# Informal Complete Metric Space and Fixed Point Theorems

## Abstract

**:**

## 1. Introduction

## 2. Informal Vector Spaces

**Example**

**1.**

**Example**

**2.**

**Definition**

**1.**

**Example**

**3.**

**Definition**

**2.**

- $1x=x$ for any $x\in X$;
- $x=y$ implies $x\oplus z=y\oplus z$ and $\alpha x=\alpha y$ for any $x,y,z\in X$ and $\alpha \in \mathbb{F}$;
- The commutative and associative laws for vector addition hold true in X; that is, $x\oplus y=y\oplus x$ and $(x\oplus y)\oplus z=x\oplus (y\oplus z)$ for any $x,y,z\in X$.

**Definition**

**3.**

- $x=y$;
- There exists $\omega \in \mathsf{\Omega}$ such that $x=y\oplus \omega $ or $x\oplus \omega =y$;
- There exists ${\omega}_{1},{\omega}_{2}\in \mathsf{\Omega}$ such that $x\oplus {\omega}_{1}=y\oplus {\omega}_{2}$.

**Remark**

**1.**

**Proposition**

**1.**

**Proof.**

## 3. Informal Metric Spaces

**Definition**

**4.**

- (i)
- $d(x,y)=0$ if and only if $x\stackrel{\mathsf{\Omega}}{=}y$ for all $x,y\in X$;
- (i′)
- $d(x,y)=0$ if and only if $x=y$ for all $x,y\in X$;
- (ii)
- $d(x,y)=d(y,x)$ for all $x,y\in X$;
- (iii)
- $d(x,y)\le d(x,z)+d(z,y)$ for all $x,y,z\in X$.

- The pair $(X,d)$ is called a pseudo-metric space if and only if d satisfies conditions (ii) and (iii).
- The pair $(X,d)$ is called a metric space if and only if d satisfies conditions (${i}^{\prime}$), (ii), and (iii).
- The pair $(X,d)$ is called a informal metric space if and only if d satisfies conditions (i), (ii), and (iii).

- (iv)
- We say that d satisfies the null super-inequality if and only if, for any ${\omega}_{1},{\omega}_{2}\in \mathsf{\Omega}$ and $x,y,z\in X$, we have$$d(x\oplus {\omega}_{1},y\oplus {\omega}_{2})\ge d(x,y),\phantom{\rule{1.em}{0ex}}d(x\oplus {\omega}_{1},y)\ge d(x,y)andd(x,y\oplus {\omega}_{2})\ge d(x,y).$$
- (iv′)
- We say that d satisfies the null sub-inequality if and only if, for any ${\omega}_{1},{\omega}_{2}\in \mathsf{\Omega}$ and $x,y\in X$, we have$$d(x\oplus {\omega}_{1},y\oplus {\omega}_{2})\le d(x,y),\phantom{\rule{1.em}{0ex}}d(x\oplus {\omega}_{1},y)\le d(x,y)andd(x,y\oplus {\omega}_{2})\le d(x,y).$$
- (iv″)
- We say that d satisfies the null equality if and only if, for any ${\omega}_{1},{\omega}_{2}\in \mathsf{\Omega}$ and $x,y\in X$, we have$$\begin{array}{ccc}\hfill d(x\oplus {\omega}_{1},y\oplus {\omega}_{2})=d(x,y),& d(x\oplus {\omega}_{1},y)=d(x,y)& and\phantom{\rule{8.53581pt}{0ex}}d(x,y\oplus {\omega}_{2})=d(x,y).\hfill \end{array}$$

**Example**

**4.**

## 4. Cauchy Sequences

**Definition**

**5.**

- Suppose that $(X,d)$ is a metric space. By condition (${i}^{\prime}$) in Definition 4, we see that $x=y$. This shows the uniqueness.
- Suppose that $(X,d)$ is an informal metric space. By condition (i) in Definition 4, we see that $x\stackrel{\mathsf{\Omega}}{=}y$. Recall that if $\mathsf{\Omega}$ is closed under the vector addition, then we can consider the equivalence classes. In this case, we also see that y is in the equivalence class $\left[x\right]$.

**Definition**

**6.**

**Proposition**

**2.**

**Proof.**

**Definition**

**7.**

- A sequence ${\left\{{x}_{n}\right\}}_{n=1}^{\infty}$ in X is called a Cauchy sequence if and only if, given any $\u03f5>0$, $N\in \mathbb{N}$ exists, such that $d({x}_{n},{x}_{m})<\u03f5$ for all $n>N$ and $m>N$.
- A subset M of X is said to be complete if and only if every Cauchy sequence in M is convergent to some element in M.

**Proposition**

**3.**

**Example**

**5.**

## 5. Near Fixed Point Theorems

**Definition**

**8.**

**Example**

**6.**

- $T\left([{a}^{L},{a}^{U}]\right)\oplus [-{k}_{1},{k}_{1}]=[{a}^{L},{a}^{U}]$;
- $T\left([{a}^{L},{a}^{U}]\right)=[{a}^{L},{a}^{U}]\oplus [-{k}_{1},{k}_{1}]$;
- $T\left([{a}^{L},{a}^{U}]\right)\oplus [-{k}_{1},{k}_{1}]=[{a}^{L},{a}^{U}]\oplus [-{k}_{2},{k}_{2}]$,

**Remark**

**2.**

- By definition, we see that $T\left(x\right)\stackrel{\mathsf{\Omega}}{=}x$ if and only if ${\omega}_{1},{\omega}_{2}\in \mathsf{\Omega}$ exist, such that $T\left(x\right)=x$, $T\left(x\right)\oplus {\omega}_{1}=x$, or $T\left(x\right)=x\oplus {\omega}_{1}$ or $T\left(x\right)\oplus {\omega}_{1}=x\oplus {\omega}_{2}$.
- If the informal vector space X owns a zero element θ, then the (conventional) fixed point is also a near fixed point.
- If the informal vector space X turns into a (conventional) vector space over $\mathbb{F}$, then the concepts of near fixed point and (conventional) fixed point are equivalent.

**Definition**

**9.**

**Example**

**7.**

**Theorem**

**1.**

- There is a unique equivalence class $\left[x\right]$ satisfying that if $\overline{x}\notin \left[x\right]$, then $\overline{x}$ cannot be a near fixed point, which shows the sense of uniqueness.
- Suppose that $\overline{x}\in \left[x\right]$. Then $\overline{x}$ is also a near fixed point of T satisfying $T\left(\overline{x}\right)\stackrel{\mathsf{\Omega}}{=}\overline{x}$ and $\left[\overline{x}\right]=\left[x\right]$.
- Suppose that $\overline{x}$ is a near fixed point of T. Then $\overline{x}\in \left[x\right]$; i.e., $\left[\overline{x}\right]=\left[x\right]$. In other words, if x and $\overline{x}$ are the near fixed points of T, then $x\stackrel{\mathsf{\Omega}}{=}\overline{x}$.

**Proof.**

**Example**

**8.**

**Definition**

**10.**

- $x\stackrel{\mathsf{\Omega}}{=}y$, i.e., $\left[x\right]=\left[y\right]$ implies $d\left(T\right(x),T(y\left)\right)=0$;
- $x\stackrel{\mathsf{\Omega}}{=}y$, i.e., $\left[x\right]\ne \left[y\right]$ implies $d\left(T\right(x),T(y\left)\right)<d(x,y)$.

**Theorem**

**2.**

- There is a unique equivalence class $\left[x\right]$ satisfying that if $\overline{x}\notin \left[x\right]$ then $\overline{x}$ cannot be a near fixed point, which shows the sense of uniqueness.
- Suppose that $\overline{x}\in \left[x\right]$. Then $\overline{x}$ is also a near fixed point of T, satisfying $T\left(\overline{x}\right)\stackrel{\mathsf{\Omega}}{=}\overline{x}$ and $\left[\overline{x}\right]=\left[x\right]$.
- Suppose that $\overline{x}$ is a near fixed point of T. Then $\overline{x}\in \left[x\right]$; i.e., $\left[\overline{x}\right]=\left[x\right]$. In other words, if x and $\overline{x}$ are the near fixed points of T, then $x\stackrel{\mathsf{\Omega}}{=}\overline{x}$.

**Proof.**

- Suppose that ${T}^{n}\left({x}_{0}\right)\stackrel{\mathsf{\Omega}}{=}x$. Since T is a weakly strict contraction of X, it follows that$$d({T}^{n+1}\left({x}_{0}\right),T\left(x\right))=0<\u03f5.$$
- Suppose that ${T}^{n}\left({x}_{0}\right)\stackrel{\mathsf{\Omega}}{\ne}x$. Since T is a weakly strict contraction of X, we have$$d({T}^{n+1}\left({x}_{0}\right),T\left(x\right))<d({T}^{n}\left({x}_{0}\right),x)<\u03f5\mathrm{for}n\ge N.$$

**Definition**

**11.**

- $x\stackrel{\mathsf{\Omega}}{=}y$, i.e., $\left[x\right]=\left[y\right]$ implies $d\left(T\right(x),T(y\left)\right)=0$;
- given any $\u03f5>0$, $\delta >0$ exists, such that $\u03f5\le d(x,y)<\u03f5+\delta $ implies $d\left(T\right(x),T(y\left)\right)<\u03f5$ for any $x\stackrel{\mathsf{\Omega}}{\ne}y$, i.e., $\left[x\right]\ne \left[y\right]$.

**Remark**

**3.**

**Lemma**

**1.**

**Proof.**

- Suppose that $\left[{x}_{n-1}\right]\ne \left[{x}_{n}\right]$. By Remark 3, we have$${c}_{n}=d({x}_{n},{x}_{n+1})=d({T}^{n}\left(x\right),{T}^{n+1}\left(x\right))<d({T}^{n-1}\left(x\right),{T}^{n}\left(x\right))=d({x}_{n-1},{x}_{n})={c}_{n-1}.$$
- Suppose that $\left[{x}_{n-1}\right]=\left[{x}_{n}\right]$. Then, by the first condition of Definition 11,$${c}_{n}=d({T}^{n}\left(x\right),{T}^{n+1}\left(x\right))=d(T\left({x}_{n-1}\right),T\left({x}_{n}\right))=0<{c}_{n-1}.$$

- Let m be the first index in the sequence ${\left\{{x}_{n}\right\}}_{n=1}^{\infty}$ such that $\left[{x}_{m-1}\right]=\left[{x}_{m}\right]$. Then, we can show that ${c}_{m-1}={c}_{m}={c}_{m+1}=\cdots =0$. Since ${x}_{m-1}\stackrel{\mathsf{\Omega}}{=}{x}_{m}$, we have ${c}_{m-1}=d({x}_{m-1},{x}_{m})=0$. The first condition of Definition 11 says that$$0=d(T\left({x}_{m-1}\right),T\left({x}_{m}\right))=d({T}^{m}\left(x\right),{T}^{m+1}\left(x\right))=d({x}_{m},{x}_{m+1})={c}_{m},$$
- Suppose that $\left[{x}_{m+1}\right]\ne \left[{x}_{m}\right]$ for all $m\ge 1$. Since the sequence ${\left\{{c}_{n}\right\}}_{n=1}^{\infty}$ is decreasing, we can assume that ${c}_{n}\downarrow \u03f5>0$, i.e., ${c}_{n}\ge \u03f5>0$ for all n, which says that $\delta >0$ exists, such that $\u03f5\le {c}_{m}<\u03f5+\delta $ for some m, i.e., $\u03f5\le d({x}_{m},{x}_{m+1})<\u03f5+\delta $. The second condition of Definition 11 says that$${c}_{m+1}=d({x}_{m+1},{x}_{m+2})=d({T}^{m+1}\left(x\right),{T}^{m+2}\left(x\right))=d(T\left({x}_{m}\right),T\left({x}_{m+1}\right))<\u03f5,$$

**Theorem**

**3.**

- There is a unique equivalence class $\left[x\right]$ satisfying that if $\overline{x}\notin \left[x\right]$, then $\overline{x}$ cannot be a near fixed point, which shows the sense of uniqueness.
- Suppose that $\overline{x}\in \left[x\right]$. Then $\overline{x}$ is also a near fixed point of T satisfying $T\left(\overline{x}\right)\stackrel{\mathsf{\Omega}}{=}\overline{x}$ and $\left[\overline{x}\right]=\left[x\right]$.
- Suppose that $\overline{x}$ is a near fixed point of T. Then $\overline{x}\in \left[x\right]$; i.e., $\left[\overline{x}\right]=\left[x\right]$. In other words, if x and $\overline{x}$ are the near fixed points of T, then $x\stackrel{\mathsf{\Omega}}{=}\overline{x}$.

**Proof.**

## Funding

## Conflicts of Interest

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**MDPI and ACS Style**

Wu, H.-C.
Informal Complete Metric Space and Fixed Point Theorems. *Axioms* **2019**, *8*, 126.
https://doi.org/10.3390/axioms8040126

**AMA Style**

Wu H-C.
Informal Complete Metric Space and Fixed Point Theorems. *Axioms*. 2019; 8(4):126.
https://doi.org/10.3390/axioms8040126

**Chicago/Turabian Style**

Wu, Hsien-Chung.
2019. "Informal Complete Metric Space and Fixed Point Theorems" *Axioms* 8, no. 4: 126.
https://doi.org/10.3390/axioms8040126