Fundamental Characteristics of the Product-Operated Metric Spaces

Abstract

In the field of metric fixed point theory, there are several generalized or modified types of metric space. One such type is called multiplicative metric space. It was initially introduced as a modified version of the metric space, but was later found to be equivalent to the metric space. However, it allows the researchers to view the concept of metric space from a different perspective. Consequently, the idea of product-operated metric space is introduced in this article, which is obtained by removing the slackness of the multiplicative metric space. This article also presents some fundamental characteristics of the product-operated metric space and investigates the existence of fixed points for self-maps in product-operated metric spaces.

1. Introduction and Preliminaries

The concept of non-Newtonian calculus, also known as multiplicative calculus, was presented by Grossman and Katz [1]. The idea used to establish this non-Newtonian calculus was to interchange the roles of subtraction and addition in basic calculus with division and multiplication, respectively. Following the ideology of Grossman and Katz [1], Bashirov et al. [2] presented the notion of multiplicative metric spaces in the following way:

Definition 1.

A mapping $d_m:W\times W\to [1,\infty )$ is said to be a multiplicative metric [2] on a nonempty set W if for all $x,y,z\in W$, $d_m$ satisfies the following conditions:

(m1)

$d_m(x,y)>1$ for all $x,y\in W$ with $x\ne y$ and $d_m(x,y)=1$ if and only if $x=y$;

(m2)

$d_m(x,y)=d_m(y,x)$ for all $x,y\in W$;

(m3)

$d_m(x,z)\le d_m(x,y)·d_m(y,z)$ for all $x,y,z\in W$.

Several researchers have contributed to the establishment of the fixed point theory using multiplicative metric spaces. For instance, Özavsar and Cevikel [3] investigated the topological properties of multiplicative metric spaces along with some fixed-point theorems for the multiplicative contraction mappings of multiplicative metric spaces; He et al. [4] established common fixed point results for weak commutative mappings in multiplicative metric spaces; Abbas et al. [5] studied common fixed point results for mapping satisfying generalized rational-type contraction conditions in multiplicative metric spaces; Yamaod and Sintunavarat [6] established fixed-point results for generalized contraction mappings with cyclic $(\alpha ,\beta )$-admissibility in multiplicative metric spaces; Gu and Cho [7] established common fixed point results for four maps satisfying $\varphi$-contractive condition; Mongkolkeha and Sintunavarat [8] studied the existence of best proximity points for multiplicative proximal contraction mapping.

Later, many researchers, including Abodayeh et al. [9]; Agarwal et al. [10]; Došenović and Radenović [11] published remarks on multiplicative metric spaces, stating that the concept of multiplicative metric spaces is equivalent to the concept of metric spaces. A detailed survey of multiplicative metric spaces and related fixed point results was conducted by Došenović et al. [12].

The literature on fixed point theory contains several generalized versions of metric spaces and fixed point theorems in these generalized versions: for example, partial metric space [13], vector-valued metric space [14], b-metric space [15], vector-valued b-metric space [16], K-metric space [17], orthogonal m-metric spaces [18] etc. One of the most recent generalized version of metric space was presented in [19] called Czerwik vector-valued $\mathcal{R}$-metric space. Readers can find a comprehensive study of metric fixed point theory and generalized forms of metric spaces in [20,21,22].

Discussion on Multiplicative Metric Spaces

We know that 0 is the additive identity, and 1 is the multiplicative identity in the set of real numbers under the usual operations of addition and multiplication. To define the multiplicative metric spaces, Bashirov et al. [2] suggested two modifications in the concept of metric spaces: first, interchange the addition operation with the multiplication operation in the triangle inequality, and second, interchange additive identity with multiplicative identity. One can think about the following question: What was the need to interchange additive identity with multiplicative identity? A possible reason was the axiom $\left(m_3\right)$ of Definition 1. Because when

$$d_m(k,l)=0\mathrm{if}\mathrm{and}\mathrm{only}\mathrm{if}k=l$$

and

$$d_m(k,q)\le d_m(k,l)·d_m(l,q)\forall k,l,q\in W.$$

Then, for $k=l$ and $l\ne q$, we get

$$d_m(k,q)\le d_m(k,l)·d_m(l,q)=0\forall k,q\in W.$$

That is

$$d_m(l,q)\le 0\forall l\ne q.$$

2. Main Results

In this section, we introduce the concept of product-operated metric spaces, which involves a multiplication operation in the triangle inequality rather than an addition operation. It is also important to note that this concept is different from multiplicative metric spaces because we use additive identity and define the triangle inequality in such a way that avoids the above-mentioned issue—that is, $d_m(l,q)\le 0\forall l\ne q.$

Definition 2.

A mapping $d_M:W\times W\to \mathbb{R}$ is called a product-operated metric on W, if for all $w_1,w_2,w_3\in W$ the following conditions are satisfied:

$\left(c_1\right)$

$d_M(w_1,w_2)\ge 0$;

$d_M(w_1,w_2)=0$ if and only if $w_1=w_2$;

$\left(c_2\right)$

$d_M(w_1,w_2)=d_M(w_2,w_1)$;

$\left(c_3\right)$

$d_M(w_1,w_3)\le d_M(w_1,w_2)·d_M(w_2,w_3)$, provided $w_1\ne w_2$ and $w_2\ne w_3$.

Then, the pair $(W,d_M)$ is called a product-operated metric space.

Example 1.

Consider $W=\{a,b\}$. Define a product-operated metric on W by

$$d_M(w,q)=\left\{\begin{array}{c}0.8,if w\ne q\hfill \\ 0,if w=q.\hfill \end{array}\right.$$

It is important to note that the above-defined function is a product-operated metric on W but not a multiplicative metric on W.

Example 2.

Consider $W=\mathbb{N}$ as the set of all natural numbers. Define a product-operated metric on W by

$$d_M(w,q)=\left\{\begin{array}{c}0,if w=q\hfill \\ 1,if w,q are even numbers\hfill \\ 9,if w,q are odd numbers\hfill \\ 3,otherwise.\hfill \end{array}\right.$$

The axioms $\left(c_1\right)$ and $\left(c_2\right)$ are trivially held. Now, we discuss $\left(c_3\right)$ by the following cases:

• Case 1: Consider $w_1=e_1$, $w_2=e_2$, and $w_3=e_3$, where $e_1,e_2,e_3$ are even numbers with $e_1\ne e_2$ and $e_2\ne e_3$. Then $d_M(e_1,e_3)=1or0$, $d_M(e_1,e_2)=1$, and $d_M(e_2,e_3)=1$.
• Case 2: Consider $w_1=e_1$, $w_2=e_2$, and $w_3=o_3$, where $e_1,e_2$ are even numbers with $e_1\ne e_2$ and $o_3$ is an odd number. Then $d_M(e_1,o_3)=3$, $d_M(e_1,e_2)=1$, and $d_M(e_2,o_3)=3$.
• Case 3: Consider $w_1=e_1$, $w_2=o_2$, and $w_3=o_3$, where $e_1$ is an even number and $o_2,o_3$ are odd numbers with $o_2\ne o_3$. Then $d_M(e_1,o_3)=3$, $d_M(e_1,o_2)=3$, and $d_M(o_2,o_3)=9$.
• Case 4: Consider $w_1=o_1$, $w_2=o_2$, and $w_3=o_3$, where $o_1,o_2,o_3$ are odd numbers with $o_1\ne o_2$ and $o_2\ne o_3$. Then $d_M(o_1,o_3)=9or0$, $d_M(o_1,o_2)=9$, and $d_M(o_2,o_3)=9$.
• Case 5: Consider $w_1=o_1$, $w_2=e_2$, and $w_3=o_3$, where $o_1,o_3$ are odd numbers and $e_2$ is an even number. Then $d_M(o_1,o_3)=9or0$, $d_M(o_1,e_2)=3$, and $d_M(e_2,o_3)=3$.
• Hence, we get $d_M(w_1,w_3)\le d_M(w_1,w_2)·d_M(w_2,w_3)$, provided $w_1\ne w_2$ and $w_2\ne w_3$.

It is important to note that the above-defined function is neither a metric, since $d_M(3,5)=9\nleqq d_M(3,4)+d_M(4,5)=3+3$, nor a multiplicative metric, $d_M(3,3)=0\ne 1$, on W.

Example 3.

Consider $W=\mathbb{N}$ is the set of all natural numbers. Define a product-operated metric on W by

$$d_M(w,q)=\left\{\begin{array}{c}{2}^{|w-q|},ifw\ne q\hfill \\ 0,ifw=q.\hfill \end{array}\right.$$

It is clear that axioms $c_1$ and $c_2$ hold. If $a,b>0$ and $a\le b$, then $2^a\le 2^b$. As $|w_1-w_3|\le |w_1-w_2|+|w_2-w_3|$, then, $2^{|w_1-w_3|}\le 2^{|w_1-w_2|+|w_2-w_3|}=2^{|w_1-w_2|}·2^{|w_2-w_3|}$. Thus, we say that $d_M(w_1,w_3)\le d_M(w_1,w_2)·d_M(w_2,w_3)$, provided $w_1\ne w_2$ and $w_2\ne w_3$. Therefore, $(W,d_M)$ is a product-operated metric space.

Note that it is not a metric space. To check consider $w_1=2$, $w_2=4$ and $w_3=6$, then $d_M(2,4)=2^2=4$, $d_M(2,6)=2^4=16$, $d_M(4,6)=2^2=4$. Now we have

$$d_M(2,6)=2^4=16>d_M(2,4)+d_M(4,6)=4+4.$$

Further, it is not a multiplicative metric space. Since $d_M(2,2)=0\ne 1$.

A few other examples of product-operated metric spaces are given below.

Example 4.

(a)

Consider $W=[1,\infty )$. A product-operated metric on W is defined by

$$d_M(w,q)=\left\{\begin{array}{c}wq,ifw\ne q\hfill \\ 0,ifw=q.\hfill \end{array}\right.$$

(b)

Consider $W=(0,1)$. A product-operated metric on W is defined by

$$d_M(w,q)=\left\{\begin{array}{c}\frac{1}{wq},ifw\ne q\hfill \\ 0,ifw=q.\hfill \end{array}\right.$$

(c)

Consider $W=[1,\infty )$. A product-operated metric on W is defined by

$$d_M(w,q)=\left\{\begin{array}{c}max\{w,q\},ifw\ne q\hfill \\ 0,ifw=q.\hfill \end{array}\right.$$

Definition 3.

The concepts of convergent and Cauchy sequences in a product-operated metric space, $(W,d_M)$, are defined as follows:

(1)

A sequence $\left(w_n\right)$ in W is convergent to $w\in W$, if ${lim}_{n\to \infty }{d}_M(w_n,w)=0$.

(2)

A sequence $\left(w_n\right)$ in W is Cauchy if ${lim}_{n,m\to \infty }{d}_M(w_n,w_m)=0$.

Definition 4.

A product-operated metric space is said to be complete if each Cauchy sequence in it is convergent.

2.1. Motivation

Consider the fission process of a cell at different stages, as defined in the following table.

$$\begin{array}{ccc}& Stage1& 1cell\hfill \\ & Stage2& 2cells\hfill \\ & Stage3& 4cells\hfill \\ & Stage4& 8cells\hfill \\ & Stage5& 16cells\hfill \\ & Stage6& 32cells\hfill \\ & Stage7& 64cells\hfill \\ & & ⋮\hfill \end{array}$$

Now, we are interested in finding a function that describes how many cells are generated by one cell between two stages in the above-listed fission process.

A required function for Stage i and Stage j can be defined as

$$CG(S_i,S_j)=CG(S_j,S_i)=\left\{\begin{array}{c}{2}^{|i-j|},ifi\ne j\hfill \\ 0,ifi=j.\hfill \end{array}\right.$$

For instance, consider the following points:

(i)

For Stage 1 and Stage 7, we say that a single cell of Stage 1 is converted into 64 cells of Stage 7. Alternatively, we say that 64 cells of Stage 7 are created by a single cell of Stage 1. Whereas, $CG(S_1,S_7)=CG(S_7,S_1)=2^{|1-7|}=2^{\left|6\right|}=64$.

(ii)

For Stage 2 and Stage 7, we say that 2 cells of Stage 2 are converted into 64 cells of Stage 7; that is, each cell of Stage 2 creates 32 cells in Stage 7. Whereas, $CG(S_2,S_7)=2^{|2-7|}=2^{\left|5\right|}=32$.

(iii)

For Stage 3 and Stage 7, we say that 4 cells of Stage 3 are converted into 64 cells of Stage 7; that is, each cell of Stage 3 creates 16 cells in Stage 7. Whereas, $CG(S_3,S_7)=2^{|3-7|}=2^{\left|4\right|}=16$.

(iv)

If both stages are the same, then no generation occurs; that is, $CG(S_i,S_i)=0$ means each cell in stage i generates no cell in stage i.

Remark 1.

Note that the defined function $CG$ for the above-listed fission process is a product-operated metric. Hence, we can define the distance between two stages of the fission process by the number of cells generated by one cell between the two stages.

2.2. Basic Results for Product-Operated Metric Spaces

In this section, we discuss a few results on the characteristics of product-operated metric spaces, which are also required to derive the results of the next section.

Theorem 1.

Every convergent sequence in a product-operated metric space, $(W,d_M)$, is Cauchy.

Proof. 

Suppose that $w_n\to a$ as $n\to \infty$.

If $\left(w_n\right)$ contains a constant tail, that is, $w_n=a$$\forall n>k$ for some $k\in \mathbb{N}$, then it is obvious that $d_M(w_n,w_m)=0$$\forall n,m>k$.

If $\left(w_n\right)$ does not contain a constant tail, then $w_n\ne a$ for infinitely many n. Using $\left(c_3\right)$, we get

$$d_M(w_n,w_m)\le d_M(w_n,a)·d_M(a,w_m)\forall n,m\in \mathbb{N}$$

provided that $d_M(w_n,a)\ne 0$ and $d_M(w_m,a)\ne 0$. By applying the limit $n,m\to \infty$, we get $d_M(w_n,w_m)\to 0$. □

Theorem 2.

Let $(W,d_M)$ be a product-operated metric space, and let $\left(w_n\right)$ be a convergent sequence in W. Then the limit point of $\left(w_n\right)$ is unique.

Proof. 

Suppose that $w_n\to a$ and $w_n\to b$ as $n\to \infty$. We have to show that $a=b$.

First, suppose that $\left(w_n\right)$ has a constant tail. That is, there exists $n_1\in \mathbb{N}$ such that $w_n=c$$\forall n\ge n_1$ for some c in W. Then $d_M(w_n,a)=d_M(c,a)$$\forall n\ge n_1$, this implies $d_M(c,a)=0$ since $w_n\to a$. Similarly, $d_M(c,b)=0$ since $w_n\to b$. Thus, we get $c=a=b$. Hence, the limit point is unique in this case.

Next, suppose that $\left(w_n\right)$ has no constant tail. Then, for infinitely many n, we have $d_M(w_n,a)\ne 0$ and $d_M(w_n,b)\ne 0$. By the facts $w_n\to a$ and $w_n\to b$ as $n\to \infty$, we say that $d_M(w_n,a)\to 0$ and $d_M(w_n,b)\to 0$ as $n\to \infty$. Thus, we get $d_M(w_n,a)·d_M(w_n,b)\to 0$ as $n\to \infty .$ By $\left(c_3\right)$, we get

$$d_M(a,b)\le d_M(a,w_n)·d_M(w_n,b)$$

provided $d_M(w_n,a)\ne 0$ and $d_M(w_n,b)\ne 0$. By applying the limit, we get $d_M(a,b)=0$.

Hence, the limit point of a convergent sequence is unique. □

Theorem 3.

If a sequence $\left(w_n\right)$ is convergent to a in a product-operated metric space, $(W,d_M)$, then each subsequence of $\left(w_n\right)$ is also convergent to a.

Proof. 

Given that $w_n\to a$. On the contrary, assume that $\left(w_{n_k}\right)$ is a subsequence of $\left(w_n\right)$ such that $w_{n_k}\to b$ with $b\ne a$.

First, we assume that $\left(w_{n_k}\right)$ has a constant tail. That is, $w_{n_k}=b$$\forall n_k\ge n_{k_0}$. Then, by $\left(c_3\right)$, we get

$$d_M(b,w_n)\le d_M(b,a)·d_M(a,w_n)\forall n\in \mathbb{N}$$

provided that $d_M(w_n,a)\ne 0$. This implies that $w_n\to b$ as $n\to \infty$. Since the limit point of a convergent sequence is unique, our assumption $w_{n_k}\to b$ with $b\ne a$ is wrong when $\left(w_{n_k}\right)$ has a constant tail.

Next, if we assume that $\left(w_{n_k}\right)$ has no constant tail, then $w_{n_k}\ne b$ for infinitely many terms. Using $\left(c_3\right)$, we get

$$d_M(a,w_{n_k})\le d_M(a,b)·d_M(b,w_{n_k})\forall n_k$$

provided that $d_M(w_{n_k},b)\ne 0$. This implies that $w_{n_k}\to a$ as $k\to \infty$, which contradicts our assumption.

Hence, if $w_n\to a$, then each subsequence $\left(w_{n_k}\right)$ of $\left(w_n\right)$ converges to a. □

The direct proof of the above theorem is given below.

On the contrary, assume that $\left(w_{n_k}\right)$ is a subsequence of $\left(w_n\right)$ such that $w_{n_k}\to b$ with $b\ne a$. Then, by $\left(c_3\right)$, we get

$$d_M(b,w_n)\le d_M(b,a)·d_M(a,w_n)\forall n\in \mathbb{N}$$

provided that $d_M(w_n,a)\ne 0$. This implies that $w_n\to b$ as $n\to \infty$, which is a contradiction to our assumption, that is, $b\ne a$, since the limit point of a convergent sequence is unique. Hence, $b=a$. That is, every subsequence $\left(w_{n_k}\right)$ of $\left(w_n\right)$ converges to a.

Theorem 4.

Let $(W,d_M)$ be a product-operated metric space, and let $w_n\to a$ as $n\to \infty$. Then, $d_M(w_n,b)\to d_M(a,b)$ for each $b\in W$.

Proof. 

Consider $b\in W$ to be an arbitrary element. We discuss the proof of this theorem in the following two cases:

Case 1: When $a=b$, the result is trivial.

Case 2: When $a\ne b$.

If the sequence $\left(w_n\right)$ contains a constant tail. That is, $w_n=a$$\forall n\ge m$. Then, $d_M(w_n,b)=d_M(a,b)\forall n\ge m$. Thus, ${lim}_{n\to \infty }{d}_M(w_n,b)=d_M(a,b)$.

If the sequence $\left(w_n\right)$ does not contain a constant tail, then $d_M(w_n,b)>0$ for infinitely many n. Now, suppose on the contrary that $d_M(w_n,b)\nrightarrow d_M(a,b)$. Then, for some constant $c>0$, we get $|d_M(w_n,b)-d_M(a,b)|>c>0$ for infinitely many sufficiently large values of n. That is, either $d_M(w_n,b)<d_M(a,b)$ or $d_M(w_n,b)>d_M(a,b)$ for infinitely many sufficiently large n.

Case 2a: Consider $d_M(w_n,b)>d_M(a,b)$ for infinitely many n. Then, by $\left(c_3\right)$, we get

$$d_M(a,b)<d_M(w_n,b)\le d_M(w_n,a)·d_M(a,b)$$

provided $d_M(w_n,a)\ne 0$. By applying the limit $n\to \infty$, we get $d_M(a,b)\le 0$.

Case 2b: Consider $d_M(w_n,b)<d_M(a,b)$ for infinitely many n. Then we get

$$d_M(w_n,a)·d_M(w_n,b)\le d_M(w_n,a)·d_M(a,b).$$

That is,

$$d_M(a,b)\le d_M(w_n,a)·d_M(w_n,b)\le d_M(w_n,a)·d_M(a,b)$$

provided $d_M(w_n,a)\ne 0$ and $d_M(w_n,b)\ne 0$. Again, by applying the limit $n\to \infty$, we get $d_M(a,b)\le 0$.

Hence, Cases 2a and 2b imply that $d_M(a,b)\le 0$, but in Case 2, we have $d_M(a,b)>0$, which means our assumption that $|d_M(w_n,b)-d_M(a,b)|>0$ for infinitely many n is wrong. Therefore, we conclude that $d_M(w_n,b)\to d_M(a,b)$ as $n\to \infty$. □

Theorem 5.

Let $(W,d_M)$ be a product-operated metric space, and let $\left(w_n\right)$ and $\left(q_n\right)$ be two sequences in W. If $w_n\to a$ and $q_n\to b$ as $n\to \infty$, then $d_M(w_n,q_n)\to d_M(a,b)$ as $n\to \infty$.

Proof. 

Since $w_m\to a$ as $m\to \infty$, using Theorem 4, we get

$$d_M(w_m,q_n)\to d_M(a,q_n)\mathrm{for}\mathrm{each}{q}_n\mathrm{as}m\to \infty .$$

Since $q_n\to b$ as $n\to \infty$, using Theorem 4, we get

$$d_M(a,q_n)\to d_M(a,b)\mathrm{as}n\to \infty .$$

Therefore, we conclude that

$$d_M(w_n,q_n)\to d_M(a,b)\mathrm{as}n\to \infty .$$

□

Definition 5.

Let $(W,d_M)$ be a product-operated metric space. Then

(1)

An open ball of radius r with center $w_0$ is defined by

$$B_{d_M}(w_0,r)=\{w\in W:d_M(w,w_0)<r\}.$$

(2)

An open set in W is a set containing an open ball about each of its points.

Theorem 6.

An open ball $B_{d_M}(w_0,r)$ in a product-operated metric space, $(W,d_M)$, is an open set.

Proof. 

Let $q\in B_{d_M}(w_0,r)$ be an arbitrary element with $q\ne w_0$. If $q=w_0$, then we say that $B_{d_M}(w_0,r)\subseteq B_{d_M}(w_0,r)$. Let $\rho =\frac{r}{1+d_M(q,w_0)}$ and $z\in B_{d_M}(q,\rho )$ be an arbitrary element, that is, $d_M(q,z)<\rho =\frac{r}{1+d_M(q,w_0)}$. If we consider that $q=z$, then trivially we have $d_M(w_0,z)=d_M(w_0,q)<r$. If $q\ne z$, then by $\left(c_3\right)$, we get

$$\begin{array}{ccc}\hfill d_M(w_0,z)& \le & d_M(w_0,q)·d_M(q,z)\hfill \\ & <& d_M(w_0,q)·\rho \le r.\hfill \end{array}$$

Thus, we get $B_{d_M}(q,\rho )\subseteq B_{d_M}(w_0,r)$ for each $q\in B_{d_M}(w_0,r)$. Hence, an open ball is an open set. □

Note that the set $\{B_{d_M}(w,r):w\in W,r>0\}$ defines a neighborhood system for the topology $\tau$ on W induced by the product-operated metric space.

2.3. Fixed Point Results on Product-Operated Metric Spaces

In this section, we will derive the results on product-operated metric spaces to ensure the existence of fixed points for self-mapping.

Lemma 1.

Let $(W,d_M)$ be a product-operated metric space, and let $\left(w_n\right)$ be any iterative sequence for a map $R:W\to W$ such that $w_n=R{w}_{n-1}$, $w_{n-1}\ne w_n$$\forall n\in \mathbb{N}$, and

$$d_M(w_n,w_{n+1})\le r^n{d}_M(w_0,w_1)\forall n\in \mathbb{N}.$$

(1)

Then, $w_k\ne w_{k+p}$ for each $k,p\in \mathbb{N}$.

Proof.

On the contrary, suppose that there are two natural numbers k and p such that $w_k=w_{k+p}$. Then, we have $R{w}_k=R{w}_{k+p}$, that is, $w_{k+1}=w_{k+p+1}$. Continuing in this way, we get

$$\begin{array}{ccc}& & w_{k+1}=w_{k+p+1}\hfill \\ & & w_{k+2}=w_{k+p+2}\hfill \\ & & w_{k+3}=w_{k+p+3}\hfill \\ & & ⋮\hfill \\ & & w_{k+p}=w_{k+p+p}⟹w_k=w_{k+2p}\hfill \\ & & w_{k+p+1}=w_{k+2p+1}⟹w_{k+1}=w_{k+2p+1}\hfill \\ & & w_{k+p+2}=w_{k+2p+2}⟹w_{k+2}=w_{k+2p+2}\hfill \\ & & w_{k+p+3}=w_{k+2p+3}⟹w_{k+3}=w_{k+2p+3}\hfill \\ & & ⋮\hfill \\ & & w_{k+2p}=w_{k+3p}⟹w_k=w_{k+3p}\hfill \\ & & w_{k+2p+1}=w_{k+3p+1}⟹w_{k+1}=w_{k+3p+1}\hfill \\ & & w_{k+2p+2}=w_{k+3p+2}⟹w_{k+2}=w_{k+2p+2}\hfill \\ & & ⋮\hfill \end{array}$$

Hence, a sequence $\left(w_n\right)$ is of the form

$$(w_0,w_1,w_2,\cdots ,w_{k+p-1},w_k,w_{k+1},w_{k+2},\cdots ,w_{k+2p-1},w_k,w_{k+1},w_{k+2},\cdots ,w_{k+3p-1},\cdots ).$$

By considering the above sequence and (1), we get

$$d_M(w_k,w_{k+1})=d_M(w_{k+p},w_{k+p+1})\le r^{k+p}{d}_M(w_0,w_1).$$

$$d_M(w_k,w_{k+1})=d_M(w_{k+2p},w_{k+2p+1})\le r^{k+2p}{d}_M(w_0,w_1).$$

$$\begin{array}{c}{d}_M(w_k,w_{k+1})=d_M(w_{k+3p},w_{k+3p+1})\le r^{k+3p}{d}_M(w_0,w_1)\\ ⋮\end{array}$$

Thus, we conclude that

$$d_M(w_k,w_{k+1})\le r^{k+lp}{d}_M(w_0,w_1)\forall l\in \mathbb{N}.$$

The above inequality holds only if $w_k=w_{k+1}$, and it contradicts the given hypothesis of the lemma. Hence, our assumption is wrong, and there are no k and p for which $w_k=w_{k+p}$.□

We now state and prove some results regarding the existence of fixed points for a self-map using a product-operated metric space.

Theorem 7.

Let $(W,d_M)$ be a complete product-operated metric space, and let $R:W\to W$ and $\alpha :W\times W\to [0,\infty )$ be two mappings such that for each $w,l\in W$:

$$\begin{array}{ccc}\hfill & & \alpha (w,l)\ge 1⟹\hfill \\ & & d_M(Rw,Rl)\le rmax\left\{d_M(w,l),\sqrt{d_M(w,Rw)·d_M(l,Rl)},\sqrt{d_M(w,Rl)·d_M(l,Rw)}\right\}\hfill \end{array}$$

(2)

where $r\in (0,1)$. Also, consider the following axioms:

(i)

there exists $w_0\in W$ with $\alpha (w_0,R{w}_0)\ge 1$;

(ii)

for each $w,l\in W$ with $\alpha (w,l)\ge 1$, we have $\alpha (Rw,Rl)\ge 1$;

(iii)

for each $\left(w_n\right)$ in W with $\alpha (w_n,w_{n+1})\ge 1\forall n\in \mathbb{N}$ and $w_n\to w\in W$, we have $\alpha (w_n,w)\ge 1$$\forall n\in \mathbb{N}$.

Then, R has a fixed point.

Proof. 

Axiom (i) implies the existence of an element $w_0\in W$ with $\alpha (w_0,R{w}_0)\ge 1$. Using this expression and Axiom (ii), we reach the fact that $\alpha (R^n{w}_0,R^{n+1}{w}_0)\ge 1$$\forall n\in \mathbb{N}$. Thus, by defining a sequence $\left(w_n\right)$ such that $w_n=R{w}_{n-1}=R^n{w}_0\forall n\in \mathbb{N}$, we get $\alpha (w_n,R{w}_n)\ge 1$$\forall n\in \mathbb{N}\cup \left\{0\right\}$. If there is some $n_0$ such that $w_{n_0}=w_{n_0+1}$, then $w_{n_0}$ is a fixed point of R, which completes the proof. Hence, we assume that $w_n\ne w_{n+1}$$\forall n\in \mathbb{N}\cup \left\{0\right\}$. Then, by (2), we get

$$\begin{array}{ccc}\hfill d_M(R{w}_n,R{w}_{n+1})& \le & rmax\{d_M(w_n,w_{n+1}),\sqrt{d_M(w_n,R{w}_n)·d_M(w_{n+1},R{w}_{n+1})},\hfill \\ & & \sqrt{d_M(w_n,R{w}_{n+1})·d_M(w_{n+1},R{w}_n)}\}\forall n\in \mathbb{N}\cup \left\{0\right\}.\hfill \end{array}$$

That is,

$$\begin{array}{ccc}\hfill d_M(w_{n+1},w_{n+2})& \le & rmax\{d_M(w_n,w_{n+1}),\sqrt{d_M(w_n,w_{n+1})·d_M(w_{n+1},w_{n+2})},\hfill \\ & & \sqrt{d_M(w_n,w_{n+2})·d_M(w_{n+1},w_{n+1})}\}\hfill \\ & \le & rmax\left\{d_M(w_n,w_{n+1}),d_M(w_{n+1},w_{n+2})\right\}\forall n\in \mathbb{N}\cup \left\{0\right\}.\hfill \end{array}$$

(3)

If the $max\{d_M(w_n,w_{n+1}),d_M(w_{n+1},w_{n+2})\}=d_M(w_{n+1},w_{n+2})$ for some n, then (3), yields that $d_M(w_{n+1},w_{n+2})=0$, which is impossible. Thus,

$$\begin{array}{c}\hfill d_M(w_{n+1},w_{n+2})\le r{d}_M(w_n,w_{n+1})\forall n\in \mathbb{N}\cup \left\{0\right\}.\end{array}$$

(4)

Using induction and the above inequality, we get

$$\begin{array}{c}\hfill d_M(w_n,w_{n+1})\le r^n{d}_M(w_0,w_1)\mathrm{for}\mathrm{each}n\in \mathbb{N}.\end{array}$$

(5)

As $\left(w_n\right)$ is of the form $w_n=R{w}_{n-1}$, $w_{n-1}\ne w_n$$\forall n\in \mathbb{N}$ and also satisfying (5), by Lemma 1, we get $w_n\ne w_{n+p}$ for each $n,p\in \mathbb{N}$. Thus, $\left(c_3\right)$ and (5) imply the following inequalities:

$$\begin{array}{ccc}\hfill d_M(w_n,w_m)& \le & d_M(w_n,w_{n+1})·d_M(w_{n+1},w_{n+2})·\cdots ·d_M(w_{m-1},w_m)\hfill \\ & \le & [r^n·r^{n+1}·\cdots ·r^{m-1}]d_M(w_0,w_1)\hfill \\ & <& r^n{d}_M(w_0,w_1)\forall m>n.\hfill \end{array}$$

This implies that $d_M(w_n,w_m)\to 0$ as $n,m\to \infty$, since $r<1$, that is, $\left(w_n\right)$ is a Cauchy sequence in $(W,d_M)$. Hence, there exists $w_{*}\in W$ such that $w_n\to w_{*}$, since $(W,d_M)$ is complete. Using Axiom (iii), we get $\alpha (w_n,w_{*})\ge 1$$\forall n\in \mathbb{N}$. Then, by (2), we get

$$\begin{array}{ccc}\hfill d_M(R{w}_n,R{w}_{*})& \le & rmax\{d_M(w_n,w_{*}),\sqrt{d_M(w_n,R{w}_n)·d_M(w_{*},R{w}_{*})},\hfill \\ & & \sqrt{d_M(w_n,R{w}_{*})·d_M(w_{*},R{w}_n)}\}\forall n\in \mathbb{N}.\hfill \end{array}$$

Letting $n\to \infty$ in the above inequality, we conclude $d_M(w_{*},R{w}_{*})=0$. Hence, $w_{*}$ is a fixed point of R. □

Corollary 1.

Let $(W,d_M)$ be a complete product-operated metric space, and let $R:W\to W$ be a mapping that satisfies one of the following inequalities:

$$d_M(Rw,Rl)\le r{d}_M(w,l)\forall w,l\in W$$

$$d_M(Rw,Rl)\le r\sqrt{d_M(w,Rw)·d_M(l,Rl))}\forall w,l\in W$$

$$d_M(Rw,Rl)\le r\sqrt{d_M(w,Rl)·d_M(l,Rw)}\forall w,l\in W$$

where $r\in (0,1)$. Then, R has a fixed point.

Example 5.

Consider $W=\mathbb{N}$ as the set of all natural numbers. Define a product-operated metric on W by

$$d_M(w,q)=\left\{\begin{array}{c}0,if w=q\hfill \\ 1,if w,q are even numbers\hfill \\ 9,if w,q are odd numbers\hfill \\ 3,otherwise.\hfill \end{array}\right.$$

Define a function $R:W\to W$ by

$$Rw=\left\{\begin{array}{c}2w,if w is an odd number\hfill \\ 2,if w is an even number.\hfill \end{array}\right.$$

Now, we show that the above-defined function satisfies the first inequality of the above corollary.

• Case 1: If $w=e_1$ and $l=o_1$, where $e_1$ is an even number and $o_1$ is an odd number. Then $d_M(e_1,0_1)=3$ and $d_M(R{e}_1,R{o}_1)=1,or0$.
• Case 2: If $w=o_1$ and $l=o_2$, where $o_1,o_2$ are two odd numbers. Then either $d_M(o_1,o_2)=9$ and $d_M(R{o}_1,R{o}_2)=1$, provided $o_1\ne o_2$, or $d_M(o_1,o_2)=0$ and $d_M(R{o}_1,R{o}_2)=0$, provided $o_1=o_2$.
• Case 3: If $w=e_1$ and $l=e_2$, where $e_1,e_2$ are two even numbers. Then either $d_M(e_1,e_2)=1$ and $d_M(R{e}_1,R{e}_2)=0$, provided $e_1\ne e_2$, or $d_M(e_1,e_2)=0$ and $d_M(R{e}_1,R{e}_2)=0$, provided $e_1=e_2$.

Hence,

$$d_M(Rw,Rl)\le (1/2)d_M(w,l)\forall w,l\in W.$$

Therefore, Corollary 1 ensures the existence of a fixed point.

Remark 2.

It is important to note that the function $R:\mathbb{N}\to \mathbb{N}$ defined by

$$Rw=\left\{\begin{array}{c}2w,if w is an odd number\hfill \\ 2,if w is an even number\hfill \end{array}\right.$$

does not satisfy the Banach fixed point theorem with respect to a usual metric on $\mathbb{N}$, that is, $d(n,m)=|n-m|$. Since, for $w=2$ and $l=3$, we have

$$d(Rw,Rl)=|2-6|=4>|2-3|=d(w,l).$$

Example 6.

Consider $W=[0,\infty )$ and define a product-operated metric on W by

$$d_M(w,q)=\left\{\begin{array}{c}{2}^{|w-q|},ifw\ne q\hfill \\ 0,ifw=q.\hfill \end{array}\right.$$

Define a map $R:W\to W$ by

$$Rw=\left\{\begin{array}{c}\frac{w+5}{2},w>1\hfill \\ 1,0\le w\le 1\hfill \end{array}\right.$$

and $\alpha :W\times W\to [0,\infty )$ by

$$\alpha (w,l)=\left\{\begin{array}{c}1,w,l\le 1\hfill \\ 0,otherwise.\hfill \end{array}\right.$$

The reader can easily calculate that

$$d_M(Rw,Rl)\le \frac{1}{2}{d}_M(w,l)$$

for all $w,l\in W$ with $\alpha (w,l)=1$. Also, the rest of the axioms for Theorem 7 are true. Therefore, R has a fixed point.

Remark 3.

Note that in the above-defined example, Corollary 1 is not applicable. For instance, consider $w=2$ and $l=0$. Then, we get

$$d_M(R2,R0)=d_M(7/2,1)=2^{|5/2|}>2^{\left|2\right|}=d_M(2,0)$$

$$d_M(R2,R0)=d_M(7/2,1)=2^{|5/2|}>\sqrt{2^{|2-7/2|}{2}^{|0-1|}}=\sqrt{d_M(2,R2)·d_M(0,R0))}$$

and

$$d_M(R2,R0)=d_M(7/2,1)=2^{|5/2|}>\sqrt{2^{|2-1|}{2}^{|0-7/2|}}=\sqrt{d_M(2,R0)·d_M(0,R2)}.$$

Example 7.

Consider $W=[0,\infty )$ and define a product-operated metric on W by

$$d_M(w,q)=\left\{\begin{array}{c}{2}^{|w-q|},ifw\ne q\hfill \\ 0,ifw=q.\hfill \end{array}\right.$$

Define a map $R:W\to W$ by

$$Rw=\left\{\begin{array}{c}{w}^2-4w+6,w>1\hfill \\ 3,0\le w\le 1\hfill \end{array}\right.$$

and $\alpha :W\times W\to [0,\infty )$ by

$$\alpha (w,l)=\left\{\begin{array}{c}1,w,l\in [0,1]\cup \left\{3\right\}\hfill \\ 0,otherwise.\hfill \end{array}\right.$$

The reader can easily calculate that

$$d_M(Rw,Rl)\le \frac{1}{3}{d}_M(w,l)$$

for all $w,l\in W$ with $\alpha (w,l)=1$. Also, the rest of the axioms for Theorem 7 are true. Therefore, R has a fixed point.

Theorem 8.

Let $(W,d_M)$ be a complete product-operated metric space, and let $R:W\to W$ and $\alpha :W\times W\to [0,\infty )$ be two mappings such that for each $w\in W$ with $\alpha (w,Rw)\ge 1$, the following inequality is satisfied:

$$d_M(Rw,R^{2}w)\le rmax\left\{d_M(w,Rw),\sqrt{d_M(w,Rw)·d_M(Rw,R^{2}w)},\sqrt{d_M(w,R^{2}w)}\right\}$$

(6)

where $r\in (0,1)$. Also, consider the following axioms:

(i)

there exists $w_0\in W$ with $\alpha (w_0,R{w}_0)\ge 1$;

(ii)

for each $w\in W$ with $\alpha (w,Rw)\ge 1$, we have $\alpha (Rw,R^{2}w)\ge 1$;

(iii)

for each $\left(w_n\right)$ in W with $w_n\to w\in W$, we have $R{w}_n\to Rw$.

Then, R has a fixed point.

Proof. 

Using axioms (i) and (ii), we get $\alpha (w_n,w_{n+1})=\alpha (w_n,R{w}_n)\ge 1$$\forall n\in \mathbb{N}\cup \left\{0\right\}$, where $w_n=R{w}_{n-1}=R^n{w}_0\forall n\in \mathbb{N}$, and $w_n\ne w_{n+1}$$\forall n\in \mathbb{N}\cup \left\{0\right\}$; otherwise, we have a fixed point of R. Then, by (6), we get

$$\begin{array}{ccc}\hfill d_M(R{w}_n,R^2{w}_n)& \le & rmax\{d_M(w_n,R{w}_n),\sqrt{d_M(w_n,R{w}_n)·d_M(R{w}_n,R^2{w}_n)},\hfill \\ & & \sqrt{d_M(w_n,R^2{w}_n)}\}.\hfill \end{array}$$

That is,

$$\begin{array}{ccc}\hfill d_M(w_{n+1},w_{n+2})& \le & rmax\{d_M(w_n,w_{n+1}),\sqrt{d_M(w_n,w_{n+1})·d_M(w_{n+1},w_{n+2})},\hfill \\ & & \sqrt{d_M(w_n,w_{n+2})}\}\hfill \\ & \le & rmax\{d_M(w_n,w_{n+1}),\sqrt{d_M(w_n,w_{n+1})·d_M(w_{n+1},w_{n+2})},\hfill \\ & & \sqrt{d_M(w_n,w_{n+1})·d_M(w_{n+1},w_{n+2})}\}\hfill \\ & \le & rmax\left\{d_M(w_n,w_{n+1}),d_M(w_{n+1},w_{n+2})\right\}\hfill \\ & =& r{d}_M(w_n,w_{n+1})\forall n\in \mathbb{N}\cup \left\{0\right\}\hfill \end{array}$$

(7)

otherwise, we have a contradiction to our supposition. Thus, we get

$$\begin{array}{c}\hfill d_M(w_n,w_{n+1})\le r^n{d}_M(w_0,w_1)\mathrm{for}\mathrm{each}n\in \mathbb{N}.\end{array}$$

(8)

Using $\left(c_3\right)$ and (8), we get

$$\begin{array}{ccc}\hfill d_M(w_n,w_m)& \le & d_M(w_n,w_{n+1})·d_M(w_{n+1},w_{n+2})·\cdots ·d_M(w_{m-1},w_m)\hfill \\ & \le & [r^n·r^{n+1}·\cdots ·r^{m-1}]d_M(w_0,w_1)\hfill \\ & <& r^n{d}_M(w_0,w_1)\forall m>n.\hfill \end{array}$$

This implies that $d_M(w_n,w_m)\to 0$ as $n,m\to \infty$, since $r<1$, that is $\left(w_n\right)$ is a Cauchy sequence in $(W,d_M)$. Hence, there exists $w_{*}\in W$ such that $w_n\to w_{*}$. As $w_n\to w_{*}$, we get $R{w}_n\to R{w}_{*}$ by Axiom (iii). That is, ${lim}_{n\to \infty }{d}_M(w_{n+1},R{w}_{*})={lim}_{n\to \infty }{d}_M(R{w}_n,R{w}_{*})=0$. Since the limit point of a convergent sequence is unique, $w_{*}=R{w}_{*}$. □

3. Conclusions

This article presents the notion of product-operated metric spaces. The fundamental characteristics of the product-operated metric spaces are also presented to establish a basic literature. A few results to ensure the existence of fixed points for self-mappings are derived in product-operated metric spaces. We now leave the researchers with an open problem: find some contractive-type inequalities in product-operated metric spaces to investigate the existence of fixed points for single-valued and multi-valued mappings.