Approximate Best Proximity Pairs in Metric Space

Abstract

Let 𝐴 and 𝐵 be nonempty subsets of a metric space 𝑋 and also 𝑇∶𝐴∪𝐵→𝐴∪𝐵 and 𝑇(𝐴)⊆𝐵, 𝑇(𝐵)⊆𝐴. We are going to consider element 𝑥∈𝐴 such that 𝑑(𝑥,𝑇𝑥)≤𝑑(𝐴,𝐵)+𝜖 for some 𝜖>0. We call pair (𝐴,𝐵) an approximate best proximity pair. In this paper, definitions of approximate best proximity pair for a map and two maps, their diameters, 𝑇-minimizing a sequence are given in a metric space.

1. Introduction

Let 𝑋 be a metric space and 𝐴 and 𝐵 nonempty subsets of 𝑋, and 𝑑(𝐴,𝐵) is distance of 𝐴 and 𝐵. If 𝑑(𝑥₀,𝑦₀)=𝑑(𝐴,𝐵), then the pair (𝑥₀,𝑦₀) is called a best proximity pair for 𝐴 and 𝐵 and put prox(𝐴,𝐵)∶={(𝑥,𝑦)∈𝐴×𝐵∶𝑑(𝑥,𝑦)=𝑑(𝐴,𝐵)}(1.1) as the set of all best proximity pair (𝐴,𝐵). Best proximity pair evolves as a generalization of the concept of best approximation. That reader can find some important result of it in [1–4].

Now, as in [5] (see also [4, 6–11]), we can find the best proximity points of the sets 𝐴 and 𝐵, by considering a map 𝑇∶𝐴∪𝐵→𝐴∪𝐵 such that 𝑇(𝐴)⊆𝐵 and 𝑇(𝐵)⊆𝐴. Best proximity pair also evolves as a generalization of the concept of fixed point of mappings. Because if 𝐴∩𝐵≠∅, every best proximity point is a fixed point of 𝑇.

We say that the point 𝑥∈𝐴∪𝐵 is an approximate best proximity point of the pair (𝐴,𝐵), if 𝑑(𝑥,𝑇𝑥)≤𝑑(𝐴,𝐵)+𝜖, for some 𝜖>0.

In the following, we introduce a concept of approximate proximity pair that is stronger than proximity pair.

Definition 1.1. Let 𝐴 and 𝐵 be nonempty subsets of a metric space 𝑋 and 𝑇∶𝐴∪𝐵→𝐴∪𝐵 a map such that 𝑇(𝐴)⊆𝐵, 𝑇(𝐵)⊆𝐴. put 𝑃^𝑎𝑇(𝐴,𝐵)={𝑥∈𝐴∪𝐵∶𝑑(𝑥,𝑇𝑥)≤𝑑(𝐴,𝐵)+𝜖forsome𝜖>0}.(1.2) We say that the pair (𝐴,𝐵) is an approximate best proximity pair if 𝑃^𝑎𝑇(𝐴,𝐵)≠∅.

Example 1.2. Suppose that 𝑋=𝐑², 𝐴={(𝑥,𝑦)∈𝑋∶(𝑥−𝑦)²+𝑦²≤1}, and 𝐵={(x,𝑦)∈𝑋∶(𝑥+𝑦)²+𝑦²≤1} with 𝑇(𝑥,𝑦)=(−𝑥,𝑦) for (𝑥,𝑦)∈𝑋. Then 𝑑((𝑥,𝑦),𝑇(𝑥,𝑦))≤𝑑(𝐴,𝐵)+𝜖 for some 𝜖>0. Hence 𝑃^𝑎𝑇(𝐴,𝐵)≠∅.

2. Approximate Best Proximity

In this section, we will consider the existence of approximate best proximity points for the map 𝑇∶𝐴∪𝐵→𝐴∪𝐵, such that 𝑇(𝐴)⊆𝐵, 𝑇(𝐵)⊆𝐴, and its diameter.

Theorem 2.1. Let 𝐴 and 𝐵 be nonempty subsets of a metric space 𝑋. Suppose that the mapping 𝑇∶𝐴∪𝐵→𝐴∪𝐵 is satisfying 𝑇(𝐴)⊆𝐵, 𝑇(𝐵)⊆𝐴, and lim_𝑛→∞𝑑𝑇^𝑛𝑥,𝑇^𝑛+1𝑥=𝑑(𝐴,𝐵)forsome𝑥∈𝐴∪𝐵.(2.1) Then the pair (𝐴,𝐵) is an approximate best proximity pair.

Proof. Let 𝜖>0 be given and 𝑥∈𝐴∪𝐵 such that lim_𝑛→∞𝑑(𝑇^𝑛𝑥,𝑇^𝑛+1𝑥)=𝑑(𝐴,𝐵); then there exists 𝑁₀>0 such that ∀𝑛≥𝑁₀∶𝑑𝑇^𝑛𝑥,𝑇^𝑛+1𝑥<𝑑(𝐴,𝐵)+𝜖.(2.2) If 𝑛=𝑁₀, then 𝑑(𝑇^𝑁0(𝑥),𝑇(𝑇^𝑁0(𝑥)))<𝑑(𝐴,𝐵)+𝜖, and 𝑇^𝑁0(𝑥)∈𝑃^𝑎𝑇(𝐴,𝐵) and 𝑃^𝑎𝑇(𝐴,𝐵)≠∅.

Theorem 2.2. Let 𝐴 and 𝐵 be nonempty subsets of a metric space 𝑋. Suppose that the mapping 𝑇∶𝐴∪𝐵→𝐴∪𝐵 is satisfying 𝑇(𝐴)⊆𝐵, 𝑇(𝐵)⊆𝐴 and 𝑑(𝑇𝑥,𝑇𝑦)≤𝛼𝑑(𝑥,𝑦)+𝛽[𝑑(𝑥,𝑇𝑥)+𝑑(𝑦,𝑇𝑦)]+𝛾𝑑(𝐴,𝐵)(2.3) for all 𝑥,𝑦∈𝐴∪𝐵, where 𝛼,𝛽,𝛾≥0 and 𝛼+2𝛽+𝛾<1. Then the pair (𝐴,𝐵) is an approximate best proximity pair.

Proof. If 𝑥∈𝐴∪𝐵, then 𝑑𝑇𝑥,𝑇²𝑥≤𝛼𝑑(𝑥,𝑇𝑥)+𝛽𝑑(𝑥,𝑇𝑥)+𝑑𝑇𝑥,𝑇²𝑥+𝛾𝑑(𝐴,𝐵).(2.4) Therefore, 𝑑𝑇𝑥,𝑇²𝑥≤𝛼+𝛽1−𝛽𝑑(𝑥,𝑇𝑥)+1−𝛾𝛽𝑑(𝐴,𝐵).(2.5) Now if 𝑘=(𝛼+𝛽)/(1−𝛽), then 𝑑𝑇𝑥,𝑇²𝑥≤𝑘𝑑(𝑥,𝑇𝑥)+(1−𝑘)𝑑(𝐴,𝐵)(2.6) also 𝑑𝑇²𝑥,𝑇³𝑥≤𝑘²𝑑(𝑥,𝑇𝑥)+1−𝑘²𝑑(𝐴,𝐵).(2.7) Therefore, 𝑑𝑇^𝑛𝑥,𝑇^𝑛+1𝑥≤𝑘^𝑛𝑑(𝑥,𝑇𝑥)+(1−𝑘^𝑛)𝑑(𝐴,𝐵),(2.8) and so 𝑑𝑇^𝑛𝑥,𝑇^𝑛+1𝑥⟶𝑑(𝐴,𝐵),as𝑛⟶∞.(2.9) Therefore, by Theorem 2.1, 𝑃^𝑎𝑇(𝐴,𝐵)≠∅; then pair (𝐴,𝐵) is an approximate best proximity pair.

Definition 2.3. Let 𝐴 and 𝐵 be nonempty subsets of a metric space 𝑋. Suppose that the mapping 𝑇∶𝐴∪𝐵→𝐴∪𝐵 is satisfying 𝑇(𝐴)⊆𝐵, 𝑇(𝐵)⊆𝐴. We say that the sequence {𝑧_𝑛}⊆𝐴∪𝐵 is T-minimizing if lim_𝑛→∞𝑑𝑧_𝑛,𝑇𝑧_𝑛=𝑑(𝐴,𝐵).(2.10)

Theorem 2.4. Let 𝐴 and 𝐵 be nonempty subsets of a metric space 𝑋, suppose that the mapping 𝑇∶𝐴∪𝐵→𝐴∪𝐵 is satisfying 𝑇(𝐴)⊆𝐵, 𝑇(𝐵)⊆𝐴. If {𝑇^𝑛𝑥} is a T-minimizing for some 𝑥∈𝐴∪𝐵, then (𝐴,𝐵) is an approximate best pair proximity.

Proof. Since lim_𝑛→∞𝑑𝑇^𝑛𝑥,𝑇^𝑛+1𝑥=𝑑(𝐴,𝐵)forsome𝑥∈𝐴∪𝐵,(2.11) therefore, by Theorem 2.1, 𝑃^𝑎𝑇(𝐴,𝐵)≠∅; then pair (𝐴,𝐵) is an approximate best proximity pair.

Theorem 2.5. Let 𝐴 and 𝐵 be nonempty subsets of a normed space 𝑋 such that 𝐴∪𝐵 is compact. Suppose that the mapping 𝑇∶𝐴∪𝐵→𝐴∪𝐵 is satisfying 𝑇(𝐴)⊆𝐵, 𝑇(𝐵)⊆𝐴, 𝑇 is continuous and ‖𝑇𝑥−𝑇𝑦‖≤‖𝑥−𝑦‖,(2.12) where (𝑥,𝑦)∈𝐴×𝐵. Then 𝑃^𝑎𝑇(𝐴,𝐵) is nonempty and compact.

Proof. Since 𝐴∪𝐵 compact, there exists a 𝑧₀∈𝐴∪𝐵 such that ‖‖𝑧₀−𝑇𝑧₀‖‖=inf_{𝑧∈𝐴∪𝐵}‖𝑧−𝑇𝑧‖⋅(∗) If ‖𝑧₀−𝑇𝑧₀‖>𝑑(𝐴,𝐵), then ‖𝑇𝑧₀−𝑇²𝑧₀‖<‖𝑧₀−𝑇𝑧₀‖ which contradict to the definition of 𝑧₀, (𝑇𝑧₀∈𝐴∪𝐵 and by (*) ‖𝑇𝑧₀−𝑇(𝑇𝑧₀)‖≥‖𝑧₀−𝑇𝑧₀‖). Therefore, ‖𝑧₀−𝑇𝑧₀‖=𝑑(𝐴,𝐵)≤𝑑(𝐴,𝐵)+𝜖 for some 𝜖>0 and 𝑧₀∈𝑃^𝑎𝑇(𝐴,𝐵). Therefore, 𝑃^𝑎𝑇(𝐴,𝐵) is nonempty.
Also, if {𝑧_𝑛}⊆𝑃^𝜖𝑇(𝐴,𝐵), then ‖𝑧_𝑛−𝑇𝑧_𝑛‖<𝑑(𝐴,𝐵)+𝜖, for some 𝜖>0, and by compactness of 𝐴∪𝐵, there exists a subsequence 𝑧_𝑛𝑘 and a 𝑧₀∈𝐴∪𝐵 such that 𝑧_𝑛𝑘→𝑧₀ and so ‖‖𝑧₀−𝑇𝑧₀‖‖=lim_𝑘→∞‖‖𝑧_𝑛𝑘−𝑇𝑧_𝑛𝑘‖‖<𝑑(𝐴,𝐵)+𝜖(2.13) for some 𝜖>0, hence 𝑃^𝑎𝑇(𝐴,𝐵) is compact.

Example 2.6. If 𝐴=[−3,−1],𝐵=[1,3], and 𝑇∶𝐴∪𝐵→𝐴∪𝐵 such that 𝑇(𝑥)=⎧⎪⎨⎪⎩1−1−2−𝑥,𝑥𝑥∈𝐴,2,𝑥∈𝐵,(2.14) then 𝑃^𝑎𝑇(𝐴,𝐵) is compact, and we have 𝑃^𝑎𝑇(𝐴,𝐵)==={{{𝑥1𝑥,∈∈−1𝐴𝐴}∪∪.𝐵𝐵∶∶𝑑𝑑((𝑥𝑥,,𝑇𝑇𝑥𝑥))<<𝑑2+(𝐴𝜖,𝐵fo)r+so𝜖mfoer𝜖so>m0e}𝜖>0}(2.15) That is compact.

In the following, by diam(𝑃^𝑎𝑇(𝐴,𝐵)) for a set 𝑃^𝑎𝑇(𝐴,𝐵)≠∅, we will understand the diameter of the set 𝑃^𝑎𝑇(𝐴,𝐵).

Definition 2.7. Let 𝑇∶𝐴∪𝐵→𝐴∪𝐵 be a continuous map such that 𝑇(𝐴)⊆𝐵, 𝑇(𝐵)⊆𝐴 and 𝜖>0. We define diameter 𝑃^𝑎𝑇(𝐴,𝐵) by diam𝑃^𝑎𝑇(𝐴,𝐵)=sup𝑑(𝑥,𝑦)∶𝑥,𝑦∈𝑃^𝑎𝑇(𝐴,𝐵).(2.16)

Theorem 2.8. Let 𝑇∶𝐴∪𝐵→𝐴∪𝐵, such that 𝑇(𝐴)⊆𝐵, 𝑇(𝐵)⊆𝐴 and 𝜖>0. If there exists an 𝛼∈[0,1] such that for all (𝑥,𝑦)∈𝐴×𝐵𝑑(𝑇𝑥,𝑇𝑦)≤𝛼𝑑(𝑥,𝑦),(2.17) then diam𝑃^𝑎𝑇(𝐴,𝐵)≤2𝜖1−𝛼+2𝑑1(−𝐴,𝛼𝐵).(2.18)

Proof. If 𝑥,𝑦∈𝑃^𝑎𝑇(𝐴,𝐵), then 𝑑(𝑥,𝑦)≤≤𝑑𝜖₁(𝑥,𝑇𝑥)+𝑑(𝑇𝑥,𝑇𝑦)+𝑑(𝑇𝑦,𝑦)+𝛼𝑑(𝑥,𝑦)+2𝑑(𝐴,𝐵)+𝜖₂.(2.19) Put 𝜖=Max{𝜖₁,𝜖₂}, therefore, 𝑑(𝑥,𝑦)≤2𝜖/(1−𝛼)+(2𝑑(𝐴,𝐵))/(1−𝛼). Hence diam(𝑃^𝑎𝑇(𝐴,𝐵))≤2𝜖/(1−𝛼)+(2𝑑(𝐴,𝐵))/(1−𝛼).

3. Approximate Best Proximity for Two Maps

In this section, we will consider the existence of approximate best proximity points for two maps 𝑇∶𝐴∪𝐵→𝐴∪𝐵 and 𝑆∶𝐴∪𝐵→𝐴∪𝐵, and its diameter.

Definition 3.1. Let 𝐴 and 𝐵 be nonempty subsets of a metric space (𝑋,𝑑) and let 𝑇∶𝐴∪𝐵→𝐴∪𝐵𝑆∶𝐴∪𝐵→𝐴∪𝐵 two maps such that 𝑇(𝐴)⊆𝐵, 𝑆(𝐵)⊆𝐴. A point (𝑥,𝑦) in 𝐴×𝐵 is said to be an approximate-pair fixed point for (𝑇,𝑆) in 𝑋 if there exists 𝜖>0𝑑(𝑇𝑥,𝑆𝑦)≤𝑑(𝐴,𝐵)+𝜖.(3.1) We say that the pair (𝑇,𝑆) has the approximate-pair fixed property in 𝑋 if 𝑃^{𝑎(𝑇,𝑆)}(𝐴,𝐵)≠∅, where 𝑃^{𝑎(𝑇,𝑆)}(𝐴,𝐵)={(𝑥,𝑦)∈𝐴×𝐵∶𝑑(𝑇𝑥,𝑆𝑦)≤𝑑(𝐴,𝐵)+𝜖forsome𝜖>0}.(3.2)

Theorem 3.2. Let 𝐴 and 𝐵 be nonempty subsets of a metric space (𝑋,𝑑) and let 𝑇∶𝐴∪𝐵→𝐴∪𝐵 and 𝑆∶𝐴∪𝐵→𝐴∪𝐵 be two maps such that 𝑇(𝐴)⊆𝐵, 𝑆(𝐵)⊆𝐴. If, for every (𝑥,𝑦)∈𝐴×𝐵, 𝑑(𝑇^𝑛(𝑥),𝑆^𝑛(𝑦))⟶𝑑(𝐴,𝐵),(3.3) then (𝑇,𝑆) has the approximate-pair fixed property.

Proof. For 𝜖>0, Suppose (𝑥,𝑦)∈𝐴×𝐵. Since ∃𝑛₀𝑑(𝑇^𝑛(𝑥),𝑆^𝑛(𝑦))⟶𝑑(𝐴,𝐵),>0s.t.∀𝑛≥𝑛₀∶𝑑(𝑇^𝑛(𝑥),𝑆^𝑛(𝑦))<𝑑(𝐴,𝐵)+𝜖,(3.4) then 𝑑(𝑇(𝑇^𝑛−1(𝑥),𝑆(𝑆^𝑛−1(𝑦))<𝑑(𝐴,𝐵)+𝜖 for every 𝑛≥𝑛₀. Put 𝑥₀=𝑇^𝑛0−1(𝑥) and 𝑦₀=𝑆^𝑛0−1(𝑦)). Hence 𝑑(𝑇(𝑥₀),𝑆(𝑦₀))≤𝑑(𝐴,𝐵)+𝜖 and 𝑃^{𝑎(𝑇,𝑆)}(𝐴,𝐵)≠∅.

Theorem 3.3. Let 𝐴 and 𝐵 be nonempty subsets of a metric space (𝑋,𝑑) and let 𝑇∶𝐴∪𝐵→𝐴∪𝐵 and 𝑆∶𝐴∪𝐵→𝐴∪𝐵 be two maps such that 𝑇(𝐴)⊆𝐵, 𝑆(𝐵)⊆𝐴 and, for every (𝑥,𝑦)∈𝐴×𝐵, 𝑑(𝑇x,𝑆𝑦)≤𝛼𝑑(𝑥,𝑦)+𝛽[𝑑(𝑥,𝑇𝑥)+𝑑(𝑦,𝑆𝑦)]+𝛾𝑑(𝐴,𝐵),(3.5) where 𝛼,𝛽,𝛾≥0 and 𝛼+2𝛽+𝛾<1. Then if 𝑥 is an approximate fixed point for 𝑇, or 𝑦 is an approximate fixed point for 𝑆, then 𝑃^{𝑎(𝑇,𝑆)}(𝐴,𝐵)≠∅.

Proof. If (𝑥,𝑦)∈𝐴×𝐵, then 𝑑(𝑇𝑥,𝑆(𝑇𝑥))≤𝛼𝑑(𝑥,𝑇𝑥)+𝛽[𝑑(𝑥,𝑇𝑥)+𝑑(𝑇𝑥,𝑆(𝑇𝑥))]+𝛾𝑑(𝐴,𝐵).(3.6) Therefore, 𝑑(𝑇𝑥,𝑆(𝑇𝑥))≤𝛼+𝛽1−𝛽𝑑(𝑥,𝑇𝑥)+1−𝛾𝛽𝑑(𝐴,𝐵).(3.7) Now if 𝑘=(𝛼+𝛽)/(1−𝛽), then 𝑑(𝑇𝑥,𝑆(𝑇𝑥))≤𝑘𝑑(𝑥,𝑇𝑥)+(1−𝑘)𝑑(𝐴,𝐵)(∗) also 𝑑(𝑆𝑦,𝑇(𝑆𝑦))≤𝑘𝑑(𝑦,𝑆𝑦)+(1−𝑘)𝑑(𝐴,𝐵).(∗∗) If 𝑥 is an approximate fixed point for 𝑇, then there exists a 𝜖>0 and by (*) 𝑑(𝑇𝑥,𝑆(𝑇𝑥))≤≤=<𝑘𝑘𝑑𝑑𝑑(((𝑑𝐴𝐴(𝑥(,,𝐴,𝐵𝐵𝑇,))𝑥𝐵++))+𝑘𝜖+.𝜖(𝜖1)−+𝑘()1𝑑−(𝐴𝑘),𝐵𝑑)(𝐴,𝐵)(3.8) And (𝑥,𝑇𝑥)∈𝑃^{𝑎(𝑇,𝑆)}(𝐴,𝐵); also if 𝑦 is an approximate fixed point for 𝑆, then there exists a 𝜖>0 and by (**) 𝑑(𝑆𝑦,𝑇(𝑆𝑦))≤≤=<𝑘𝑘𝑑𝑑𝑑(((𝑑𝐴𝐴(𝑦(,,,𝐴𝐵𝐵𝑆,))𝑦𝐵++))+𝑘𝜖+.𝜖(𝜖1)−+𝑘()1𝑑−(𝐴𝑘),𝐵𝑑)(𝐴,𝐵)(3.9) And (𝑦,𝑆𝑦)∈𝑃^{𝑎(𝑇,𝑆)}(𝐴,𝐵). Therefore, 𝑃^{𝑎(𝑇,𝑆)}(𝐴,𝐵)≠∅.

Theorem 3.4. Let 𝐴 and 𝐵 be nonempty subsets of a metric space (𝑋,𝑑) and let 𝑇∶𝐴∪𝐵→𝐴∪𝐵 and 𝑆∶𝐴∪𝐵→𝐴∪𝐵 be two continuous maps such that 𝑇(𝐴)⊆𝐵, 𝑆(𝐵)⊆𝐴. If, for every (𝑥,𝑦)∈𝐴×𝐵, 𝑑(𝑇𝑥,𝑆𝑦)≤𝛼𝑑(𝑥,𝑦)+𝛾𝑑(𝐴,𝐵),(3.10) where 𝛼,𝛾≥0 and 𝛼+𝛾=1, also let {𝑥_𝑛} and {𝑦_𝑛} be as follows: 𝑥_𝑛+1=𝑆𝑦_𝑛,𝑦_𝑛+1=𝑇𝑥_𝑛forsome𝑥₁,𝑦₁∈𝐴×𝐵,𝑛∈𝑁.(3.11) If {𝑥_𝑛} has a convergent subsequence in 𝐴, then there exists a 𝑥₀∈𝐴 such that 𝑑(𝑥₀,𝑇𝑥₀)=𝑑(𝐴,𝐵).

Proof. We have 𝑑𝑥_𝑛+1,𝑦_𝑛+1≤=𝛼𝑑𝑑𝑇𝑥𝑥_𝑛𝑛,𝑆𝑦_𝑛≤≤𝛼⋯_𝑛+1,𝑦_𝑛+𝛾(𝑑(𝐴,𝐵)𝑑𝑥₀,𝑦₀+(1+𝛼+⋯+𝛼^𝑛)𝛾𝑑(𝐴,𝐵).(3.12) If {𝑥_𝑛𝑘}_𝑘≥1 converges to 𝑥₁∈𝐴, that is, 𝑥_𝑛𝑘→𝑥₁, then 𝑑𝑥_𝑛𝐾+1,𝑦_𝑛𝑘+1≤𝛼^𝑛𝑘+1𝑑𝑥₀,𝑦₀+1+𝛼+⋯+𝛼^𝑛𝑘𝛾𝑑(𝐴,𝐵).(3.13) Since 𝑇 is continuous, then 𝑑𝑥_𝑛𝑘+1,𝑇𝑥_𝑛𝑘⟶𝛾1−𝛼𝑑(𝐴,𝐵)=𝑑(𝐴,𝐵).(3.14) Therefore, 𝑑(𝑥₁,𝑇𝑥₁)=𝑑(𝐴,𝐵).

Definition 3.5. Let 𝑇∶𝐴∪𝐵→𝐴∪𝐵 and 𝑆∶𝐴∪𝐵→𝐴∪𝐵 be continues maps such that 𝑇(𝐴)⊆𝐵 and 𝑆(𝐵)⊆𝐴. We define diameter 𝑃^{𝑎(𝑇,𝑆)}(𝐴,𝐵) by diam𝑃^{𝑎(𝑇,𝑆)}(𝐴,𝐵)=sup{𝑑(𝑥,𝑦)∶𝑑(𝑇𝑥,𝑇𝑦)≤𝜖+𝑑(𝐴,𝐵)forsome𝜖>0}.(3.15)

Example 3.6. Suppose 𝐴={(𝑥,0)∶0≤𝑥≤1}, 𝐵={(𝑥,1)∶0≤𝑥≤1}, 𝑇(𝑥,0)=𝑇(𝑥,1)=(1/2,1), and 𝑆(𝑥,1)=𝑆(𝑥,0)=(1/2,0). Then 𝑑(𝑇(𝑥,0),𝑆(𝑦,1))=1 and diam(𝑃^{𝑎(𝑇,𝑆)}(𝐴,𝐵))=diam(𝐴×𝐵)=√2.

Theorem 3.7. Let 𝑇∶𝐴∪𝐵→𝐴∪𝐵 and 𝑆∶𝐴∪𝐵→𝐴∪𝐵 be continuous maps such that 𝑇(𝐴)⊆𝐵, 𝑆(𝐵)⊆𝐴. If there exists 𝑎  𝑘∈[0,1], 𝑑(𝑥,𝑇𝑥)+𝑑(𝑆𝑦,𝑦)≤𝑘𝑑(𝑥,𝑦),(3.16) then diam𝑃^{𝑎(𝑇,𝑆)}(𝐴,𝐵)≤𝜖1−𝑘+𝑑1(𝐴−,𝑘𝐵)forsome𝜖>0.(3.17)

Proof. If (𝑥,𝑦)∈𝑃^{𝑎(𝑇,𝑆)}(𝐴,𝐵), then 𝑑(𝑥,𝑦)≤≤𝜖𝑑+(𝑥𝑘,𝑇𝑑𝑥()𝑥+,𝑦𝑑)+(𝑇𝑑𝑥,(𝑆𝐴𝑦,)𝐵+).𝑑(𝑆𝑦,𝑦)(3.18) Therefore, 𝑑(𝑥,𝑦)≤𝜖/(1−𝑘)+(𝑑(𝐴,𝐵))/(1−𝑘). Then diam(𝑃^{𝑎(𝑇,𝑆)}(𝐴,𝐵))≤𝜖/(1−𝑘)+(𝑑(𝐴,𝐵))/(1−𝑘).