A short proof of the metrizability of $\mathcal{F}$-metric spaces

TL;DR

This paper presents a concise proof demonstrating that $\

Contribution

It offers a simplified proof of the metrizability of $\\mathcal{F}$-metric spaces, expanding understanding of their structure.

Findings

$\\mathcal{F}$-metric spaces are metrizable.

The proof simplifies previous approaches.

Enhances theoretical understanding of generalized metric spaces.

Abstract

The main purpose of this manuscript is to provide a short proof of the metrizability of $\mathcal{F}$-metric spaces introduced by Jleli and Samet in \cite[\, Jleli, M. and Samet, B., On a new generalization of metric spaces, J. Fixed Point Theory Appl. (2018) 20:128]{JS1}.

Equations

$$\underset{n\to +\infty }{\lim }{t}_n=0⟺\underset{n\to +\infty }{\lim }f(t_n)=-\infty .$$

$$D(x,y)>0⟹f(D(x,y))\le f\left(\sum _{i=1}^{N-1}D(u_i,u_{i+1})\right)+\alpha .$$

$$d(x,y)=\text{mbox}inf\left\{\sum _{i=1}^{N-1}D(u_i,u_{i+1}):N\in \mathbb{N},N\ge 2,\{u_i{\}}_{i=1}^N\subseteq X\text{mbox}with(u_1,u_N)=(x,y)\right\}.$$

$$f(D(x,z))\le f(D(x,y)+D(y,z))+\alpha .$$

$$f(D(x,z))<f(\epsilon )$$

$$\Rightarrow D(x,z)<\epsilon .$$

Full text

2010 Mathematics Subject Classification. $54$E$35$, $54$H$99$.

1. Introduction

Recently, Jleli and Samet [3] proposed a new generalization of usual metric space concept. By means of a certain class of functions, the authors defined the notion of an $\mathcal{F}$-metric space. Firstly, we will recall the definition of such kind of spaces. Let $\mathcal{F}$ denote the class of functions $f:(0,\infty)\rightarrow\mathbb{R}$ which satisfy the following conditions :

($\mathcal{F}_{1}$) $f$ is non-decreasing, i.e., $0<s<t\Rightarrow f(s)\leq f(t)$.

($\mathcal{F}_{2}$) For every sequence $\{t_{n}\}_{n\in\mathbb{N}}\subseteq(0,+\infty)$, we have

[TABLE]

Now due to Jleli and Samet, the definition of an $\mathcal{F}$-metric space is as follows:

Definition 1.1**.**

[3] Let $X$ be a non-empty set and $D:X\times X\rightarrow[0,\infty)$ be a given mapping. Suppose there exists $(f,\alpha)\in\mathcal{F}\times[0,\infty)$ such that

1. (D1)

$D(x,y)=0\Longleftrightarrow x=y~{}\forall~{}(x,y)\in X\times X$. 2. (D2)

$D(x,y)=D(y,x),~{}$ $\forall~{}(x,y)\in X\times X$. 3. (D3)

For every $(x,y)\in X\times X$, for each $N\in\mathbb{N},N\geq 2$ and for every $(u_{i})_{i=1}^{N}\subseteq X$ with $(u_{1},u_{N})=(x,y)$, we have

[TABLE]

Then $D$ is said to be an $\mathcal{F}$-metric on $X$ and the pair $(X,D)$ is said to be an $\mathcal{F}$-metric space.

In our earlier manuscript [4], we have already proved that this new generalization of metric space is indeed metrizable by a metric $d:X\times X\rightarrow\mathbb{R}$ defined as follows

[TABLE]

In that proof, we showed that $\tau=\tau_{\mathcal{F}}$ where $\tau$ denotes the topology generated by $d$ in 1.1 and $\tau_{\mathcal{F}}$ denotes the topology generated by the $\mathcal{F}$-metric $D.$ We also showed that the notions of Cauchy sequence, completeness, Banach contraction principle are equivalent to that of usual metric spaces. In this manuscript, we will give a very short proof of the metrizability of $\mathcal{F}$-metric spaces. We use Chittenden’s metrization theorem [1] in our proof. Further, for more details one can see [2].

Before proceeding to our main result, we first recall the metrization result due to Chittenden [1]. Let $X$ be a topological space and $F:X\times X\rightarrow[0,\infty)$ be a distance function on $X$. If the distance function $F$ satisfies the following conditions:

1. (i)

$F(x,y)=0\Longleftrightarrow x=y~{}\forall~{}(x,y)\in X\times X$. 2. (ii)

$F(x,y)=F(y,x),~{}$ $\forall~{}(x,y)\in X\times X$. 3. (iii)

(Uniformly regular) For every $\varepsilon>0$ and $x,y,z\in X$ there exists $\phi(\varepsilon)>0$ such that if $F(x,y)<\phi(\varepsilon)$ and $F(y,z)<\phi(\varepsilon)$ then $F(x,z)<\varepsilon$

then the topological space $X$ is metrizable.

2. Main Result

In this section, we will provide a short proof of the metrizability of $\mathcal{F}$-metric spaces.

Theorem 2.1**.**

Let $X$ be an $\mathcal{F}$-metric space with $(f,\alpha)\in\mathcal{F}\times[0,\infty)$. Then $X$ is metrizable.

Proof.

Let $X$ be an $\mathcal{F}$-metric space with $(f,\alpha)\in\mathcal{F}\times[0,\infty)$. By the definition of an $\mathcal{F}$-metric space, the distance function $D:X\times X\rightarrow[0,\infty)$ satisfies the first two conditions of Chittenden’s metrization result i.e,

1. (i)

$D(x,y)=0\Longleftrightarrow x=y~{}\forall~{}(x,y)\in X\times X$; 2. (ii)

$D(x,y)=D(y,x),~{}$ $\forall~{}(x,y)\in X\times X$.

Now we will prove the third condition i.e., the ‘uniformly regular’ condition. Let $\varepsilon>0$ and $x,y,z\in X.$ If $x=z$, then $D(x,z)=0$. So in this case $\phi(\varepsilon)=c$ where $c$ is any positive real number will serve the purpose. Now let $x\neq z.$ Then $D(x,z)>0.$ So by the definition of an $\mathcal{F}$-metric space we have,

[TABLE]

Now by the $\mathcal{F}_{2}$ condition, for $(f(\varepsilon)-\alpha)\in\mathbb{R}$ there exists $\delta>0$ such that $0<t<\delta\Rightarrow f(t)<f(\varepsilon)-\alpha.$ Now let us choose $\phi(\varepsilon)=\frac{\delta}{2}.$ If $D(x,y)<\frac{\delta}{2}$ and $D(y,z)<\frac{\delta}{2}$ then $D(x,y)+D(y,z)<\delta.$ So by the equation 2.1, we have

[TABLE]

[TABLE]

This shows that the distance function $D$ of an $\mathcal{F}$-metric space satisfies the uniformly regular condition. Consequently, by Chittenden’s metrization result we can conclude that the $\mathcal{F}$-metric space $X$ is metrizable. ∎

Acknowledgement**.**

The Research is funded by the Council of Scientific and Industrial Research (CSIR), Government of India under the Grant Number: $25(0285)/18/EMR-II$.