Subsequential Continuity in Neutrosophic Metric Space with Applications

Abstract

This paper introduces two concepts, subcompatibility and subsequential continuity, which are, respectively, weaker than the existing concepts of occasionally weak compatibility and reciprocal continuity. These concepts are studied within the framework of neutrosophic metric spaces. Using these ideas, a common fixed point theorem is developed for a system involving four maps. Furthermore, the results are applied to solve the Volterra integral equation, demonstrating the practical use of these findings in neutrosophic metric spaces.

1. Introduction

Zadeh introduced FSs in 1965 [1], and the concept of FMSs has been extensively explored by numerous researchers through various approaches. Kramosil and Michalek initially introduced FMS in 1975 [2], laying the foundation for further developments. Later, in 1994, George and Veeramani refined this concept by introducing significant modifications and investigating the Hausdorff topology for these spaces [3]. Over the years, researchers have expanded on these ideas, introducing statistical FMS, fuzzy pseudo metric spaces, and other related notions [2,4,5,6]. Various studies have contributed to the advancement of fuzzy metric concepts, drawing from different perspectives [5,7,8,9,10,11].

Several researchers, including [2,12,13,14,15,16,17,18,19], have contributed to the study of commuting mappings and FPTs for compatible and subcompatible maps in FMSs [20,21]. These theorems serve as a significant extension and generalization of various FPTs originally formulated for contractive-type mappings in metric spaces and similar structures. Furthermore, the work of [22] was instrumental in introducing the concept of type $\left(\beta \right)$ compatibility within FMS. Their research established links between compatible maps and those classified under types $\left(\alpha \right)$ and $\left(\beta \right)$, while also formulating a series of FPTs specifically tailored for type $\left(\beta \right)$ compatible maps in FMS.

Grabiec [19], in 1988, expanded the classical FPTs initially proposed by Banach and later refined by Edelstein in 1962 [23], extending their applicability to FMS as outlined by the definitions provided by Kramosil and Michalek.

In 1983, Krassimir Atanassov [24,25] introduced the concept of IFSs as an extension of Lotfi Zadeh’s FS theory, broadening the conventional understanding of set structures. Later, in 2004, Park [26] extended FMSs, originally formulated by George and Veeramani in 1994 [3], by proposing IFM-spaces. This development provided a natural progression in the field, linking these spaces to existing results in metric spaces (MS) [24,25,26,27].

Further advancements followed in 2006 when Gregori et al. [28] analyzed the convergence of topologies induced by fuzzy metrics and IFM. Expanding on this, Saadati et al. [29] gave the idea of modified IFMSs. After that Saadati, Sedghi, and Shobe [29] refined the IFMS framework of Park’s work [26] and the foundational ideas of George and Veeramani [3], leading to the development of modified IFMSs. Numerous researchers have since explored various aspects of IFMS, employing diverse methodologies [30,31]. Several other researchers [32] have explored additional fixed point results within the context of NMSs.

A major advancement in this field came with Smarandache’s work in 1999, which introduced the concept of measuring dilemmas while also generalizing existing mathematical structures that represent imprecision. Many researchers have contributed to the development of neutrosophic theory, including those who formulated neutrosophic logic [33] and the Dombi neutrosophic graph [34]. Ongoing research continues to expand the notion of NSs and their wide-ranging applications.

In neutrosophic set theory, each event is characterized by three key components: truthfulness (T), indeterminacy (I), and falsity (F), as originally introduced by Smarandache in 2003. NMSs further extend these ideas by utilizing continuous triangular norms (CTN) and continuous triangular conorms (CTC) to facilitate their study and application. In the present paper, Section 2 presents the idea of NS, and NMS [35] is then defined with the help of CTC and CTN. Additionally, a definition of Cauchy sequences in NMS is provided.

Moving to Section 3, some definitions and examples of weakly commuting, compatible mappings, as well as occasionally weakly compatible, subsequentially continuous, and subcompatible mappings in NMS, are introduced. In Section 4, FPT is established by using subsequentially continuous and subcompatible mappings in NMS. Section 5 focuses on illustrating the practical applications of the results, presenting examples to validate the findings. Section 6 provides the conclusion of the study. Lastly, the abbreviations and references used in the paper are given.

2. Preliminaries

In this section, some preliminaries are given that will be used later on.

Definition 1

([7]). A CTN is a binary operator $\uplus :[0,1]\times [0,1]\mapsto [0,1]$ if ⊎ satisfies the following criteria:

For $p,l,f,t\in [0,1]$,

1. 

⊎ is continuous,

2. 

if $p⩽f$ and $l⩽t$, then $p\uplus l⩽f\uplus t$,

3. 

$p\uplus 1=p$, and

4. 

⊎ is associative and commutative.

Example 1

([7]). (i) $\iota \uplus \nu =\iota \nu$ (ii) $\iota \uplus \nu =min(\iota ,\nu )$∀$\iota ,\nu \in [0,1].$

Definition 2

([7]). A CTC is a binary operator $\odot :[0,1]\times [0,1]\mapsto [0,1]$ if ⊙ satisfies the following criteria:

For $p,l,f,t\in [0,1]$,

1. 

⊙ is continuous,

2. 

If $p⩽f$ and $l⩽t$, then $p\odot l⩽f\odot t$,

3. 

$p\odot 0=p$,

4. 

⊙ is associative and commutative.

Example 2

([7]). (i) $\mathsf{g}\odot \mathsf{h}=\mathsf{g}+\mathsf{h}-\mathsf{g}\mathsf{h}$ (ii) $\mathsf{g}\odot \mathsf{h}=max(\mathsf{g},\mathsf{h})$ for all $\mathsf{g},\mathsf{h}\in [0,1]$

Definition 3

([35]). Let $\mathfrak{A}$ be an arbitrary set. Then $\aleph =<\mathsf{l},\mathsf{P}\left(\mathsf{l}\right),\mathsf{Q}\left(\mathsf{l}\right),\mathsf{R}\left(\mathsf{l}\right)>:\mathsf{l}\in \mathfrak{A}$ is an NS such that $\aleph :\mathfrak{A}\times \mathfrak{A}\times {\mathbb{R}}^{+}\mapsto [\mathsf{0},\mathsf{1}]$. Let ⊎ be a CTN and let ⊙ be the CTC. Then $(\mathfrak{A},\aleph ,\uplus ,\odot )$ is NMS if the following criteria are satisfied. ∀$\mathsf{l},\mathsf{m},v\in \mathfrak{A}$,

1. 

$0⩽\mathsf{P}(\mathsf{l},\mathsf{m},\mathfrak{t})⩽1,0⩽\mathsf{Q}(\mathsf{l},\mathsf{m},\mathfrak{t})⩽1,0⩽\mathsf{R}(\mathsf{l},\mathsf{m},\mathfrak{t})⩽1$, for all $\mathfrak{t}\in {\mathbb{R}}^{+}$,

2. 

$\mathsf{P}(\mathsf{l},\mathsf{m},\mathfrak{t})+\mathsf{Q}(\mathsf{l},\mathsf{m},\mathfrak{t})+\mathsf{R}(\mathsf{l},\mathsf{m},\mathfrak{t})⩽3$, (for $\mathfrak{t}\in {\mathbb{R}}^{+}$),

3. 

$\mathsf{P}(\mathsf{l},\mathsf{m},\mathfrak{t})=1$ (for $\mathfrak{t}>0)$ iff $\mathsf{l}=\mathsf{m}$,

4. 

$\mathsf{P}(\mathsf{l},\mathsf{m},\mathfrak{t})=\mathsf{P}(\mathsf{m},\mathsf{l},\mathfrak{t})$ (for $\mathfrak{t}>0$),

5. 

$\mathsf{P}(\mathsf{l},\mathsf{m},\mathfrak{t})\uplus \mathsf{P}(\mathsf{m},v,s)⩽\mathsf{P}(\mathsf{l},v,\mathfrak{t}+s)(\forall \mathfrak{t},s>0)$,

6. 

$\mathsf{P}(\mathsf{l},\mathsf{m},.)$:$[0,\infty )\mapsto [0,1]$ is continuous,

7. 

$li{m}_{\mathfrak{t}\mapsto \infty }\mathsf{P}(\mathsf{l},\mathsf{m},\mathfrak{t})=1(\forall \mathfrak{t}>0)$,

8. 

$\mathsf{Q}(\mathsf{l},\mathsf{m},\mathfrak{t})=0$ (for $\mathfrak{t}>0)$ iff $\mathsf{l}=\mathsf{m}$,

9. 

$\mathsf{Q}(\mathsf{l},\mathsf{m},\mathfrak{t})=\mathsf{Q}(\mathsf{m},\mathsf{l},\mathfrak{t})$ (for $\mathfrak{t}>0$),

10. 

$\mathsf{Q}(\mathsf{l},\mathsf{m},\mathfrak{t})\odot \mathsf{Q}(\mathsf{m},v,s)⩾\mathsf{Q}(\mathsf{l},v,\mathfrak{t}+s)(\forall \mathfrak{t},s>0)$,

11. 

$\mathsf{Q}(\mathsf{l},\mathsf{m},.)$:$[0,\infty )\mapsto [0,1]$ is continuous,

12. 

$li{m}_{\mathfrak{t}\mapsto \infty }\mathsf{Q}(\mathsf{l},\mathsf{m},\mathfrak{t})=0(\forall \mathfrak{t}>0)$,

13. 

$\mathsf{R}(\mathsf{l},\mathsf{m},\mathfrak{t})=0$ (for $\mathfrak{t}>0)$ iff $\mathsf{l}=\mathsf{m}$,

14. 

$\mathsf{R}(\mathsf{l},\mathsf{m},\mathfrak{t})=\mathsf{R}(\mathsf{m},\mathsf{l},\mathfrak{t})$ (for $\mathfrak{t}>0$),

15. 

$\mathsf{R}(\mathsf{l},\mathsf{m},\mathfrak{t})\odot \mathsf{R}(\mathsf{m},v,s)⩾B(\mathsf{l},v,\mathfrak{t}+s)(\forall \mathfrak{t},s>0)$,

16. 

$\mathsf{R}(\mathsf{l},\mathsf{m},.)$:$[0,\infty )\mapsto [0,1]$ is continuous,

17. 

$li{m}_{\mathfrak{t}\mapsto \infty }\mathsf{R}(\mathsf{l},\mathsf{m},\mathfrak{t})=\mathsf{0}$ (for all $\mathfrak{t}>\mathsf{0}$), and

18. 

if $\mathfrak{t}⩽\mathsf{0}$, then $\mathsf{P}(\mathsf{l},\mathsf{m},\mathfrak{t})=\mathsf{0},\mathsf{Q}(\mathsf{l},\mathsf{m},\mathfrak{t})=\mathsf{1}$ and $\mathsf{R}(\mathsf{l},\mathsf{m},\mathfrak{t})=\mathsf{1}$.

Thus, $\aleph =(\mathsf{P},\mathsf{Q},\mathsf{R})$ is called neutrosophic metric (NM) on $\mathfrak{A}$.

The functions $\mathsf{P}(\mathsf{l},\mathsf{m},\mathfrak{t}),\mathsf{Q}(\mathsf{l},\mathsf{m},\mathfrak{t}),\mathsf{R}(\mathsf{l},\mathsf{m},\mathfrak{t})$ denote the nearness degree, neutralness degree, and non-nearness degree between $\mathsf{l}$ and $\mathsf{m}$ with respect to $\mathfrak{t}$, respectively.

Definition 4

([35]). Consider an NMS $(\mathfrak{A},\aleph ,\uplus ,\odot )$. A sequence $\left\{{\mathsf{a}}_{\mathsf{n}}\right\}$ cgs $\mathsf{a}\in \mathfrak{A}$ iff, for $\epsilon >0$ and $l\in (\mathsf{0},\mathsf{1})$, ∃$\mathsf{m}\in \mathbb{N}$ such that ∀$\mathsf{n}\ge \mathsf{m}$

$$\begin{array}{cc}\hfill \mathsf{P}({\mathsf{a}}_{\mathsf{n}},\mathsf{a},\epsilon )& >1-l,\hfill \\ \hfill \mathsf{Q}({\mathsf{a}}_{\mathsf{n}},\mathsf{a},\epsilon )& <l,\hfill \\ \hfill \mathsf{Q}({\mathsf{a}}_{\mathsf{n}},\mathsf{a},\epsilon )& <l\hfill \end{array}$$

or alternatively,

$$\begin{array}{cc}\hfill \underset{\mathsf{n}\to \infty }{lim}\mathsf{P}({\mathsf{a}}_{\mathsf{n}},{\mathsf{a}}_{\mathsf{m}},\epsilon )& =1,\hfill \\ \hfill \underset{\mathsf{n}\to \infty }{lim}\mathsf{Q}({\mathsf{a}}_{\mathsf{n}},{\mathsf{a}}_{\mathsf{m}},\epsilon )& =0,\hfill \\ \hfill \underset{\mathsf{n}\to \infty }{lim}\mathsf{Q}({\mathsf{a}}_{\mathsf{n}},{\mathsf{a}}_{\mathsf{m}},\epsilon )& =0\hfill \end{array}$$

as $\epsilon \mapsto \infty$.

Definition 5

([35]). Consider an NMS $(\mathfrak{A},\aleph ,\uplus ,\odot )$. A sequence $\left\{a_n\right\}$ in $\mathfrak{A}$ is termed as a Cauchy sequence, if for a given $l>\mathsf{0}$ and $\epsilon >\mathsf{0}$, ∃$n_0\in \mathbb{N}$ such that

$$\mathsf{P}(a_l,a_r,\epsilon )>1-l,\mathsf{Q}(a_l,a_r,\epsilon )<l,\mathsf{R}(a_l,a_r,\epsilon )<l,$$

for all $l,r\ge n_0$.

Definition 6

([35]). If every Cauchy sequence is convergent, then an NMS $(\mathfrak{A},\aleph ,\uplus ,\odot )$ is complete.

In the next section, key definitions and illustrative examples of subcompatibility and subsequential continuity within the framework of NMS are provided. These definitions and examples serve as the groundwork for the theoretical development and will be utilized in later sections to establish fixed point results and demonstrate their applications in solving practical problems within the framework of NMS.

3. Subcompatibility and Subsequential Continuity in NMS

Definition 7.

Let ζ and υ be two maps within NMS $(\mathfrak{A},\aleph ,\uplus ,\odot )$. We define ζ and υ as weakly commuting when the following conditions holds: $\mathsf{P}(\zeta \upsilon b,\upsilon \zeta b,\varrho )\ge \mathsf{P}(\zeta b,\upsilon b,\varrho ),\mathsf{Q}(\zeta \upsilon b,\upsilon \zeta b,\varrho )\le \mathsf{Q}(\zeta b,\upsilon b,\varrho ),\mathsf{R}(\zeta \upsilon b,\upsilon \zeta b,\varrho )\le \mathsf{R}(\zeta b,\upsilon b,\varrho )$, for all $b\in \mathfrak{A}$ and $\varrho >0.$

Example 3.

Let $\mathfrak{A}=\mathbb{R}$ and consider the NMS $(\mathfrak{A},\aleph ,\uplus ,\odot )$, where $\aleph =(\mathsf{P},\mathsf{Q},\mathsf{R})$ represents the neutrosophic metric functions given as $\mathsf{P}(\mathfrak{x},\mathfrak{y},\varrho )={\displaystyle \frac{\varrho }{1+|\mathfrak{x}-\mathfrak{y}|}}$, $\mathsf{Q}(\mathfrak{x},\mathfrak{y},\varrho )={\displaystyle \frac{|\mathfrak{x}-\mathfrak{y}|}{1+|\mathfrak{x}-\mathfrak{y}|}}$, and $\mathsf{R}(\mathfrak{x},\mathfrak{y},\varrho )=1-\mathsf{P}(\mathfrak{x},\mathfrak{y},\varrho )-\mathsf{Q}(\mathfrak{x},\mathfrak{y},\varrho )$. Define the mappings $\zeta ,\upsilon :A\to A$ as $\zeta \left(b\right)=b+1$ and $\upsilon \left(b\right)=2b$

Now, the following conditions are verified: AS $\zeta \upsilon \left(b\right)=\zeta \left(2b\right)=2b+1$ and $\upsilon \zeta \left(b\right)=\upsilon (b+1)=2(b+1)=2b+2$. Thus,

$$\mathsf{P}(\zeta \upsilon b,\upsilon \zeta b,\varrho )={\displaystyle \frac{\varrho }{1+\left|\right(2b+1)-(2b+2\left)\right|}}={\displaystyle \frac{\varrho }{1+1}}={\displaystyle \frac{\varrho }{2}},$$

and

$$\mathsf{P}(\zeta b,\upsilon b,\varrho )={\displaystyle \frac{\varrho }{1+\left|\right(b+1)-(2b\left)\right|}}={\displaystyle \frac{\varrho }{1+|b-1|}}.$$

Clearly, $\mathsf{P}(\zeta \upsilon b,\upsilon \zeta b,\varrho )\ge \mathsf{P}(\zeta b,\upsilon b,\varrho )$ holds.

$$\mathsf{Q}(\zeta \upsilon b,\upsilon \zeta b,\varrho )={\displaystyle \frac{\left|\right(2b+1)-(2b+2\left)\right|}{1+\left|\right(2b+1)-(2b+2\left)\right|}}={\displaystyle \frac{1}{2}},$$

and

$$\mathsf{Q}(\zeta b,\upsilon b,\varrho )={\displaystyle \frac{\left|\right(b+1)-(2b\left)\right|}{1+\left|\right(b+1)-(2b\left)\right|}}={\displaystyle \frac{|b-1|}{1+|b-1|}}.$$

Clearly, $\mathsf{Q}(\zeta \upsilon b,\upsilon \zeta b,\varrho )\le \mathsf{Q}(\zeta b,\upsilon b,\varrho )$ holds.

Since $\mathsf{R}(\mathfrak{x},\mathfrak{y},\varrho )=1-\mathsf{P}(\mathfrak{x},\mathfrak{y},\varrho )-\mathsf{Q}(\mathfrak{x},\mathfrak{y},\varrho )$, it can be verified that

$$\mathsf{R}(\zeta \upsilon b,\upsilon \zeta b,\varrho )\le \mathsf{R}(\zeta b,\upsilon b,\varrho ).$$

Thus, ζ and υ satisfy all the conditions.

Definition 8.

Assume an NMS $(\mathfrak{A},\aleph ,\uplus ,\odot )$. Assume ζ and υ are maps. These maps are compatible, if, for any $\varrho >0$, the following axioms are fulfilled:

$$\begin{array}{c}\hfill \underset{n\to \infty }{lim}\mathsf{P}(\zeta \upsilon a_n,\upsilon \zeta a_n,\varrho )=1,\underset{n\to \infty }{lim}\mathsf{Q}(\zeta \upsilon a_n,\upsilon \zeta a_n,\varrho )=0\end{array}$$

and

$$\begin{array}{c}\hfill \underset{n\to \infty }{lim}\mathsf{R}(\zeta \upsilon a_n,\upsilon \zeta a_n,\varrho )=0,\end{array}$$

whenever a sequence $\left\{a_n\right\}$ in $\mathfrak{A}$ such that both ${lim}_{n\to \infty }\zeta a_n$ and ${lim}_{n\to \infty }\upsilon a_n$ converge to the same point $\mathsf{b}\in \mathfrak{A}$.

Example 4.

Assume $\mathfrak{A}=\mathbb{R}$ and the maps $\zeta \left(\mathfrak{x}\right)=2\mathfrak{x}$ and $\upsilon \left(\mathfrak{x}\right)={\displaystyle \frac{\mathfrak{x}}{2}}$ for $\mathfrak{x}\in \mathbb{R}$. Consider a sequence $\left\{a_n\right\}$ defined by $a_n={\displaystyle \frac{1}{n}}$. Then for $\varrho >0$,

$$\begin{array}{c}\hfill \mathsf{P}(\zeta \upsilon a_n,\upsilon \zeta a_n,\varrho )\to 1,\mathsf{Q}(\zeta \upsilon a_n,\upsilon \zeta a_n,\varrho )\to 0,\mathit{and}\mathsf{R}(\zeta \upsilon a_n,\upsilon \zeta a_n,\varrho )\to 0,\mathit{as}n\to \infty .\end{array}$$

Also,

$$\begin{array}{c}\hfill \underset{n\to \infty }{lim}\zeta a_n=\underset{n\to \infty }{lim}2·{\displaystyle \frac{1}{n}}=0,\underset{n\to \infty }{lim}\upsilon a_n=\underset{n\to \infty }{lim}{\displaystyle \frac{1}{2n}}=0.\end{array}$$

Both mappings $\zeta a_n$ and $\upsilon a_n$ converge to 0. Thus, these maps are compatible in NMS.

Definition 9.

Two mappings ζ and υ of an NMS $(\mathfrak{A},\aleph ,\uplus ,\odot )$ are called reciprocal continuous if $\zeta \upsilon a_n\to \zeta b$ and $\upsilon \zeta a_n\to \upsilon b,$ whenever $\{a_n\}$ is a sequence such that $\zeta a_n,\upsilon a_n\to b$ for some b in $\mathfrak{A}$.

Example 5.

Let $\mathfrak{A}=\mathbb{R}$ and consider the NMS $(\mathfrak{A},\aleph ,\uplus ,\odot )$, where

$\aleph =(\mathsf{P},\mathsf{Q},\mathsf{R})$ represents the neutrosophic metric functions given as

$\mathsf{P}(\mathfrak{x},\mathfrak{y},\varrho )={\displaystyle \frac{\varrho }{1+|\mathfrak{x}-\mathfrak{y}|}}$, $\mathsf{Q}(\mathfrak{x},\mathfrak{y},\varrho )={\displaystyle \frac{|\mathfrak{x}-\mathfrak{y}|}{1+|\mathfrak{x}-\mathfrak{y}|}}$, and $\mathsf{R}(\mathfrak{x},\mathfrak{y},\varrho )=1-\mathsf{P}(\mathfrak{x},\mathfrak{y},\varrho )-\mathsf{Q}(\mathfrak{x},\mathfrak{y},\varrho )$.

Define the mappings $\zeta ,\upsilon :\mathfrak{A}\to \mathfrak{A}$ as $\zeta \left(b\right)=b+1$ and $\upsilon \left(b\right)=2b$

Now, let $\left\{a_n\right\}$ be a sequence in $\mathfrak{A}$ such that

$$a_n={\displaystyle \frac{1}{n}},\mathit{so}\mathit{that}{a}_n\to 0\mathit{as}n\to \infty .$$

Also,

$$\zeta a_n=a_n+1={\displaystyle \frac{1}{n}}+1,\upsilon a_n=2a_n={\displaystyle \frac{2}{n}}.$$

As $n\to \infty$, both $\zeta a_n\to 1$ and $\upsilon a_n\to 0$.

Now, the condition $\zeta a_n,\upsilon a_n\to b$ is verified for some $b\in \mathfrak{A}$: Let $b=0$. For the given sequence $\left\{a_n\right\}$, $\zeta a_n\to b+1=1$, and $\upsilon a_n\to 2b=0$, which satisfies the condition.

Now,

$$\zeta \upsilon a_n=\zeta \left(2a_n\right)=2a_n+1={\displaystyle \frac{2}{n}}+1,\upsilon \zeta a_n=\upsilon (a_n+1)=2(a_n+1)={\displaystyle \frac{2}{n}}+2.$$

As $n\to \infty$, $\zeta \upsilon a_n\to \zeta b=1$ and $\upsilon \zeta a_n\to \upsilon b=0$.

Since both limits are consistent, ζ and υ are reciprocal continuous.

Remark 1.

When both ζ and υ are continuous, it is evident that they exhibit reciprocal continuity. However, it is important to note that this reciprocity does not guarantee continuity in return. Moreover, within the context of the typical FPT applied to a compatible pair of mappings that satisfy contractive conditions, the continuity of one of the mappings either ζ or υ implies their reciprocal continuity, but the converse is not true.

Example 6.

Let $\mathfrak{A}=\mathbb{R}$, and define NMS $(\mathfrak{A},\aleph ,\uplus ,\odot )$ where $\mathsf{P}(\mathfrak{x},\mathfrak{y},\varrho )={\displaystyle \frac{\varrho }{1+|\mathfrak{x}-\mathfrak{y}|}}$, $\mathsf{Q}(\mathfrak{x},\mathfrak{y},\varrho )={\displaystyle \frac{|\mathfrak{x}-\mathfrak{y}|}{1+|\mathfrak{x}-\mathfrak{y}|}}$, and $\mathsf{R}(\mathfrak{x},\mathfrak{y},\varrho )=1-\mathsf{P}(\mathfrak{x},\mathfrak{y},\varrho )-\mathsf{Q}(\mathfrak{x},\mathfrak{y},\varrho )$. Define $\zeta ,\upsilon :\mathfrak{A}\to \mathfrak{A}$ mappings as

$\zeta \left(\mathfrak{x}\right)=\mathfrak{x}+1$ and $\upsilon \left(\mathfrak{x}\right)=2\mathfrak{x}$.

Both ζ and υ are continuous.

Since both mappings ζ and υ are continuous, they will exhibit reciprocal continuity. That is, for any sequence $\left\{a_n\right\}$ such that $\zeta a_n\to b$ and $\upsilon a_n\to b$ as $n\to \infty$, we will have

$$\zeta \upsilon a_n\to \zeta b\mathit{and}\upsilon \zeta a_n\to \upsilon b.$$

For example, if $a_n={\displaystyle \frac{1}{n}}$, then $\zeta a_n={\displaystyle \frac{1}{n}}+1\to 1$ and $\upsilon a_n={\displaystyle \frac{2}{n}}\to 0$.

Thus, we have $\zeta \upsilon a_n=\zeta \left({\displaystyle \frac{2}{n}}\right)={\displaystyle \frac{2}{n}}+1\to 1$ and $\upsilon \zeta a_n=\upsilon \left({\displaystyle \frac{1}{n}}+1\right)={\displaystyle \frac{2}{n}}+2\to 2$.

Therefore, the mappings satisfy reciprocal continuity.

However, ζ and υ are continuous, and this does not guarantee that they will be reciprocal continuous in every case without additional conditions.

Thus, the continuity of ζ and υ ensures their reciprocal continuity, but reciprocal continuity does not necessarily imply continuity for these mappings.

Definition 10.

Consider an NMS $(\mathfrak{A},\aleph ,\uplus ,\odot )$. Let ζ and υ be self-maps on the set $\mathfrak{A}$. A point a within $\mathfrak{A}$ is defined as a CP of ζ and υ if and only if $\zeta \left(a\right)=\upsilon \left(a\right)$. In such a case, the value $w=\zeta \left(a\right)=\upsilon \left(a\right)$ is known as a coincidence point for both ζ and $\upsilon .$

Definition 11.

Consider an NMS denoted as $(\mathfrak{A},\aleph ,\uplus ,\odot )$. A pair of self-mappings denoted as $(\zeta ,\upsilon )$ is said to be WC if it exhibits commutativity at the point of coincidence. In other words, if ∃ a point $\mathsf{u}$ in $\mathfrak{A}$ such that $\zeta \left(\mathsf{u}\right)=\upsilon \left(\mathsf{u}\right)$, then $\zeta \left(\upsilon \right(\mathsf{u}\left)\right)=\upsilon \left(\zeta \right(\mathsf{u}\left)\right)$.

Example 7.

Let $\mathfrak{A}=\mathbb{R}$, and define an NMS $(\mathfrak{A},\aleph ,\uplus ,\odot )$ where

$\mathsf{P}(\mathfrak{x},\mathfrak{y},\varrho )={\displaystyle \frac{\varrho }{1+|\mathfrak{x}-\mathfrak{y}|}}$, $\mathsf{Q}(\mathfrak{x},\mathfrak{y},\varrho )={\displaystyle \frac{|\mathfrak{x}-\mathfrak{y}|}{1+|\mathfrak{x}-\mathfrak{y}|}}$, and $\mathsf{R}(\mathfrak{x},\mathfrak{y},\varrho )=1-\mathsf{P}(\mathfrak{x},\mathfrak{y},\varrho )-\mathsf{Q}(\mathfrak{x},\mathfrak{y},\varrho )$.

Define two mappings $\zeta ,\upsilon :\mathfrak{A}\to \mathfrak{A}$ as $\zeta \left(\mathfrak{x}\right)=\mathfrak{x}+2$ and $\upsilon \left(\mathfrak{x}\right)=\mathfrak{x}+2$. To check if $(\zeta ,\upsilon )$ are weakly compatible, we need to verify if ∃ a point $\mathfrak{u}\in \mathfrak{A}$ such that $\zeta \left(\mathfrak{u}\right)=\upsilon \left(\mathfrak{u}\right)$, and if so, whether $\zeta \left(\upsilon \right(\mathfrak{u}\left)\right)=\upsilon \left(\zeta \right(\mathfrak{u}\left)\right)$. Thus, to find $\mathfrak{u}$ where $\zeta \left(\mathfrak{u}\right)=\upsilon \left(\mathfrak{u}\right)$

$$\zeta \left(\mathfrak{u}\right)=\mathfrak{u}+2,\upsilon \left(\mathfrak{u}\right)=\mathfrak{u}+2.$$

Clearly, $\zeta \left(\mathfrak{u}\right)=\upsilon \left(\mathfrak{u}\right)$ for all $\mathfrak{u}\in \mathfrak{A}$, because both mappings are identical.

Now commutativity can be verified at $\mathfrak{u}$. For any $\mathfrak{u}\in \mathfrak{A}$,

$$\zeta \left(\upsilon \right(\mathfrak{u}\left)\right)=\zeta (\mathfrak{u}+2)=(\mathfrak{u}+2)+2=\mathfrak{u}+4,$$

and

$$\upsilon \left(\zeta \right(\mathfrak{u}\left)\right)=\upsilon (\mathfrak{u}+2)=(\mathfrak{u}+2)+2=\mathfrak{u}+4.$$

Thus, $\zeta \left(\upsilon \right(\mathfrak{u}\left)\right)=\upsilon \left(\zeta \right(\mathfrak{u}\left)\right)$ holds for all $\mathfrak{u}\in \mathfrak{A}$.

Since $\zeta \left(\mathfrak{u}\right)=\upsilon \left(\mathfrak{u}\right)$ for all $\mathfrak{u}\in \mathfrak{A}$, and $\zeta \left(\upsilon \right(\mathfrak{u}\left)\right)=\upsilon \left(\zeta \right(\mathfrak{u}\left)\right)$ for all $\mathfrak{u}\in \mathfrak{A}$, the pair $(\zeta ,\upsilon )$ is weakly compatible.

Thus, the mappings $\zeta \left(\mathfrak{x}\right)=\mathfrak{x}+2$ and $\upsilon \left(\mathfrak{x}\right)=\mathfrak{x}+2$ are weakly compatible because they satisfy the condition of commutativity at the point of coincidence. Specifically, since $\zeta \left(\mathfrak{u}\right)=\upsilon \left(\mathfrak{u}\right)$ for all $\mathfrak{u}$, they exhibit commutativity at every point, satisfying the definition of weak compatibility in an NMS.

Remark 2.

Two maps that are compatible also satisfy the condition of being weakly compatible, but the reverse statement is not necessarily true.

Definition 12.

Two self-maps denoted as ζ and υ within the framework of an NMS represented as $(\mathfrak{A},\aleph ,\uplus ,\odot )$ are termed occasionally weakly compatible (OWC) if ∃ a point $a\in \mathfrak{A}$ that is a coincidence point for both ζ and υ, and at this point both ζ and υ exhibit commutativity.

Example 8.

Let $\mathfrak{A}=\mathbb{R}$, and define an NMS $(\mathfrak{A},\aleph ,\uplus ,\odot )$, where

$\mathsf{P}(\mathfrak{x},\mathfrak{y},\varrho )={\displaystyle \frac{\varrho }{1+|\mathfrak{x}-\mathfrak{y}|}}$, $\mathsf{Q}(\mathfrak{x},\mathfrak{y},\varrho )={\displaystyle \frac{|\mathfrak{x}-\mathfrak{y}|}{1+|\mathfrak{x}-\mathfrak{y}|}}$, and $\mathsf{R}(\mathfrak{x},\mathfrak{y},\varrho )=1-\mathsf{P}(\mathfrak{x},\mathfrak{y},\varrho )-\mathsf{Q}(\mathfrak{x},\mathfrak{y},\varrho )$.

Define two mappings $\zeta ,\upsilon :\mathfrak{A}\to \mathfrak{A}$ as $\zeta \left(\mathfrak{x}\right)=\mathfrak{x}+1$ and $\upsilon \left(\mathfrak{x}\right)=\mathfrak{x}+1$.

Now,

$$\zeta \left(a\right)=a+1,\upsilon \left(a\right)=a+1.$$

Clearly, $\zeta \left(a\right)=\upsilon \left(a\right)$ for all $a\in \mathfrak{A}$, so every $a\in \mathfrak{A}$ is a coincidence point.

Again, for any $a\in \mathfrak{A}$,

$$\zeta \left(\upsilon \right(a\left)\right)=\zeta (a+1)=(a+1)+1=a+2,$$

and

$$\upsilon \left(\zeta \right(a\left)\right)=\upsilon (a+1)=(a+1)+1=a+2.$$

Since $\zeta \left(\upsilon \right(a\left)\right)=\upsilon \left(\zeta \right(a\left)\right)$, ζ and υ exhibit commutativity at every coincidence point.

Thus, both maps are OWC.

Definition 13.

Consider an NMS $(\mathfrak{A},\aleph ,\uplus ,\odot )$, self-maps ζ and υ on the set $\mathfrak{A}$ are defined to be subcompatible if ∃ a sequence $\{a_n\}$ in $\mathfrak{A}$ such that ${lim}_{n\to \infty }\zeta a_n={lim}_{n\to \infty }\upsilon a_n=b,b\in \mathfrak{A}$ and satisfy ${lim}_{n\to \infty }\mathsf{P}(\zeta \upsilon a_n,\upsilon \zeta a_n,\varrho )=1,{lim}_{n\to \infty }\mathsf{Q}(\zeta \upsilon a_n,\upsilon \zeta a_n,\varrho )=0,{lim}_{n\to \infty }\mathsf{R}(\zeta \upsilon a_n,\upsilon \zeta a_n,\varrho )=0$.

Example 9.

Consider the set $\mathfrak{A}=[0,1]$, and define the neutrosophic metric functions $\mathsf{P},\mathsf{Q},\mathsf{R}:\mathfrak{A}\times \mathfrak{A}\times [0,\infty )\to [0,1]$ as follows:

$\mathsf{P}(\mathfrak{x},\mathfrak{y},\varrho )=1-|\mathfrak{x}-\mathfrak{y}|$, $\mathsf{Q}(\mathfrak{x},\mathfrak{y},\varrho )=|\mathfrak{x}-\mathfrak{y}|$, $\mathsf{R}(\mathfrak{x},\mathfrak{y},\varrho )=|\mathfrak{x}-\mathfrak{y}|$.

Let the operations ⊎ and ⊙ be defined as the standard addition and multiplication on $[0,1]$ modulo 1, respectively.

Define the self-maps ζ and υ on A as $\zeta \left(\mathfrak{x}\right)={\displaystyle \frac{\mathfrak{x}}{2}}$ and $\upsilon \left(\mathfrak{x}\right)={\displaystyle \frac{\mathfrak{x}}{3}}$.

Now, take the sequence $\left\{a_n\right\}$ in $\mathfrak{A}$ as $a_n={\displaystyle \frac{1}{n}}$ for $n\ge 1$.

Now, convergence of $\zeta a_n$ and $\upsilon a_n$ can be verified: $\zeta \left(a_n\right)={\displaystyle \frac{1}{2n}}$ and $\upsilon \left(a_n\right)={\displaystyle \frac{1}{3n}}$.

As $n\to \infty$, ${lim}_{n\to \infty }\zeta \left(a_n\right)=0$ and ${lim}_{n\to \infty }\upsilon \left(a_n\right)=0$.

Thus, ${lim}_{n\to \infty }\zeta \left(a_n\right)={lim}_{n\to \infty }\upsilon \left(a_n\right)=b=0$, where $b\in \mathfrak{A}$.

Now, subcompatibility conditions can be evaluated:

$\zeta \upsilon \left(a_n\right)=\zeta \left({\displaystyle \frac{1}{3n}}\right)={\displaystyle \frac{1}{6n}}$ and $\upsilon \zeta \left(a_n\right)=\upsilon \left({\displaystyle \frac{1}{2n}}\right)={\displaystyle \frac{1}{6n}}$.

Thus,

${lim}_{n\to \infty }\mathsf{P}(\zeta \upsilon \left(a_n\right),\upsilon \zeta \left(a_n\right),\varrho )=1-\left|{\displaystyle \frac{1}{6n}}-{\displaystyle \frac{1}{6n}}\right|=1$, ${lim}_{n\to \infty }\mathsf{Q}(\zeta \upsilon \left(a_n\right),\upsilon \zeta \left(a_n\right),\varrho )=\left|{\displaystyle \frac{1}{6n}}-{\displaystyle \frac{1}{6n}}\right|=0$, and ${lim}_{n\to \infty }\mathsf{R}(\zeta \upsilon \left(a_n\right),\upsilon \zeta \left(a_n\right),\varrho )=\left|{\displaystyle \frac{1}{6n}}-{\displaystyle \frac{1}{6n}}\right|=0$.

Thus, the maps ζ and υ satisfy the conditions for subcompatibility in the neutrosophic metric space $(\mathfrak{A},\aleph ,\uplus ,\odot )$ with the given sequence $\left\{a_n\right\}$.

Thus, it is evident that two occasionally weakly compatible maps are also subcompatible, but the reverse statement is not always true. The example provided below illustrates that there can be instances of subcompatible maps that do not qualify as occasionally weakly compatible (OWC).

Example 10.

Consider an MS $(\mathfrak{A},d)$, where d is defined as $d(\varkappa ,\varsigma )=|\varkappa -\varsigma |$, ϰ, ς∈$A=[0,\infty )$ and, for any positive value $\varrho \in (0,\infty )$, we have the operators ⊎ and ⊙ defined as follows: The $TN$ operator is denoted by ⊎ with its default operation being the minimum, i.e., $TN\mathfrak{x}\uplus \mathfrak{y}=min(\mathfrak{x},\mathfrak{y})$, and the $TC$ operator is denoted by ⊙ with its default operation being the maximum, i.e., $TC\mathfrak{x}\odot \mathfrak{y}=max(\mathfrak{x},\mathfrak{y})$. Define $\mathsf{P}(\varkappa ,\varsigma ,\varrho )=$$\begin{array}{cc}\{& \begin{array}{cc}{\displaystyle \frac{\varrho }{\varrho +|\varkappa -\varsigma |}}& \varrho >0\\ 0& \varrho =0\end{array}\end{array}$, $\mathsf{Q}(\varkappa ,\varsigma ,\varrho )=$$\begin{array}{cc}\{& \begin{array}{cc}{\displaystyle \frac{\varkappa -\varsigma }{\varrho +|\varkappa -\varsigma |}}& \varrho >0\\ 1& \varrho =1\end{array}\end{array}$ and $\mathsf{R}(\varkappa ,\varsigma ,\varrho )=$ $\begin{array}{cc}\{& \begin{array}{cc}{\displaystyle \frac{|\varkappa -\varsigma |}{\varrho }}& \varrho >0\\ 0& \varrho =0\end{array}\end{array}$. Clearly, $(\mathfrak{A},\aleph ,\uplus ,\odot )$ is a complete NMS.

Define ζ and υ as follows: $\zeta \left(\varkappa \right)={\varkappa }^2$ and $\upsilon \left(\varkappa \right)=$ $\begin{array}{cc}\{& \begin{array}{cc}\varkappa +2& \varkappa \in [0,4]\uplus (9,\infty )\\ \varkappa +12& \varkappa \in (4,9]\end{array}\end{array}$.  Consider the sequence $\{{\varkappa }_n\}$ in set $\mathfrak{A}$ defined as ${\varkappa }_n=2+{\displaystyle \frac{1}{n}}$ for $n=1,2,3,\dots$.

As n approaches infinity, the limits are as follows:

• ${lim}_{n\to \infty }\zeta {\varkappa }_n={lim}_{n\to \infty }\upsilon {\varkappa }_n=4$, so 4 is an element of set $\mathfrak{A}$.
• Additionally, $\zeta \upsilon {\varkappa }_n$ approaches 16 and $\upsilon \zeta$ also tends to 16 as n approaches infinity.

Thus, ${lim}_{n\to \infty }\mathsf{P}(\zeta \upsilon {\varkappa }_n,\upsilon \zeta {\varkappa }_n,\varrho )=1,{lim}_{n\to \infty }\mathsf{Q}(\zeta \upsilon {\varkappa }_n,\upsilon \zeta {\varkappa }_n,\varrho )=0$ and

${lim}_{n\to \infty }\mathsf{R}(\zeta \upsilon {\varkappa }_n,\upsilon \zeta {\varkappa }_n,\varrho )=0$.

Thus, the subcompatibility of ζ and υ is proved. However, $\zeta \varkappa =\upsilon \varkappa$ if and only if $\varkappa =2$ and $\zeta \upsilon \left(2\right)\ne \upsilon \zeta \left(2\right)$. Therefore, ζ and υ do not satisfy the OWC property.

Our next goal is to introduce the concept of subsequential continuity within the framework of NMS. This concept is more relaxed than the previously established notion of reciprocal continuity, which was firstly given by Pant [36]. A similar notion was also introduced by H. Boudhadjera [20] within the framework of MS.

Definition 14.

Let $(\mathfrak{A},\aleph ,\uplus ,\odot )$ be NMS. Self-maps ζ and υ defined on the set $\mathfrak{A}$ are called subsequentially continuous iff ∃ a sequence $\{a_n\}$ in $\mathfrak{A}$ such that ${lim}_{n\to \infty }\zeta a_n={lim}_{n\to \infty }\upsilon a_n=b,b\in \mathfrak{A}$ and satisfy ${lim}_{n\to \infty }\zeta \upsilon a_n=\zeta b,{lim}_{n\to \infty }\upsilon \zeta a_n=\upsilon b.$

It is evident that, if both $\zeta$ and $\upsilon$ are continuous or reciprocally continuous, then they are subsequentially continuous. However, the subsequent example demonstrates that ∃ pairs of maps that are subsequentially continuous, but do not meet the criteria of being continuous or reciprocally continuous.

Example 11.

Consider an MS $(\mathfrak{A},d)$, where the metric d is defined as $d(\varkappa ,\varsigma )=|\varkappa -\varsigma |$. For any elements ϰ and ς in $\mathfrak{A}=[0,\infty )$ and for any positive value $\varrho \in (0,\infty )$, we have the operators ⊎ and ⊙ defined as follows:

• $TN\mathsf{x}\uplus \mathsf{y}=min.(\mathsf{x},\mathsf{y})$ and $TC\mathsf{x}\odot \mathsf{y}=max.(\mathsf{x},\mathsf{y})$. Define $\mathsf{P}(\varkappa ,\varsigma ,\varrho )=$$\begin{array}{cc}\{& \begin{array}{cc}{\displaystyle \frac{\varrho }{\varrho +|\varkappa -\varsigma |}}& \varrho >0\\ 0& \varrho =0\end{array}\end{array}$, $\mathsf{Q}(\varkappa ,\varsigma ,\varrho )=$$\begin{array}{cc}\{& \begin{array}{cc}{\displaystyle \frac{\varkappa -\varsigma }{\varrho +|\varkappa -\varsigma |}}& \varrho >0\\ 1& \varrho =1\end{array}\end{array}$ and $\mathsf{R}(\varkappa ,\varsigma ,\varrho )=$ $\begin{array}{cc}\{& \begin{array}{cc}{\displaystyle \frac{|\varkappa -\varsigma |}{\varrho }}& \varrho >0\\ 0& \varrho =0\end{array}\end{array}$. Clearly, $(\mathfrak{A},\aleph ,\uplus ,\odot )$ is a complete NMS.

Define ζ and υ as follows: $\zeta \left(\varkappa \right)=$ $\begin{array}{cc}\{& \begin{array}{cc}1+\varkappa & \varkappa \in [0,1]\\ 2\varkappa -1& \varkappa \in [1,\infty ).\end{array}\end{array}$ and $\upsilon \left(\varkappa \right)=$ $\begin{array}{cc}\{& \begin{array}{cc}1-\varkappa & \varkappa \in [0,1]\\ 3\varkappa -2& \varkappa \in [1,\infty ).\end{array}\end{array}$. Clearly ζ and υ are discontinuous at $\varkappa =1$. Consider the sequence $\{{\varkappa }_n\}$ in set $\mathfrak{A}$, defined as ${\varkappa }_n={\displaystyle \frac{1}{n}}$ for $n=1,2,3,\dots$.

As n approaches infinity, the limits are as follows:

• ${lim}_{n\to \infty }\zeta {\varkappa }_n={lim}_{n\to \infty }\upsilon {\varkappa }_n=1$, so 1 is an element of set $\mathfrak{A}$.
• Additionally, $\zeta \upsilon {\varkappa }_n$ approaches 2 (which is $\zeta \left(1\right)$) and $\upsilon \zeta$ also tends to 1 (which is $\upsilon \left(1\right)$) as n approaches infinity.

Therefore, ζ and υ exhibit subsequential continuity.

Now, consider the sequence $\{{\varkappa }_n\}$ in set $\mathfrak{A}$ defined as ${\varkappa }_n=1+{\displaystyle \frac{1}{n}}$ for $n=1,2,3,...$. In this case, as n approaches infinity for the sequence $\{{\varkappa }_n\}$ in set $\mathfrak{A}$, defined as ${\varkappa }_n=1+{\displaystyle \frac{1}{n}}$ for $n=1,2,3,\dots$:

${lim}_{n\to \infty }\zeta {\varkappa }_n={lim}_{n\to \infty }\upsilon {\varkappa }_n=1$, so 1 is an element of set $\mathfrak{A}$. However, $\zeta \upsilon {\varkappa }_n$ approaches 1 (which is $\zeta \left(1\right)$), not 2.

As a result, when n approaches infinity, ζ and υ do not exhibit reciprocal continuity as $\zeta \upsilon {\varkappa }_n$ does not converge to $\zeta \left(1\right)$.

Lemma 1.

Consider the sequence $\{u_n\}$ within the NMS $(\mathfrak{A},\aleph ,\uplus ,\odot )$. If ∃ a constant k in the open interval $(0,1)$ such that

$$\begin{array}{c}\hfill \mathsf{P}(u_n,u_{n+1},k\varrho )\ge \mathsf{P}(u_{n-1},u_n,\varrho ),\mathsf{Q}(u_n,u_{n+1},k\varrho )\le \mathsf{Q}(u_{n-1},u_n,\varrho )\end{array}$$

and

$$\begin{array}{c}\hfill \mathsf{R}(u_n,u_{n+1},k\varrho )\le \mathsf{R}\left(u_{n-1,u_n,\varrho }\right),\end{array}$$

$\forall \varrho >0$ and $n=1,2,3...$. Then $\{u_n\}$ is a Cauchy sequence in $\mathfrak{A}$.

Proof. 

Consider the following conditions: $\mathsf{P}(u_n,u_{n+1},k\varrho )\ge \mathsf{P}(u_{n-1},u_n,\varrho )$, $\mathsf{Q}(u_n,u_{n+1},k\varrho )\le \mathsf{Q}(u_{n-1},u_n,\varrho )$, and $\mathsf{R}(u_n,u_{n+1},k\varrho )\le \mathsf{R}(u_{n-1},u_n,\varrho )$.

These imply that

$\left\{\mathsf{P}(u_n,u_{n+1},k\varrho )\right\}$ is a non-decreasing sequence bounded above by 1, and

$\left\{\mathsf{Q}(u_n,u_{n+1},k\varrho )\right\}$ and $\left\{\mathsf{R}(u_n,u_{n+1},k\varrho )\right\}$ are non-increasing sequences bounded below by 0.

By the monotone convergence theorem,

$$\mathsf{P}(u_n,u_{n+1},k\varrho )\to l_{\mathsf{P}},\mathsf{Q}(u_n,u_{n+1},k\varrho )\to l_{\mathsf{Q}},\mathsf{R}(u_n,u_{n+1},k\varrho )\to l_{\mathsf{R}},$$

where $l_{\mathsf{P}}\in [0,1]$, $l_{\mathsf{Q}}\in [0,1]$, and $l_{\mathsf{R}}\in [0,1]$.

Additionally, for $m>n$, using the properties of the neutrosophic metric, we have

$$\mathsf{P}(u_m,u_n,\varrho )\ge \mathsf{P}(u_{m-1},u_m,k^{m-n}\varrho ),$$

$$\mathsf{Q}(u_m,u_n,\varrho )\le \mathsf{Q}(u_{m-1},u_m,k^{m-n}\varrho ),$$

$$\mathsf{R}(u_m,u_n,\varrho )\le \mathsf{R}(u_{m-1},u_m,k^{m-n}\varrho ).$$

⇒

$$\mathsf{P}(u_{m-1},u_m,k^{m-n}\varrho )\to l_{\mathsf{P}},\mathsf{Q}(u_{m-1},u_m,k^{m-n}\varrho )\to l_{\mathsf{Q}},\mathsf{R}(u_{m-1},u_m,k^{m-n}\varrho )\to l_{\mathsf{R}},$$

as $m,n\to \infty$.

Additionally, for $ϵ>0$, N is chosen such that, for all $m,n\ge N$,

$$\mathsf{P}(u_{m-1},u_m,k^{m-n}\varrho )>1-ϵ,\mathsf{Q}(u_{m-1},u_m,k^{m-n}\varrho )<ϵ,\mathsf{R}(u_{m-1},u_m,k^{m-n}\varrho )<ϵ.$$

By the triangle inequality,

$$\mathsf{P}(u_m,u_n,\varrho )\to 1,\mathsf{Q}(u_m,u_n,\varrho )\to 0,\mathsf{R}(u_m,u_n,\varrho )\to 0.$$

Thus, $\left\{u_n\right\}$ is a Cauchy sequence in $\mathfrak{A}$. □

Lemma 2.

Consider the NMS denoted as $(\mathfrak{A},\aleph ,\uplus ,\odot )$. For all elements $\mathfrak{a}$ and $\mathfrak{b}$ belonging to $\mathfrak{A}$ as well as for any positive number ϱ, if ∃ a constant k in $(0,1),$ such that

$$\begin{array}{c}\hfill \mathsf{P}(\mathfrak{a},\mathfrak{b},k\varrho )\ge \mathsf{P}(\mathfrak{a},\mathfrak{b},\varrho ),\mathsf{Q}(\mathfrak{a},\mathfrak{b},k\varrho )\le \mathsf{Q}(\mathfrak{a},\mathfrak{b},\varrho ),\mathsf{R}(\mathfrak{a},\mathfrak{b},k\varrho )\le \mathsf{R}(\mathfrak{a},\mathfrak{b},\varrho ),\end{array}$$

then $\mathfrak{a}=\mathfrak{b}$.

Proof. 

If $\mathsf{P}(\mathfrak{a},\mathfrak{b},k\varrho )\ge \mathsf{P}(\mathfrak{a},\mathfrak{b},\varrho )$, $\mathsf{Q}(\mathfrak{a},\mathfrak{b},k\varrho )\le \mathsf{Q}(\mathfrak{a},\mathfrak{b},\varrho )$ and $\mathsf{R}(\mathfrak{a},\mathfrak{b},k\varrho )\le \mathsf{R}(\mathfrak{a},\mathfrak{b},\varrho )$ for all $t>0$ and some constant $0<k<1$, then

$$\mathsf{P}(\mathfrak{a},\mathfrak{b},\mathfrak{s})\ge \mathsf{P}\left(\mathfrak{a},\mathfrak{b},{\displaystyle \frac{\mathfrak{s}}{k}}\right)\ge \mathsf{P}\left(\mathfrak{a},\mathfrak{b},{\displaystyle \frac{\mathfrak{s}}{k^2}}\right)\ge \cdots \ge \mathsf{P}\left(\mathfrak{a},\mathfrak{b},{\displaystyle \frac{\mathfrak{s}}{k^n}}\right),$$

$$\mathsf{Q}(\mathfrak{a},\mathfrak{b},\mathfrak{s})\le \mathsf{Q}\left(\mathfrak{a},\mathfrak{b},{\displaystyle \frac{\mathfrak{s}}{k}}\right)\le \mathsf{Q}\left(\mathfrak{a},\mathfrak{b},{\displaystyle \frac{\mathfrak{s}}{k^2}}\right)\le \cdots \le \mathsf{Q}\left(\mathfrak{a},\mathfrak{b},{\displaystyle \frac{\mathfrak{s}}{k^n}}\right)$$

and

$$\mathsf{R}(\mathfrak{a},\mathfrak{b},\mathfrak{s})\le \mathsf{R}\left(\mathfrak{a},\mathfrak{b},{\displaystyle \frac{\mathfrak{s}}{k}}\right)\le \mathsf{R}\left(\mathfrak{a},\mathfrak{b},{\displaystyle \frac{\mathfrak{s}}{k^2}}\right)\le \cdots \le \mathsf{R}\left(\mathfrak{a},\mathfrak{b},{\displaystyle \frac{\mathfrak{s}}{k^n}}\right)$$

for all $\mathfrak{s}>0$ and $\mathfrak{a},\mathfrak{b}\in \mathfrak{A}$. Letting $n\to \infty$, we have $\mathsf{P}(\mathfrak{a},\mathfrak{b},\mathfrak{s})=1$, $\mathsf{Q}(\mathfrak{a},\mathfrak{b},\mathfrak{s})=0$, and $\mathsf{R}(\mathfrak{a},\mathfrak{b},\mathfrak{s})=0$, so

$$\mathfrak{a}=\mathfrak{b}.$$

□

Now, we move forward to formulate the main theorem by leveraging the concepts of subcompatible and subsequentially continuous maps. These definitions, introduced earlier, form the basis for deriving significant fixed point results in NMS.

4. Result and Discussion

Theorem 1.

Consider four self-maps $\zeta ,\psi ,\upsilon$ and τ of NMS $(\mathfrak{A},\aleph ,\uplus ,\odot )$ with continuous TN ⊎ and continuous TC ⊙ defined by $\varrho \uplus \varrho \ge \varrho$ and $(1-\varrho )\odot (1-\varrho )\le (1-\varrho )$$\forall \varrho \in [0,1]$. If the pairs $(\zeta ,\upsilon )$ and $(\psi ,\tau )$ are both compatible and subsequentially continuous, then the following holds:

1. 

ζ and υ possess a CP.

2. 

ψ and τ possess a CP.

3. 

Let $\forall a,b\in A$, with $k\in (0,1)$ and $\varrho >0$,

$$\begin{array}{cc}\hfill \mathsf{P}(\zeta a,\psi b,k\varrho )& \ge \mathsf{P}(\upsilon a,\tau b,\varrho )\uplus \mathsf{P}(\zeta a,\upsilon a,\varrho )\uplus \mathsf{P}(\psi b,\tau b,\varrho )\uplus \mathsf{P}(\psi b,\upsilon a,2\varrho )\hfill \\ \hfill & \uplus \mathsf{P}(\zeta a,\tau b,\varrho ),\hfill \\ \hfill \mathsf{Q}(\zeta a,\psi b,k\varrho )& \le \mathsf{Q}(\upsilon a,\tau b,\varrho )\odot \mathsf{Q}(\zeta a,\upsilon a,\varrho )\odot \mathsf{Q}(\psi b,\tau b,\varrho )\odot \mathsf{Q}(\psi b,\upsilon a,2\varrho )\hfill \\ \hfill & \odot \mathsf{Q}(\zeta a,\tau b,\varrho ),\hfill \\ \hfill \mathsf{R}(\zeta a,\psi b,k\varrho )& \le \mathsf{R}(\upsilon a,\tau b,\varrho )\odot \mathsf{R}(\zeta a,\upsilon a,\varrho )\odot \mathsf{R}(\psi b,\tau b,\varrho )\odot \mathsf{R}(\psi b,\upsilon a,2\varrho )\hfill \\ \hfill & \odot \mathsf{R}(\zeta a,\tau b,\varrho ).\hfill \end{array}$$

Thus, $\zeta ,\psi ,\tau$ and υ have a unique CFP.

Proof. 

Given that $(\zeta ,\upsilon )$ and $(\psi ,\tau )$ are both subcompatible and subsequentially continuous, it follows that there are two sequences denoted as $\{a_n\}$ and $\{b_n\}$ within the set $\mathfrak{A}$ such that

${lim}_{n\to \infty }\zeta a_n={lim}_{n\to \infty }\upsilon a_n=c,c\in A,$ which satisfies

$$\begin{array}{c}\hfill \underset{n\to \infty }{lim}\mathsf{P}(\zeta \upsilon a_n,\upsilon \zeta a_n,\varrho )=\mathsf{P}(\zeta c,\upsilon c,\varrho )=1,\\ \hfill \underset{n\to \infty }{lim}\mathsf{Q}(\zeta \upsilon a_n,\upsilon \zeta a_n,\varrho )=\mathsf{Q}(\zeta c,\upsilon c,\varrho )=0\end{array}$$

and

$$\begin{array}{c}\hfill \underset{n\to \infty }{lim}\mathsf{R}(\zeta \upsilon a_n,\upsilon \zeta a_n,\varrho )=\mathsf{R}(\zeta c,\upsilon c,\varrho )=0.\end{array}$$

Additionally, ${lim}_{n\to \infty }\psi b_n={lim}_{n\to \infty }\tau b_n=c^{\prime },c^{\prime }\in A$, which satisfies

$$\begin{array}{c}\hfill \underset{n\to \infty }{lim}\mathsf{P}(\psi \tau b_n,\tau \psi b_n,\varrho )=\mathsf{P}(\psi c^{\prime },\tau c^{\prime },\varrho )=1,\\ \hfill \underset{n\to \infty }{lim}\mathsf{Q}(\psi \tau b_n,\tau \psi b_n,\varrho )=\mathsf{Q}(\psi c^{\prime },\tau c^{\prime },\varrho )=0\end{array}$$

and

$$\begin{array}{c}\hfill \underset{n\to \infty }{lim}\mathsf{R}(\psi \tau b_n,\tau \psi b_n,\varrho )=\mathsf{R}(\psi c^{\prime },\tau c^{\prime },\varrho )=0.\end{array}$$

Therefore, $\zeta c=\upsilon c$ and $\psi c^{\prime }=\tau c^{\prime }$, i.e., c is a CP of $\zeta$ and $\upsilon$ and $c^{\prime }$ is a CP of $\psi$ and $\tau$.

Now, we will prove that $c=c^{\prime }$. By condition (3), we have taken $a=a_n$ and $b=b_n$,

$$\begin{array}{cc}\hfill \mathsf{P}(\zeta a_n,\psi b_n,k\varrho )& \ge \mathsf{P}(\upsilon a_n,\tau b_n,\varrho )\uplus \mathsf{P}(\zeta a_n,\upsilon a_n,\varrho )\uplus \mathsf{P}(\psi b_n,\tau b_n,\varrho )\uplus \mathsf{P}(\psi b_n,\upsilon a_n,2\varrho )\hfill \\ \hfill & \uplus \mathsf{P}(\zeta a_n,\tau b_n,\varrho ),\hfill \\ \hfill \mathsf{Q}(\zeta a_n,\psi b_n,k\varrho )& \le \mathsf{Q}(\upsilon a_n,\tau b_n,\varrho )\odot \mathsf{Q}(\zeta a_n,\upsilon a_n,\varrho )\odot \mathsf{Q}(\psi b_n,\tau b_n,\varrho )\odot \mathsf{Q}(\psi b_n,\upsilon a_n,2\varrho )\hfill \\ \hfill & \odot \mathsf{Q}(\zeta a_n,\tau b_n,\varrho )\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill \mathsf{R}(\zeta a_n,\psi b_n,k\varrho )& \le \mathsf{R}(\upsilon a_n,\tau b_n,\varrho )\odot \mathsf{R}(\zeta a_n,\upsilon a_n,\varrho )\odot \mathsf{R}(\psi b_n,\tau b_n,\varrho )\odot \mathsf{R}(\psi b_n,\upsilon a_n,2\varrho )\hfill \\ \hfill & \odot \mathsf{R}(\zeta a_n,\tau b_n,\varrho ).\hfill \end{array}$$

Assuming the limit is $n\to \infty$, we have

$$\begin{array}{c}\hfill \mathsf{P}(c,c^{\prime },k\varrho )\ge \mathsf{P}(c,c^{\prime },\varrho )\uplus \mathsf{P}(c,c,\varrho )\uplus \mathsf{P}(c^{\prime },c^{\prime },\varrho )\uplus \mathsf{P}(c^{\prime },c,2\varrho )\uplus \mathsf{P}(c,c^{\prime },\varrho ),\\ \hfill \mathsf{Q}(c,c^{\prime },k\varrho )\le \mathsf{Q}(c,c^{\prime },\varrho )\odot \mathsf{Q}(c,c,\varrho )\odot \mathsf{Q}(c^{\prime },c^{\prime },\varrho )\odot \mathsf{Q}(c^{\prime },c,2\varrho )\odot \mathsf{Q}(c,c^{\prime },\varrho )\end{array}$$

and

$$\begin{array}{c}\hfill \mathsf{R}(c,c^{\prime },k\varrho )\le \mathsf{R}(c,c^{\prime },\varrho )\odot \mathsf{R}(c,c,\varrho )\odot \mathsf{R}(c^{\prime },c^{\prime },\varrho )\odot \mathsf{R}(c^{\prime },c,2\varrho )\odot \mathsf{R}(c,c^{\prime },\varrho ).\end{array}$$

$$\begin{array}{c}\hfill \Rightarrow \mathsf{P}(c,c^{\prime },k\varrho )\ge \mathsf{P}(c,c^{\prime },\varrho ),\mathsf{Q}(c,c^{\prime },k\varrho )\le \mathsf{Q}(c,c^{\prime },\varrho )\end{array}$$

and

$$\begin{array}{c}\hfill \mathsf{R}(c,c^{\prime },k\varrho )\le \mathsf{R}(c,c^{\prime },\varrho ).\end{array}$$

By Lemma 2, we have $c=c^{\prime }$.

Furthermore, we will prove that $\zeta c=c$. To establish this claim, we utilize inequality (3) by selecting $a=c$ and $b=b_n$. This gives

$$\begin{array}{cc}\hfill \mathsf{P}(\zeta c,\psi b_n,k\varrho )& \ge \mathsf{P}(\upsilon c,\tau b_n,\varrho )\uplus \mathsf{P}(\zeta c,\upsilon c,\varrho )\uplus \mathsf{P}(\psi b_n,\tau b_n,\varrho )\uplus \mathsf{P}(\psi b_n,\upsilon c,2\varrho )\hfill \\ \hfill & \uplus \mathsf{P}(\zeta c,\tau b_n,\varrho ),\hfill \\ \hfill \mathsf{Q}(\zeta c,\psi b_n,k\varrho )& \le \mathsf{Q}(\upsilon c,\tau b_n,\varrho )\odot \mathsf{Q}(\zeta c,\upsilon c,\varrho )\odot \mathsf{Q}(\psi b_n,\tau b_n,\varrho )\odot \mathsf{Q}(\psi b_n,\upsilon c,2\varrho )\hfill \\ \hfill & \odot \mathsf{Q}(\zeta c,\tau b_n,\varrho )\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill \mathsf{R}(\zeta c,\psi b_n,k\varrho )& \le \mathsf{R}(\upsilon c,\tau b_n,\varrho )\odot \mathsf{R}(\zeta c,\upsilon c,\varrho )\odot \mathsf{R}(\psi b_n,\tau b_n,\varrho )\odot \mathsf{R}(\psi b_n,\upsilon c,2\varrho )\hfill \\ \hfill & \odot \mathsf{R}(\zeta c,\tau b_n,\varrho ).\hfill \end{array}$$

Assuming the limit is $n\to \infty$, one can obtain

$$\begin{array}{cc}\hfill \mathsf{P}(\zeta c,c^{\prime },k\varrho )& \ge \mathsf{P}(\zeta c,c^{\prime },\varrho )\uplus \mathsf{P}(\zeta c,\zeta c,\varrho )\uplus \mathsf{P}(c^{\prime },c^{\prime },\varrho )\uplus \mathsf{P}(c^{\prime },\zeta c,2\varrho )\hfill \\ \hfill & \uplus \mathsf{P}(\zeta c,c^{\prime },\varrho ),\hfill \\ \hfill \mathsf{Q}(\zeta c,c^{\prime },k\varrho )& \le \mathsf{Q}(\zeta c,c^{\prime },\varrho )\odot \mathsf{Q}(\zeta c,\zeta c,\varrho )\odot \mathsf{Q}(c^{\prime },c^{\prime },\varrho )\odot \mathsf{Q}(c^{\prime },\zeta c,2\varrho )\hfill \\ \hfill & \odot \mathsf{Q}(\zeta c,c^{\prime },\varrho )\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill \mathsf{R}(\zeta c,c^{\prime },k\varrho )& \le \mathsf{R}(\zeta c,c^{\prime },\varrho )\odot \mathsf{R}(\zeta c,\zeta c,\varrho )\odot \mathsf{R}(c^{\prime },c^{\prime },\varrho )\odot \mathsf{R}(c^{\prime },\zeta c,2\varrho )\hfill \\ \hfill & \odot \mathsf{R}(\zeta c,c^{\prime },\varrho ).\hfill \end{array}$$

$$\begin{array}{c}\hfill \Rightarrow \mathsf{P}(\zeta c,c^{\prime },k\varrho )\ge \mathsf{P}(\zeta c,c^{\prime },\varrho ),\mathsf{Q}(\zeta c,c^{\prime },k\varrho )\le \mathsf{Q}(\zeta c,c^{\prime },\varrho ),\end{array}$$

and

$$\begin{array}{c}\hfill \mathsf{R}(\zeta c,c^{\prime },k\varrho )\le \mathsf{R}(\zeta c,c^{\prime },\varrho ).\end{array}$$

By Lemma 2, we have $\zeta c=c^{\prime }=c$.

Again, we claim that $\psi c=c$. By inequality (3) of Theorem 1, assuming $a=c$ and $b=c$, we obtain

$$\begin{array}{cc}\hfill \mathsf{P}(\zeta c,\psi c,k\varrho )& \ge \mathsf{P}(\upsilon c,\tau c,\varrho )\uplus \mathsf{P}(\zeta c,\upsilon c,\varrho )\uplus \mathsf{P}(\psi c,\tau c,\varrho )\uplus \mathsf{P}(\psi c,\upsilon c,2\varrho )\hfill \\ \hfill & \uplus \mathsf{P}(\zeta c,\tau c,\varrho ),\hfill \\ \hfill \mathsf{Q}(\zeta c,\psi c,k\varrho )& \le \mathsf{Q}(\upsilon c,\tau c,\varrho )\odot \mathsf{Q}(\zeta c,\upsilon c,\varrho )\odot \mathsf{Q}(\psi c,\tau c,\varrho )\odot \mathsf{Q}(\psi c,\upsilon c,2\varrho )\hfill \\ \hfill & \odot \mathsf{Q}(\zeta c,\tau c,\varrho )\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill \mathsf{R}(\zeta c,\psi c,k\varrho )& \le \mathsf{R}(\upsilon c,\tau c,\varrho )\odot \mathsf{R}(\zeta c,\upsilon c,\varrho )\odot \mathsf{R}(\psi c,\tau c,\varrho )\odot \mathsf{R}(\psi c,\upsilon c,2\varrho )\hfill \\ \hfill & \odot \mathsf{R}(\zeta c,\tau c,\varrho ).\hfill \end{array}$$

$$\begin{array}{cc}\hfill \Rightarrow \mathsf{P}(c,\psi c,k\varrho )& \ge \mathsf{P}(c,\psi c,\varrho )\uplus \mathsf{P}(\zeta c,\zeta c,\varrho )\uplus \mathsf{P}(\psi c,\psi c,\varrho )\uplus \mathsf{P}(\psi c,c,2\varrho )\hfill \\ \hfill & \uplus \mathsf{P}(c,\psi c,\varrho ),\hfill \\ \hfill \mathsf{Q}(c,\psi c,k\varrho )& \le \mathsf{Q}(c,\psi c,\varrho )\odot \mathsf{Q}(\zeta c,\zeta c,\varrho )\odot \mathsf{Q}(\psi c,\psi c,\varrho )\odot \mathsf{Q}(\psi c,c,2\varrho )\hfill \\ \hfill & \odot \mathsf{Q}(c,\psi c,\varrho )\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill \mathsf{R}(c,\psi c,k\varrho )& \le \mathsf{R}(c,\psi c,\varrho )\odot \mathsf{R}(\zeta c,\zeta c,\varrho )\odot \mathsf{R}(\psi c,\psi c,\varrho )\odot \mathsf{R}(\psi c,c,2\varrho )\hfill \\ \hfill & \odot \mathsf{R}(c,\psi c,\varrho ).\hfill \end{array}$$

By Lemma 2, we have $c=\psi c=\tau c$.

Therefore, $c=\zeta c=\psi c=\upsilon c=\tau c;$ i.e., c is a common FP of $\zeta ,\psi ,\upsilon$ and $\tau .$

Uniqueness: Now suppose ∃ another FP w of $\zeta ,\psi ,\upsilon$ and $\tau .$

Under condition (3) of Theorem 1, if we set $a=c$ and $b=w$, we obtain,

$$\begin{array}{cc}\hfill \mathsf{P}(\zeta c,\psi w,k\varrho )& \ge \mathsf{P}(\upsilon c,\tau w,\varrho )\uplus \mathsf{P}(\zeta c,\upsilon c,\varrho )\uplus \mathsf{P}(\psi w,\tau w,\varrho )\uplus \mathsf{P}(\psi w,\upsilon c,2\varrho )\hfill \\ \hfill & \uplus \mathsf{P}(\zeta c,\tau w,\varrho ),\hfill \\ \hfill \mathsf{Q}(\zeta c,\psi w,k\varrho )& \le \mathsf{Q}(\upsilon c,\tau w,\varrho )\odot \mathsf{Q}(\zeta c,\upsilon c,\varrho )\odot \mathsf{Q}(\psi w,\tau w,\varrho )\odot \mathsf{Q}(\psi w,\upsilon c,2\varrho )\hfill \\ \hfill & \odot \mathsf{Q}(\zeta c,\tau w,\varrho )\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill \mathsf{R}(\zeta c,\psi w,k\varrho )& \le \mathsf{R}(\upsilon c,\tau w,\varrho )\odot \mathsf{R}(\zeta c,\upsilon c,\varrho )\odot \mathsf{R}(\psi w,\tau w,\varrho )\odot \mathsf{R}(\psi w,\upsilon c,2\varrho )\hfill \\ \hfill & \odot \mathsf{R}(\zeta c,\tau w,\varrho ).\hfill \end{array}$$

$$\begin{array}{c}\hfill \Rightarrow \mathsf{P}(c,w,k\varrho )\ge \mathsf{P}(c,w,\varrho ),\mathsf{Q}(c,w,k\varrho )\le \mathsf{Q}(c,w,\varrho )\end{array}$$

and

$$\begin{array}{c}\hfill \mathsf{R}(c,w,k\varrho )\le \mathsf{R}(c,w,\varrho ).\end{array}$$

Therefore, in accordance with Lemma 2, it can be concluded that $c=w$.

When we substitute $\upsilon =\tau$ in the above theorem, the following result emerges. □

Corollary 1.

Let us consider three self-maps $\zeta ,\psi$, and υ of NMS $(\mathfrak{A},\aleph ,\uplus ,\odot )$ with continuous TN ⊎ and continuous TC ⊙ defined by $\varrho \uplus \varrho \ge \varrho$ and $(1-\varrho )\odot (1-\varrho )\le (1-\varrho )$$\forall \varrho \in [0,1]$. If the pairs $(\zeta ,\upsilon )$ and $(\psi ,\upsilon )$ are compatible and subsequentially continuous, then

1. 

ζ and υ have a CP, and

2. 

ψ and υ have a CP.

3. 

Further, let $\forall a,b\in \mathfrak{A},k\in (0,1),\varrho >0$,

$$\begin{array}{cc}\hfill \mathsf{P}(\zeta a,\psi b,k\varrho )& \ge \mathsf{P}(\upsilon a,\upsilon b,\varrho )\uplus \mathsf{P}(\zeta a,\upsilon a,\varrho )\uplus \mathsf{P}(\psi b,\upsilon b,\varrho )\uplus \mathsf{P}(\psi b,\upsilon a,2\varrho )\hfill \\ \hfill & \uplus \mathsf{P}(\zeta a,\upsilon b,\varrho ),\hfill \\ \hfill \mathsf{Q}(\zeta a,\psi b,k\varrho )& \le \mathsf{Q}(\upsilon a,\upsilon b,\varrho )\odot \mathsf{Q}(\zeta a,\upsilon a,\varrho )\odot \mathsf{Q}(\psi b,\upsilon b,\varrho )\odot \mathsf{Q}(\psi b,\upsilon a,2\varrho )\hfill \\ \hfill & \odot \mathsf{Q}(\zeta a,\upsilon b,\varrho )\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill \mathsf{R}(\zeta a,\psi b,k\varrho )& \le \mathsf{R}(\upsilon a,\upsilon b,\varrho )\odot \mathsf{R}(\zeta a,\upsilon a,\varrho )\odot \mathsf{R}(\psi b,\upsilon b,\varrho )\odot \mathsf{R}(\psi b,\upsilon a,2\varrho )\hfill \\ \hfill & \odot \mathsf{R}(\zeta a,\upsilon b,\varrho ).\hfill \end{array}$$

Then, $\zeta ,\psi$ and υ have a unique common FP.

Proof. 

The proof is the same as in the above theorem. If $\upsilon =\tau$, the desired result will be obtained. □

Remark 3.

We can also establish the validity of this theorem by substituting subsequential continuity of the pairs with reciprocal continuity of the pairs.

Remark 4.

The conclusions of the previously mentioned theorems remain valid when we replace compatibility with subcompatibility and subsequential continuity with reciprocal continuity while maintaining all other hypotheses unchanged.

5. Application

Let us take an example of volterra I.E and apply FPT to find the solution.

Consider the following I.E, which is a Volterra I.E of the first kind:

$$u\left(\mathfrak{z}\right)=\mathfrak{z}+\varrho {\int }_0^{\mathfrak{z}}{e}^{-(\mathfrak{z}-\varpi )}u\left(\varpi \right)d\varpi ,$$

where •$u\left(\mathfrak{z}\right)$ is the unknown function,

• $\varrho$ is a given constant and
• $0\le \mathfrak{z}\le 1$.

We want to prove that this I.E has a unique solution on the interval $[0,1]$.

Step 1: 

Define the Operator:

We define an operator $\mathfrak{M}$ as follows:

$$\mathfrak{M}\left(u\right)\left(\mathfrak{z}\right)=\mathfrak{z}+\varrho {\int }_0^{\mathfrak{z}}{e}^{-(\mathfrak{z}-\varpi )}u\left(\varpi \right)d\varpi .$$

To prove that $\mathfrak{M}$ is continuous, we will show that, if a sequence of functions $\left\{u_n\right\}$ converges to some function u (pointwise or in some other suitable norm), then the sequence of transformed functions $\mathfrak{M}\left(u_n\right)$ converges to $\mathfrak{M}\left(u\right)$.

Let us consider a sequence of functions $\left\{u_n\right\}$ that converges to u pointwise:

$$u_n\left(\mathfrak{z}\right)\to u\left(\mathfrak{z}\right)\mathrm{as}n\to \infty .$$

We will show that

$$\mathfrak{M}\left(u_n\right)\left(\mathfrak{z}\right)\to \mathfrak{M}\left(u\right)\left(\mathfrak{z}\right)\mathrm{as}n\to \infty .$$

For this, we can break down $\mathfrak{M}\left(u_n\right)\left(\mathfrak{z}\right)$:

$$\mathfrak{M}\left(u_n\right)\left(\mathfrak{z}\right)=\mathfrak{z}+\varrho {\int }_0^{\mathfrak{z}}{e}^{-(\mathfrak{z}-\varpi )}{u}_n\left(\varpi \right)d\varpi .$$

Now, we will use the concept of dominated convergence. For this we will show that

• $u_n\left(\varpi \right)$ converges pointwise to $u\left(\varpi \right)$ as $n\to \infty$ for each $\varpi$ in the interval $[0,\mathfrak{z}]$.
• ∃ a dominating function $g\left(\mathfrak{z}\right)$ such that $|u_n\left(\varpi \right)|\le g\left(\mathfrak{z}\right)$∀n and $\varpi$ in $[0,\mathfrak{z}]$ and $g\left(\mathfrak{z}\right)$ is integrable on $[0,\mathfrak{z}]$.
• The dominated convergence theorem (DCT) is used to justify interchanging the limit and the integral sign in $\mathfrak{M}\left(u_n\right)\left(\mathfrak{z}\right)$.

By the DCT, we can conclude that

$$\mathfrak{M}\left(u_n\right)\left(\mathfrak{z}\right)\to \mathfrak{M}\left(u\right)\left(\mathfrak{z}\right)\mathrm{as}n\to \infty .$$

This establishes the subsequential continuity of the operator $\mathfrak{M}$ when sequence of functions $\left\{u_n\right\}$ converges pointwise to u.

Step 2: 

Show $\mathfrak{M}$ is a Contraction Mapping:

To apply the FPT, we have to show that $\mathfrak{M}$ is a contraction mapping on a suitable MS. For this purpose, we can use the $L^{\infty }$ norm, which measures the supremum of the absolute values of the function over the interval $[0,1]$.

Now, to prove that $\mathfrak{M}$ is a contraction mapping, we will prove that ∃ a constant $0\le k<1$ such that

$${|\mathfrak{M}\left(u\right)-\mathfrak{M}\left(v\right)|}_{\infty }\le k{|u-v|}_{\infty },$$

where ${\left|u\right|}_{\infty }={sup}_{0\le \mathfrak{z}\le 1}\left|u\left(\mathfrak{z}\right)\right|$ is the $L^{\infty }$ norm.

By analyzing $\left|\mathfrak{M}\right(u\left)\right(\mathfrak{z})-\mathfrak{M}(v\left)\right(\mathfrak{z}\left)\right|$, we can derive an expression involving the difference $\left|u\right(\mathfrak{z})-v(\mathfrak{z}\left)\right|$. By using the Mean Value Theorem for integrals, we can find a k such that the contraction condition holds.

Step 3: 

Apply the Fixed-Point Theorem:

Once we have established that $\mathfrak{M}$ is a contraction mapping, we can apply the FPT. This theorem guarantees that ∃ a unique function u in the MS of continuous functions on $[0,1]$ such that $\mathfrak{M}\left(u\right)=u$. This function u is the unique solution to the original integral equation.

This method shows that the I.E has a unique solution on the interval $[0,1]$ by applying the FPT and using the concept of a contraction mapping with subcompatibility or subsequential continuity.

Implementation by taking a mathematical example

Step 1: 

Defining the Operator:

Consider the sequence of functions $\left\{u_n\left(\mathfrak{z}\right)\right\}={\displaystyle \frac{{\mathfrak{z}}^n}{n}}$ for $n\ge 1$ on the interval $[0,1]$. We aim to show that $u_n\left(\mathfrak{z}\right)$ converges pointwise to $u\left(\mathfrak{z}\right)=0$ on $[0,1]$.

(i) Pointwise Convergence:

For any fixed $\mathfrak{z}$ in $[0,1]$, as n approaches infinity, ${\displaystyle \frac{{\mathfrak{z}}^n}{n}}$ converges to 0. Since the numerator ${\mathfrak{z}}^n$ grows slower than the denominator, n goes to infinity. Therefore, $u_n\left(\mathfrak{z}\right)$ converges pointwise to $u\left(\mathfrak{z}\right)=0$.

(ii) Existence of Dominating Function:

We will find a dominating function $g\left(\mathfrak{z}\right)$ such that $|u_n\left(\mathfrak{z}\right)|\le g\left(\mathfrak{z}\right)$ for all n and $\mathfrak{z}$ in $[0,1]$.

Since $u_n\left(\mathfrak{z}\right)={\displaystyle \frac{{\mathfrak{z}}^n}{n}}$ and $0\le \mathfrak{z}\le 1$, we have $|u_n\left(\mathfrak{z}\right)|\le {\displaystyle \frac{1}{n}}$ for all n.

Now, choose $g\left(\mathfrak{z}\right)=1$. Then $|u_n\left(\mathfrak{z}\right)|\le {\displaystyle \frac{1}{n}}\le 1$ for all n and $\mathfrak{z}$ in $[0,1]$. Hence, $g\left(\mathfrak{z}\right)=1$ is a dominating function.

(iii) Dominated Convergence Theorem (DCT):

With $g\left(\mathfrak{z}\right)=1$, which is integrable over $[0,1]$, we can apply the DCT to justify the interchange of the limit and the integral sign.

Thus,

$$\mathfrak{M}\left(u_n\right)\left(\mathfrak{z}\right)=\mathfrak{z}+\varrho {\int }_0^{\mathfrak{z}}{e}^{-(\mathfrak{z}-\varpi )}{u}_n\left(\varpi \right)d\varpi =\mathfrak{z}+\varrho {\int }_0^{\mathfrak{z}}{e}^{-(\mathfrak{z}-\varpi )}{\displaystyle \frac{{\varpi }^n}{n}}d\varpi .$$

Since $e^{-(\mathfrak{z}-\varpi )}$ is bounded on $[0,1]$, we can replace $u_n\left(\varpi \right)$ with $u\left(\varpi \right)=0$ by the DCT. Therefore, $\mathfrak{M}\left(u_n\right)\left(\mathfrak{z}\right)$ converges to $\mathfrak{M}\left(u\right)\left(\mathfrak{z}\right)$ as $n\mapsto \infty$.

This establishes the subsequential continuity of the operator $\mathfrak{M}$ when functions $u_n$ converges pointwise to u.

Step 2: 

Show $\mathfrak{M}$ is a Contraction Mapping:

To prove that $\mathfrak{M}$ is a contraction mapping, we need to find a constant $0\le k<1$ such that for any two functions $u\left(\mathfrak{z}\right)$ and $v\left(\mathfrak{z}\right)$ in the function space, the operator $\mathfrak{M}$ satisfies the contraction condition:

$${|\mathfrak{M}\left(u\right)-\mathfrak{M}\left(v\right)|}_{\infty }\le k{|u-v|}_{\infty },$$

where ${\left|f\right|}_{\infty }={sup}_{\mathfrak{z}\in [0,1]}\left|f\left(\mathfrak{z}\right)\right|$ represents the supremum norm over the interval $[0,1]$.

Let $\mathfrak{M}\left(u\right)$ and $\mathfrak{M}\left(v\right)$ be

$$\mathfrak{M}\left(u\right)\left(\mathfrak{z}\right)=\mathfrak{z}+\varrho {\int }_0^{\mathfrak{z}}{e}^{-(\mathfrak{z}-\varpi )}u\left(\varpi \right)d\varpi .$$

$$\mathfrak{M}\left(v\right)\left(\mathfrak{z}\right)=\mathfrak{z}+\varrho {\int }_0^{\mathfrak{z}}{e}^{-(\mathfrak{z}-\varpi )}v\left(\varpi \right)d\varpi .$$

Thus, the difference $\mathfrak{M}\left(u\right)\left(\mathfrak{z}\right)-\mathfrak{M}\left(v\right)\left(\mathfrak{z}\right)$ is

$$\mathfrak{M}\left(u\right)\left(\mathfrak{z}\right)-\mathfrak{M}\left(v\right)\left(\mathfrak{z}\right)=\varrho {\int }_0^{\mathfrak{z}}{e}^{-(\mathfrak{z}-\varpi )}(u\left(\varpi \right)-v\left(\varpi \right))d\varpi .$$

Now, ${|\mathfrak{M}\left(u\right)-\mathfrak{M}\left(v\right)|}_{\infty }$ is

$${|\mathfrak{M}\left(u\right)-\mathfrak{M}\left(v\right)|}_{\infty }=\underset{\mathfrak{z}\in [0,1]}{sup}|\mathfrak{M}\left(u\right)\left(\mathfrak{z}\right)-\mathfrak{M}\left(v\right)\left(\mathfrak{z}\right)|$$

$$=\underset{\mathfrak{z}\in [0,1]}{sup}\left|\varrho {\int }_0^{\mathfrak{z}}{e}^{-(\mathfrak{z}-\varpi )}(u\left(\varpi \right)-v\left(\varpi \right))d\varpi \right|.$$

Substituting the expression for $\mathfrak{M}\left(u\right)\left(\mathfrak{z}\right)-\mathfrak{M}\left(v\right)\left(\mathfrak{z}\right)$,

$${|\mathfrak{M}\left(u\right)-\mathfrak{M}\left(v\right)|}_{\infty }=\underset{\mathfrak{z}\in [0,1]}{sup}\left|\varrho {\int }_0^{\mathfrak{z}}{e}^{-(\mathfrak{z}-\varpi )}(u\left(\varpi \right)-v\left(\varpi \right))d\varpi \right|.$$

The absolute value of the integral can be bounded by the integral of the absolute value:

$${|\mathfrak{M}\left(u\right)-\mathfrak{M}\left(v\right)|}_{\infty }\le \underset{\mathfrak{z}\in [0,1]}{sup}\varrho {\int }_0^{\mathfrak{z}}{e}^{-(\mathfrak{z}-\varpi )}|u\left(\varpi \right)-v\left(\varpi \right)|d\varpi .$$

Since $e^{-(\mathfrak{z}-\varpi )}\le 1$ for all $\mathfrak{z},\varpi \in [0,1]$, we have

$${|\mathfrak{M}\left(u\right)-\mathfrak{M}\left(v\right)|}_{\infty }\le \underset{\mathfrak{z}\in [0,1]}{sup}\varrho {\int }_0^{\mathfrak{z}}|u\left(\varpi \right)-v\left(\varpi \right)|d\varpi .$$

By the definition of the supremum norm,

$$|u\left(\varpi \right)-v\left(\varpi \right)|\le {|u-v|}_{\infty },\forall \varpi \in [0,1].$$

Thus,

$${|\mathfrak{M}\left(u\right)-\mathfrak{M}\left(v\right)|}_{\infty }\le \underset{\mathfrak{z}\in [0,1]}{sup}\varrho {\int }_0^{\mathfrak{z}}{|u-v|}_{\infty }d\varpi .$$

Since ${|u-v|}_{\infty }$ is independent of $\varpi$, it can be factored out:

$${|\mathfrak{M}\left(u\right)-\mathfrak{M}\left(v\right)|}_{\infty }\le \varrho {|u-v|}_{\infty }\underset{\mathfrak{z}\in [0,1]}{sup}{\int }_0^{\mathfrak{z}}1d\varpi .$$

Simplifying the integral ${\int }_0^{\mathfrak{z}}1d\varpi =\mathfrak{z}$, it follows that

$${|\mathfrak{M}\left(u\right)-\mathfrak{M}\left(v\right)|}_{\infty }\le \varrho {|u-v|}_{\infty }\underset{\mathfrak{z}\in [0,1]}{sup}\mathfrak{z}.$$

Since $\mathfrak{z}\le 1$ for $\mathfrak{z}\in [0,1]$, we get

$${|\mathfrak{M}\left(u\right)-\mathfrak{M}\left(v\right)|}_{\infty }\le \varrho {|u-v|}_{\infty }.$$

If $\varrho <1$, we can set $k=\varrho$, where $0\le k<1$. Thus, $\mathfrak{M}$ satisfies the contraction condition:

$${|\mathfrak{M}\left(u\right)-\mathfrak{M}\left(v\right)|}_{\infty }\le k{|u-v|}_{\infty }.$$

Step 3: 

Apply the Fixed-Point Theorem:

The Fixed-Point Theorem states:

If Y is a complete metric space and $\mathfrak{M}:Y\to Y$ is a contraction mapping with contraction constant $k<1$, then $\mathfrak{M}$ has a unique FP $u^{*}$ in Y such that $\mathfrak{M}\left(u^{*}\right)=u^{*}$.

Here, Y is the space of continuous functions equipped with the supremum norm ${\left|f\right|}_{\infty }$ over the interval $[0,1]$, and $\mathfrak{M}:Y\to Y$ is the operator defined as

$$\mathfrak{M}\left(u\right)\left(\mathfrak{z}\right)=\mathfrak{z}+\varrho {\int }_0^{\mathfrak{z}}{e}^{-(\mathfrak{z}-\varpi )}u\left(\varpi \right)d\varpi .$$

We have already shown that $\mathfrak{M}$ is a contraction mapping with contraction constant $k=\varrho$, which satisfies $0\le \varrho <1$.

Now, as $\mathfrak{M}$ is a contraction mapping on a complete MS, ∃ a unique FP $u^{*}$ such that $\mathfrak{M}\left(u^{*}\right)=u^{*}$. In other words, ∃ a unique function $u\left(\mathfrak{z}\right)$ satisfying the I.E:

$$u\left(\mathfrak{z}\right)=\mathfrak{z}+\varrho {\int }_0^{\mathfrak{z}}{e}^{-(\mathfrak{z}-\varpi )}u\left(\varpi \right)d\varpi .$$

To find the solution, let $u_0\left(\mathfrak{z}\right)=0$ by the Picard iteration method. Find $u_1\left(\mathfrak{z}\right)$:

$$u_{n+1}\left(\mathfrak{z}\right)=\mathfrak{z}+\varrho {\int }_0^{\mathfrak{z}}{e}^{-(\mathfrak{z}-\varpi )}{u}_n\left(\varpi \right)d\varpi$$

Since $u_0\left(\mathfrak{z}\right)=0$, the integral term becomes zero, and we have

$$u_1\left(\mathfrak{z}\right)=\mathfrak{z}+\varrho {\int }_0^{\mathfrak{z}}{e}^{-(\mathfrak{z}-\varpi )}·0d\varpi$$

$$u_1\left(\mathfrak{z}\right)=\mathfrak{z}$$

Therefore, $u_1\left(\mathfrak{z}\right)=\mathfrak{z}$ is the first approximation in the Picard iteration method when the initial guess is $u_0\left(\mathfrak{z}\right)=0$.

To find $u_2\left(\mathfrak{z}\right)$, the second approximation in the Picard iteration method, we use $u_1\left(\mathfrak{z}\right)=\mathfrak{z}$ obtained from the previous iteration as the input function in the iteration formula:

$$u_{n+1}\left(\mathfrak{z}\right)=\mathfrak{z}+\varrho {\int }_0^{\mathfrak{z}}{e}^{-(\mathfrak{z}-\varpi )}{u}_n\left(\varpi \right)d\varpi$$

$$u_2\left(\mathfrak{z}\right)=\mathfrak{z}+\varrho {\int }_0^{\mathfrak{z}}{e}^{-(\mathfrak{z}-\varpi )}·\varpi d\varpi$$

We will compute this integral to find $u_2\left(\mathfrak{z}\right)$:

$$u_2\left(\mathfrak{z}\right)=\mathfrak{z}+\varrho \left[-{\displaystyle \frac{1}{e^{\mathfrak{z}-\varpi }}}·\varpi {|}_0^{\mathfrak{z}}+{\int }_0^{\mathfrak{z}}{\displaystyle \frac{1}{e^{\mathfrak{z}-\varpi }}}·1d\varpi \right]$$

$$u_2\left(\mathfrak{z}\right)=\mathfrak{z}+\varrho \left[-{\displaystyle \frac{1}{e^{\mathfrak{z}-\mathfrak{z}}}}·\mathfrak{z}+{\int }_0^{\mathfrak{z}}{\displaystyle \frac{1}{e^{\mathfrak{z}-\varpi }}}d\varpi \right]$$

$$u_2\left(\mathfrak{z}\right)=\mathfrak{z}+\varrho \left[-{\displaystyle \frac{\mathfrak{z}}{1}}+\left(-{\displaystyle \frac{1}{e^{\mathfrak{z}-\varpi }}}{|}_0^{\mathfrak{z}}\right)\right]$$

$$u_2\left(\mathfrak{z}\right)=\mathfrak{z}+\varrho \left[-\mathfrak{z}+1-{\displaystyle \frac{1}{e^{\mathfrak{z}}}}\right]$$

$$u_2\left(\mathfrak{z}\right)=(1-\varrho )\mathfrak{z}+\varrho -\varrho {\displaystyle \frac{1}{e^{\mathfrak{z}}}}$$

Therefore, $u_2\left(\mathfrak{z}\right)=(1-\varrho )\mathfrak{z}+\varrho -\varrho {\displaystyle \frac{1}{e^{\mathfrak{z}}}}$. This is the second approximation in the Picard iteration method.

By using the boundary conditions, we can calculate the $u\left(\mathfrak{z}\right)$, which is the unique solution to the I.E on $[0,1]$.

Therefore, by applying the FPT, we have established the existence and uniqueness of the solution to the I.E. This completes the proof that the given Volterra I.E has a unique solution on the interval $[0,1]$.

Now we are giving an example in decision theory.

Example 12.

Application in Decision Theory:

In decision-making problems involving uncertainty, neutrosophic metric spaces (NMSs) provide a robust framework to evaluate alternatives by considering degrees of membership (T), non-membership (F), and indeterminacy (I). This framework is particularly useful in situations where the available information is incomplete, inconsistent, or vague, such as selecting suppliers for a sustainable supply chain.

For example, a company needs to choose the best supplier from four candidates ($S_1,S_2,S_3,S_4$) based on three criteria:

1. 

cost efficiency (f),

2. 

delivery time (g), and

3. 

environmental sustainability (h).

Each supplier’s performance on these criteria is evaluated in the neutrosophic metric space, where

• T represents the degree of membership (e.g., how well a supplier meets the criterion),
• F represents the degree of non-membership (e.g., how poorly a supplier performs), and
• I represents the degree of indeterminacy (e.g., uncertainty or ambiguity in the evaluation).

The overall performance of a supplier is mapped using k, which combines the three criteria into a single score.

Now, the mappings f (cost efficiency) and g (delivery time) are subcompatible if their evaluations align sufficiently, meaning there is no strong contradiction between their outputs. For example, if a supplier scores high on cost efficiency (f) but moderately on delivery time (g), their neutrosophic evaluations should overlap in a way that allows meaningful aggregation into k (overall performance). Subcompatibility ensures that combining f, g, and h into k results in a coherent overall evaluation. Again, if a sequence of evaluations for a supplier (e.g., periodic assessments over time) converges to a certain neutrosophic value under mappings f and g, then k should reflect this consistency; i.e., gradual changes in criterion evaluations should lead to gradual changes in the overall performance k, avoiding abrupt or discontinuous shifts.

Numerical Example

Consider supplier $S_1$ evaluated as follows:

• $f\left(S_1\right)=(0.8,0.1,0.1)$ for cost efficiency,
• $g\left(S_1\right)=(0.7,0.2,0.1)$ for delivery time, and
• $h\left(S_1\right)=(0.6,0.2,0.2)$ for environmental sustainability.

The overall performance $k\left(S_1\right)$ is calculated as

$$k\left(S_1\right)=max\{f\left(S_1\right),g\left(S_1\right),h\left(S_1\right)\}.$$

Using the component-wise maximum,

$$k\left(S_1\right)=(max(0.8,0.7,0.6),max(0.1,0.2,0.2),max(0.1,0.1,0.2)),$$

$$k\left(S_1\right)=(0.8,0.2,0.2).$$

Here,

• $T=0.8$: the supplier $S_1$ has a high overall membership degree;
• $F=0.2$: there is a moderate degree of non-membership;
• $I=0.2$: some indeterminacy exists in the evaluation.

Now, subcompatibility and sequential continuity are used:

1. 

Subcompatibility: $f\left(S_1\right)$ and $g\left(S_1\right)$ are subcompatible if their neutrosophic evaluations do not strongly contradict each other. For example, $f\left(S_1\right)=(0.8,0.1,0.1)$ and $g\left(S_1\right)=(0.7,0.2,0.1)$ align well because their T, F, and I values are close and compatible.

2. 

Sequential Continuity: If the evaluations of $f\left(S_1\right)$, $g\left(S_1\right)$, and $h\left(S_1\right)$ change gradually over time (e.g., $f\left(S_1\right)\to (0.85,0.05,0.1)$), then $k\left(S_1\right)$ should also change gradually, ensuring that the overall performance mapping remains consistent.

Thus, in NMSs, the decision-making process accounts for uncertainty and inconsistency in supplier evaluations. Subcompatibility ensures the coherent aggregation of criteria into overall performance, while sequential continuity maintains consistency in evaluations over time, making NMS an effective tool for solving complex, uncertain problems such as supplier selection.

6. Conclusions

Within this study, some innovative concepts, such as specific subcompatibility and subsequential continuity, are presented. They offer a less restrictive framework compared to OWC and reciprocal continuity within the context of NMS. Leveraging these concepts, a common fixed point theorem applicable to four mappings is formulated. Our findings are further extended within the NMS framework. To showcase the practical relevance of this study, illustrative examples were provided to validate the new results and demonstrate their applicability in integral equations.