# Coupled Fixed Point Theorems in Topological Spaces with Size Function Topology

Authors: Gun Sunyeekhan
Journal: Abstract and Applied Analysis
 (2022)

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## Abstract

Regarding the concept of size function topology, this allows us to view the topological spaceX as a metric-like space. We prove the existence and uniqueness for a coupled fixed point of the map F:X×X⟶X satisfying some certain contractive conditions.


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## Body

## 1. Introduction

The Banach contraction mapping theorem is the classical fixed point theorem saying that for a self-mapT:X⟶X, if X,d is a complete metric space and there exists λ∈0,1 such that
(1)dTx,Ty≤λdx,y,forallx,y∈X,then T admits a unique fixed point. This classical theorem has been applied to many problems in physics, optimization, and economics. There is no surprise why many researchers have attempted to extend the theorem to the general setting. One way is to replace the contraction condition the weaker contractive conditions (see [1–3]). Another approach is to give the alternative metric structure on the space (see [4–6]).Later, Bhaskar and Lakshmikantham [7] established some coupled fixed point theorems for a map F:X×X⟶X. They proved that if X is a partially ordered complete metric space and a map F has a mixed monotone property satisfying a certain contractive condition, then F admits a unique coupled fixed point. After this framework, many researchers have attempted to generalize their results (see [8–15]).In 2017, Robdera [16] introduced the concept of size function topology. This concept allows us to view the topological space as a metric-like space. Our main goal is to establish some coupled fixed point theorems on the topological space equipped with the size function topology. First, we recall some standard notations and definitions for size function topologies given by Robdera.Definition 1.
LetX,τ be a topological space. A function δ:τ⟶0,+∞ is called a topology size function or simply τ-size function if
(i)
U=∅, then δU=0(ii)
maxδU,δV≤δU∪V≤δU+δV for all U,V∈τHis idea behind this definition has arisen from the diameter of an open interval in the set of real numbers.Example 2.
Letℝ be the set of real number. For any open set U⊆ℝ, U can be uniquely decomposed as a disjoint union of open intervals aα,bα, U=∐αaα,bα. Define
(2)δU=supdiamaα,bαforallα,where diamA=supx−yx,y∈A for any A⊆ℝ.Then,δ is a topology size function on X.Definition 3.
LetX,τ be a topological space and δ be a τ-size function. For a∈X and r>0. We define a δ-ball centered at a of radius r to be
(3)Bδa;r=⋃U∈τa∈UandδU<r.
We can verify that the collection of allδ-balls forms a basis for a topology on X. The topology generated by this basis is denoted by τδ. Clearly, τδ⊆τ.In the rest of this paper, we always assume thatX,τ is a topological space, δ is a τ-size function, and ℕ denote the set of positive integers.Definition 4.
(i)
Aτ-size function δ is said to be k-uniform if for all a∈X and r>0, we have that δBδa;r≤rk for some k>1. For example, the topology size function δ, given in Example 2, is a 2-uniform(ii)
A sequencexn in X is said to δ converge to x∈X if for every ε>0, there exists N∈ℕ such that for n≥N, xn∈Bδx;ε. A point x is called a δ-limit of the sequence xn and simply denoted by limn→∞xn=x(iii)
A sequencexn in X is said to be δ-Cauchy if for every ε>0, there exists N∈ℕ such that for n,m≥N, xn∈Bδxm;ε(iv)
A subsetA of X is said to be δ complete if every δ-convergent sequence in Aδ converges to a δ-limit in A. Denote the set of all δ-limits of A by A¯Remark 5 (see [16]).
(i)
Everyδ-convergent sequence is a δ-Cauchy sequence(ii)
Clearly,A⊆A¯, and we obtain that A is δ-complete if and only if A=A¯ if and only if every δ-Cauchy sequence δ converges in A


## 2. Main Results

### 2.1. Fixed Point Theorem

For a given topological spaceX,τ and a τ-size function δ. Let T:X⟶X be a self-map.Definition 6.
A pointx∈X is said to be a fixed point of T if Tx=x.
We denote thenth iteration of T by
(4)T∘T∘T∘⋯∘T⏟n−copiesx=Tnxforallx∈X.A functionψ:0,∞⟶0,∞ is called a contractant if it is increasing and limn→∞ψnt=0 for all t∈0,∞. Note that if ψ is a contractant, then ψ0=0 and ψt<t for all t∈0,∞.For a subsetA of X, define δ∗A=supδUU∈τandU⊂A.Definition 7.
A mapT:X⟶X is called a δ-contraction if there exists a contractant ψ:0,∞⟶0,∞ such that
(5)δ∗TU¯<ψδU,for all U∈τ, δU is finite.The uniqueness for the fixed point of the mapT is relied heavily on the following property of the τ-size function δ.Definition 8.
Aτ-size function δ is said to separate points in X or simply X is δ separated if any two distinct points x and y in X, there exist rx,ry>0 such that Bδx;rx∩Bδy;ry=∅.Example 9.
Consider the discrete spaceX,PX where PX is the power set of X. Define a topology size function δ:PX⟶0,∞ by
(6)δA=AforallA∈PX,where A is the number of elements in A. Clearly, X is δ separated.Observe that if we considerX,PX as the discrete metric space, then we have that diamx=0, but δBδx;2=δx=1 for all x∈X.Robdera proved the extended Matkowski fixed point theorem as follows.Theorem 10 (Robdera M. Theorem 3.1).
LetX,τ be a topological space and δ be a k-uniform τ-size function. Let T:X⟶X be a δ-contraction. Assume that X is δ complete, δ separated, and δX is finite. Then, T admits a unique fixed point.Based on the work of Robdera, the fixed point results were obtained by the fact that the size of the domainδX is finite. In this work, we weaken this condition by introducing the notion of locally finite size as follows.Definition 11.
LetX,τ be a topological space and δ be a τ-size function. We say that X is δ locally finite size if for each x∈X, there exists U∈τ such that x∈U and δU is finite.We can verify that ifδX is finite, then X is δ locally finite size. The discrete space X,PX in Example 9 is an example of a topological size function space which is locally finite size.Theorem 12.
LetX,τ be a topological space and δ be a k-uniform τ-size function. Let T:X⟶X be a δ-contraction. Assume that X is δ complete, δ separated, and δ locally finite size. Then, T admits a unique fixed point.Proof.
Letψ:0,∞⟶0,∞ be a contractant and x0 be an arbitrary point in X. Then, we define a sequence xn in X by setting xn=Tnx0.
First, we will show that the sequencexn is δ-Cauchy. Since X is δ locally finite size, there exists U∈τ such that x0,Tx0∈U and δU is finite, which says δU=δ0.
Hence, we have thatTx0,T2x0∈TU and δ∗TU≤δ∗TU¯<ψδU=ψδ0. This implies that there exists U1∈τ such that Tx0,T2x0∈U1⊆TU and δU1<ψδ0.
Then,T2x0,T3x0∈TU1⊆T2U and δ∗TU1≤δ∗TU1¯<ψ2δU=ψ2δ0.
Continuing in this way, we obtain that forn∈ℕ, Tnx0,Tn+1x0∈TnU and δ∗TnU<ψnδ0.
Givenε>0, then ε>ψε>0. Since ψnδ0⟶0 as n⟶∞, there exists N∈ℕ such that for n≥N, we have δ∗TnU<ε−ψε and ψnδ0<ψε.
Hence, forn,m≥N, we have that Tnx0 and Tmx0 are in TnU∪TmU and
(7)δ∗TnU∪TmU≤δ∗TnU+δ∗TmU≤ψnδ0+ε−ψε<ε.
This implies thatTmx0∈BδTnx0;ε. Thus, xn=Tnx0 is δ-Cauchy. Since X is δ complete, there exists x∞∈X such that Tnx0δ converges to x∞.
Next, we will show thatTx∞=x∞. Given ε>0, since T is continuous (see [11]), there exists r>0 such that x∈Bδx∞;r implies that Tx∈BδTx∞;ε/2. Setting η=minε/2,r/2. Since Tnx∞⟶x∞ as n⟶∞, there exists M∈ℕ such that for n≥M, Tnx∞∈Bδx∞;η.
In particular,TMx∞∈Bδx∞;η implies that there exists U′∈τ such that TMx∞,x∞∈U′ and δU′<η. By the continuity of T, we have that TM+1x∞,Tx∞∈TU′ and δ∗TU′<ε/2.
Hence,x∞,Tx∞∈U′∪TU′ and
(8)δ∗U′∪TU′≤δ∗U′+δ∗TU′≤η+ε2<ε.
Therefore,Tx∞∈Bδx∞;ε for every ε>0. Hence, Tx∞=x∞.
Finally, we will show thatx∞ is the unique fixed point of T. Assume that there exists y∞∈X such that Ty∞=y∞. Suppose that x∞≠y∞. Since X is δ separated, there exists ε>0 such that Bδx∞;ε∩Bδy∞;ε=∅. Obviously, y∞∉Bδx∞;ε. Observe that Tx∞,Ty∞∈TBδx∞;ε/2k∪TBδy∞;ε/2k and
(9)δ∗TBδx∞;ε2k∪TBδy∞;ε2k≤δ∗TBδx∞;ε2k+δ∗TBδy∞;ε2k<ψδBδx∞;ε2k+ψδBδy∞;ε2k<ψε2+ψε2<ε.
Hence,y∞=Ty∞∈Bδx∞;ε which is a contradiction. Therefore, x∞=y∞. The proof is complete.Notice that the mapψ:0,∞⟶0,∞ is defined by ψt=αt, where α∈0,1 is a contractant. Immediately, we obtain the extension of the Banach contraction mapping theorem as follows.Corollary 13.
LetX,τ be a topological space and δ be a k-uniform τ-size function. Assume that there exists α∈0,1 such that
(10)δ∗TU¯<αδU,for all U∈τ, δU is finite. If X is δ complete, δ separated, and δ locally finite size, then T admits a unique fixed point.


### 2.2. Coupled Fixed Point Theorem

LetX,τ be a topological space and δ be a τ-size function. Let F:X×X⟶X be a map.Definition 14.
(i)
A pointx,y∈X×X is called a coupled fixed point of F if Fx,y=x and Fy,x=y.(ii)
A pointx,x∈X×X is called a fixed point of F if Fx,x=x. Clearly, a fixed point of F is a coupled fixed point of .Definition 15.
A mapF:X×X⟶X is said to satisfy the product contraction I if there exists a contractant ψ:0,∞⟶0,∞ such that
(11)δ∗FU×V¯<ψminδU,δV,for all U,V∈τ, δU and δV are finite.Example 16.
LetX=0,∞⊂ℝ. Let τ be the topology on X given by
(12)τ=∅,X,0,aa>0.Define the topology size functionδ by δ∅=0, δX=∞, and δ0,a=a.LetF:0,∞×0,∞⟶0,∞ be defined by
(13)Fx,y=1,for all x,y∈0,∞.Then, fora,b>0, we have that for any contractant ψ:0,∞⟶0,∞(14)0=δ∗F0,a×0,b¯<ψminδ0,a,δ0,b=ψmina,b.Therefore,F satisfies the product contraction I. Observe that F admits a unique coupled fixed point 1,1.Remark 17.
Observe that we can naturally define the product topology onX×X inherited from X. More precisely, the basis element is of the form U×V where U,V∈τδ. But it is still questionable to define the size function corresponding to this product topology in term of δ, because the union of two basis elements might not be a product form. Nevertheless, the following lemma can be obtained by the product contraction I.Lemma 18.
LetX,τ be a topological space and δ be a τ-size function. Let F:X×X⟶X be a map satisfying the product contraction I. Then, for every x,y∈X×X and ε>0, there exists r>0 such that for U,V∈τ, if x∈U,y∈V and minδU,δV<r, then δ∗FU×V<ε.Proof.
Letx,y∈X×X and ε>0. Choose r=ε. Then, for U,V∈τ, if x∈U,y∈V and minδU,δV<r, we have that Fx,y∈FU×V and
(15)δ∗FU×V≤δ∗FU×V¯<ψminδU,δV<ψε<ε.Theorem 19.
LetX,τ be a topological space and δ be a k-uniform τ-size function. Let F:X×X⟶X be a map satisfying the product contraction I. Assume that X is δ complete, δ separated, and δ locally finite size. Then, F has a unique coupled fixed point.Proof.
Letψ:0,∞⟶0,∞ be a contractant and choose an arbitrary point x0,y0∈X×X. Since X is δ locally finite size, there exist U0,V0∈τ such that x0∈U0,y0∈V0,δU0 and δV0 are finite. Setting δU0=α0 and δV0=β0.
We inductively define two sequences inX as follows:
(16)x1=Fx0,y0,y1=Fy0,x0,xn=Fxn−1,yn−1,yn=Fyn−1,xn−1,foralln≥2.
Next, we will show that bothxn and yn are δ-Cauchy. Clearly, x1∈FU0×V0 and δ∗FU0×V0<ψα0. Similarly, y1∈FV0×U0 and δ∗FV0×U0<ψβ0.
Hence, there existU1,V1∈τ such that
(17)x1∈U1⊆FU0×V0,δU1<ψα0,y1∈V1⊆FV0×U0,δV1<ψβ0.
We obtain thatx2=Fx1,y1∈FU1×V1 and
(18)δ∗FU1×V1<ψminδU1,δV1<ψδU1<ψ2α0.
Similarly,y2∈FV1×U1 and δ∗FV1×U1<ψ2β0.
Inductively, we obtain that forn∈ℕ there exist Un−1,Vn−1∈τ such that
(19)xn∈FUn−1×Vn−1,δ∗FUn−1×Vn−1<ψnα0,yn∈FVn−1×Un−1,
δ∗FVn−1×Un−1<ψnβ0.
Givenε>0, then ψε<ε or ε−ψε>0. Since limn⟶∞ψnα0=0, there exist N1,N2∈ℕ such that for n≥N1 implies that ψnα0<ψε and for n≥N2 implies that δ∗FUn−1×Vn−1<ε−ψε.
Forn,m≥N=maxN1,N2, we have that xn,xm∈FUn−1×Vn−1∪FUm−1×Vm−1 and
(20)δ∗FUn−1×Vn−1∪FUm−1×Vm−1≤δ∗FUn−1×Vn−1+δ∗FUm−1×Vm−1<ε−ψε+ψmα0<ε.
This means thatxn∈Bδxm;ε, so xn is δ-Cauchy. By the analogous argument, we can show that yn is also δ-Cauchy.
By theδ-completeness of X, there exist x∞ and y∞ in X such that limn⟶∞xn=x∞ and limn⟶∞yn=y∞.
Next, we will show thatx∞,y∞ is a coupled fixed point or Fx∞,y∞=x∞ and Fy∞,x∞=y∞.
Givenε>0. By Lemma 18, there is r>0 such that for U,V∈τ if x∞,y∞∈U×V and minδU,δV<r, then δ∗FU×V≤ε/2.
Sincelimn⟶∞xn=x∞ and limn⟶∞yn=y∞, there exist M1,M2∈ℕ such that
(21)n>M1impliesthatxn∈Bδx∞;η,n>M2impliesthatyn∈Bδy∞;η,where η=minε/2,r/2.
In particular, there existU′,V′∈τ such that xM1,x∞∈U′, yM2,y∞∈V′, and δU′,δV′<η≤r/2. Then, minδU′,δV′<r.
Hence, we have thatFx∞,y∞∈FU′×V′ and δ∗FU′×V′<ε/2.
Thus,x∞,Fx∞,y∞∈U′∪FU′×V′ and
(22)δ∗U′∪FU′×V′≤δ∗U′+δ∗FU′×V′<η+ε2<ε.
That isFx∞,y∞∈Bδx∞;ε for all ε>0. Hence, Fx∞,y∞=x∞. Similarly, we can show that Fy∞,x∞=y∞.
Finally, we will show thatx∞,y∞ is the unique coupled fixed point of F. Suppose that there is x′∞,y′∞∈X×X such that Fx∞′,y∞′=x′∞ and Fy∞′,x∞′=y′∞. Suppose that x∞,y∞≠x∞′,y∞′. Then, x∞≠x′∞ or y∞≠y′∞.
Assume thatx∞≠x′∞. Since X is δ separated and k-uniform, there exists ε>0 such that Bδx∞;ε∩Bδx′∞;ε=∅. Clearly, x′∞∉Bδx∞;ε. Since X is δ locally finite size, there exist V,V′∈τ such that y∞∈V,y′∞∈V′ and δV,δV′ are finite. We have that Fx∞,y∞,Fx∞′,y∞′∈FBδx∞;ε/2k×V∪FBδx∞′;ε/2k×V′ and
(23)δ∗FBδx∞;ε2k×V∪FBδx∞′;ε2k×V′≤δ∗FBδx∞;ε2k×V+δ∗FBδx∞′;ε2k×V′<ψε2+ψε2<ε.
This implies thatx′∞=Fx′∞,y′∞∈Bδx∞;ε which is a contradiction.
Similarly, if we assume thaty∞≠y′∞, then it will also lead to the contradiction. Therefore, x∞,y∞=x′∞,y′∞. The proof is complete.Remark 20.
Observe that in the proof of Theorem19, we can replace the product contraction I by the weaker condition:
(24)δ∗FU×V¯<12ψδU+ψδVforallU,V∈τ,if the contractant ψ is assumed to be a linear function. By the Cauchy functional equation, this implies that ψt=λt for some 0<λ<1.Definition 21.
A mapF:X×X⟶X is said to satisfy the product contraction II if there exists λ∈0,1 such that
(25)δ∗FU×V¯≤λ2δU+δV,for all U,V∈τ, δU and δV are finite.Example 22.
LetX=0,∞⊂ℝ. Let τ be the topology on X given by
(26)τ=∅,X,0,aa>0.Define the topology size functionδ by δ∅=0,δX=∞ and δ0,a=a.LetF:0,∞×0,∞⟶0,∞ be defined by
(27)Fx,y=x+y6,for all x,y∈0,∞.Then, fora,b>0,
(28)a+b6=δ∗F0,a×0,b¯≤λ2δ0,a+δ0,b=λ2a+b,where λ=1/3.Therefore,F satisfies the product contraction II. Observe that F has a unique coupled fixed point 0,0.Before we prove the next main theorem, the following lemma is required.Lemma 23.
LetX,τ be a topological space and δ be a τ-size function. Let F:X×X⟶X be a map satisfying the product contraction II. Then, for every x,y∈X×X and ε>0, there exists r>0 such that for U,V∈τ, if x∈U,y∈V and δU+δV<r, then δ∗FU×V<ε.Proof.
Letx,y∈X×X and ε>0. Choose r=2ε/λ. Then, for U,V∈τ, if x∈U,y∈V and δU+δV<r, we have that Fx,y∈FU×V and
(29)δ∗FU×V≤δ∗FU×V¯<λ2δU+δV<λ22ελ=ε.Theorem 24.
LetX,τ be a topological space and δ be a k-uniform τ-size function. Let F:X×X⟶X be a map satisfying the product contraction II. If X is δ complete, δ separated, and δ locally finite size. Then, F admits a unique coupled fixed point.Proof.
Letx0,y0∈X×X be an arbitrary point. Since X is δ locally finite size, there exist U0,V0∈τ such that x0∈U0,y0∈V0,δU0 and δV0 are finite. Setting δU0=α0 and δV0=β0 and let α0+β0=γ0.
We inductively define two sequences inX as follows:
(30)x1=Fx0,y0,y1=Fy0,x0,xn=Fxn−1,yn−1,yn=Fyn−1,xn−1foralln≥2.
Next, we will show that bothxn and yn are δ-Cauchy. Clearly, x1∈FU0×V0 and δ∗FU0×V0<λ/2γ0. Similarly, y1∈FV0×U0 and δ∗FV0×U0<λ/2γ0.
Hence, there existU1,V1∈τ such that
(31)x1∈U1⊆FU0×V0,δU1<λ2γ0,y1∈V1⊆FV0×U0,δV1<λ2γ0.
We obtain thatx2=Fx1,y1∈FU1×V1 and
(32)δ∗FU1×V1≤δ∗FU1×V1¯=λ/22γ0.
Similarly,y2∈FV1×U1 and δ∗FV1×U1<λ/22γ0.
Continuing in this manner, we obtain that forn∈ℕ there exist Un−1,Vn−1∈τ such that
(33)xn∈FUn−1×Vn−1,δ∗FUn−1×Vn−1<λ2nγ0,yn∈FVn−1×Un−1,δ∗FVn−1×Un−1<λ2nγ0.
Givenε>0, then λε<ε or ε1−λ>0. Since limn⟶∞λ/2nγ0=0, there exist N1,N2∈ℕ such that for n≥N1 implies that λ/2nγ0<λε and for n≥N2 implies that δ∗FUn−1×Vn−1<ε1−λ.
Forn,m≥N=maxN1,N2, we have that xn,xm∈FUn−1×Vn−1∪FUm−1×Vm−1 and
(34)δ∗FUn−1×Vn−1∪FUm−1×Vm−1≤δ∗FUn−1×Vn−1+δ∗FUm−1×Vm−1<ε1−λ+λ2mγ0<ε.
This meansxn∈Bδxm;ε, so xn is δ-Cauchy. By the similar argument, we can show that yn is also δ-Cauchy.
By theδ-completeness of X, there exist x∞ and y∞ in X such that limn⟶∞xn=x∞ and limn⟶∞yn=y∞.
Next, we will show thatx∞,y∞ is a coupled fixed point or Fx∞,y∞=x∞ and Fy∞,x∞=y∞.
Givenε>0, by Lemma 23, there is r>0 such that for U,V∈τ if x∞,y∞∈U×V and δU+δV<r, then δ∗FU×V<ε/2.
Sincelimn⟶∞xn=x∞ and limn⟶∞yn=y∞, there exist M1,M2∈ℕ such that
(35)n>M1impliesthatxn∈Bδx∞;η,n>M2impliesthatyn∈Bδy∞;η,where η=minε/2,r/2.
In particular, there existU′,V′∈τ such that xM1,x∞∈U′, yM2,y∞∈V′, and δU′,δV′<η≤r/2. Then, δU′+δV′<r.
Hence, we have thatFx∞,y∞∈FU′×V′ and δ∗FU′×V′<ε/2.
Thus,x∞,Fx∞,y∞∈U′∪FU′×V′ and
(36)δ∗U′∪FU′×V′≤δ∗U′+δ∗FU′×V′<η+ε2<ε.
That isFx∞,y∞∈Bδx∞;ε for all ε>0. Hence, Fx∞,y∞=x∞. Similarly, we can show that Fy∞,x∞=y∞.
Finally, we will show thatx∞,y∞ is the unique coupled fixed point of F. Suppose that there is x′∞,y′∞∈X×X such that Fx′∞,y′∞=x′∞ and Fy′∞,x′∞=y′∞. Suppose that x∞,y∞≠x′∞,y′∞. Then, x∞≠x′∞ or y∞≠y′∞.
Assume thatx∞≠x′∞. Since X is δ separated and k-uniform, there exists ε1>0 such that Bδx∞;ε1∩Bδx′∞;ε1=∅. Clearly, x′∞∉Bδx∞;ε1.Case 1.
Ify∞≠y′∞, then there exists ε2>0 such that Bδy∞;ε2∩Bδy′∞;ε2=∅. Setting ε3=minε1,ε2. We have that Fx∞,y∞,Fx∞′,y∞′∈FBδx∞;ε3/4k×Bδy∞;ε3/4k∪FBδx∞′;ε3/4k×Bδy∞′;ε3/4k and
(37)δ∗FBδx∞;ε34k×Bδy∞;ε34k∪FBδx∞′;ε32k×Bδy∞′;ε34k≤δ∗FBδx∞;ε34k×Bδy∞;ε34k+δ∗FBδx∞′;ε34k×Bδy∞′;ε34k<λ2δBδx∞;ε34k+δBδy∞;ε34k+λ2δBδx∞′;ε34k+δBδy∞′;ε34k<λ2ε32+ε32<ε3.This implies thatx′∞=Fx∞′,y∞′∈Bδx∞;ε3⊂Bδx∞;ε1 which is a contradiction.Case 2.
Ify∞=y′∞, then there exists V∈τ such that y∞=y′∞∈V and δV is finite. Setting ε4=minε1,δV. We have that Fx∞,y∞,Fx∞′,y∞′∈FBδx∞;ε4/4k×Bδy∞;ε4/4k∪FBδx∞′;ε4/4k×Bδy∞′;ε4/4k and we can similarly show that
(38)δ∗FBδx∞;ε44k×Bδy∞;ε44k∪FBδx∞′;ε44k×Bδy∞′;ε44k<ε4.
This implies thatx′∞=Fx∞′,y∞′∈Bδx∞;ε4⊂Bδx∞;ε1 which is a contradiction.Similarly, if we assume thaty∞≠y′∞, then it will also lead to the contradiction. Hence, x∞,y∞=x∞′,y∞′. The proof is complete.If spaceX failed to be separated by the topology size function δ, then the map F might not have a unique coupled fixed point.Example 25.
Consider the closed interval0,1. Let τ be the topology on 0,1 given by
(39)τ=∅,0,1,0,aa<1.
Define the topology size functionδ by δ∅=0, δ0,1=1, and δ0,a=a. Clearly, δ does not separate the points on 0,1.
LetF:0,1×0,1⟶0,1 be defined by
(40)Fx,y=x2+y2,for all x,y∈0,1. We obtain that F satisfies the product contraction II. The coupled fixed points of F are 0,0 and 1/2,1/2.


## 2.1. Fixed Point Theorem

For a given topological spaceX,τ and a τ-size function δ. Let T:X⟶X be a self-map.Definition 6.
A pointx∈X is said to be a fixed point of T if Tx=x.
We denote thenth iteration of T by
(4)T∘T∘T∘⋯∘T⏟n−copiesx=Tnxforallx∈X.A functionψ:0,∞⟶0,∞ is called a contractant if it is increasing and limn→∞ψnt=0 for all t∈0,∞. Note that if ψ is a contractant, then ψ0=0 and ψt<t for all t∈0,∞.For a subsetA of X, define δ∗A=supδUU∈τandU⊂A.Definition 7.
A mapT:X⟶X is called a δ-contraction if there exists a contractant ψ:0,∞⟶0,∞ such that
(5)δ∗TU¯<ψδU,for all U∈τ, δU is finite.The uniqueness for the fixed point of the mapT is relied heavily on the following property of the τ-size function δ.Definition 8.
Aτ-size function δ is said to separate points in X or simply X is δ separated if any two distinct points x and y in X, there exist rx,ry>0 such that Bδx;rx∩Bδy;ry=∅.Example 9.
Consider the discrete spaceX,PX where PX is the power set of X. Define a topology size function δ:PX⟶0,∞ by
(6)δA=AforallA∈PX,where A is the number of elements in A. Clearly, X is δ separated.Observe that if we considerX,PX as the discrete metric space, then we have that diamx=0, but δBδx;2=δx=1 for all x∈X.Robdera proved the extended Matkowski fixed point theorem as follows.Theorem 10 (Robdera M. Theorem 3.1).
LetX,τ be a topological space and δ be a k-uniform τ-size function. Let T:X⟶X be a δ-contraction. Assume that X is δ complete, δ separated, and δX is finite. Then, T admits a unique fixed point.Based on the work of Robdera, the fixed point results were obtained by the fact that the size of the domainδX is finite. In this work, we weaken this condition by introducing the notion of locally finite size as follows.Definition 11.
LetX,τ be a topological space and δ be a τ-size function. We say that X is δ locally finite size if for each x∈X, there exists U∈τ such that x∈U and δU is finite.We can verify that ifδX is finite, then X is δ locally finite size. The discrete space X,PX in Example 9 is an example of a topological size function space which is locally finite size.Theorem 12.
LetX,τ be a topological space and δ be a k-uniform τ-size function. Let T:X⟶X be a δ-contraction. Assume that X is δ complete, δ separated, and δ locally finite size. Then, T admits a unique fixed point.Proof.
Letψ:0,∞⟶0,∞ be a contractant and x0 be an arbitrary point in X. Then, we define a sequence xn in X by setting xn=Tnx0.
First, we will show that the sequencexn is δ-Cauchy. Since X is δ locally finite size, there exists U∈τ such that x0,Tx0∈U and δU is finite, which says δU=δ0.
Hence, we have thatTx0,T2x0∈TU and δ∗TU≤δ∗TU¯<ψδU=ψδ0. This implies that there exists U1∈τ such that Tx0,T2x0∈U1⊆TU and δU1<ψδ0.
Then,T2x0,T3x0∈TU1⊆T2U and δ∗TU1≤δ∗TU1¯<ψ2δU=ψ2δ0.
Continuing in this way, we obtain that forn∈ℕ, Tnx0,Tn+1x0∈TnU and δ∗TnU<ψnδ0.
Givenε>0, then ε>ψε>0. Since ψnδ0⟶0 as n⟶∞, there exists N∈ℕ such that for n≥N, we have δ∗TnU<ε−ψε and ψnδ0<ψε.
Hence, forn,m≥N, we have that Tnx0 and Tmx0 are in TnU∪TmU and
(7)δ∗TnU∪TmU≤δ∗TnU+δ∗TmU≤ψnδ0+ε−ψε<ε.
This implies thatTmx0∈BδTnx0;ε. Thus, xn=Tnx0 is δ-Cauchy. Since X is δ complete, there exists x∞∈X such that Tnx0δ converges to x∞.
Next, we will show thatTx∞=x∞. Given ε>0, since T is continuous (see [11]), there exists r>0 such that x∈Bδx∞;r implies that Tx∈BδTx∞;ε/2. Setting η=minε/2,r/2. Since Tnx∞⟶x∞ as n⟶∞, there exists M∈ℕ such that for n≥M, Tnx∞∈Bδx∞;η.
In particular,TMx∞∈Bδx∞;η implies that there exists U′∈τ such that TMx∞,x∞∈U′ and δU′<η. By the continuity of T, we have that TM+1x∞,Tx∞∈TU′ and δ∗TU′<ε/2.
Hence,x∞,Tx∞∈U′∪TU′ and
(8)δ∗U′∪TU′≤δ∗U′+δ∗TU′≤η+ε2<ε.
Therefore,Tx∞∈Bδx∞;ε for every ε>0. Hence, Tx∞=x∞.
Finally, we will show thatx∞ is the unique fixed point of T. Assume that there exists y∞∈X such that Ty∞=y∞. Suppose that x∞≠y∞. Since X is δ separated, there exists ε>0 such that Bδx∞;ε∩Bδy∞;ε=∅. Obviously, y∞∉Bδx∞;ε. Observe that Tx∞,Ty∞∈TBδx∞;ε/2k∪TBδy∞;ε/2k and
(9)δ∗TBδx∞;ε2k∪TBδy∞;ε2k≤δ∗TBδx∞;ε2k+δ∗TBδy∞;ε2k<ψδBδx∞;ε2k+ψδBδy∞;ε2k<ψε2+ψε2<ε.
Hence,y∞=Ty∞∈Bδx∞;ε which is a contradiction. Therefore, x∞=y∞. The proof is complete.Notice that the mapψ:0,∞⟶0,∞ is defined by ψt=αt, where α∈0,1 is a contractant. Immediately, we obtain the extension of the Banach contraction mapping theorem as follows.Corollary 13.
LetX,τ be a topological space and δ be a k-uniform τ-size function. Assume that there exists α∈0,1 such that
(10)δ∗TU¯<αδU,for all U∈τ, δU is finite. If X is δ complete, δ separated, and δ locally finite size, then T admits a unique fixed point.


## 2.2. Coupled Fixed Point Theorem

LetX,τ be a topological space and δ be a τ-size function. Let F:X×X⟶X be a map.Definition 14.
(i)
A pointx,y∈X×X is called a coupled fixed point of F if Fx,y=x and Fy,x=y.(ii)
A pointx,x∈X×X is called a fixed point of F if Fx,x=x. Clearly, a fixed point of F is a coupled fixed point of .Definition 15.
A mapF:X×X⟶X is said to satisfy the product contraction I if there exists a contractant ψ:0,∞⟶0,∞ such that
(11)δ∗FU×V¯<ψminδU,δV,for all U,V∈τ, δU and δV are finite.Example 16.
LetX=0,∞⊂ℝ. Let τ be the topology on X given by
(12)τ=∅,X,0,aa>0.Define the topology size functionδ by δ∅=0, δX=∞, and δ0,a=a.LetF:0,∞×0,∞⟶0,∞ be defined by
(13)Fx,y=1,for all x,y∈0,∞.Then, fora,b>0, we have that for any contractant ψ:0,∞⟶0,∞(14)0=δ∗F0,a×0,b¯<ψminδ0,a,δ0,b=ψmina,b.Therefore,F satisfies the product contraction I. Observe that F admits a unique coupled fixed point 1,1.Remark 17.
Observe that we can naturally define the product topology onX×X inherited from X. More precisely, the basis element is of the form U×V where U,V∈τδ. But it is still questionable to define the size function corresponding to this product topology in term of δ, because the union of two basis elements might not be a product form. Nevertheless, the following lemma can be obtained by the product contraction I.Lemma 18.
LetX,τ be a topological space and δ be a τ-size function. Let F:X×X⟶X be a map satisfying the product contraction I. Then, for every x,y∈X×X and ε>0, there exists r>0 such that for U,V∈τ, if x∈U,y∈V and minδU,δV<r, then δ∗FU×V<ε.Proof.
Letx,y∈X×X and ε>0. Choose r=ε. Then, for U,V∈τ, if x∈U,y∈V and minδU,δV<r, we have that Fx,y∈FU×V and
(15)δ∗FU×V≤δ∗FU×V¯<ψminδU,δV<ψε<ε.Theorem 19.
LetX,τ be a topological space and δ be a k-uniform τ-size function. Let F:X×X⟶X be a map satisfying the product contraction I. Assume that X is δ complete, δ separated, and δ locally finite size. Then, F has a unique coupled fixed point.Proof.
Letψ:0,∞⟶0,∞ be a contractant and choose an arbitrary point x0,y0∈X×X. Since X is δ locally finite size, there exist U0,V0∈τ such that x0∈U0,y0∈V0,δU0 and δV0 are finite. Setting δU0=α0 and δV0=β0.
We inductively define two sequences inX as follows:
(16)x1=Fx0,y0,y1=Fy0,x0,xn=Fxn−1,yn−1,yn=Fyn−1,xn−1,foralln≥2.
Next, we will show that bothxn and yn are δ-Cauchy. Clearly, x1∈FU0×V0 and δ∗FU0×V0<ψα0. Similarly, y1∈FV0×U0 and δ∗FV0×U0<ψβ0.
Hence, there existU1,V1∈τ such that
(17)x1∈U1⊆FU0×V0,δU1<ψα0,y1∈V1⊆FV0×U0,δV1<ψβ0.
We obtain thatx2=Fx1,y1∈FU1×V1 and
(18)δ∗FU1×V1<ψminδU1,δV1<ψδU1<ψ2α0.
Similarly,y2∈FV1×U1 and δ∗FV1×U1<ψ2β0.
Inductively, we obtain that forn∈ℕ there exist Un−1,Vn−1∈τ such that
(19)xn∈FUn−1×Vn−1,δ∗FUn−1×Vn−1<ψnα0,yn∈FVn−1×Un−1,
δ∗FVn−1×Un−1<ψnβ0.
Givenε>0, then ψε<ε or ε−ψε>0. Since limn⟶∞ψnα0=0, there exist N1,N2∈ℕ such that for n≥N1 implies that ψnα0<ψε and for n≥N2 implies that δ∗FUn−1×Vn−1<ε−ψε.
Forn,m≥N=maxN1,N2, we have that xn,xm∈FUn−1×Vn−1∪FUm−1×Vm−1 and
(20)δ∗FUn−1×Vn−1∪FUm−1×Vm−1≤δ∗FUn−1×Vn−1+δ∗FUm−1×Vm−1<ε−ψε+ψmα0<ε.
This means thatxn∈Bδxm;ε, so xn is δ-Cauchy. By the analogous argument, we can show that yn is also δ-Cauchy.
By theδ-completeness of X, there exist x∞ and y∞ in X such that limn⟶∞xn=x∞ and limn⟶∞yn=y∞.
Next, we will show thatx∞,y∞ is a coupled fixed point or Fx∞,y∞=x∞ and Fy∞,x∞=y∞.
Givenε>0. By Lemma 18, there is r>0 such that for U,V∈τ if x∞,y∞∈U×V and minδU,δV<r, then δ∗FU×V≤ε/2.
Sincelimn⟶∞xn=x∞ and limn⟶∞yn=y∞, there exist M1,M2∈ℕ such that
(21)n>M1impliesthatxn∈Bδx∞;η,n>M2impliesthatyn∈Bδy∞;η,where η=minε/2,r/2.
In particular, there existU′,V′∈τ such that xM1,x∞∈U′, yM2,y∞∈V′, and δU′,δV′<η≤r/2. Then, minδU′,δV′<r.
Hence, we have thatFx∞,y∞∈FU′×V′ and δ∗FU′×V′<ε/2.
Thus,x∞,Fx∞,y∞∈U′∪FU′×V′ and
(22)δ∗U′∪FU′×V′≤δ∗U′+δ∗FU′×V′<η+ε2<ε.
That isFx∞,y∞∈Bδx∞;ε for all ε>0. Hence, Fx∞,y∞=x∞. Similarly, we can show that Fy∞,x∞=y∞.
Finally, we will show thatx∞,y∞ is the unique coupled fixed point of F. Suppose that there is x′∞,y′∞∈X×X such that Fx∞′,y∞′=x′∞ and Fy∞′,x∞′=y′∞. Suppose that x∞,y∞≠x∞′,y∞′. Then, x∞≠x′∞ or y∞≠y′∞.
Assume thatx∞≠x′∞. Since X is δ separated and k-uniform, there exists ε>0 such that Bδx∞;ε∩Bδx′∞;ε=∅. Clearly, x′∞∉Bδx∞;ε. Since X is δ locally finite size, there exist V,V′∈τ such that y∞∈V,y′∞∈V′ and δV,δV′ are finite. We have that Fx∞,y∞,Fx∞′,y∞′∈FBδx∞;ε/2k×V∪FBδx∞′;ε/2k×V′ and
(23)δ∗FBδx∞;ε2k×V∪FBδx∞′;ε2k×V′≤δ∗FBδx∞;ε2k×V+δ∗FBδx∞′;ε2k×V′<ψε2+ψε2<ε.
This implies thatx′∞=Fx′∞,y′∞∈Bδx∞;ε which is a contradiction.
Similarly, if we assume thaty∞≠y′∞, then it will also lead to the contradiction. Therefore, x∞,y∞=x′∞,y′∞. The proof is complete.Remark 20.
Observe that in the proof of Theorem19, we can replace the product contraction I by the weaker condition:
(24)δ∗FU×V¯<12ψδU+ψδVforallU,V∈τ,if the contractant ψ is assumed to be a linear function. By the Cauchy functional equation, this implies that ψt=λt for some 0<λ<1.Definition 21.
A mapF:X×X⟶X is said to satisfy the product contraction II if there exists λ∈0,1 such that
(25)δ∗FU×V¯≤λ2δU+δV,for all U,V∈τ, δU and δV are finite.Example 22.
LetX=0,∞⊂ℝ. Let τ be the topology on X given by
(26)τ=∅,X,0,aa>0.Define the topology size functionδ by δ∅=0,δX=∞ and δ0,a=a.LetF:0,∞×0,∞⟶0,∞ be defined by
(27)Fx,y=x+y6,for all x,y∈0,∞.Then, fora,b>0,
(28)a+b6=δ∗F0,a×0,b¯≤λ2δ0,a+δ0,b=λ2a+b,where λ=1/3.Therefore,F satisfies the product contraction II. Observe that F has a unique coupled fixed point 0,0.Before we prove the next main theorem, the following lemma is required.Lemma 23.
LetX,τ be a topological space and δ be a τ-size function. Let F:X×X⟶X be a map satisfying the product contraction II. Then, for every x,y∈X×X and ε>0, there exists r>0 such that for U,V∈τ, if x∈U,y∈V and δU+δV<r, then δ∗FU×V<ε.Proof.
Letx,y∈X×X and ε>0. Choose r=2ε/λ. Then, for U,V∈τ, if x∈U,y∈V and δU+δV<r, we have that Fx,y∈FU×V and
(29)δ∗FU×V≤δ∗FU×V¯<λ2δU+δV<λ22ελ=ε.Theorem 24.
LetX,τ be a topological space and δ be a k-uniform τ-size function. Let F:X×X⟶X be a map satisfying the product contraction II. If X is δ complete, δ separated, and δ locally finite size. Then, F admits a unique coupled fixed point.Proof.
Letx0,y0∈X×X be an arbitrary point. Since X is δ locally finite size, there exist U0,V0∈τ such that x0∈U0,y0∈V0,δU0 and δV0 are finite. Setting δU0=α0 and δV0=β0 and let α0+β0=γ0.
We inductively define two sequences inX as follows:
(30)x1=Fx0,y0,y1=Fy0,x0,xn=Fxn−1,yn−1,yn=Fyn−1,xn−1foralln≥2.
Next, we will show that bothxn and yn are δ-Cauchy. Clearly, x1∈FU0×V0 and δ∗FU0×V0<λ/2γ0. Similarly, y1∈FV0×U0 and δ∗FV0×U0<λ/2γ0.
Hence, there existU1,V1∈τ such that
(31)x1∈U1⊆FU0×V0,δU1<λ2γ0,y1∈V1⊆FV0×U0,δV1<λ2γ0.
We obtain thatx2=Fx1,y1∈FU1×V1 and
(32)δ∗FU1×V1≤δ∗FU1×V1¯=λ/22γ0.
Similarly,y2∈FV1×U1 and δ∗FV1×U1<λ/22γ0.
Continuing in this manner, we obtain that forn∈ℕ there exist Un−1,Vn−1∈τ such that
(33)xn∈FUn−1×Vn−1,δ∗FUn−1×Vn−1<λ2nγ0,yn∈FVn−1×Un−1,δ∗FVn−1×Un−1<λ2nγ0.
Givenε>0, then λε<ε or ε1−λ>0. Since limn⟶∞λ/2nγ0=0, there exist N1,N2∈ℕ such that for n≥N1 implies that λ/2nγ0<λε and for n≥N2 implies that δ∗FUn−1×Vn−1<ε1−λ.
Forn,m≥N=maxN1,N2, we have that xn,xm∈FUn−1×Vn−1∪FUm−1×Vm−1 and
(34)δ∗FUn−1×Vn−1∪FUm−1×Vm−1≤δ∗FUn−1×Vn−1+δ∗FUm−1×Vm−1<ε1−λ+λ2mγ0<ε.
This meansxn∈Bδxm;ε, so xn is δ-Cauchy. By the similar argument, we can show that yn is also δ-Cauchy.
By theδ-completeness of X, there exist x∞ and y∞ in X such that limn⟶∞xn=x∞ and limn⟶∞yn=y∞.
Next, we will show thatx∞,y∞ is a coupled fixed point or Fx∞,y∞=x∞ and Fy∞,x∞=y∞.
Givenε>0, by Lemma 23, there is r>0 such that for U,V∈τ if x∞,y∞∈U×V and δU+δV<r, then δ∗FU×V<ε/2.
Sincelimn⟶∞xn=x∞ and limn⟶∞yn=y∞, there exist M1,M2∈ℕ such that
(35)n>M1impliesthatxn∈Bδx∞;η,n>M2impliesthatyn∈Bδy∞;η,where η=minε/2,r/2.
In particular, there existU′,V′∈τ such that xM1,x∞∈U′, yM2,y∞∈V′, and δU′,δV′<η≤r/2. Then, δU′+δV′<r.
Hence, we have thatFx∞,y∞∈FU′×V′ and δ∗FU′×V′<ε/2.
Thus,x∞,Fx∞,y∞∈U′∪FU′×V′ and
(36)δ∗U′∪FU′×V′≤δ∗U′+δ∗FU′×V′<η+ε2<ε.
That isFx∞,y∞∈Bδx∞;ε for all ε>0. Hence, Fx∞,y∞=x∞. Similarly, we can show that Fy∞,x∞=y∞.
Finally, we will show thatx∞,y∞ is the unique coupled fixed point of F. Suppose that there is x′∞,y′∞∈X×X such that Fx′∞,y′∞=x′∞ and Fy′∞,x′∞=y′∞. Suppose that x∞,y∞≠x′∞,y′∞. Then, x∞≠x′∞ or y∞≠y′∞.
Assume thatx∞≠x′∞. Since X is δ separated and k-uniform, there exists ε1>0 such that Bδx∞;ε1∩Bδx′∞;ε1=∅. Clearly, x′∞∉Bδx∞;ε1.Case 1.
Ify∞≠y′∞, then there exists ε2>0 such that Bδy∞;ε2∩Bδy′∞;ε2=∅. Setting ε3=minε1,ε2. We have that Fx∞,y∞,Fx∞′,y∞′∈FBδx∞;ε3/4k×Bδy∞;ε3/4k∪FBδx∞′;ε3/4k×Bδy∞′;ε3/4k and
(37)δ∗FBδx∞;ε34k×Bδy∞;ε34k∪FBδx∞′;ε32k×Bδy∞′;ε34k≤δ∗FBδx∞;ε34k×Bδy∞;ε34k+δ∗FBδx∞′;ε34k×Bδy∞′;ε34k<λ2δBδx∞;ε34k+δBδy∞;ε34k+λ2δBδx∞′;ε34k+δBδy∞′;ε34k<λ2ε32+ε32<ε3.This implies thatx′∞=Fx∞′,y∞′∈Bδx∞;ε3⊂Bδx∞;ε1 which is a contradiction.Case 2.
Ify∞=y′∞, then there exists V∈τ such that y∞=y′∞∈V and δV is finite. Setting ε4=minε1,δV. We have that Fx∞,y∞,Fx∞′,y∞′∈FBδx∞;ε4/4k×Bδy∞;ε4/4k∪FBδx∞′;ε4/4k×Bδy∞′;ε4/4k and we can similarly show that
(38)δ∗FBδx∞;ε44k×Bδy∞;ε44k∪FBδx∞′;ε44k×Bδy∞′;ε44k<ε4.
This implies thatx′∞=Fx∞′,y∞′∈Bδx∞;ε4⊂Bδx∞;ε1 which is a contradiction.Similarly, if we assume thaty∞≠y′∞, then it will also lead to the contradiction. Hence, x∞,y∞=x∞′,y∞′. The proof is complete.If spaceX failed to be separated by the topology size function δ, then the map F might not have a unique coupled fixed point.Example 25.
Consider the closed interval0,1. Let τ be the topology on 0,1 given by
(39)τ=∅,0,1,0,aa<1.
Define the topology size functionδ by δ∅=0, δ0,1=1, and δ0,a=a. Clearly, δ does not separate the points on 0,1.
LetF:0,1×0,1⟶0,1 be defined by
(40)Fx,y=x2+y2,for all x,y∈0,1. We obtain that F satisfies the product contraction II. The coupled fixed points of F are 0,0 and 1/2,1/2.



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