# Rectangular Metric-Like Type Spaces and Related Fixed Points

Authors: N. Mlaiki; K. Abodayeh; H. Aydi; T. Abdeljawad; M. Abuloha
Journal: Journal of Mathematics
 (2018)

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## Abstract

In this paper, we introduce the concept of a rectangular metric-like space, along with its topology, and we prove some fixed point theorems for different contraction types. We also introduce the concept of modified metric-like spaces and we prove some topological and convergence properties under the symmetric convergence. Some examples are given to illustrate the new introduced metric type spaces.


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## Body

## 1. Introduction

The generalization of Banach contraction principle, which has many applications in different branches of science and engineering, depends on either generalizing the metric type space or the contractive type mapping (see [1] and the references therein). The generalization of a metric space is based on reducing or modifying the metric axioms; for example, we cite quasi-metrics, partial metrics, m-metrics, Sp metrics, rectangular metrics, b-metrics. For more details, see [2–15]. Note that losing or weakening some of the metric axioms causes the loss of some topological properties, hence bringing obstacles in proving some fixed point theorems. These obstacles force researchers to develop new techniques in the development of fixed point theory in order to resolve more real concrete applications. In this article, we restrict ourselves on developing metric-like spaces by introducing modified metric-like spaces, rectangular metric-like spaces, and rectangular modified metric-like spaces. We shall prove some fixed point theorems in rectangular metric-like spaces. Examples will be given to support the given concepts. The notion of symmetric convergence will be also studied in the setting of modified metric-like spaces.Definition 1 ([13] (partial metric space)).
LetX be a nonempty set. A mapping p:X×X→R+ is said to be a partial metric on X, if, for any x,y,z∈X, it satisfies the following conditions:(p1)
x =y if and only if p(x,y)=p(x,x)=p(y,y);(p2)
p(x,x)≤p(x,y);(p3)
p(x,y)=p(y,x);(p4)
p(x,y)≤p(x,z)+p(z,y)-p(z,z).
In this case, the pair(X,p) is called a partial metric (PM) space.Definition 2 ([9] (rectangular (or Branciari) metric space)).
LetX be a nonempty set. A mapping d:X×X→R+ is said to be a rectangular metric on X if, for any x,y∈X and all distinct points u,v∈X∖{x,y}, it satisfies the following conditions:(R1)
x=y if and only if d(x,y)=0;(R2)
d(x,y)=d(y,x);(R3)
d(x,y)≤d(x,u)+d(u,v)+d(v,y).
In this case, the pair(X,d) is called a rectangular metric (RM) space.In [15], the notion of a rectangular metric space was extended to rectangular partial metric spaces as follows.Definition 3 ([15] (rectangular partial metric space)).
LetX be a nonempty set. A mapping ρ:X×X→R+ is said to be a rectangular metric on X if, for any x,y∈X and all distinct points u,v∈X∖{x,y}, it satisfies the following conditions:(RP1)
x=y if and only if ρ(x,y)=ρ(x,x)=ρ(y,y);(RP2)
ρ(x,x)≤ρ(x,y);(RP3)
ρ(x,y)=ρ(y,x);(RP4)
ρ(x,y)≤ρ(x,u)+ρ(u,v)+ρ(v,y)-ρ(u,u)-ρ(v,v).
In this case, the pair(X,ρ) is called a rectangular partial metric (RPM) space.It is clear that every rectangular metric space is a rectangular partial metric space, but the converse is not true.Example 4 (see [15]).
LetX=[0,a] and α≥a≥3. Define the mapping ρ:X×X→R+ by (1)ρx,y=xifx=y3α+x+y2ifx,y∈1,2,x≠yα+x+y2otherwise.Then (X,ρ) is a rectangular partial metric space, but it is not a rectangular metric space, because, for any x>0, we have ρ(x,x)=x≠0.For convergence, completeness, and examples of RM, PM, and RPM spaces, we refer to [9, 13, 15]. See also the papers [16, 17].Definition 5 (see [11]).
LetX be a nonempty set. A mapping σ:X×X→R+ is said to be a metric-like on X if, for any x,y,z∈X, it satisfies the following conditions:(σ1)
σ(x,y)=0 implies x=y;(σ2)
σ(x,y)=σ(y,x);(σ3)
σ(x,y)≤σ(x,z)+σ(z,y).
In this case, the pair(X,σ) is called a metric-like space (ML-space).Every metric-like space is a topological space whose topology is generated by the base consisting of the openσ-balls (2)Bσx,δ=y∈X:σx,y-σx,x<δ,x∈X,δ>0. Note the difference between the balls Bσ(x,δ) and the balls Bp(x,δ), which is due to the absence of the smallness of the self distance condition (p2) from the metric-like. Also, since the self distance need not be zero in metric-like spaces, then convergence and completeness in metric-like spaces still resemble the case of partial metric spaces. Indeed, a sequence {xn} in a metric-like space converges to a point x∈X if and only if limn→∞σ(xn,x)=σ(x,x) and the sequence {xn} is called σ-Cauchy if limm,n→∞σ(xn,xm) exists and is finite. The metric-like space (X,σ) is called complete if for each σ-Cauchy sequence {xn} there exists x∈X such that (3)limn→∞⁡σxn,x=σx,x=limm,n→∞⁡σxn,xm.Remark 6.
Metric-like spaces lose some topological and convergence properties as known for metric spaces. We state the following. For example, limits are not unique inML-spaces. Take X={a,b} and let σ(x,y)=1 for any x,y∈X. Then, clearly the constant sequence {xn=1} converges to both a and b. Notice that σ(a,a)=σ(b,b)=1≠0. However, if xn→x and xn→y such that x,y∈Λ={z∈X:σ(z,z)=0}, then σ(x,y)≤σ(xn,x)+σ(xn,y). Letting n→∞, we conclude that σ(x,y)=0, and hence x=y.


## 2. Main Results

Upon Remark6 above, we define the following modified metric-like (mML) space.Definition 7 (modified metric-like space).
LetX be a nonempty set. A mapping σm:X×X→R+ is said to be a modified metric-like on X, if for any x,y,z∈X, it satisfies the following conditions:(mσ1)
σm(x,y)=0 implies x=y;(mσ2)
σm(x,y)=σm(y,x);(mσ3)
σm(x,y)≤σm(x,z)+σm(z,y)-σm(z,z).
The pair(X,σm) is called a modified metric-like space (mML-space).It is clear that every partial metric space is anmML-space and every mML-space is an ML-space.Example 8.
LetX={1,2,3,4,5} and define the mapping d:X×X→R+ such that d(x,y)=2 for all x≠y, d(x,x)=0 for all x≠1, and d(1,1)=3. It is clear that the conditions (σ1) and (σ2) in Definition 5 are satisfied. We need to verify (σ3). If x≠y, then we have d(x,u)+d(u,y)=2+2≥d(x,y). Also, if x=y=1, then d(1,u)+d(u,1)=2+2≥d(1,1). Finally, if x=y≠1, d(x,u)+d(u,x)≥0=d(x,x). Therefore, (X,d) is a ML-space, but it is not a mML-space, because d(2,1)+d(1,3)-d(1,1)=2+2-3=1≤d(2,3).As in the case of metric-like spaces, the openσm-balls in modified metric-like spaces are given as (4)Bσmx,δ=y∈X:σmx,y-σmx,x<δ,x∈X,δ>0. Also, note that a sequence {xn} in a modified metric-like space (X,σm) converges to a point x∈X if and only if limn→∞σm(xn,x)=σm(x,x) and the sequence {xn} is called σm-Cauchy if limn,l→∞σm(xn,xl) exists and is finite.Definition 9 (symmetrical convergence in modified metric-like spaces).
We shall say that a sequence{xn} in a modified metric-like space (X,σm) is symmetrically convergent to x∈X if, for every ϵ>0, there exists n0∈N such that, for each n≥n0, we have (5)xn∈Bσmx,ϵandx∈Bσmxn,ϵ.Equivalently,(6)limn→∞⁡σmxn,x=limn→∞⁡σmxn,xn=σmx,x.We shall denotexn→sx for symmetrical convergence. It is clear that symmetrical convergence implies σm-convergence. Our first result is as follows.Theorem 10.
Let(X,σm) be an mML-space. Then one has the following: (1)
Ifxn→sx, then {xn} is σm-Cauchy.(2)
If{xn} is σm-Cauchy and has a subsequence {xni} such that xni→sx, then xn→sx.(3)
If{xn} and {yn} are σm-Cauchy sequences, then limn→∞σm(xn,yn) exists.(4)
Ifxn→sx and yn→sy, then limn→∞σm(xn,yn)=σ(x,y).Proof.
(1)
Assumexn→sx. Then limn→∞σm(xn.yn)=σ(x,y). By (mσ3), for each l,n∈N we have (7)σmxn,xl≤σmxn,x+σmx,xl-σmx,x and(8)σmx,x≤σmx,xn+σmxn,xl+σmxl,x-σmxn,xn-σmxl,xl. Letl,n→∞. Then σm(x,x)≤limn,l→∞σm(xl,xn)≤σm(x,x). Hence limn,l→∞σm(xl,xn)=σm(x,x), and so {xn} is σm-Cauchy. (2)
Let{xn} be a σm-Cauchy and have a subsequence {xni} such that xni→sx. Then (9)limi→∞⁡σmxni,x=limi→∞⁡σmxni,xni=σmx,x. Since{xn} is σm-Cauchy, there exists r>0 such that liml,n→∞σm(xn,xl)=r. It is clear that σm(x,x)=r as well. On the other hand, by (mσ3), we have (10)σmxn,x≤σmxn,xni+σmxni,x-σmxni,xni, and(11)σmxni,x≤σmxni,xn+σmxn,x-σmxn,xn. Therefore,(12)σmxni,x-σmxni,xn+σmxn,xn≤σmxn,x≤σmxn,xni+σmxni,x-σmxni,xni. If we taken,i→∞, then σm(x,x)-r+r≤limn,l→∞σm(xn,xl)≤r+σm(x,x)-σm(x,x). It follows that limn→∞σm(xn,x)=limn→∞σm(xn,xn)=σm(x,x), and hence xn→sx. (3)
Assume that{xn} and {yn} are σm-Cauchy sequences in X. Then there exist r1,r2>0 such that limn,l→∞σm(xn,xl)=r and limn,l→∞σm(yn,yl)=r2. It is sufficient to prove that the sequence {σm(xn,yn)} is Cauchy in R. By (mσ3), for each n,l∈N, we have (13)σmxn,yn≤σmxn,xl+σmxl,yl+σmyl,xn-σmxl,xl-σmyl,yl, and(14)σmxl,yl≤σmxl,xn+σmxn,yn+σmyn,yl-σmxn,xn-σmyn,yn. From what is mentioned, it follows that(15)σmxn,xn+σmyn,yn-σmxl,xn-σmyn,yl≤σmxn,yn-σmxl,yl≤σmxn,xl+σmyl,yn-σmxl,xl-σmyl,yl. Letn,l→∞, then (16)r1+r2-r1-r2≤limn,l→∞⁡σmxn,yn-σmxl,yl≤r1+r2-r1-r2. Hence|σm(xn,yn)-σm(xl,yl)|=0, and so {σm(xn,yn)} is Cauchy in R. (4)
Assume thatxn→sx and yn→sy. Then (17)limn→∞⁡σmxn,x=limn→∞⁡σmxn,xn=σmx,x, and(18)limn→∞⁡σmyn,y=limn→∞⁡σmyn,yn=σmy,y. Now, for eachn∈N, we have (19)σmxn,yn≤σmxn,x+σmx,y+σmy,yn-σmx,x-σmy,y, and(20)σmx,y≤σmx,xn+σmxn,yn+σmyn,y-σmxn,xn-σmyn,yn. Finally, lettingn→∞ yields that (21)limn→∞σmxn,yn≤σmx,y≤limn→∞σmxn,yn, and thuslimn→∞σ(xn,yn)=σ(x,y).Now, we introduce the concepts of rectangular metric-like and rectangular modified metric-like spaces.Definition 11.
LetX be a nonempty set and ρr:X2→[0,∞) be a function. If the following conditions are satisfied for all x,y in X, (1)
ρr(x,y)=0⇒x=y;(2)
ρr(x,y)=ρr(y,x);(3)
ρr(x,y)≤ρr(x,u)+ρr(u,v)+ρr(v,y), for all distinct u,v∈X∖{x,y},
then the pair(X,ρr) is called a rectangular metric-like (RML) space.Definition 12.
LetX be a nonempty set and ρmr:X2→[0,∞) be a function. If the following conditions are satisfied for all x,y in X, (1)
ρrm(x,y)=0⇒x=y;(2)
ρrm(x,y)=ρrm(y,x);(3)
ρrm(x,y)≤ρrm(x,u)+ρrm(u,v)+ρrm(v,y)-ρrm(u,u)-ρrm(v,v), for all distinct u,v∈X∖{x,y},
then the pair(X,ρrm) is called a rectangular modified metric-like (RMML) space.Example 13.
LetX={1,2,3,4,5} and define the mapping ρr:X2→[0,∞) by (22)ρrx,y=2.5forx≠y5ifx=y=10otherwise.It is clear that conditions (1) and (2) in Definition 11 are satisfied. We need to verify the last condition. For all distinct u,v∈X∖{x,y}, we have ρr(x,u)+ρr(u,v)+ρr(v,y)=2.5+ρr(u,v)+2.5=5+ρr(u,v)≥ρr(x,y), for all x,y∈X. Therefore, (X,ρr) is a RML-space, but it is not a RMML-space. Indeed, ρr(2,1)+ρr(1,1)+ρr(1,3)-ρr(1,1)-ρr(1,1)=2.5+5+2.5-5-5=0≤ρr(2,3)=2.5. Moreover, the space (X,ρr) is not a rectangular partial metric, because the condition (RP4) of Definition 2 does not hold.Example 14.
Let(X,ρr) be a RML-space. Consider ρrm(x,y)=ρr(x,y)+a for a>0. Then the space (X,ρrm) is a RMML-space.Example 15.
LetX=(0,1) and define the mapping ρmr:X2→[0,∞) by ρmr(x,y)=|x-y|+2. Then (X,ρrm) is a RMML-space. Indeed, for any x,y∈X and distinct u,v∈X∖{x,y}, we have ρmr(x,u)+ρmr(u,v)+ρmr(v,y)-ρmr(u,u)-ρmr(v,v)=ρr(x,u)+ρr(u,v)+ρr(v,y)-ρr(u,u)-ρr(v,v)+α≤ρr(x,y)-ρr(u,u)-ρr(v,v)+α≤ρr(x,y)+α=ρrm(x,y).Definition 16.
(1)
A sequence{xn} is called ρr-convergent (resp., ρrm-convergent) in a rectangular metric-like space (X,ρr) (resp., a rectangular modified metric-like space (X,ρrm)), if there exists x∈X such that limn→∞ρr(xn,x)=ρr(x,x) (resp.limn→∞ρrm(xn,x)=ρrm(x,x)). (2)
A sequence{xn} is called ρr-Cauchy if and only if limn,m→∞ρr(xn,xm) (resp.,limn,p→∞ρrm(xn,xp)) exists and is finite. (3)
A rectangular metric-like space(X,ρr) is called ρr-complete (resp., ρrm-complete) if, for every ρr-Cauchy (resp., ρrm-Cauchy) sequence {xn} in X, there exists x∈X, such that limn→∞ρr(xn,x)=limn,m→∞ρr(xn,xm)=ρr(x,x) (resp.,limn→∞ρrm(xn,x)=limn,p→∞ρrm(xn,xp)=ρrm(x,x)).Remark 17.
The convergence defined in Definition16 is the convergence obtained in the sense of the topology generated by the open balls Bϱ(x,δ)=y∈X:ϱx,y-ϱx,x<δ, x∈X, ϱ∈ρr,ρrm. This convergence is weaker than the symmetric convergence discussed before.Definition 18 (continuity of maps).
Let(X,ϱ1) and (Y,ϱ2) be two metric type (such as ML, MML, PM, RPM, RML, RMML) spaces. The mapping f:X→Y is said (sequentially) continuous at x∈X if and only if f(xn)→ϱ2f(x), whenever {xn} is a sequence in X such that xn→ϱ1x.Our second main result concerns an existence and uniqueness theorem on rectangular metric-like spaces.Theorem 19.
Let(X,ρr) be a ρr-complete rectangular metric-like space and T be a self mapping on X. If there exists 0<k<1 such that(23)ρrTx,Ty≤kρrx,yfor  allx,y∈X,then T has a unique fixed point u in X such that ρr(u,u)=0.Proof.
Letx in X be arbitrary. Using (23), we have(24)ρrTnx,Tn+1x≤kρrTn-1x,Tnx≤⋯≤knρrx,Tx,for all n≥1. We distinguish two cases.
Case 1. Let Tnx=Tmx for some integers n≠m. For example, take m>n. We have Tm-n(Tnx)=Tnx. Choose y=Tnx and p=m-n. Then(25)Tpy=y, that is, y is a periodic point of T. By (23) and (24), we have (26)ρry,Ty=ρrTpy,Tp+1y≤kpρry,Ty. Since k∈(0,1), we get ρr(y,Ty)=0, so y=Ty; that is, y is a fixed point of T.
Case 2. Suppose that Tnx≠Tmx for all integers n≠m.
We rewrite (24) as(27)ρrTnx,Tn+1x≤knρrx,Tx≤kn1-kρrx,Tx.Similarly, by (23), we have(28)ρrTnx,Tn+2x≤kρrTn-1x,Tn+1x≤⋯≤knρrx,T2x≤kn1-kρrx,T2x.Now, if m>2 is odd, then consider m=2p+1 with p≥1. By (23) and (27), we have (29)ρrTnx,Tn+mx≤ρrTnx,Tn+1x+ρrTn+1x,Tn+2x+⋯+ρrTn+2px,Tn+2p+1x≤knρrx,Tx+kn+1ρrx,Tx+⋯+kn+2pρrx,Tx=knρrx,Tx1+k+k2+⋯+k2p≤kn1-kρrx,Tx.On the other hand, if m>2 is even, then consider m=2p with p≥2. Again, by (23), (27), and (28), (30)ρrTnx,Tn+mx≤ρrTnx,Tn+2x+ρrTn+2x,Tn+3x+⋯+ρrTn+2p-1x,Tn+2px≤knρrx,T2x+kn+2ρrx,Tx+kn+3ρrx,Tx+⋯+kn+2p-1ρrx,Tx≤knρrx,T2x+kn+21-kρrx,Tx≤knρrx,T2x+kn1-kρrx,Tx. We deduce from all cases that(31)ρrTnx,Tn+mx≤knρrx,T2x+kn1-kρrx,Txfor  alln,m≥0.The right-hand side tends to 0 as n→∞, so the sequence {Tnx} is ρr-Cauchy in the ρr-complete rectangular metric-like space (X,ρr). Due to Definition 16, there exists some u∈X such that (32)limn→∞⁡ρrTnx,u=limn,m→∞⁡ρrTnx,Tmx=ρru,u. In view of (31), we get(33)ρru,u=limn→∞ρrTnx,u=limn,m→∞ρrTnx,Tmx=0.We shall prove that Tu=u. Mention that we are still in case 2, that is, Tnx≠Tmx for all integers n≠m. Now, we distinguish three subcases.
Subcase 1. If, for all n≥0, Tnx∉{u,Tu}, the rectangular inequality implies that (34)ρru,Tu≤ρru,Tnx+ρrTnx,Tn+1x+ρrTn+1x,Tu≤ρru,Tnx+ρrTnx,Tn+1x+kρrTnx,u. Taking limit as n→∞ and using (27) and (33), we get ρr(u,Tu)=0; that is, Tu=u.
Subcase 2. If there exists an integer N such that TNx=u, due to case 2, Tnx≠u for all n>N. Similarly, Tnx≠Tu for all n>N. We reach subcase 1, so u is a fixed point of T.
Subcase 3. If there exists an integer N such that TNx=Tu, again, necessarily Tnx≠u and Tnx≠Tu for all n>N. Similarly, we get Tu=u.
We deduce thatu is a fixed point of T. To show the uniqueness of the fixed point u, assume that T has another fixed point v. By (23), (35)ρru,v=ρrTu,Tv≤kρru,v, which holds unless ρr(u,v)=0, so u=v.Next, we present the following example.Example 20.
LetX={1,2,3,4,5} and define the mapping ρr:X2→[0,∞) by (36)ρrx,y=2.5forx≠y≠15ifx=y=13ifx=1ory=10otherwise. It is not difficult to see that (X,ρr) is a complete rectangular metric-like space. Let T be a self mapping on X defined by T2=T3=T4=T5=2 and T1=3. Note that T satisfies the contraction of Theorem 19 with k=29/30, and 2 is the unique fixed point of T.Our third main result is as follows.Theorem 21.
Let(X,ρr) be a ρr-complete rectangular metric-like space and T be a self mapping on X. If there exists 0<k<1 such that(37)ρrTx,Ty≤kmax⁡ρrx,Tx,ρry,Tyfor  allx,y∈X,then T has a unique fixed point u in X such that ρr(u,u)=0.Proof.
Fixx in X. If ρr(Tnx,Tn+1x)=0 for some n, then xn is a fixed point of T. The proof is completed. From now on, we assume that ρr(Tnx,Tn+1x)>0, so Tnx≠Tn+1x for all n. By (37), we have (38)0<ρrTnx,Tn+1x≤kmax⁡ρrTn-1x,Tnx,ρrTnx,Tn+1x,for all n≥1. If, for some n≥1, max{ρr(Tn-1x,Tnx),ρr(Tnx,Tn+1x}=ρr(Tnx,Tn+1x}, then(39)0<ρrTnx,Tn+1x≤kρrTnx,Tn+1x, which is a contradiction. Thus,(40)0<ρrTnx,Tn+1x≤kρrTn-1x,Tnx≤knρrx,Tx,for  alln≥1.Here, the sequence {ρr(Tnx,Tn+1x)} is nonincreasing. As the proof of Theorem 19, we distinguish two cases.
Case 1. Let Tnx=Tmx for some integers n≠m (take m-n≥2). Let y=Tnx=Tpy, where p=m-n. Again, using (40), (41)ρry,Ty=ρrTpy,Tp+1y≤kpρry,Ty. Since k∈(0,1), we get ρr(y,Ty)=0, so y is a fixed point of T.
Case 2. Suppose that Tnx≠Tmx for all integers n≠m.
We rewrite (40) as(42)ρrTnx,Tn+1x≤knρrx,Tx≤kn1-kρrx,Tx.Similarly, by (37) and (42), we have (43)ρrTnx,Tn+2x≤kmax⁡ρrTn-1x,Tnx,ρrTn+1x,Tn+2x≤kmax⁡kn-1ρrx,Tx,kn+1ρrx,x=knρrx,Tx≤kn1-kρrx,Tx.Now, if m>2 is odd, then consider m=2p+1 with p≥1. By (37) and (42), we have (44)ρrTnx,Tn+mx≤ρrTnx,Tn+1x+ρrTn+1x,Tn+2x+⋯+ρrTn+2px,Tn+2p+1x≤knρrx,Tx1+k+k2+⋯+k2p≤kn1-kρrx,Tx.On the other hand, if m>2 is even, then consider m=2p with p≥2. Again, by (37), (42), (45)ρrTnx,Tn+mx≤ρrTnx,Tn+2x+ρrTn+2x,Tn+3x+⋯+ρrTn+2p-1x,Tn+2px≤knρrx,Tx+kn+2ρrx,Tx+kn+3ρrx,Tx+⋯+kn+2p-1ρrx,Tx≤kn1-kρrx,Tx. We deduce from all cases that(46)ρrTnx,Tn+mx≤kn1-kρrx,Txfor  alln,m≥0.The right-hand side tends to 0 as n→∞, so the sequence {Tnx} is ρr-Cauchy in the ρr-complete rectangular metric-like space (X,ρr). Due to Definition 16, there exists some u∈X such that, in view of (46),(47)ρru,u=limn→∞ρrTnx,u=limn,m→∞ρrTnx,Tmx=0.We shall prove that Tu=u. Without loss of generality, we may assume that, for all n≥0, Tnx∉{u,Tu}. By the rectangular inequality, we have (48)ρru,Tu≤ρru,Tnx+ρrTnx,Tn+1x+ρrTn+1x,Tu≤ρru,Tnx+ρrTnx,Tn+1x+kmax⁡ρrTnx,Tn+1x,ρu,Tu.Taking limit as n→∞ and using (42) and (47), we get ρr(u,Tu)=0, that is, Tu=u.
Now, letv be a fixed point of T. From (37), (49)ρru,v=ρrTu,Tv≤kρru,v.It is true unless ρr(v,v)=0, so u is the unique fixed point of T.



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