Coupled Fixed Point Theorems in Topological Spaces with Size Function Topology

Abstract

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

1. Introduction

The Banach contraction mapping theorem is the classical fixed point theorem saying that for a self-map , if is a complete metric space and there exists such that then 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 . They proved that if is a partially ordered complete metric space and a map has a mixed monotone property satisfying a certain contractive condition, then 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. Let be a topological space. A function is called a topology size function or simply -size function if (i), then (ii)max for all

His idea behind this definition has arisen from the diameter of an open interval in the set of real numbers.

Example 2. Let be the set of real number. For any open set , can be uniquely decomposed as a disjoint union of open intervals , . Define where for any .

Then, is a topology size function on .

Definition 3. Let be a topological space and be a -size function. For and . We define a -ball centered at a of radius to be We can verify that the collection of all -balls forms a basis for a topology on . The topology generated by this basis is denoted by . Clearly, .

In the rest of this paper, we always assume that is a topological space, is a -size function, and denote the set of positive integers.

Definition 4. (i)A -size function is said to be -uniform if for all and , we have that for some . For example, the topology size function , given in Example 2, is a -uniform(ii)A sequence in is said to converge to if for every , there exists such that for , . A point is called a -limit of the sequence and simply denoted by (iii)A sequence in is said to be -Cauchy if for every , there exists such that for , (iv)A subset of is said to be complete if every -convergent sequence in converges to a -limit in . Denote the set of all -limits of by

Remark 5 (see [16]). (i)Every -convergent sequence is a -Cauchy sequence(ii)Clearly, , and we obtain that is -complete if and only if if and only if every -Cauchy sequence converges in

2. Main Results

2.1. Fixed Point Theorem

For a given topological space and a -size function . Let be a self-map.

Definition 6. A point is said to be a fixed point of if .
We denote the th iteration of by

A function is called a contractant if it is increasing and for all . Note that if is a contractant, then and for all .

For a subset of , define .

Definition 7. A map is called a -contraction if there exists a contractant such that for all , is finite.

The uniqueness for the fixed point of the map is relied heavily on the following property of the -size function .

Definition 8. A -size function is said to separate points in or simply is separated if any two distinct points and in , there exist such that .

Example 9. Consider the discrete space where is the power set of . Define a topology size function by where is the number of elements in . Clearly, is separated.

Observe that if we consider as the discrete metric space, then we have that , but for all .

Robdera proved the extended Matkowski fixed point theorem as follows.

Theorem 10 (Robdera M. Theorem 3.1). Let be a topological space and be a -uniform -size function. Let be a -contraction. Assume that is complete, separated, and is finite. Then, 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 domain is finite. In this work, we weaken this condition by introducing the notion of locally finite size as follows.

Definition 11. Let be a topological space and be a -size function. We say that is locally finite size if for each , there exists such that and is finite.

We can verify that if is finite, then is locally finite size. The discrete space in Example 9 is an example of a topological size function space which is locally finite size.

Theorem 12. Let be a topological space and be a -uniform -size function. Let be a -contraction. Assume that is complete, separated, and locally finite size. Then, admits a unique fixed point.

Proof. Let be a contractant and be an arbitrary point in . Then, we define a sequence in by setting .
First, we will show that the sequence is -Cauchy. Since is locally finite size, there exists such that and is finite, which says .
Hence, we have that and . This implies that there exists such that and .
Then, and .
Continuing in this way, we obtain that for , and .
Given , then . Since as , there exists such that for , we have and .
Hence, for , we have that and are in and This implies that . Thus, is -Cauchy. Since is complete, there exists such that converges to .
Next, we will show that . Given , since is continuous (see [11]), there exists such that implies that . Setting . Since as , there exists such that for , .
In particular, implies that there exists such that and . By the continuity of , we have that and .
Hence, and Therefore, for every . Hence, .
Finally, we will show that is the unique fixed point of . Assume that there exists such that . Suppose that . Since is separated, there exists such that . Obviously, . Observe that and Hence, which is a contradiction. Therefore, . The proof is complete.

Notice that the map is defined by , where is a contractant. Immediately, we obtain the extension of the Banach contraction mapping theorem as follows.

Corollary 13. Let be a topological space and be a -uniform -size function. Assume that there exists such that for all , is finite. If is complete, separated, and locally finite size, then admits a unique fixed point.

2.2. Coupled Fixed Point Theorem

Let be a topological space and be a -size function. Let be a map.

Definition 14. (i)A point is called a coupled fixed point of if and .(ii)A point is called a fixed point of if . Clearly, a fixed point of is a coupled fixed point of .

Definition 15. A map is said to satisfy the product contraction if there exists a contractant such that for all , and are finite.

Example 16. Let . Let be the topology on given by

Define the topology size function by , , and .

Let be defined by for all .

Then, for , we have that for any contractant

Therefore, satisfies the product contraction . Observe that admits a unique coupled fixed point .

Remark 17. Observe that we can naturally define the product topology on inherited from . More precisely, the basis element is of the form where . But it is still questionable to define the size function corresponding to this product topology in term of , because the union of two basis elements might not be a product form. Nevertheless, the following lemma can be obtained by the product contraction .

Lemma 18. Let be a topological space and be a -size function. Let be a map satisfying the product contraction . Then, for every and , there exists such that for , if and , then .

Proof. Let and . Choose . Then, for , if and , we have that and

Theorem 19. Let be a topological space and be a -uniform -size function. Let be a map satisfying the product contraction . Assume that is complete, separated, and locally finite size. Then, has a unique coupled fixed point.

Proof. Let be a contractant and choose an arbitrary point . Since is locally finite size, there exist such that and are finite. Setting and .
We inductively define two sequences in as follows: Next, we will show that both and are -Cauchy. Clearly, and . Similarly, and .
Hence, there exist such that We obtain that and Similarly, and .
Inductively, we obtain that for there exist such that .
Given , then or . Since , there exist such that for implies that and for implies that .
For , we have that and This means that , so is -Cauchy. By the analogous argument, we can show that is also -Cauchy.
By the -completeness of , there exist and in such that and .
Next, we will show that is a coupled fixed point or and .
Given . By Lemma 18, there is such that for if and , then .
Since and , there exist such that where .
In particular, there exist such that , , and Then, .
Hence, we have that and .
Thus, and That is for all . Hence, . Similarly, we can show that .
Finally, we will show that is the unique coupled fixed point of . Suppose that there is such that and . Suppose that . Then, or .
Assume that . Since is separated and -uniform, there exists such that . Clearly, . Since is locally finite size, there exist such that and are finite. We have that and This implies that which is a contradiction.
Similarly, if we assume that , then it will also lead to the contradiction. Therefore, . The proof is complete.

Remark 20. Observe that in the proof of Theorem 19, we can replace the product contraction by the weaker condition: if the contractant is assumed to be a linear function. By the Cauchy functional equation, this implies that for some .

Definition 21. A map is said to satisfy the product contraction if there exists such that for all , and are finite.

Example 22. Let . Let be the topology on given by

Define the topology size function by and .

Let be defined by for all .

Then, for , where

Therefore, satisfies the product contraction . Observe that has a unique coupled fixed point .

Before we prove the next main theorem, the following lemma is required.

Lemma 23. Let be a topological space and be a -size function. Let be a map satisfying the product contraction . Then, for every and , there exists such that for , if and , then .

Proof. Let and . Choose . Then, for , if and , we have that and

Theorem 24. Let be a topological space and be a -uniform -size function. Let be a map satisfying the product contraction . If is complete, separated, and locally finite size. Then, admits a unique coupled fixed point.

Proof. Let be an arbitrary point. Since is locally finite size, there exist such that and are finite. Setting and and let .
We inductively define two sequences in as follows: Next, we will show that both and are -Cauchy. Clearly, and . Similarly, and .
Hence, there exist such that We obtain that and Similarly, and .
Continuing in this manner, we obtain that for there exist such that Given , then or . Since , there exist such that for implies that and for implies that .
For , we have that and This means , so is -Cauchy. By the similar argument, we can show that is also -Cauchy.
By the -completeness of , there exist and in such that and .
Next, we will show that is a coupled fixed point or and .
Given , by Lemma 23, there is such that for if and , then .
Since and , there exist such that where .
In particular, there exist such that , , and Then, .
Hence, we have that and .
Thus, and That is for all . Hence, . Similarly, we can show that .
Finally, we will show that is the unique coupled fixed point of . Suppose that there is such that and . Suppose that . Then, or .
Assume that . Since is separated and -uniform, there exists such that . Clearly, .

Case 1. If , then there exists such that . Setting . We have that and

This implies that which is a contradiction.

Case 2. If , then there exists such that and is finite. Setting . We have that and we can similarly show that This implies that which is a contradiction.

Similarly, if we assume that , then it will also lead to the contradiction. Hence, . The proof is complete.

If space failed to be separated by the topology size function , then the map might not have a unique coupled fixed point.

Example 25. Consider the closed interval . Let be the topology on given by Define the topology size function by , , and . Clearly, does not separate the points on .
Let be defined by for all . We obtain that satisfies the product contraction . The coupled fixed points of are and .

Data Availability

No data were used to support this study.

Conflicts of Interest

The author declares that there are no conflicts of interest.