AAA Abstract and Applied Analysis 1687-0409 1085-3375 Hindawi Publishing Corporation 10.1155/2014/458098 458098 Research Article Generalized Metric Spaces Do Not Have the Compatible Topology http://orcid.org/0000-0002-2524-6045 Suzuki Tomonari 1, 2 Du Wei-Shih 1 Department of Basic Sciences Faculty of Engineering Kyushu Institute of Technology Tobata Kitakyushu 804-8550 Japan kyutech.ac.jp 2 Department of Mathematics Faculty of Science King Abdulaziz University Jeddah Saudi Arabia kau.edu.sa 2014 482014 2014 12 05 2014 09 07 2014 4 8 2014 2014 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.

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  introduced a very interesting concept whose name is “ν-generalized metric space.”

Definition 1 (see Branciari [<xref ref-type="bibr" rid="B2">1</xref>]).

Let X be a set, let d be a function from X×X into [0,), and let νN. Then (X,d) is said to be a ν-generalized metric space if the following hold:

d(x,y)=0 if and only if x=y for any x,yX;

d(x,y)=d(y,x) for any x,yX;

d(x,y)d(x,u1)+d(u1,u2)++d(uν,y) for any x,u1,u2,,uν, yX such that x,u1,u2,,uν,y are all different.

Example 2.

Every metric space (X,d) is a 1-generalized metric space.

A 2-generalized metric space is also said to be a generalized metric space.

Definition 3 (see Branciari [<xref ref-type="bibr" rid="B2">1</xref>]).

Let X be a set and let d be a function from X×X into [0,). Then (X,d) is said to be ageneralized metric space if the following hold:

d(x,y)=0 if and only if x=y for any x,yX.

d(x,y)=d(y,x) for any x,yX.

d(x,y)d(x,u)+d(u,v)+d(v,y) for any x,u,v,yX such that x,u,v,y 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 X does not necessarily have the topology which is compatible with d; 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. <inline-formula><mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML" id="M36"><mml:mrow><mml:mi>ν</mml:mi></mml:mrow></mml:math></inline-formula>-Generalized Metric Space

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

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

Lemma 4.

Let (X,ρ) be a bounded metric space and let M be a real number satisfying (1)sup{ρ(x,y):x,yX}M. Let A and B be two subsets of X with X=AB and AB=. Define a function d from X×X into [0,) by (2)d(x,x)=0d(x,y)=d(y,x)=ρ(x,y)ifxA,yBd(x,y)=Motherwise. Then (X,d) is a generalized metric space.

Proof.

(N1) and (N2) are obvious. Let us prove (N3). Let x,y,u,vX be all different. Put (3)t=d(x,u)+d(u,v)+d(v,y). In the case where tM, (N3) holds because d(x,y)M. In the other case, where t<M, without loss of generality, we may assume xA. Then we have vA and u,yB from the definition of d. Hence, (4)d(x,y)=ρ(x,y)ρ(x,u)+ρ(u,v)+ρ(v,y)=d(x,u)+d(u,v)+d(v,y). Thus (N3) holds.

Definition 5.

Let (X,d) be a ν-generalized metric space. Then a net {xα} is said to converge to x if and only if limαd(x,xα)=0.

Definition 6.

Let X be a topological space with topology τ. Let d be a function from X×X into [0,) satisfying (N1)–(N3) with some νN. Then τ is compatible with d if and only if the following are equivalent for any net {xα} in X and xX:

limαd(x,xα)=0.

{xα} converges to x in τ.

The following is a very important example.

Example 7.

Let (5)X={(0,0)}((0,1]×[0,1]). Define a function d from X×X into [0,) by (6)d(x,x)=0d((0,0),(s,0))=d((s,0),(0,0))=s,ifs(0,1]d((s,0),(p,q))=d((p,q),(s,0))=|s-p|+q,ifs,p,q(0,1]d(x,y)=3,otherwise. Then the following hold:

(X,d) is not a metric space;

(X,d) is a generalized metric space;

X does not have a topology which is compatible with d.

Proof.

Since (7)d((0,0),(1,0))+d((1,0),(1,1))=1+1=2<3=d((0,0),(1,1)),(X,d) is not a metric space. Define a metric ρ on X by (8)ρ((s,t),(p,q))=|s-p|+|t-q|, for (s,t),(p,q)X. Put(9)A={(0,0)}((0,1]×(0,1]),B=(0,1]×{0}. Then d is equal to the d defined by Lemma 4 with M=3. Therefore, (X,d) is a generalized metric space. In order to show (iii), we will show that the following does not hold.

If a net {xα}αD converges to x and for every αD a net {x(α,β)}βEα converges to xα, then {x(α,γ)}(α,γ)D×{Eα:αD} has a subnet converging to x; see [4, page 77].

We have that {(1/l,0)}l converges to (0,0) and {(1/l,1/m)}m converges to (1/l,0) for every lN. However, since d((0,0),(1/l,1/m))=3 for (l,m)N2, a net {(1/l,1/γ(l))}(l,γ) does not converge to (0,0). Therefore there does not exist a topology which is compatible with d.

Remark 8.

For (α,γ)D×{Eα:αD}, x(α,γ)=x(α,γ(α)). For (α1,γ1),(α2,γ2)D×{Eα:αD}, (α1,γ1)(α2,γ2) if and only if α1α2 and γ1(α)γ2(α) for any αD.

Remark 9.

Indeed, let τ be the topology induced by a subbase: (10){S(x,r):xX,r>0}, where S(x,r)={yX:d(x,y)<r}. Since (11)S((0,0),2)S((1,0),2)=([0,1]×{0})({(0,0),(1,0)}((0,1]×(0,1]))={(0,0),(1,0)}, we have (12)S((0,0),2)S((1,0),1)={(1,0)}. Hence {(1,0)} is an open neighborhood of (1,0). So a sequence {(1,1/n)} does not converge to (1,0) in τ. Since limnd((1,0),(1,1/n))=0, τ is not compatible with d.

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

Example 10.

Put X=N and let νN satisfy ν2. Define a function d from X×X into [0,) by (13)d(x,x)=0,d(1,s)=d(s,1)=ν+1,ifsN{1,2},d(x,y)=1,otherwise. Then the following hold:

(X,d) is not a μ-generalized metric space for μN with μ<ν;

(X,d) is a μ-generalized metric space for μN with μν.

Proof.

(N1) and (N2) obviously hold. Let μN satisfy μ<ν. Since (14)j=1μ+1d(j,j+1)=μ+1<ν+1=d(1,μ+2), (N3) does not hold. So (X,d) is not a μ-generalized metric space. Let μN satisfy μν. Let x,u1,u2,uμ,yX be all different. Then we have (15)d(x,y)ν+1μ+1d(x,u1)+d(u1,u2)++d(uμ,y). Thus (N3) holds. Hence (X,d) is a μ-generalized metric space.

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

Definition 11.

Let (X,d) be a ν-generalized metric space.

A sequence {xj} is said to be Cauchy if and only if limjsupm>jd(xj,xm)=0.

X is said to be complete if and only if every Cauchy sequence converges to some point in X.

X is said to be Hausdorff if and only if limjd(x,xj)=limjd(y,xj)=0 implies x=y.

Lemma 12.

Let (X,d) be a ν-generalized metric space and let x,u1,,uν,yX such that x,u1,,uν are all different and u1,,uν,y are all different. Then (16)d(x,y)d(x,u1)+d(u1,u2)++d(uν,y) holds.

Proof.

In the case where x=y, the conclusion obviously holds from (N1). In the other case, where xy, the conclusion obviously holds from (N3).

3. The CJM Fixed Point Theorem

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

Theorem 13.

Let (X,d) be a complete ν-generalized metric space and let T be a CJM contraction on X; that is, the following hold:

for every ɛ>0, there exists δ>0 such that d(x,y)<ɛ+δ implies d(Tx,Ty)ɛ for any x,yX;

xy implies d(Tx,Ty)<d(x,y) for any x,yX.

Then T has a unique fixed point z of T. Moreover, limjd(Tjx,z)=0 for any xX.

Proof.

We first note that T is nonexpansive by (ii); that is (17)d(Tx,Ty)d(x,y) for any x,yX. Fix uX and define a sequence {uj} in X by uj=Tju for jN. We next show that {uj} converges to a fixed point of T, dividing the following three cases:

there exists nN such that un+1=un;

uj+1uj for all jN and there exist m,nN such that m+2n and um=un;

u1,u2, are all different.

In the first case, un is a fixed point of T. By (N1), {uj} converges to un. In the second case, from (ii), we have {d(uj,uj+1)} is strictly decreasing. So, since um+1=un+1, we have (18)d(um,um+1)=d(un,un+1)<d(um,um+1). This is a contradiction. Thus, the second case cannot be possible. In the third case, from (ii), we have {d(uj,uj+k)} is strictly decreasing for any kN. So {d(uj,uj+k)} converges to some ε10. Then we note that d(uj,uj+k)>ε1 for every jN. Arguing by contradiction, we assume ε1>0. From (i), there exists δ1>0 such that (19)d(x,y)<ε1+δ1impliesd(Tx,Ty)ε1.

From the definition of ε1, there exists nN such that d(un,un+k)<ε1+δ1. Then we have d(un+1,un+k+1)ε1. This is a contradiction. Therefore we obtain ε1=0. That is, limjd(uj,uj+k)=0 holds for any kN. Thus (20)limjmax{d(uj,uj+k):k=1,2,,ν+1}=0 holds. Fix ε2>0. Then, by (i), there exists δ2(0,ε2) such that (21)d(x,y)<ε2+2νδ2impliesd(Tx,Ty)ε2.

Let lN such that (22)max{d(uj,uj+k):k=1,2,,ν+1}<δ2, for all jN with jl. We will show (23)d(ul,ul+m)<ε2+νδ2, for mN by induction. For m=1,2,,ν+1, we have (24)d(ul,ul+m)<δ2<ε2+νδ2, and, thus, (23) holds. We assume (23) holds for some mN with m>ν. We have, by (N3), (25)d(ul+ν,ul+m)j=1νd(ul+j,ul+j-1)+d(ul,ul+m)<νδ2+ε2+νδ2=ε2+2νδ2. Hence d(ul+ν+1,ul+m+1)ε2. We put (26)α={d(ul,ul+ν+1)ifν=1d(ul,ul+1)+d(ul+1,ul+ν+1)ifν=2j=ll+ν-2d(uj,uj+1)+d(ul+ν-1,ul+ν+1)ifν>2. We note α<νδ2. By (N3), we have (27)d(ul,ul+m+1)α+d(ul+ν+1,ul+m+1)<νδ2+ε2. Thus, (23) holds for mm+1. So, by induction, (23) holds for every mN. Therefore we have shown (28)limlsupl<md(ul,um)ε2+νδ2<(ν+1)ε2. Since ε2>0 is arbitrary, we obtain that {uj} is Cauchy. Since X is complete, {uj} converges to some point zX. We have by Lemma 12 and the nonexpansiveness of T(29)d(z,Tz)(d(z,um+1)+j=1ν-1d(um+j,um+j+1)+d(um+ν,Tz))(d(z,um+1)+j=1ν-1d(um+j,um+j+1)+d(um+ν-1,z)), for sufficiently large mN. As m tends to , we obtain d(z,Tz)=0. Thus, z is a fixed point of T. The uniqueness of the fixed point is obviously followed by (ii).

Remark 14.

In , 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 [<xref ref-type="bibr" rid="B2">1</xref>]).

Let (X,d) be a complete ν-generalized metric space and let T be a contraction on X; that is, there exists r[0,1) such that (30)d(Tx,Ty)rd(x,y), for any x,yX. Then T has a unique fixed point z of T. Moreover, limjd(Tjx,z)=0 for any xX.

Remark 16.

The authors in  stated the proof in  is incorrect and gave a proof under the assumption that (X,d) is Hausdorff and ν=2. See also .

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

Proposition 17.

Let (X,d) be a ν-generalized metric space and let T be a mapping on X. Assume that there exist functions φ,ψ from [0,) into [0,) such that the following hold:

ψ(d(Tx,Ty))ψ(d(x,y))-φ(d(x,y)) for any x,yX;

ψ is nondecreasing;

infφ([s,t])>0 for any s,t(0,) with s<t.

Then T is a CJM contraction.

Proof.

Since φ(t)>0 for any t(0,), (ii) of the definition of CJM contraction obviously holds. We will show (i) of the definition of CJM contraction. Fix ɛ>0. From (iii), we can put (31)ηinf{φ(t):ɛtɛ+1}>0. We choose δ(0,1) such that (32)ψ(ɛ+δ)<limtɛ+0ψ(t)+η. Let x,yX satisfy d(x,y)<ɛ+δ. In the case where d(x,y)=0, we have d(Tx,Ty)=0 because x=y. In the case where 0<d(x,y)ɛ, we have (33)ψ(d(Tx,Ty))ψ(d(x,y))-φ(d(x,y))<ψ(d(x,y))ψ(ɛ), which implies d(Tx,Ty)<ɛ. In the other case, where d(x,y)>ɛ, we have (34)ψ(d(Tx,Ty))ψ(d(x,y))-φ(d(x,y))ψ(ɛ+δ)-η<limtɛ+0ψ(t)+η-η=limtɛ+0ψ(t), which implies d(Tx,Ty)ɛ. Hence we have d(Tx,Ty)ɛ in all cases. Therefore T 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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