# Fixed Point Theorem for Neutrosophic Triplet Partial Metric Space

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## Abstract

**:**

## 1. Introduction

## 2. Preliminaries

**Definition 1**

**(**[19]

**).**Let A be nonempty set. If the function ${p}_{m}$:AxA → ${\mathbb{R}}^{+}$ satisfies the conditions given below; p is called a PM. ⩝a, b, c ∈ A;

- (i)
- ${p}_{m}$(a, a) =${p}_{m}$(b, b) =${p}_{m}$(a, b) =${p}_{m}$(b, a)$\u27fa$a = b;
- (ii)
- ${p}_{m}$(a, a)$\le $${p}_{m}$(a, b);
- (iii)
- ${p}_{m}$(a, b) =${p}_{m}$(b, a);
- (iv)
- ${p}_{m}$(a, c)$\le $${p}_{m}$(a, b) +${p}_{m}$(b, c) − ${p}_{m}$(b, b);

**Definition 2**

**(**[12]

**).**Let N be a nonempty and # be a binary operation. Then, N is called a NT if the given below conditions are satisfied.

- (i)
- There is neutral element (neut(x)) for x ∈ N such thatx*neut(x) = neut(x)* x = x.
- (ii)
- There is anti element (anti(x)) for x ∈ N such thatx*anti(x) = anti(x)* x = neut(x).

**Definition 3**

**(**[15]

**).**Let (M, #) be a NTS and a#b ∊ N, ⩝a, b ∊ M. NTM is a map${d}_{T}$:MxM →${\mathbb{R}}^{+}$∪ {0} such that ⩝a, b, c ∈ M,

- (a)
- ${d}_{T}$(a, b) ≥ 0
- (b)
- If a = b, then${d}_{T}$(a, b) = 0
- (c)
- ${d}_{T}$(a, b) =${d}_{T}$(a, b)
- (d)
- If there exists any element c ∊ M such that${d}_{T}$(a, c) ≤${d}_{T}$(a, c*neut(b)), then${d}_{T}$(a, c*neut(b)) ≤${d}_{T}$(a, b) +${d}_{T}$(b, c).

## 3. Neutrosophic Triplet Partial Metric Space

**Definition**

**4.**

- (i)
- 0$\le $${p}_{N}$(a, a)$\le {p}_{N}$(a, b)
- (ii)
- If${p}_{N}$(a, a) =${p}_{N}$(a, b) =${p}_{N}$(b, b) = 0, then there exits any a, b such that a = b.
- (iii)
- ${p}_{N}$(a, b) =${p}_{N}$(a, b)
- (iv)
- If there exists any element b ∊ A such that${p}_{N}$(a, c) ≤${p}_{N}$(a, c#neut(b)), then${p}_{N}$(a, c#neut(b)) ≤${p}_{N}$(a, b) +${p}_{N}$(b, c) −${p}_{N}$(b, b)

**Example**

**1.**

- (i), (ii) and (iii) are apparent.
- (iv) Let ∅ be empty element of P(X). Then,${p}_{N}$(X, Y) =${p}_{N}$(X, Y ∪ ∅) since for${p}_{N}$(X, Y ∪ ∅) =${p}_{N}$(X, Y) = max{m(X), m(Y)}. Also, it is clear that
- max{m(X), m(Y)} ≤ max{m(X), m(Z)}+ max{m(Z), m(Y)} – max {m(∅), m(∅)}.

**Corollary**

**1.**

**Corollary**

**2.**

**Theorem**

**1.**

**Proof.**

- (i)
- ${p}_{N}$(X, X) = $\frac{d(X,\text{}X)+m(X)+m(X)}{2}$ = m(X) $\le $ $\frac{d(X,\text{}Y)+m(X)+m(Y)}{2}$ = ${p}_{N}$(X, Y), since for d(X,X) = 0. Thus; 0 $\le $ ${p}_{N}$(X, X) $\le {p}_{N}$(X, Y) for X, Y ∈ P(A).
- (ii)
- If ${p}_{N}$(X, X) = ${p}_{N}$(X, Y) = ${p}_{N}$(Y, Y) = 0, then
- (iii)
- $\frac{d(X,\text{}X)+m(X)+m(X)}{2}$ = $\frac{d(X,\text{}Y)+m(X)+m(Y)}{2}$ = $\frac{d(Y,\text{}Y)+m(Y)+m(Y)}{2}$ = 0 and $d(X,\text{}Y)+m(X)+m(Y$) = 0. Where, m(X) = 0, m(Y) = 0 and $d(X,\text{}Y)$ = 0. Thus, X = Y = $\varnothing $ (empty set).
- (iv)
- ${p}_{N}$(X, Y) = $\frac{d(X,\text{}Y)+m(X)+m(Y)}{2}$ = $\frac{d(Y,\text{}X)+m(Y)+m(X)}{2}$ = ${p}_{N}$(Y, X), since for $d$(X, Y)= $d$(Y, X).
- (v)
- We suppose that there exists any Z ∈ P(A) such that $m(Y\#\mathrm{neut}(Z))=$ m(Y) and ${p}_{N}(X,\text{}Y)$ ≤ ${p}_{N}(X,\text{}Y\#\mathrm{neut}(Z))$. Thus,$$\frac{d(X,\text{}Y)+m(X)+m(Y)}{2}\le \frac{d(X,\text{}Y\#\mathrm{neut}(Z))+m(X)+m(Y\#\mathrm{neut}(Z))}{2}$$From (1), $d(X,\text{}Y)$ ≤ $d(X,\text{}Y\#\mathrm{neut}(Z))$. Since (P(A), #), d) is a NTMS,$$d(X,\text{}Y\#\mathrm{neut}(Z))\le d(X,\text{}Z)+d(X,\text{}Z)$$From (1), (2)$$\frac{d(X,\text{}Y)+m(X)+m(Y)}{2}\le \frac{d(X,\text{}\mathrm{Y}\#\mathrm{neut}(\mathrm{Z}))+m(X)+m(\mathrm{Y}\#\mathrm{neut}(\mathrm{Z}))\text{}}{2}\le \frac{d(X,\text{}Z)+\mathrm{d}(Z,\text{}Y)+m(X)+m(Y)+m(Z)}{2}=\phantom{\rule{0ex}{0ex}}\frac{d(X,\text{}Z)+m(X)+m(Z)}{2}+\frac{d(Z,\text{}Y)+m(Z)+m(Y)}{2}-m(Z).\text{}\mathrm{Where},\text{}{p}_{N}(Z,\text{}Z)=m(Z).$$

**Theorem**

**2.**

**Proof.**

- (i)
- Since for ${d}_{T}$(a, a) = 0, 0 $\le $ ${p}_{N}$(a, a) = ${d}_{T}$(a, a) + k = k $\le $ ${p}_{N}$(a, b) = ${d}_{T}$(a, b) + k. Thus;
- (ii)
- 0 $\le $ ${p}_{N}$(a, a) $\le {p}_{N}$(a, b).
- (iii)
- There do not exists a, b ∈ A such that ${p}_{N}$(a, a) = ${p}_{N}$(a, b) = ${p}_{N}$(b, b) = 0 since for k ∈ ${\mathbb{R}}^{+}$ and ${d}_{T}$(a, a) = 0.
- (iv)
- ${p}_{N}$(a, b) = ${d}_{T}$(a, b) + k = ${d}_{T}$(b, a) + k, since for ${d}_{T}$(a, b) = ${d}_{T}$(b, a).
- (v)
- Suppose that there exists any element c ∊ A such that ${p}_{N}$(a, b) ≤ ${p}_{N}$(a, b#neut(c)). Then ${d}_{T}$(a, b) + k ≤ ${d}_{T}$(a, b#neut(c)) + k. Thus,$${d}_{T}(a,\text{}b)\le {d}_{T}(a,\text{}b\#\mathrm{neut}(c))$$

**Corollary**

**3.**

**Definition**

**5.**

**Definition**

**6.**

**Theorem**

**3.**

**Proof.**

**Definition**

**7.**

**Definition**

**8.**

- (i)
- There exists any element c ∊ A such that${p}_{N}$(a, b) ≤${p}_{N}$(a, b*neut(c)); ⩝a, b ∊ A.
- (ii)
- There exists k in [0, 1) such that${p}_{N}$(m(a), m(b))$\le $k.${p}_{N}$(a, b); ⩝a, b ∊ A.

**Example**

**2.**

- ${p}_{N}$(m(∅), m(∅)) =${p}_{N}$({x}, {x}) = 1$\le $0, 2.${p}_{N}$(∅, ∅) = 1, 5
- ${p}_{N}$(m(∅), m({x})) =${p}_{N}$({x}, {x, y}) = 1$\le $0, 2.${p}_{N}$(∅, {x}) = 1, 5
- ${p}_{N}$(m(∅), m({x, y})) =${p}_{N}$({x}, {x, y}) = 1$\le $0, 2.${p}_{N}$(∅, {x, y}) = 1, 5
- ${p}_{N}$(m({x}), m({x})) =${p}_{N}$({x, y}, {x, y}) = 0$\le $0, 2.${p}_{N}$({x}, {x}) = 0, 5
- ${p}_{N}$(m({x}), m({x, y})) =${p}_{N}$({x, y}, {x, y}) = 0$\le $0, 2.${p}_{N}$({x}, {x,y}) = 0, 5
- ${p}_{N}$(m({x, y}), m({x, y})) =${p}_{N}$({x, y}, {x, y}) = 0$\le $0, 2.${p}_{N}$({x, y}, {x, y}) = 0, 5
- Thus, m is a contraction for ((A,$\cap $),${p}_{N}$)

**Theorem**

**4.**

**Proof.**

- ${p}_{N}$(${x}_{2}$, ${x}_{1}$) = ${p}_{N}$(m(${x}_{1}$), m(${x}_{0}$)) $\le $ c. ${p}_{N}$(${x}_{1}$, ${x}_{0}$) and
- ${p}_{N}$(${x}_{3}$, ${x}_{2}$) = ${p}_{N}$(m(${x}_{2}$), m(${x}_{1}$)) $\le $ c. ${p}_{N}$(${x}_{2}$, ${x}_{1}$) $\le $ ${c}^{2}$. ${p}_{N}$(${x}_{1}$, ${x}_{0}$). From mathematical induction, n $\ge $ m;
- ${p}_{N}$(${x}_{m+1}$, ${x}_{m}$) = ${p}_{N}$(m(${x}_{m}$), m(${x}_{m-1}$))$\le $ c.$\text{}{p}_{N}$(${x}_{m}$, ${x}_{m-1}$) $\le $ ${c}^{m}$. ${p}_{N}$(${x}_{1}$, ${x}_{0}$). Thus; from (8) and definition of NTPMS,$$\begin{array}{cc}\hfill {p}_{N}({x}_{n},{x}_{m})\text{}\le {p}_{N}({x}_{n}{x}_{m}*\mathrm{neut}({x}_{n-1}))& \le {p}_{N}({x}_{n},{x}_{n-1})+{p}_{N}({x}_{n-1},{x}_{m})-{p}_{N}({x}_{n-1},{x}_{n-1})\hfill \\ & \le {c}^{n-1}.{p}_{N}({x}_{1},{x}_{0})+{p}_{N}({x}_{n-1},{x}_{m})-{p}_{N}({x}_{n-1},{x}_{n-1})\hfill \\ & \le \text{}{c}^{n-1}.\text{}{p}_{N}({x}_{1},\text{}{x}_{0})\text{}+\text{}{p}_{N}({x}_{n-1},\text{}{x}_{n-2})\text{}+\text{}\dots \text{}+\text{}{p}_{N}({x}_{m},\text{}{x}_{m-1})\hfill \\ & \le \text{}({c}^{n-1}+\text{}{c}^{n-2}+\dots +\text{}{c}^{m-1}+{c}^{m}).\text{}{p}_{N}({x}_{1},\text{}{x}_{0})\text{}-\text{}{{\displaystyle \sum}}_{i=m\text{}}^{n-1}{p}_{N}({x}_{i},\text{}{x}_{i})\hfill \\ & \le \text{}{{\displaystyle \sum}}_{i=m\text{}}^{n-1}{c}^{i}.{p}_{N}({x}_{1},\text{}{x}_{0})\text{}-\text{}{{\displaystyle \sum}}_{i=m\text{}}^{n-1}{p}_{N}({x}_{i},\text{}{x}_{i})\hfill \\ & \le \text{}{{\displaystyle \sum}}_{i=m\text{}}^{n-1}{c}^{i}.{p}_{N}({x}_{1},\text{}{x}_{0})\text{}+\text{}{p}_{N}({x}_{0},\text{}{x}_{0})\hfill \\ & =\text{}{{\displaystyle \sum}}_{i=m\text{}}^{n-1}{c}^{i}.{p}_{N}({x}_{1},\text{}{x}_{0})\text{}+\text{}{p}_{N}({x}_{0},\text{}{x}_{0})(\mathrm{For}\text{}n,\text{}m\to \infty )\hfill \\ & =\text{}\frac{{c}^{m}}{1-c}\text{}{p}_{N}({x}_{1},\text{}{x}_{0})+\text{}{p}_{N}({x}_{0},\text{}{x}_{0})\text{}\to \text{}{p}_{N}({x}_{0},\text{}{x}_{0})\hfill \end{array}$$

## 4. Conclusions

## Author Contributions

## Funding

## Conflicts of Interest

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**MDPI and ACS Style**

Şahin, M.; Kargın, A.; Çoban, M.A.
Fixed Point Theorem for Neutrosophic Triplet Partial Metric Space. *Symmetry* **2018**, *10*, 240.
https://doi.org/10.3390/sym10070240

**AMA Style**

Şahin M, Kargın A, Çoban MA.
Fixed Point Theorem for Neutrosophic Triplet Partial Metric Space. *Symmetry*. 2018; 10(7):240.
https://doi.org/10.3390/sym10070240

**Chicago/Turabian Style**

Şahin, Memet, Abdullah Kargın, and Mehmet Ali Çoban.
2018. "Fixed Point Theorem for Neutrosophic Triplet Partial Metric Space" *Symmetry* 10, no. 7: 240.
https://doi.org/10.3390/sym10070240