0. Then x=0.Letm∈N and yn=0∀n. Then d(yn,m)=d(o,m)=1/m.Solim⁡d(yn,m)=d(o,m).Ifzn≠0∀n,andlim⁡(1/zn)=0 the d(zn,m)=|1/zn-1/m| for n>mso that lim⁡d(zn,m)=1/m=d(0,m).Thus ifm∈N andlim⁡d(xn,x)=0 then lim⁡d(xn,m)=d(x,m).Clearly this holds whenm=∞ or m=0 as well.HenceC4 holds.C
2 does not hold: let xn=x,yn=n+1,andx=0:
(44)d(xn,x)=d(n,0)=1n,hencelim⁡d(xn,x)=0,d(yn,x)=d(n+1,0)=1n+1,hencelim⁡d(yn,x)=0,lim⁡d(xn,yn)=lim⁡d(n,n+1)⟹lim⁡d(xn,yn)≠0.C3 does not hold:
(45)xn=n,yn=n+2,zn=xn,d(xn,yn)=d(n,n+2)=|1n+2-1n|=1n-1n+2,d(yn,zn)=d(xn,yn)=1n-1n+2,lim⁡d(xn,zn)=lim⁡d(n,n)=1,lim⁡d(xn,yn)=lim⁡d(yn,zn)=0butlim⁡d(xn,zn)=1.C5 holds since C4⇒C5.Remarks. From this example we can conclude that(1)
C5 does not imply C2 as otherwise, since C4⇒C5 it would follow that C4⇒C2 which does not hold as is evident from the above example,(2)
in ad-symmetric space, convergent sequences are necessarily Cauchy sequences.Example 10.
Ad-symmetric space (X,d) in which C4 holds but C2, C3 fail to hold.LetX=N∪{0}. Define d on X×X as follows:
(46)d(x,y)=d(y,x)=1foreveryx,y∈X,d(2m,0)=1,d(2m-1,0)=12m-1∀m,d(0,0)=0,d(m,n)={1m+1nifm+nisevenor|m-n|=1,1ifm+nisoddand|m-n|>2.
Clearly (X,d) is a d-symmetric space.We first characterize all convergent sequences in(X,d).Suppose thatlim⁡d(xn,x)=0. We show that x=0.Ifx is odd and xn is even d(xn,x)=1ifxn>x+2.Solim⁡d(xn,x)≠0. Thus xn is even for at most finitely many n.We may thus assume thatxn is odd ∀n.Thed(xn,x)=1/xn+1/x so that d(xn,x)≥1/x>0.Hencex cannot be odd. Now suppose that x>0 and x is even.Thend(xn,x)=1ifxn=0 if xn is odd and |xn-x|>2while d(xn,x)=1/xn+1/xifxn+x  is  even  or  |xn-x|=1. In all cases lim⁡d(xn,x)≠0.Hence the only possibility isx=0.We now show that the following are equivalent.(a)
lim⁡d(xn,x)=0in R,(b)
there exists a positive integerN such that xn is positive and even, only if n<N.Assumption (b):xn is odd or zero if n≥N so that lim⁡d(xn,0)=lim⁡(1/xn)=0.Hence (b)⇒(a).Assumption (a): sinced(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≥N.ConsequentlyC5 holds.C
1 does not hold: let xn=2n+1,yn=2n and x=0;
(47)lim⁡d(xn,x)=lim⁡12n+1=0,lim⁡d(xn,yn)=lim⁡12n+1+12n=0.
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).Thenx=0 and then there exists N such that xn is “0” or odd and yn=0 or odd for n≥N and lim⁡(1/xn)=lim⁡(1/yn)=0.Ifxn=yn=0, d(xn,yn)=0.Ifxn=0, yn is odd, d(xn,yn)=1/yn.Ifyn=0, xn is odd, d(xn,yn)=1/xn.Ifxn is odd and yn is odd, d(xn,yn)=1/xn+1/yn.Consequentlylim⁡d(xn,yn)=0.C
3 does not hold: let xn=0, yn=2n+1, and zn=2n:
(48)d(xn,yn)=12n+1,d(yn,zn)=12n+1+12n,d(xn,zn)=1
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:(49)lim⁡d(xn,0)=lim⁡12n+1=0,lim⁡d(xn,3)=1,lim⁡d(0,3)=13.Example 11.
The following example shows that there exist symmetric spaces in whichC does not hold.LetX={0,1/2,1/3,1/4,…}.Defined(x,x)=0, d(x,y)=d(y,x)(50)d(1n,0)=d(0,1n)=1n∀ninN,d(1n,1m)=1∀n,minN.
Then (X,d) is a symmetric space; {1/n} converges to 0 but is not a Cauchy sequence.



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*Source: 745031-2014-10-14.xml*