# Completeness ofb∼Metric Spaces and the Fixed Points of Generalized Multivalued Quasicontractions

Authors: Basit Ali; Mujahid Abbas; Manuel de la Sen
Journal: Discrete Dynamics in Nature and Society
 (2020)

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## Abstract

In this article, we present a completeness characterization ofb∼metric space via existence of fixed points of generalized multivalued quasicontractions. The purpose of this paper is twofold: (a) to establish the existence of fixed points of multivalued quasicontractions in the setup of b∼ metric spaces and (b) to establish completeness of a b∼ metric space which is a topological property in nature with existence of fixed points of generalized multivalued quasicontractions. Further, a comparison of our results with comparable results shows that the results obtained herein improve and unify the existing results in the literature applicable to the case where existing results fail.


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## Body

## 1. Introduction and Preliminaries

In this paper, we consider an important problem in mathematics known as the “completeness problem.” Completeness problem is to know under what circumstances the underlying space is complete. Completeness of space guarantees the convergence of all Cauchy sequences. There are some other problems like fixed point problem and variational problem which are equivalent to the completeness problem and readers interested in equivalence of completeness and fixed point problem are referred to a survey article by Cobzaş [1]. Completeness problem in mathematics has correspondence with an important problem known as “the end problem” in behavioral sciences. The end problem is to determine where and when human dynamics defined as succession of positions that starts from an initial position and follows transitions ends somewhere. For details on the completeness problem and the end problem, we refer to [2] and references therein.On the other hand, the notion of a distance between two objects of an abstract setZ plays a vital role in mathematics and other related disciplines such as routing theory, graph theory, matching problems, and decision-making processes. Due to its significance in solving problems in various disciples, the concept of a distance function (metric) has been generalized in many ways (see [3] and the reference therein). One of such generalizations is the notion of a b∽metric introduced by Czerwik [4–6].In Section1, we fix the notations along with the provision of necessary tools to prove the main results in this paper. In Section 2, we solve fixed point problem in the context of b∼ metric spaces via more general multivalued quasicontractions. In Section 3, we obtain equivalence between completeness problem and fixed point problem in the context of b∼ metric spaces.Throughout this article, letZ be a nonempty set and ℕ, ℕ1, ℝ, and ℝ+ denote the sets of positive integers, nonnegative integers, reals, and nonnegative reals, respectively.Czerwik [4–6] introduced b∼ metric spaces.Definition 1.
A bifunctionρ:Z×Z⟶ℝ+ is a b∼ metric on Z if there exists a κ∈ℝ with κ≥1 such that for u,y,z in Z, ρ satisfies(i)
a1−ρu,y=0 if u=y(ii)
a2−ρu,y=ρy,u(iii)
a3−ρu,y≤κρu,z+κρz,y
The pairZ,ρ is known as b∼ metric space (s) (shortly as b∼ MS (s)) with b-metric constant κ. Clearly, for κ=1, Z,ρ is a metric space, but there are b∼ metrics that are not metrics (see [4, 7, 8]).
LetZ,ρ be a b∼ MS and un a sequence in Z. Then un is a Cauchy sequence if for any given ε>0, there is a Kε∈ℕ so that ρun,um<ε for all m,n≥Kε, or equivalently, un is Cauchy if(1)limn⟶∞ρun+p,un=0,for all p∈ℕ. Further, un is convergent if there is a u in Z and for any ε>0, there exists Kε∈ℕ so that ρun,u<ε for all n≥Kε, or equivalently, a sequence un in Z converges to u in Z if(2)limn⟶∞ρun,u=0,and we write un⟶u as n⟶∞. A subset F⊆Z is closed in Z if for every sequence un in F with un⟶u as n⟶∞, implies u∈F and F⊆Z is bounded if(3)supz,w∈Fρz,w<∞.
Ab∼ MS Z,ρ is complete if every Cauchy sequence in Z converges. Some topological properties have been discussed in [9]. For instance, a b-metric ρ is not continuous in both variables in general. However, if ρ is continuous in one variable, then due to the symmetry property of the b-metric ρ, it is continuous in the other variable as well. Moreover, a subset(4)Bεu0=u∈Z:ρu0,u<εin Z,ρ is not an open set (in general) but if ρ is continuous in one variable, then Bεu0 is an open subset in Z. Moreover, suppose that the b-metric ρ in this paper is continuous in one variable.
The following lemma provides an important tool to prove that a sequence that satisfies certain contractive condition is Cauchy in ab∼ MS.Lemma 1 (see [10]).
If a sequenceun in a b∼ MS Z,ρ satisfies(5)ρun+1,un+2≤rρun,un+1,∀n∈ℕ,for some(6)0≤r<1κ,then it is Cauchy sequence in Z.Recently, Suzuki [11] replaced condition (1) with 0<r<1 and proved the following important lemma in the context of b∼ MSs. It is worth mentioning that the following lemma shows that the criteria to establish Cauchyness of a sequence in b∼ metric and ordinary metric spaces are the same.Lemma 2 (see [11]).
If a sequenceun in a b∼ MS Z,ρ satisfies(7)ρun+1,un+2≤rρun,un+1,∀n∈ℕ,for some r∈0,1, then it is a Cauchy sequence in Z.LetZ,ρ be a b∼ MS and CBZ (PZ) the set of nonempty, closed, bounded subsets (nonempty subsets) of Z . For E,F∈CBZ, the mapping Hρ, defined as(8)HρE,F=maxδE,F,δF,E,is called Hausdorff metric on CBZ induced by ρ, where(9)δE,F=supu∈Eρu,F,ρu,F=infw∈Fρu,w.The following lemma encompasses some important properties ofb∼ MSs that we will use in proving the main results of this article.Lemma 3 (see [4–6, 8]).
For ab∼ MS Z,ρ, u,y∈Z and E,F∈CBZ, the following assertions hold:(c1) CBZ,Hρ is a b∼ MS.(c2) For all u∈E,ρu,F≤HρE,F.(c3) For all u,y in Z, ρu,E≤κρu,y+κρy,E.(c4) For t>1 and a∈E, there is a b∈F so that ρa,b≤tHρE,F.(c5) For every t>0 and a∈E, there is a b∈F so that ρa,b≤HρE,F+t.(c6) a∈E¯=E if and only if ρa,E=0, where E¯ is the closure of E in Z,ρ.(c7) For any sequence un in Z,(10)ρu0,un≤κρu0,u1+κ2ρu1,u2+⋯+κn−1ρun−2,un−1+κn−1ρun−1,un.Consider a self-mapping onZ, that is, h:Z⟶Z and a multivalued mapping T:Z⟶PZ. The fixed point of T or h is a point u in Z, if u∈Tu or u=hu, and we denote the fixed point set of T or h as FixT or Fixh.Regarding the existence of a unique fixed point ofh, one of the important theorems is Banach contraction principle (BCP) [12]. This principle states that if a self-mapping h on Z (a complete metric space) satisfies(11)ρhu,hy≤rρu,y,for some 0≤r<1, then Fixh is singleton. The mapping that satisfies (11) is called a Banach contraction. BCP has been generalized further in more than one directions. For example, Suzuki [13] presented a generalization of BCP that characterized the metric completeness. Kannan [14, 15] presented a fixed point result that is independent of BCP but characterizes the metric completeness (see [16]), whereas BCP does not characterize the completeness of underlying metric spaces (see [17, 18]). Among other generalizations of Banach contractions, Ciric [19] introduced a self-mapping h:Z⟶Z that satisfies(12)ρhu,hy≤rmaxρu,y,ρu,hu,ρy,hy,ρu,hy,ρy,hu,for some 0≤r<1 and termed it a quasicontraction. Further, they obtained fixed point results for quasicontractions in orbitally complete metric spaces.Nadler [20] extended the BCP for multivalued mapping on Z as given below.Theorem 1.
LetZ,ρ be a complete metric space and T:Z⟶CBZ such that(13)HρTu,Ty≤rρu,y,for all u,y∈Z and some r∈0,1; then, FixT is nonempty.Amini-Harandi [21] generalized Theorem 1 by presenting the following result.Theorem 2 (see [21]).
LetZ,ρ be a complete metric space and T:Z⟶CBZ. If(14)HρTu,Ty≤rmaxρu,y,ρu,Tu,ρy,Ty,ρu,Ty,ρy,Tu,for all u,y∈Z and some r∈0,2−1, then FixT is nonempty.Kikkawa and Suzuki [22] generalized Theorem 1 using a function ξ1:0,1⟶2−1,1 defined as(15)ξ1t=11+t.Theorem 3 (see [22]).
LetZ,ρ be a complete metric space. If T:Z⟶CBZ satisfies(16)ξ1rρu,Tu≤ρu,yimpliesHρTu,Ty≤rρu,y,for all u,y∈Z and for some r∈0,1, then FixT is nonempty.Jovanovic et al. [23] extended BCP in the framework of b∼ MSs.Theorem 4.
LetZ,ρ be a complete b∼ MS and h be a self-mapping on Z satisfying(17)ρhu,hy≤rρu,y,for all u,y∈Z and for some r∈0,1/κ. Then, Fixh is singleton.Dung and Hang [24] replaced the condition 0≤r<1/κ by 0≤r<1 in the above theorem. They obtained the following version of BCP which shows that BCP can be obtained in b∼ MSs without imposing any additional condition on r.Theorem 5.
LetZ,ρ be a complete b∼ MS and h:Z⟶Z be a self-mapping on Z satisfying(18)ρhu,hy≤rρu,y,for all u,y∈Z and for some r∈0,1. Then, Fixh is singleton.Note that there are fixed point theorems (see [7]) that cannot be transported from metric to b∽MSs as in the case of BCP.Aydi et al. [25] obtained b∼ metric version of Theorem 2.Theorem 6 (see [25]).
LetZ,ρ be a complete b∼ MS and T:Z⟶CBZ. If T satisfies(19)HρTu,Ty≤rmaxρu,y,ρu,Tu,ρy,Ty,ρu,Ty,ρy,Tu,for all u,y∈Z and for some r∈0,1 with r<1/κ2+κ, then FixT is nonempty.Using a functionξ2:0,1⟶2−1,1 defined as ξ2r=1/1+κr, Kutbi et al. [26] presented the next result in b∼ MSs.Theorem 7 (see [26]).
LetZ,ρ be a complete b∼ MS. If T:Z⟶CBZ satisfies(20)ξ2rρu,Tu≤κρu,yimpliesHρTu,Ty≤rρu,y,for all u,y∈Z and for some r∈0,1 with r<1/κ2+κ, then FixT is nonempty.
LetZ,ρ be a b∼ MS, h:Z⟶Z and T:Z⟶CBZ and let u,y be in Z, we use the notations(21)NΥu,y=maxρu,y,ρu,Υu,ρy,Υy,ρu,Υy,ρy,Υu,MΥu,y=maxρu,y,ρu,Υu,ρy,Υy,ρu,Υy,ρy,Υu,ρu,y−1ρu,Υuρy,Υy,where Υ∈h,T and(22)Λ=ξ:ℝ+×ℝ+⟶ℝ:ξs,t≤sκ−t,where κ is the b∼ metric constant (see [27]).Recently, Alolaiyan et al. [27] obtained the following theorem for quasicontractions which is a generalization of results in [7, 13, 21, 25, 28, 29].Theorem 8 (see [27]).
LetZ,ρ be a complete b∼ MS. If T:Z⟶CBZ satisfies(23)ξρu,Tu,ρu,y≤0impliesHρTu,Ty≤rNTu,y,for all u,y∈Z and for some r∈0,1 with r<1/κ2+κ, where ξ∈Λ, then FixT is nonempty.Theorem 9.
Further, in [27], by considering a class Ϝ of b∼ metrics ρ such that for any Cauchy sequence un and any u in Z, the sequence ρu,un is Cauchy in ℝ. One typical example of such b∼ metric ρ is(24)ρu,y=vu,yp,for p≥1, where v is a metric on Z (see [27]). For the class Ϝ of b∼ metrics, they have obtained the following result.Theorem 10.
LetZ,ρ be a b∼ MS such that ρ∈Ϝ. For r∈0,1 with r<1/κ2+κ, assume that Ar,η is a class of self-mappings T of Z that satisfies the following.
(a) Foru,y∈Z,(25)ηrρu,Tu≤κρu,yimpliesρTu,Ty≤rρu,y,where η:0,1⟶0,1.LetBr,η be the set of mappings T on Z that satisfies (a).(b)TZ is denumerable and every subset D⊆TZ is closed.Then, statements (i)–(iii) are equivalent.(i)
Z,ρ is complete.(ii)
FixT is nonempty for every T∈Ar,η for all r∈0,1 where r<1/κ2+κ.(iii)
FixT is nonempty for every T∈Br,η for some r∈0,1 that satisfies r<1/κ2+κ.Motivated by the work in [27], we provide the following definition.Definition 2.
LetZ,ρ be a b∼ MS. A mapping T:Z⟶CBZ is a generalized multivalued Ciric–Suzuki-type (shortly CS−type) quasicontraction if there exists r∈0,1 with r<1/2κ such that(26)ξρu,Tu,ρu,y≤0impliesHρTu,Ty≤rMTu,y,for all u,y∈Z, with u≠y, where ξ∈Λ. If T is replaced with a self-mapping h:Z⟶Z, then h is generalized CS−type quasicontraction.
In this article, we provide with some new fixed point theorems for generalized multivaluedCS−type quasicontractions in b∼ MSs. It is worth noting that the existence of fixed point sets of CS−type quasicontractions further establishes the completeness of underlying b∼ MSs. The authors in [27] obtained a completeness characterization of a specific class of b∼ MSs. We extend their result and present a completeness characterization of any b∼ MS.


## 2. Fixed Points of Generalized MultivaluedCS−Type Quasicontractions

The following is the first main result about the fixed points of generalized multivaluedCS−type quasicontractions of b∼ MSs.Theorem 11.
LetZ,ρ be a complete b∼ MS and T:Z⟶CBZ be a generalized multivalued CS−type quasicontraction. Then, FixT is nonempty.Proof.
Note that for anyu,y in Z with u≠y, we have MTu,y>0. As r<1/2κ, we can choose a positive real number α satisfying(27)r2+α=14κ,that isr+α=1212κ+r.
Ifβ=r+α, then 0<β<1 and α>0. Let u0 be any point in Z and u1∈Tu0. If u1=u0∈Tu0, u0∈FixT. So, assume that u0≠u1.(28)ξρu0,Tu0,ρu0,u1≤1κρu0,Tu0−ρu0,u1≤0.
So, by the given assumption onT, we get that(29)HρTu0,Tu1≤rMTu0,u1.
AsαMTu0,u1>0, by c5 in Lemma 3, we have u2∈Tu1 that satisfies(30)ρu1,u2≤HρTu0,Tu1+αMTu0,u1.
From (29)–(30), we obtain(31)ρu1,u2≤HρTu0,Tu1+αMTu0,u1≤rMTu0,u1+αMTu0,u1=βMTu0,u1=βmaxρu0,u1,ρu0,Tu0,ρu1,Tu1,ρu0,Tu1,ρu1,Tu0,ρu0,u1−1ρu0,Tu0ρu1,Tu1≤βmaxρu0,u1,ρu1,u2,ρu0,u2≤βmaxρu0,u1,ρu1,u2,κρu0,u1+κρu1,u2≤κβρu0,u1+ρu1,u2.
That is,(32)ρu1,u2≤κβ1−κβρu0,u1.
Continuing like this, we obtain a sequenceun in Z that satisfies(33)un+1∈Tun,ρun,un+1≤κβ1−κβρun−1,un,for all n∈ℕ1. Set αn=ρun,un+1; then, from (33), we have(34)αn≤γαn−1,where γ=κβ/1−κβ. As κ≥1 and r<1/2κ, we have(35)κβ=κ212κ+r<12,γ=κβ1−κβ<1.
That is,γ<1. By Lemma 2, un is a Cauchy sequence in Z, and hence(36)limn⟶∞ρun,u=0,for some u∈Z. Now, we show that(37)ρu,Tw≤rmaxρu,w,ρw,Tw,for all w≠u. From (36), we choose n0∈ℕ so that(38)ρun,u<13κρu,w,for all n≥n0 and for all w≠u. Hence,(39)ξρun,Tun,ρun,w≤1κρun,Tun−ρun,w≤1κρun,un+1−ρun,w≤1κκρun,u+κρu,un+1−ρun,w≤23κρu,w−ρun,w=1κρu,w−13ρu,w−ρun,w≤1κρu,w−κρun,u−ρun,w≤1κκρun,w−ρun,w=0,for all n≥n0. This further implies that(40)ρun+1,Tw≤HρTun,Tw≤rMTun,w=rmaxρun,w,ρun,Tun,ρw,Tw,ρun,Tw,ρw,Tun,ρun,w−1ρun,Tunρw,Tw≤rmaxρun,w,ρun,un+1,ρw,Tw,ρun,Tw,ρw,un+1,ρun,w−1ρun,un+1ρw,Tw,for all n≥n0. By applying the limit as n⟶∞, we obtain(41)ρu,Tw≤rmaxρu,w,ρw,Tw,ρu,Tw.
Ifmaxρu,w,ρw,Tw,ρu,Tw=ρu,Tw, then we get(42)ρu,Tw≤rρu,Tw<βρu,Tw<ρu,Tw,a contradiction. Consequently, (37) holds for all w≠u.
Now, assume thatz∉Tz for all z∈Z. As r<1/2κ, 1/2rκ−1>0. Since Tu is nonempty for all u in Z, for(43)ε=12rκ−1ρu,Tu>0,there exists a∈Tu such that(44)ρu,a<ρu,Tu+ε.
Hence, we obtain(45)2κrρu,a<ρu,Tu.
Since(46)ξρu,Tu,ρu,a≤1κρu,Tu−ρu,a≤ρu,a−ρu,a=0,we obtain(47)HρTu,Ta≤rMTu,a≤rmaxρu,a,ρu,Tu,ρa,Ta,ρu,Ta,ρa,Tu,ρu,a−1ρu,Tuρa,Ta≤rmaxρu,a,ρu,a,ρa,Ta,ρu,Ta,ρa,a,ρu,a−1ρu,aρa,Ta=rmaxρu,a,ρa,Ta,ρu,Ta.
If(48)maxρu,a,ρa,Ta,ρu,Ta=ρa,Ta,then(49)ρa,Ta≤HρTu,Ta≤rρa,Ta,which implies that either a∈Ta, that is a∈FixT, or ρa,Ta<ρa,Ta, a contradiction. Consequently, we have(50)HρTu,Ta≤rmaxρu,a,ρu,Ta.
If(51)maxρu,a,ρa,Ta,ρu,Ta=ρu,Ta,then (37)–(50) imply that(52)HρTu,Ta≤rρu,Ta≤r2maxρu,a,ρa,Ta≤rmaxρu,a,ρa,Ta.
Again ifmaxρu,a,ρa,Ta=ρa,Ta, then either a∈Ta or ρa,Ta<ρa,Ta, so(53)HρTu,Ta≤rρu,a.
Also, from (37)–(53), we obtain that(54)ρu,Ta≤rmaxρu,a,ρa,Ta≤rmaxρu,a,HρTu,Ta≤rρu,a.
From (45), (53), and (54), we have(55)ρu,Tu≤κρu,Ta+κHρTu,Ta≤κrρu,a+κrρu,a=2κrρu,a<ρu,Tu,a contradiction. Hence, u∈Tu.Remark 1.
As(56)1κ2+κ≤12κandNTu,y≤MTu,y,for all u,y∈Z, u≠y, we obtain Theorem 8 as a particular case of Theorem 11. Further, Theorem 8 is a generalization of Theorems 1–3, 6, and 7.Corollary 1.
LetZ,ρ be a complete b∼ MS and T:Z⟶CBZ. If(57)ρu,Tu≤κρu,yimpliesHρTu,Ty≤rMTu,y,for all u,y∈Z, such that u≠y, and for some r∈0,1/2κ, then FixT is nonempty.Corollary 2.
LetZ,ρ be a complete b∼ MS and T:Z⟶CBZ. If(58)ρu,Tu≤κρu,yimpliesHρTu,Ty≤rNTu,y,for all u,y∈Z and for some r∈0,1/2κ, then FixT is nonempty.The next two corollaries are the extensions of Theorem6 in b∼ MSs as (56) holds for κ≥1.Corollary 3.
LetZ,ρ be a complete b∼ MS and T:Z⟶CBZ. If(59)HρTu,Ty≤rMTu,y,for all u,y∈Z, such that u≠y, and for some r∈0,1/2κ, then FixT is nonempty.Corollary 4.
LetZ,ρ be a complete b∼ MS and T:Z⟶CBZ. If(60)HρTu,Ty≤rNTu,y,for all u,y∈Z, such that u≠y, and for some r∈0,1/2κ, then FixT is nonempty.Remark 2.
Ifκ=1 in Corollary 4, we get Theorem 2.Example 1.
LetZ=u1,u2,u3,u4,u5 and ρ:Z×Z⟶ℝ+ be defined as(61)ρu1,u2=ρu1,u3=5.28,ρu1,u4=ρu1,u5=12,ρu2,u5=ρu3,u4=ρu3,u5=9,ρu2,u4=5.72,ρu2,u3=6,ρu4,u5=3,ρu,u=0andρu,y=ρy,u,for allu,y∈Z.
As(62)ρu4,u1=12≰ρu4,u2+ρu2,u1=11,ρ is not a metric on Z. For κ≥12/11>1, ρ is a b∼ metric on Z. Let ξs,t=s/κ−t∈Λ, r=11/25. Then,(63)1κ2+κ=121276≈0.438<r=0.44<0.458≈12κ.
Define a mappingT:Z⟶CBZ by(64)Tu=u1,ifu∈u1,u2,u3,u2,ifu∈u4,u3,ifu∈u5.
For allu,y∈u1,u2,u3,(65)HρTu,Ty=0≤rMTu,y.
Note that(66)MTu1,u4=maxρu1,u4,ρu4,Tu4,ρu1,Tu1,ρu4,Tu1,ρu1,Tu4,ρu1,u4−1ρu4,Tu4ρu1,Tu1=maxρu1,u4,ρu4,u2,ρu1,u1,ρu4,u1,ρu1,u2,ρu1,u4−1ρu4,u2ρu1,u1=max12,5.72,0,12,5.28,0=12,MTu1,u5=maxρu1,u5,ρu5,Tu5,ρu1,Tu1,ρu5,Tu1,ρu1,Tu5,ρu1,u5−1ρu5,Tu5ρu1,Tu1=maxρu1,u5,ρu4,u3,ρu1,u1,ρu5,u1,ρu1,u3,ρu1,u5−1ρu5,u3ρu1,u1=max12,9,0,12,5.28,0=12.
Further(67)MTu2,u4=maxρu2,u4,ρu4,Tu4,ρu2,Tu2,ρu4,Tu2,ρu2,Tu4,ρu2,u4−1ρu4,Tu4ρu2,Tu2=maxρu2,u4,ρu4,u2,ρu2,u1,ρu4,u1,ρu2,u2,ρu2,u4−1ρu4,u2ρu2,u1=max5.72,5.72,5.28,12,0,5.28=12,MTu2,u5=maxρu2,u5,ρu5,Tu5,ρu2,Tu2,ρu5,Tu2,ρu2,Tu5,ρu2,u5−1ρu5,Tu5ρu2,Tu2=maxρu2,u5,ρu5,u3,ρu2,u1,ρu5,u1,ρu2,u3,ρu2,u5−1ρu5,u3ρu2,u1=max9,9,12,12,6,5.28=12.
Moreover,(68)MTu3,u4=maxρu3,u4,ρu4,Tu4,ρu3,Tu3,ρu4,Tu3,ρu3,Tu4,ρu3,u4−1ρu4,Tu4ρu3,Tu3=maxρu3,u4,ρu4,u2,ρu3,u1,ρu4,u1,ρu3,u2,ρu3,u4−1ρu4,u2ρu3,u1=max9,5.72,5.28,12,6,3.35=12,MTu3,u5=maxρu3,u5,ρu5,Tu5,ρu3,Tu3,ρu5,Tu3,ρu3,Tu5,ρu3,u5−1ρu5,Tu5ρu3,Tu3=maxρu3,u5,ρu5,u3,ρu3,u1,ρu5,u1,ρu3,u3,ρu3,u5−1ρu5,u3ρu3,u1=max9,9,5.28,12,0,5.28=12.
Hence, foru∈u1,u2,u3 and y∈u4,u5,(69)MTu,y=12,HρTu,Ty=5.28.
Consequently, foru∈u1,u2,u3 and y∈u4,u5, the implication(70)ξρu,Tu,ρu,y≤0impliesHρTu,Ty≤rMTu,yholds true. For u,y∈u4,u5, u≠y, we get(71)ξρu4,Tu4,ρu4,u5=κ−1ρu4,Tu4−ρu4,u5≈2.24>0,ξρu5,Tu5,ρu5,u4=κ−1ρu5,Tu5−ρu4,u5≈5.25>0.
Hence, all the assumptions in Theorem11 are satisfied.
Now based on the above calculations, we present a comparison among Theorem11 and some results from the literature.(1)
Ifu=u1, and y=u4, then(72)ξρu1,Tu1,ρu1,u4=κ−1ρu1,Tu1−ρu1,u4=−12≤0,but(73)HρTu1,Tu4=ρu1,u2=5.28≰12r=rMTu1,u4=rNTu1,u4,for anyr<0.438≈1/κ2+κ, so Theorem 8 (a main result in [27]) cannot be applied in this example. Hence, Theorem 8 is a generalization of Theorem 8.(2)
Now,HρTu4,Tu5=ρu2,u3=6 and(74)MTu4,u5=maxρu4,u5,ρu4,Tu4,ρu5,Tu5,ρu4,Tu5,ρu5,Tu4,ρu4,u5−1ρu4,Tu4ρu5,Tu5=maxρu4,u5,ρu4,u2,ρu5,u3,ρu4,u3,ρu5,u2,ρu4,u5−1ρu4,u2ρu5,u3=max3,5.72,9,9,9,5.7293=17.16,NTu4,u5=9.
On the other hand,(75)HρTu4,Tu5=6≰9r=rNTu4,u5,for any r<0.438≈1/κ2+κ. Hence, Theorem 6 is not applicable in this example. So, Theorem 11 is a proper generalization of Theorem 6 which itself is a generalization of Theorems 1 and 2 and Theorem 1 of [7].
We give another example to show the fact that Theorem11 is applicable to certain cases where several existing results are not.Example 2.
LetZ=u1,u2,u3. Suppose that bifunction ρ:Z×Z⟶ℝ+ is given by(76)ρu1,u2=4,ρu1,u3=1.4,ρu2,u3=1.6,ρu,u=0,ρu,y=ρy,u,for allu,y∈Z.
As4=ρu2,u1≰ρu2,u3+ρu3,u1=3, ρ is not a metric on Z. For κ≥4/3>1, ρ is a b∼ metric on Z. Consider T:Z⟶CBZ defined as(77)Tu=u1,u3,ifu∈u1,u3,u1,ifu∈u2.
If we considerξs,t=s/κ−t∈Λ and r=0.35, then(78)1κ2+κ=928≈0.32<r=0.35<0.375=38=12κ.
Foru,y∈u1,u3, we have(79)HρTu,Ty=0≤rMTu,y.
Hence, foru∈u1,u3 and y=u2,(80)HρTu,Ty=1.4,MTu,y=NTu,y=4asρu,Tu=0,andρy,Ty=4.
That is,(81)HρTu,Ty=1.4≤1.4=rMTu,y.
Hence,(82)ξρu,Tu,ρu,y≤0impliesHρTu,Ty≤rMTu,y,for all u,y∈Z, u≠y. That is, all the assumptions of Theorem 11 are met. If u=u2, y=u3, then(83)ξ2rρu3,Tu3=0≤κρu3,u2=2,HρTu3,Tu2=ρu1,u3=1.4.
ButHρTu3,Tu2=1.4≰1.6r=rρu3,u2 for any r<9/28=1/κ2+κ. Hence, Theorem 7 cannot be applied in this situation. So, Theorem 11 is a proper generalization of Theorem 7 which generalizes Theorems 1 and 3.Corollary 5.
LetZ,ρ be a complete b∼ MS and h:Z⟶Z be a generalized CS−type quasicontraction. Then, Fixh is singleton.Proof.
From Theorem11, there exists u0∈Fixh. To prove the uniqueness, assume u0,y0∈Fixh with u0≠y0. As(84)ξρu0,hu0,ρu0,y0≤1κρu0,hu0−ρu0,y0=−1κρu0,y0≤0,we have(85)ρu0,y0=ρhu0,hy0≤rMhu0,y0=rmaxρu0,y0,ρu0,hu0,ρy0,hy0,ρu0,hy0,ρy0,hu0,ρu0,y0−1ρu0,hu0ρy0,hy0=rρu0,y0<ρu0,y0asr<1,a contradiction. Hence, Fixh is singleton.Corollary 6.
LetZ,ρ be a complete b∼ MS and h:Z⟶Z. If(86)ρu,hu≤κρu,yimpliesρhu,hy≤rρu,y,for all u,y∈Z and for some r∈0,1/2κ, then Fixh is singleton.Corollary 7.
LetZ,ρ be a complete b∼ MS and h:Z⟶Z. If there exists ξ∈Λ and r∈0,1 with r<1/2κ such that(87)ξρu,hu,ρu,y≤0impliesρhu,hy≤rρu,y,for all u,y∈Z, then Fixh is singleton.Proof.
It follows from Corollary5.Corollary 8.
LetZ,ρ be a complete b∼MS and h:Z⟶Z. If there exists r∈0,1 with r<1/2κ such that(88)ηrρu,hu≤κρu,yimpliesρhu,hy≤rρu,y,for all u,y∈Z, where η:0,1⟶0,1, then Fixh is singleton.Proof.
Consider(89)ξs,t=ηrκs−t≤sκ−t.
Hence,ξ∈Λ. If s=ρu,hu and t=ρu,y, then(90)ξρu,hu,ρu,y=ηrκs−t≤0.
Hence, the result follows from Corollary7.


## 3. Completeness Characterization ofb∽ Metric Spaces

In this section, we discuss the completeness ofb∼ MSs via fixed point results.In 1959, Connel [17] studied some important topological properties of fixed point sets of mappings of metric spaces and presented an example of a separable and locally contractible incomplete metric space Z such that every Banach contraction on Z has a fixed point. This shows that the converse of BCP does not hold true (see [18]). For the sake of completeness and convenience, we present the example initially given in ([17], Example 3).Example 3.
Letf:Z⟶Z be any Banach contraction on Z, where(91)I0=0,0,1,0,a horizontal line segment joining0,0to1,0,In=n−1,0,n−1,1,n∈ℕ,vertical line segments,Z=∪n∈ℕ1In.
We prove thatFixf is nonempty. Assume that Fixf is empty. Let f0 be the restriction of f on I0. Define(92)π:f0I0⟶I0,h=π°f0:I0⟶I0.
Sinceh is a continuous mapping on I0, there is p,0∈Fixh.
Claim:p=1/k for some integer k. If p≠1/k for all integers k, then(93)hp,0=π°f0p,0=p,0.
Moreover,(94)f0p,0∉In,for alln∈ℕ.
Otherwise, iff0p,0∈In1 for some n1∈ℕ, then(95)f0p,0=1n1,y,for some y∈0,1. Consequently, we get(96)hp,0=πf0p,0=π1n1,y=1n1,0.
Buthp,0=p,0 implies that p=1/n1, a contradiction, as p≠1/k for all integers k. Hence, (94) holds. Thus, f0p,0∈I0 and(97)πf0p,0=f0p,0.
From (93)–(97), we have(98)f0p,0=p,0,a contradiction as f0 has no fixed point. Hence, p=1/k for some integer k. As Fixf is empty,(99)f1k,0=1k,y,for some 0<y≤1. Set(100)A=y:there is azso thatf1k,y=1k,z,z>y.
Ift=supA, there is a sequence tn in A such that tn⟶t as n⟶∞ and there is s∈0,1 such that(101)f1k,t=1k,s.
We claim thatt=s.


### 3.1. Case (I)

Ifs<t, then there exists s1 such that s<s1<t and tn∈s1,t eventually. As(102)f1k,tn⟶f1k,t=1k,sasn⟶∞,(103)f1k,tn∈1k,z:z∈0,s1,eventually. This implies(104)f1k,tn=1k,un.But0<un<s1≤tn eventually, a contradiction to tn∈A.


### 3.2. Case (II)

Ift<s, then there is s1 such that t<s1<s and tn∈t,s1 eventually. Since(105)f1k,tn⟶f1k,t=1k,sasn⟶∞,(106)f1k,tn∈1k,z:z∈s1,s,eventually and(107)f1k,tn=1k,un.Butt<tn<s1<un<s (eventually). Hence, there exists an integer n0 such that(108)f1k,tn0=1k,un0.Butt<tn0<s1<un0<s. This implies tn0∈A, a contradiction as tn0>t=supA. Hence, s=t, and(109)f1k,t=1k,t.That is,Fixf is nonempty but Z is not complete as it is not closed.Remark 3.
As every metric is ab∼ metric, Theorem 5 and Example 3 show that BCP in b∼ MSs does not characterize the completeness of underlying b∼ MS.
Park and Rhoads [30] commented on completeness characterization of metric spaces and Alolaiyan et al. [27] presented those comments in the context of b∼ MSs as follows. “Let Z,ρ be a b∼ MS and B be a class of mappings of Z such that if any map in B has a fixed point, then Z is complete. Let A be a class of mappings of Z containing B such that completeness of Z implies the existence of fixed point of any map in A.”Theorem 12 (see [27, 30]).
IfZ,ρ is a b∼ MS, then(110)Zis complete ifFixTis nonempty for any mapping inTI^A.Now, we provide a characterization of completeness ofb∼ MS.Theorem 13.
LetZ,ρ be a b∼ MS and T:Z⟶Z, such that(a)
Forx,y∈Z,(111)ηrρx,Tx≤κρx,yimpliesρTx,Ty≤rρx,y,wherer∈0,1/2κ and η:0,1⟶0,1.(b)
Forx,y∈Z,(112)ηrρx,Tx≤κρx,yimpliesρTx,Ty≤rρx,y,wherer∈0,1/κ2+κ and η:0,1⟶0,1.(c)
TZ is denumerable.(d)
EveryM⊆TZ is closed.
LetAr,η be a class of mappings T of Z that satisfies (a), Ar,η∗ be a class of mappings T of Z that satisfies (b), and Br,η be a set of mappings T of Z that satisfies (b), (c), and (d). Then, statements (1)–(4) are equivalent:(1)
b∼ MS Z,ρ is complete.(2)
FixT is nonempty for every T∈Ar,η for all r∈0,1 with r<1/2κ.(3)
FixT is nonempty for every T∈Ar,η∗ for all r∈0,1 with r<1/κ2+κ.(4)
FixT is nonempty for every T∈Br,η for some r∈0,1 with r<1/κ2+κ.Proof.
By Corollary8, (6) implies (11).
As1/κ2+κ≤1/2κ, Ar,η∗⊆Ar,η. Hence, (6) implies (29). Since Br,η⊆Ar,η∗, (29) implies (30). (30) implies (6), and assume on contrary that (30) holds but Z,ρ is incomplete. That is, there is a sequence un which is Cauchy but does not converge. Define g:Z⟶0,∞ as(113)gx=limn⟶∞supρx,un,for x∈Z. As un is Cauchy, for ε=1/κ>0, we can choose mε∈ℕ so that for all k∈ℕ,(114)ρumε,umε+k≤1κ.
Hence, for allk∈ℕ, we get(115)ρx,umε+k≤κρx,umε+κρumε,umε+k≤κρx,umε+1,which implies that the sequence ρx,un is bounded in ℝ for every x∈Z. This further implies that the function g is well defined. Further, gx>0 for all x in Z. For ε>0, there exists Kε∈ℕ such that for all k∈ℕ,(116)ρun,un+k<ε,for all n≥Kε. Hence, we get for all k∈ℕ,(117)0≤gun=limk⟶∞supρun,un+k<ε,for all n≥Kε. That is,(118)limn⟶∞gun=0.
From (118), for every x∈Z, there exists υ∈ℕ such that(119)guυ≤rηr3κ4+rκηrgx.
IfTx=uυ, then(120)gTx≤rηr3κ4+rκηrgxandTx∈un:n∈ℕ,for all x∈Z. From (120), we have gTx<gx; hence, Tx≠x for all x∈Z. That is, FixT is empty. As TZ⊂un:n∈ℕ, (c) holds. Note that (d) holds as well. Further, g satisfies(121)gx−κgy≤κρx,y,for allx,y∈Z,gy−κgx≤κρx,y,for allx,y∈Z,gx−κgTx≤κρx,Tx,for allx∈Z,ρTx,Ty≤κgTx+κgTy,for allx,y∈Z.
Now fixx,y∈Z such that(122)ηrρx,Tx≤κρx,y.
We need to show that (111) holds. Observe that(123)ρx,y≥ηrκρx,Tx≥ηrκ2gx−κgTx≥ηrκ21−rηr3κ3+rηrgx=3κηr3κ3+rηrgx.
We have two cases.Case (1) Supposegy≥2κgx; then,(124)ρTx,Ty≤κgTx+κgTy≤rηr3κ3+rηrgx+rηr3κ3+rηrgy≤r3κgx+gy+2r3κgy−2κgx=r31κgx+1κgy+2κgy−4gx≤r33κgy−3gx≤r1κgy−gx≤rρx,y.Case (2) Whenevergy<2κgx, from (123),(125)ρTx,Ty≤κgTx+κgTy≤rηr3κ3+rηrgx+rηr3κ3+rηrgy≤κrηr3κ3+rηrgx+2κrηr3κ3+rηrgx=3κrηr3κ3+rηrgx=r3κηr3κ3+rηrgx≤rρx,y.
Hence,(126)ηrρx,Tx≤κρx,yimpliesρTx,Ty≤rρx,y,for all x,y∈Z. From (30), mapping FixT is nonempty, a contradiction. Hence, Z is complete.


## 3.1. Case (I)

Ifs<t, then there exists s1 such that s<s1<t and tn∈s1,t eventually. As(102)f1k,tn⟶f1k,t=1k,sasn⟶∞,(103)f1k,tn∈1k,z:z∈0,s1,eventually. This implies(104)f1k,tn=1k,un.But0<un<s1≤tn eventually, a contradiction to tn∈A.


## 3.2. Case (II)

Ift<s, then there is s1 such that t<s1<s and tn∈t,s1 eventually. Since(105)f1k,tn⟶f1k,t=1k,sasn⟶∞,(106)f1k,tn∈1k,z:z∈s1,s,eventually and(107)f1k,tn=1k,un.Butt<tn<s1<un<s (eventually). Hence, there exists an integer n0 such that(108)f1k,tn0=1k,un0.Butt<tn0<s1<un0<s. This implies tn0∈A, a contradiction as tn0>t=supA. Hence, s=t, and(109)f1k,t=1k,t.That is,Fixf is nonempty but Z is not complete as it is not closed.Remark 3.
As every metric is ab∼ metric, Theorem 5 and Example 3 show that BCP in b∼ MSs does not characterize the completeness of underlying b∼ MS.
Park and Rhoads [30] commented on completeness characterization of metric spaces and Alolaiyan et al. [27] presented those comments in the context of b∼ MSs as follows. “Let Z,ρ be a b∼ MS and B be a class of mappings of Z such that if any map in B has a fixed point, then Z is complete. Let A be a class of mappings of Z containing B such that completeness of Z implies the existence of fixed point of any map in A.”Theorem 12 (see [27, 30]).
IfZ,ρ is a b∼ MS, then(110)Zis complete ifFixTis nonempty for any mapping inTI^A.Now, we provide a characterization of completeness ofb∼ MS.Theorem 13.
LetZ,ρ be a b∼ MS and T:Z⟶Z, such that(a)
Forx,y∈Z,(111)ηrρx,Tx≤κρx,yimpliesρTx,Ty≤rρx,y,wherer∈0,1/2κ and η:0,1⟶0,1.(b)
Forx,y∈Z,(112)ηrρx,Tx≤κρx,yimpliesρTx,Ty≤rρx,y,wherer∈0,1/κ2+κ and η:0,1⟶0,1.(c)
TZ is denumerable.(d)
EveryM⊆TZ is closed.
LetAr,η be a class of mappings T of Z that satisfies (a), Ar,η∗ be a class of mappings T of Z that satisfies (b), and Br,η be a set of mappings T of Z that satisfies (b), (c), and (d). Then, statements (1)–(4) are equivalent:(1)
b∼ MS Z,ρ is complete.(2)
FixT is nonempty for every T∈Ar,η for all r∈0,1 with r<1/2κ.(3)
FixT is nonempty for every T∈Ar,η∗ for all r∈0,1 with r<1/κ2+κ.(4)
FixT is nonempty for every T∈Br,η for some r∈0,1 with r<1/κ2+κ.Proof.
By Corollary8, (6) implies (11).
As1/κ2+κ≤1/2κ, Ar,η∗⊆Ar,η. Hence, (6) implies (29). Since Br,η⊆Ar,η∗, (29) implies (30). (30) implies (6), and assume on contrary that (30) holds but Z,ρ is incomplete. That is, there is a sequence un which is Cauchy but does not converge. Define g:Z⟶0,∞ as(113)gx=limn⟶∞supρx,un,for x∈Z. As un is Cauchy, for ε=1/κ>0, we can choose mε∈ℕ so that for all k∈ℕ,(114)ρumε,umε+k≤1κ.
Hence, for allk∈ℕ, we get(115)ρx,umε+k≤κρx,umε+κρumε,umε+k≤κρx,umε+1,which implies that the sequence ρx,un is bounded in ℝ for every x∈Z. This further implies that the function g is well defined. Further, gx>0 for all x in Z. For ε>0, there exists Kε∈ℕ such that for all k∈ℕ,(116)ρun,un+k<ε,for all n≥Kε. Hence, we get for all k∈ℕ,(117)0≤gun=limk⟶∞supρun,un+k<ε,for all n≥Kε. That is,(118)limn⟶∞gun=0.
From (118), for every x∈Z, there exists υ∈ℕ such that(119)guυ≤rηr3κ4+rκηrgx.
IfTx=uυ, then(120)gTx≤rηr3κ4+rκηrgxandTx∈un:n∈ℕ,for all x∈Z. From (120), we have gTx<gx; hence, Tx≠x for all x∈Z. That is, FixT is empty. As TZ⊂un:n∈ℕ, (c) holds. Note that (d) holds as well. Further, g satisfies(121)gx−κgy≤κρx,y,for allx,y∈Z,gy−κgx≤κρx,y,for allx,y∈Z,gx−κgTx≤κρx,Tx,for allx∈Z,ρTx,Ty≤κgTx+κgTy,for allx,y∈Z.
Now fixx,y∈Z such that(122)ηrρx,Tx≤κρx,y.
We need to show that (111) holds. Observe that(123)ρx,y≥ηrκρx,Tx≥ηrκ2gx−κgTx≥ηrκ21−rηr3κ3+rηrgx=3κηr3κ3+rηrgx.
We have two cases.Case (1) Supposegy≥2κgx; then,(124)ρTx,Ty≤κgTx+κgTy≤rηr3κ3+rηrgx+rηr3κ3+rηrgy≤r3κgx+gy+2r3κgy−2κgx=r31κgx+1κgy+2κgy−4gx≤r33κgy−3gx≤r1κgy−gx≤rρx,y.Case (2) Whenevergy<2κgx, from (123),(125)ρTx,Ty≤κgTx+κgTy≤rηr3κ3+rηrgx+rηr3κ3+rηrgy≤κrηr3κ3+rηrgx+2κrηr3κ3+rηrgx=3κrηr3κ3+rηrgx=r3κηr3κ3+rηrgx≤rρx,y.
Hence,(126)ηrρx,Tx≤κρx,yimpliesρTx,Ty≤rρx,y,for all x,y∈Z. From (30), mapping FixT is nonempty, a contradiction. Hence, Z is complete.



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*Source: 5183291-2020-09-02.xml*