A Type of Fuzzy Metric and Its Applications

Abstract

In this paper, we aim to investigate a type of lattice-valued fuzzy metric within the framework of L-topology. Firstly, we present a comprehensive construction theorem for this type of metric, utilizing the concept of L-quasi metric. Secondly, we provide an equivalent characterization through the use of C-nbd clusters, which are formed from all $B_r$: one of four types of basic spheres defined herein. Thirdly, recognizing that these four types of basic spheres serve as essential tools for characterizing various metrics, we meticulously examine the relationships among them and outline a series of topological properties associated with these metrics, which include their opening and closing characteristics, symmetrical property, and more. Finally, in addressing the corresponding symmetry problem between two types of basic spheres, namely $B_r\left(a\right)$ and $Q_r\left(a\right)$, we introduce a novel fuzzy p-metric and demonstrate tht the L-real line $\mathbf{R}\left(L\right)$ satisfies this fuzzy p-metric.

1. Introduction

Metrization represents a far-reaching and fundamental problem in general topology [1]. By the 1950s, through the collective efforts of mathematicians such as J. Nagata, M.H. Stone, R.H. Bing, Y.M. Smirnov, and C.H. Dowker, this problem was successfully resolved (see [2,3,4,5,6]). The subject is now considered largely complete. Nevertheless, some researchers continue to explore new avenues and seek innovative approaches to extend the existing theoretical framework.

The integration of fuzzy set theory [7] into topology was pioneered by C.L. Chang in his 1968 work [8]. His ideas were rapidly adopted and widely recognized as $[0,1]$-topology, which was later generalized by J.A. Goguen [9] to the L-fuzzy setting, evolving into what is now known as L-topology. Since then, lattice-valued topology has emerged as a major research field, yielding numerous creative results concerning metrics (see [10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27], etc.).

However, extending the classical concept of a metric and its associated metrization theorems from general topology to the lattice-valued context has remained a significant challenge. To date, several interpretations of fuzzy metric spaces have been proposed within lattice-valued topology (see [10,15,17,18,24,28,29,30], etc.). Among these, two types of fuzzy metric spaces have attracted considerable attention for their generalizations of classical metrics.

The first type is induced by a fuzzy metric defined as a mapping $p:Y\times Y\to [0,+\infty )$, where Y denotes the set of all fuzzy points of an underlying set X (see [11,15,16,17,24,26], etc.). In this framework, each fuzzy metric generates a corresponding fuzzy topology.

The second type, attributed to researchers such as I. Kramosil, J. Michalek, V. George, and P. Veeramani (see [18,29,31,32], etc.), features a fuzzy metric where the distance between objects is fuzzy, while the objects themselves remain crisp (see [18,22,27,29,31,33,34,35,36,37], etc.). The topology induced by this type is always fuzzifying (see Corollary 3.9 in [38]). In recent years, this form of fuzzy metric has been extensively developed by experts including V. Gregori, S. Romaguera, J. Gutiérrez, and Morillas (see [29,32,34,36,39], etc.). Additionally, related advances in fixed point theory and t-norm spaces can be found in [31,35,40,41,42] and [43,44,45,46], respectively.

Regarding the first type, besides consistently satisfying the following conditions:

(A1)

If $a\ge b$, then $p(a,b)=0$;

(A2)

$p(a,c)\le p(a,b)+p(b,c)$;

(A3)

$\forall a,b\in M$, $\exists x\nleqq a^{\prime }$ such that $p(b,x)<r$⇔$\exists y\nleqq b^{\prime }$ such that $p(a,y)<r$.

Additionally, each of these metrics satisfies one of the following four conditions:

(B1)

$p(a,b)={\displaystyle \underset{c\ll b}{\bigvee }}p(a,c)$;

(B2)

$p(a,b)={\displaystyle \underset{c\ll a}{\bigwedge }}p(c,b)$;

(B3)

$p(a,b)={\displaystyle \underset{b\ll c}{\bigwedge }}p(a,c)$;

(B4)

$p(a,b)={\displaystyle \underset{a\ll c}{\bigvee }}p(c,b)$,

where “$\ll$” denotes the way-below relation in domain theory (see Preliminaries). Based on these conditions, P. Chen categorized the first type of metrics into four distinct kinds [10], summarized as follows:

The first is the Erceg pseudo-metric, introduced by M.A. Erceg in 1979 [17], which satisfies (B1) in addition to (A1)–(A3). In 1984, J.H. Liang established a Urysohn-type metrization theorem: An L-topological space is Erceg metrizable if it is $T_1$, regular, and $C_{II}$ [20]. Liang later conjectured that these conditions were sufficient but not necessary [47,48]. In 1985, M.K. Luo constructed an example showing that an Erceg metric topology need not be $Q-C_1$ and may lack a $\sigma$-locally finite base [21], confirming Liang’s conjecture. The complexity of Erceg’s axioms led to cumbersome proofs; later, building on Peng’s work [49], P. Chen and F.G. Shi provided a significant simplification [14].

The second is Shi’s pseudo-metric (also called p.q.-metric), proposed by L.C. Yang in 1988 [26], which satisfies (B2). Yang showed that every topological molecular lattice with the $C_{II}$ property is p.q.-metrizable. This metric was further developed by F.G. Shi [24,50,51] and P. Chen [13,14].

The third is Deng’s pseudo-metric, introduced by Z.K. Deng in 1982 [15], satisfying (B3). Deng proved that a $[0,1]$-topological space that is $T_1$, regular, and $C_{II}$ is Deng metrizable [16]. Although Y.Y. Lan and F. Long attempted a related result [52], their proof contained an error, as we note. P. Chen later extended Deng’s metric to general $L^X$ [11].

The fourth is Chen’s pseudo-metric, proposed by P. Chen in 2017 [13], which satisfies (B4). Chen established several key results:

(a)

Every Deng pseudo-metric on $L^X$ is a Chen pseudo-metric;

(b)

On $I^X$, a Shi pseudo-metric is a Deng pseudo-metric iff it is a Chen pseudo-metric;

(c)

An Erceg pseudo-metric on $I^X$ is a Chen pseudo-metric if it satisfies $p(x_{{\lambda }_1},y_{{\lambda }_2})=p(y_{1-{\lambda }_2},x_{1-{\lambda }_1})$;

(d)

On $I^X$, letting C, E, D, and Y denote the classes of Chen, Erceg, Deng, and Shi pseudo-metrics, respectively, then $D=C\cap Y\cap E$.

Beyond these, many mathematicians have generalized classical metrics by relaxing certain axioms, leading to pseudo-metric, semi-metric, quasi-metric spaces [53], and more recent variants such as partial metrics [54,55], generalized metrics [56], S-metrics [57,58], $S_b$-metrics [59], b-metrics [60,61], strong b-metrics [62], and solutions to complex fuzzy matrix equations [63]. A recent generalization is the ⊕-sb-metric space [64], extending strong b-metric spaces.

Despite extensive research, the metrization problem for general L-topologies remains open. However, for $[0,1]$-topologies, Chen recently proved a significant result [11]: A $[0,1]$-topological space $(X,\delta )$ that is $T_1$, regular, and has a $\sigma$-locally finite base is metrizable by each of the Deng, Erceg, Chen, and Shi metrics.

In this paper, we study a new type of lattice-valued metric in L-topology. We present a construction theorem for this metric using L-quasi-metrics as introduced by Chen [10], and characterize it equivalently via C-neighborhood clusters formed by basic spheres $B_r$ (of four types defined herein). These spheres are crucial for describing metric topologies; we analyze their interrelations and topological properties including openness, closure, and symmetry. To address symmetry issues between specific spheres $B_r\left(a\right)$ and $Q_r\left(a\right)$, we introduce a new fuzzy p-metric and demonstrate that the L-real line $\mathbf{R}\left(L\right)$ [19] satisfies this metric.

2. Preliminaries

In this section, we present the basic definitions used in the paper. Throughout, X denotes a nonempty set and $I=[0,1]$. A fuzzy set on X is a mapping $A:X\to I$, and the family of all such sets is denoted by $I^X$. A fuzzy point $x_{\lambda }$, for $x\in X$ and $\lambda \in (0,1)$, is defined as follows:

$$x_{\lambda }\left(y\right)=\left\{\begin{array}{cc}\lambda ,\hfill & \mathrm{if} y=x;\hfill \\ 0,\hfill & \mathrm{if} y\ne x.\hfill \end{array}\right.$$

Let L be a completely distributive lattice with an order-reversing involution “′” [65,66]. The set of all L-fuzzy sets on X, denoted $L^X$ [9], inherits the lattice structure and involution from L by pointwise definitions of ∨, ∧, ′. The least and greatest elements in $L^X$ are denoted by ${\displaystyle \underline{0}}$ and ${\displaystyle \underline{1}}$, respectively.

An element $e\in L$-$\left\{{\displaystyle \underline{0}}\right\}$ is called co-prime if $e\le p\vee q$ implies $e\le p$ or $e\le q$ for any $p,q\in L$. The set of all nonzero co-prime elements in L is denoted by $M\left(L\right)$. An L-fuzzy point is defined as $x_{\lambda }$ for $x\in X$ and $\lambda \in M\left(L\right)$ [66]. The set of all such L-fuzzy points is denoted $M\left(L^X\right)$, abbreviated as M. Thus, M consists of all nonzero co-prime elements in $L^X$.

For $a,b\in L^X$, we say a is way below b, written $a\ll b$, if for every directed subset $D\subseteq L^X$ with $b\le supD$, there exists some $d\in D$ such that $a\le d$. A subset $B\subset L^X$ is called a cover (respectively, a proper cover) of a if $a\le supB$ (respectively, $a=supB$). If for every $x\in B$ there exists $y\in C$ such that $x\le y$, then B is said to refine C. A set $B,C\subset L^X$ is called a minimal set of aa if it is a proper cover of aa and refines every cover of a. Denote by $\mathcal{T}\left(a\right)$ the collection of all minimal sets of a. Note that the union of any subfamily of $\mathcal{T}\left(a\right)$ is again a minimal set of a; hence, each $a\in L^X$ has a greatest minimal set, denoted ${\beta }^{*}\left(a\right)$ [66]. Let ${\beta }^{*}\left(a\right)=\beta \left(a\right)\cap M$. Then, $x_{\lambda }\in {\beta }^{*}\left(a\right)$ if and only if $x_{\lambda }$. We also adopt the conventions $\vee \varnothing ={\displaystyle \underline{0}}$ and $\wedge \varnothing ={\displaystyle \underline{1}}$. For other unexplained terminology and further details, we refer to [1,9,66,67].

Definition 1

([10]). Mapping $p:M\times M\to [0,+\infty )$ is called an L-quasi-metric on $L^X$ if it satisfies the following:

(A1) 

If $a\ge b$, then $p(a,b)=0$;

(A2) 

$\forall a,b,c\in M$, $p(a,c)\le p(a,b)+p(b,c)$.

L-quasi-metric p is called an L-pseudo-metric on $L^X$ if it additionally satisfies the following:

(A3) 

$\forall a,b\in M$, $\exists y\nleqq b^{\prime }$ s.t. $p(a,y)<r\iff$$\exists x\nleqq a^{\prime }$ s.t. $p(b,x)<r$.

An L-pseudo-metric p is called an L-metric if it also satisfies the following:

(A4) 

If $p(a,b)=0$, then $a\ge b$.

Definition 2.

Let $p:M\times M⟶[0,\infty )$ be a mapping. For any $a,b\in M$ and $r\in [0,+\infty )$, define the following spheres:

(1) 

$D_r\left(a\right)=\bigvee \{c\in M\mid p(a,c)<r\}$;

(2) 

$B_r\left(a\right)=\bigvee \{c\in M\mid p(a,c)\le r\}$;

(3) 

$Q_r\left(b\right)=\bigvee \{c\in M\mid p(c,b)>r\}$;

(4) 

$P_r\left(b\right)=\bigvee \{c\in M\mid p(c,b)\ge r\}$.

The mapping $B_r\left(a\right):M⟶L^X$ is called the closed neighborhood mapping of a, and the family $\{B_r\left(a\right)\mid a\in M,r\in [0,+\infty )\}$ is called the closed neighborhood mapping cluster (abbreviated as C-nbd cluster) of p.

Definition 3

([13]). Let $B_r$ and $B_s$ be two closed neighborhood mappings. For any $a\in M$, define the composition $B_r\circ B_s\left(a\right)=\bigvee \left\{B_r\left(b\right)\right|b\le B_s\left(a\right),b\in M\}$.

Theorem 1

([13]). If p is an L-quasi-metric, then (1) $B_r\left(b\right)={\displaystyle \underset{r<s}{\bigwedge }}{D}_s\left(b\right)$; (2) $B_r\left(b\right)={\displaystyle \underset{r<s}{\bigwedge }}{B}_s\left(b\right)$.

Theorem 2

([13]). p is an L-pseudo-metric. Then, (1)${\displaystyle \underset{u<r}{\bigwedge }}{P}_u\left(a\right)=P_r\left(a\right)$; (2)${\displaystyle \underset{z\nleqq b^{\prime }}{\bigvee }}{P}_{\lambda }{\left(z\right)}^{\prime }\le D_{\lambda }\left(b\right)$.

Theorem 3

([13]). If mapping $p:M\times M\to [0,+\infty )$ satisfies (A3), then ${\displaystyle \underset{b\ll a}{\bigvee }}{D}_r\left(b\right)=D_r\left(a\right)$.

Theorem 4

([14]). A Shi pseudo-metric is necessarily an Erceg pseudo-metric on $L^X$, but the converse does not hold.

Theorem 5

([19,66]). The L-$fuzzy$ real line (or L-real line) $\mathbf{R}\left(L\right)$ is the set of equivalence classes of mappings $\lambda :\mathbf{R}\to L$ satisfying the following:

$${\displaystyle \underset{t\in \mathbf{R}}{\bigvee }}\lambda \left(t\right)=1,{\displaystyle \underset{t\in \mathbf{R}}{\bigwedge }}\lambda \left(t\right)=0.$$

Two mappings $\lambda ,\mu$ are equivalent if and only if for all $t\in \mathbf{R}$,

$$\lambda (t-)=\mu (t-),\lambda (t+)=\mu (t+).$$

The standard L-fuzzy topology on $\mathbf{R}\left(L\right)$ is generated by the subbase $\{l_t,r_t|t\in \mathbf{R}\}$, where

$$l_t\left(\lambda \right)=\lambda {(t-)}^{\prime },r_t\left(\lambda \right)=\lambda (t+).$$

For any $A\subset \mathbf{R}$,

$${\displaystyle \underset{t\in A}{\bigvee }}{l}_t=l_{supA},{\displaystyle \underset{t\in A}{\bigvee }}{r}_t=r_{infA}.$$

Moreover, for all $t\in \mathbf{R}$ and $s>0$,

$$l_t\le {\displaystyle r_t^{\prime }}\le l_{t+s},r_t\le {\displaystyle l_t^{\prime }}\le r_{t-s}.$$

Theorem 6

([66,68]). On $\mathbf{R}\left(L\right)$, define mappings $\epsilon ,\sigma :M\left(L^{\mathbf{R}\left(L\right)}\right)\to \mathbf{R}$ by

$$\epsilon \left(e\right)=sup\left\{t\right|e\le {\displaystyle l_t^{\prime }}\},\sigma \left(e\right)=inf\left\{t\right|e\le {\displaystyle r_t^{\prime }}\},\forall e\in M\left(L^{\mathbf{R}\left(L\right)}\right)$$

Then,

(a) 

$\epsilon \left(e\right)=max\left\{t\right|e\le {\displaystyle l_t^{\prime }}\},\sigma \left(e\right)=min\left\{t\right|e\le {\displaystyle r_t^{\prime }}\}$;

(b) 

For $a,b\in M\left(L^{\mathbf{R}\left(L\right)}\right)$ with $a\le b$, we have $\epsilon \left(a\right)\ge \epsilon \left(b\right)$ and $\sigma \left(a\right)\le \sigma \left(b\right)$;

(c) 

Define $p(b,a)=max\left\{\epsilon \right(b)-\epsilon (a),\sigma (a)-\sigma (b),0\}$. Then, p is a p.q. metric and $P_r\left(a\right)={\displaystyle l_{\epsilon \left(a\right)+r}^{\prime }}\bigvee {\displaystyle r_{\sigma \left(a\right)-r}^{\prime }}.$

3. Construction of the First Metric and Its Equivalent Characterization

Definition 4.

A mapping $p:M\times M\to [0,+\infty )$ is called a first pseudo-metric on $L^X$ if it satisfies the following:

(A1) 

if $a\ge b$, then $p(a,b)=0$.

(A2) 

$p(a,c)\le p(a,b)+p(b,c)$, for all $a,b,c\in M$.

(B1) 

$p(a,b)={\displaystyle \underset{c\ll b}{\bigvee }}p(a,c)$, for all $a,b\in M$.

(A3) 

$\forall a,b\in M$, $\exists x\nleqq a^{\prime }$ s.t. $p(b,x)<r\iff$$\exists y\nleqq b^{\prime }$ s.t. $p(a,y)<r$.

If p additionally satisfies

(A4) 

if $p(a,b)=0$, then $a\ge b$,

then it is called a first metric.

Theorem 7.

If $p_0$ is an L-quasi-metric, L-pseudo-metric, or L metric on $L^X$ and define $p:M\times M\to [0,+\infty )$ by the equation

$$p(a,b)={\displaystyle \underset{y\ll b}{\bigvee }}{\displaystyle \underset{x\ll a}{\bigwedge }}{p}_0(x,y),$$

then p is an L-quasi-metric, the first pseudo-metric, or the first metric, respectively.

Proof. 

It suffices to prove the following:

• The function p defined by $p(a,b)={\displaystyle \underset{y\ll b}{\bigvee }}{\displaystyle \underset{x\ll a}{\bigwedge }}{p}_0(x,y)$ is an L-quasi-metric.
• If $p_0$ satisfies symmetry (A3), then p also satisfies (A3).
• If $p_0$ satisfies separation (A4), then p also satisfies (A4).

(1) We prove that p is an L-quasi-metric.

(A1). Suppose $a\ge b$. For any $y\ll b$, we have $y\in {\beta }^{*}\left(a\right)$. Then, ${\displaystyle \underset{x\ll a}{\bigwedge }}{p}_0(x,y)\le p_0(y,y)=0$; it follows that $p(a,b)=0$.

(A2). Let $r=p(a,c)={\displaystyle \underset{v\ll c}{\bigvee }}{\displaystyle \underset{u\ll a}{\bigwedge }}{p}_0(u,v)$. For all $s<r$, there exists some $v_0\ll c$ such that

$${\displaystyle \underset{u\ll a}{\bigwedge }}{p}_0(u,v_0)>s.$$

Let $y_0\ll b$. Since $p_0$ satisfies the triangle inequality,

$$p_0(u,v_0)\le (p_0(u,y_0)+p_0(y_0,v_0)$$

taking the infimum over $u\ll a$ on both sides,

$${\displaystyle \underset{u\ll a}{\bigwedge }}{p}_0(u,v_0)\le {\displaystyle \underset{u\ll a}{\bigwedge }}[(p_0(u,y_0)+p_0(y_0,v_0)]$$

Since $p_0(y_0,v_0)$ is independent of u, we can write

$${\displaystyle \underset{u\ll a}{\bigwedge }}[(p_0(u,y_0)+p_0(y_0,v_0)]\le {\displaystyle \underset{y\ll b}{\bigvee }}{\displaystyle \underset{u\ll a}{\bigwedge }}{p}_0(u,y)+p_0(y_0,v_0)=p(a,b)+p_0(y_0,v_0).$$

Noticing that $y_0\ll b$ and $y_0$ is arbitrary, we have

$$s<{\displaystyle \underset{u\ll a}{\bigwedge }}{p}_0(u,v_0)\le p(a,b)+{\displaystyle \underset{y_0\ll b}{\bigwedge }}{p}_0(y_0,v_0)$$

$$\le p(a,b)+{\displaystyle \underset{v\ll c}{\bigvee }}{\displaystyle \underset{y_0\ll b}{\bigwedge }}{p}_0(y_0,v)$$

$$=p(a,b)+p(b,c).$$

Because $s<r$ and s is arbitrary, we can acquire $p(a,b)+p(b,c)\ge r=p(a,c)$.

(B1). Let $r_1=p(a,b)={\displaystyle \underset{y\ll b}{\bigvee }}{\displaystyle \underset{x\ll a}{\bigwedge }}{p}_0(x,y)$ and $r_2={\displaystyle \underset{y\ll b}{\bigvee }}{\displaystyle \underset{x\ll a}{\bigwedge }}p(x,y)$. For any $s<r_1$, because $p(a,b)={\displaystyle \underset{y\ll b}{\bigvee }}{\displaystyle \underset{x\ll a}{\bigwedge }}{p}_0(x,y)>s,$ there exists $y_s$ such that $y_s\ll b$ and satisfying ${\displaystyle \underset{x\ll a}{\bigwedge }}{p}_0(x,y_s)>s.$ By the below relationship (see Preliminaries), we can choose $z_s$ such that $y_s\ll z_s\ll b$. Therefore,

$$r_2={\displaystyle \underset{y\ll b}{\bigvee }}{\displaystyle \underset{x\ll a}{\bigwedge }}p(x,y)={\displaystyle \underset{y\ll b}{\bigvee }}{\displaystyle \underset{x\ll a}{\bigwedge }}{\displaystyle \underset{y_1\ll b_1}{\bigvee }}{\displaystyle \underset{x_1\ll x}{\bigwedge }}{p}_0(x_1,y_1)$$

$$\ge {\displaystyle \underset{x\ll a}{\bigwedge }}{\displaystyle \underset{y_1\ll z_s}{\bigvee }}{\displaystyle \underset{x_1\ll x}{\bigwedge }}{p}_0(x_1,y_1)$$

$$\ge {\displaystyle \underset{x\ll a}{\bigwedge }}{\displaystyle \underset{x_1\ll x}{\bigwedge }}{p}_0(x_1,y_s)$$

$$={\displaystyle \underset{x\ll a}{\bigwedge }}{p}_0(x,y_s)>s.$$

Because s is arbitrary, we can obtain $r_1\le r_2$.

Next, we prove $r_1\ge r_2$. Let $t<r_2$. Then, there exists $y_0$ satisfying $y_0\ll b$ such that ${\displaystyle \underset{x\ll a}{\bigwedge }}p(x,y_0)>t$. Because

$${\displaystyle \underset{x\ll a}{\bigwedge }}p(x,y_0)={\displaystyle \underset{x\ll a}{\bigwedge }}{\displaystyle \underset{y\ll y_0}{\bigvee }}{\displaystyle \underset{z\ll x}{\bigwedge }}{p}_0(z,y)>t,$$

there exists $y_x\ll y_0$ for every $x\ll a$ such that

$${\displaystyle \underset{z\ll x}{\bigwedge }}{p}_0(z,y_x)>t.$$

Thus, we can obtain

$${\displaystyle \underset{x\ll a}{\bigwedge }}{\displaystyle \underset{z\ll x}{\bigwedge }}{p}_0(z,y_x)\ge t.$$

Consequently,

$$r_1=p(a,b)={\displaystyle \underset{y\ll b}{\bigvee }}{\displaystyle \underset{x\ll a}{\bigwedge }}{p}_0(x,y)\ge {\displaystyle \underset{x\ll a}{\bigwedge }}{p}_0(x,y_0)$$

$$={\displaystyle \underset{x\ll a}{\bigwedge }}{\displaystyle \underset{z\ll x}{\bigwedge }}{p}_0(z,y_0)\ge {\displaystyle \underset{x\ll a}{\bigwedge }}{\displaystyle \underset{z\ll x}{\bigwedge }}{p}_0(z,y_x)\ge t.$$

Because t is arbitrary, we can get $r_1\ge r_2$. So,

$$p(a,b)={\displaystyle \underset{y\ll b}{\bigvee }}{\displaystyle \underset{x\ll a}{\bigwedge }}p(x,y)={\displaystyle \underset{x\ll a}{\bigvee }}{\displaystyle \underset{e\ll x}{\bigvee }}{\displaystyle \underset{y\ll b}{\bigwedge }}p(y,e)={\displaystyle \underset{x\ll a}{\bigvee }}p(b,x).$$

Therefore, p satisfies (B1).

(2). Let $a,b\in M$. Then, by $y\nleqq b^{\prime }$, there exists $y_0\ll y$ such that $y_0\nleqq b^{\prime }$. Furthermore, with the help of $b^{\prime }={\displaystyle \underset{b_0\ll b}{\bigwedge }}{\displaystyle b_0^{\prime }}$, there exists $b_0\ll b$ such that $y_0\nleqq {\displaystyle b_0^{\prime }}$. Hence, there is ${\displaystyle b_0^{\prime }}\ll b^{\prime }$ such that $y_0\nleqq {\displaystyle b_0^{\prime }}$. Because $p(a,y)={\displaystyle \underset{y_1\ll y}{\bigvee }}{\displaystyle \underset{x\ll a}{\bigwedge }}{p}_0(x,y_1)<r$ and ${\displaystyle \underset{x\ll a}{\bigwedge }}{p}_0(x,y_0)<r$, there exists $x_0$ such that $x_0\ll a$ and $p_0(x_0,y_0)<r$. Owing to $p_0$ satisfying (A3), there exists x such that $x\nleqq {\displaystyle x_0^{\prime }}$ and $p_0(b_0,x)<r$. By $x\nleqq {\displaystyle x_0^{\prime }}$ and $x_0\ll a$, we can obtain $x_0\le a\iff a^{\prime }\le {\displaystyle x_0^{\prime }}$, and then $x\nleqq a^{\prime }$. According to (A1) and (A2), when $b\ge c$, we have $p(a,c)\le p(a,b)$. Thus,

$$p(b,x)={\displaystyle \underset{v\ll x}{\bigvee }}{\displaystyle \underset{u\ll b}{\bigwedge }}{p}_0(u,v)\le {\displaystyle \underset{u\ll b}{\bigwedge }}{p}_0(u,x)\le p_0(b_0,x)<r.$$

As a result, p satisfies (A3). Therefore, p is the first pseudo-metric.

(3). According to (A1) and (A2), we have $p_0(a,b)=0$ when $a\ge b$. Therefore, by $p_0(a,c)\le p_0(a,b)+p_0(b,c)$, when $a\ge b$, $p_0(a,c)\le p_0(b,c)$ holds.

Because $p_0$ satisfies (A1) and (A2), we can get

$$p_0(a,\tilde{b})\le {\displaystyle \underset{\tilde{a}\ll a}{\bigwedge }}{p}_0(\tilde{a},\tilde{b})\left(1\right).$$

Suppose that

$$p(a,b)={\displaystyle \underset{\tilde{b}\ll b}{\bigvee }}{\displaystyle \underset{\tilde{a}\ll a}{\bigwedge }}{p}_0(\tilde{a},\tilde{b})=0.$$

Then we can deduce that ${\displaystyle \underset{\tilde{a}\ll a}{\bigwedge }}{p}_0(\tilde{a},\tilde{b})=0$ for each $\tilde{b}\ll b$. By (1), we can obtain $p_0(a,\tilde{b})\le {\displaystyle \underset{\tilde{a}\ll a}{\bigwedge }}\tilde{a}\ll a$ and $p_0(\tilde{a},\tilde{b})=0$, and then $p_0(a,\tilde{b})=0$. Noticing that $p_0$ satisfies (A4), we can acquire $a\ge \tilde{b}.$ Because $\tilde{b}\le a$ and $\tilde{b}$ is arbitrary, we have $a\ge b$. Therefore, p satisfies (A4), as desired. □

Next, we prove that $B_r\left(b\right)$ is a closed set in the first metric topological space and further provide an equivalent characterization of the first metric by the $C-nbd$ cluster as follows.

Theorem 8.

If p is the first pseudo-metric on $L^X$, then $\overline{B_r\left(b\right)}=B_r\left(b\right)$.

Proof. 

It suffices to prove $\overline{B_r\left(b\right)}\le B_r\left(b\right)$. Let $h\ll \overline{B_r\left(b\right)}={\displaystyle \underset{s>0}{\bigwedge }}{D}_s\left(B_r\left(b\right)\right)$. Then, $h\ll D_s\left(B_r\left(b\right)\right)$ for each $s>0$. Hence, there exists $a\ll B_r\left(b\right)$ such that $p(a,h)<s$. Thus, we have

$$p(b,h)\le p(b,a)+p(a,h)<s+r.$$

Therefore, $h\le D_{r+s}\left(b\right)$, and then $h\le {\displaystyle \underset{s>0}{\bigwedge }}{D}_{r+s}\left(b\right)$. According to (1) in Theorem 1, we have $h\le B_r\left(b\right)$. This implies that $\overline{B_r\left(b\right)}\le B_r\left(b\right)$, as desired. □

Theorem 9.

If mapping $p:M\times M\to [0,+\infty )$ satisfies (A1) and (A2), then $D_r\left(b\right)={\displaystyle \underset{s<r}{\bigvee }}{B}_s\left(b\right)$.

Proof. 

If $s<r$, then $B_s\left(b\right)\le D_r\left(b\right)$, and then ${\displaystyle \underset{s<r}{\bigvee }}{B}_s\left(b\right)\le D_r\left(b\right)$. Conversely, for each $c\ll D_r\left(b\right)$, we have $p(b,c)<r$. Take a number s such that $p(b,c)<s<r$. Then, $c\le B_s\left(b\right)$. Hence, $c\le {\displaystyle \underset{s<r}{\bigvee }}{B}_s\left(b\right)$. Because c is arbitrary, we can obtain $D_r\left(b\right)\le {\displaystyle \underset{s<r}{\bigvee }}{B}_s\left(b\right)$, as desired. □

Theorem 10.

Suppose that $p:M\times M\to [0,+\infty )$ is the first pseudo-metric. Then, $\left\{B_r\right|r\in [0,+\infty )\}$ satisfies the following conditions:

(R1) 

$\forall b\in M$, $\forall r\in [0,+\infty )$, $b\le B_r\left(b\right);$

(R2) 

$\forall r,s\in [0,+\infty )$, $B_r\circ B_s\le B_{r+s};$

(R3) 

$\forall b\in M$, $B_r\left(b\right)={\displaystyle \underset{r<s}{\bigwedge }}{B}_s\left(b\right);$

(R4) 

$\forall r\in [0,+\infty )$, $a,b\in M$, ${\displaystyle \underset{s<r}{\bigvee }}{B}_s\left(a\right)\nleqq b^{\prime }\iff {\displaystyle \underset{s<r}{\bigvee }}{B}_s\left(b\right)\nleqq a^{\prime }$.

Proof. 

At first, we prove the following (R0).

(R0) $c\le B_r\left(b\right)\iff p(b,c)\le r$ for all $b,c\in M$.

In fact, it suffices to prove that $c\le B_r\left(b\right)\Rightarrow p(b,c)\le r$. Let $c\le B_r\left(b\right)$. Then, for each $h\ll c$, there exists an $e\in M$ such that $e\ge h$ and $p(b,e)\le r$. So $p(b,h)\le p(b,e)\le r$. Thus, $p(b,c)={\displaystyle \underset{h\ll c}{\bigvee }}p(b,h)\le r$.

(R1) can be obtained from (R0) and (A1).

(R2) can be obtained from (R0) and (A2).

(R3) holds from the following implications:

$$\forall c,b\in M\left(L^X\right),c\le B_r\left(b\right)\iff p(b,c)\le r\iff \forall s>r,p(b,c)\le s$$

$$\iff \forall s>r,c\le B_s\left(b\right)\iff c\le {\displaystyle \underset{r<s}{\bigwedge }}{B}_s\left(b\right).$$

(R4) is straightforward by (2) in Theorem 1 and (A3), as desired. □

Theorem 11.

Let $\left\{B_r\right|B_r:M\to L^X,\forall r\in [0,+\infty )\}$ be a family of mappings satisfying (R1)–(R4). Define a mapping $p:M\times M\to [0,+\infty )$ by

$$p(b,a)=\bigwedge \left\{s\right|a\le B_s\left(b\right)\}.$$

Then, p is the first pseudo-metric, and the family of closed neighborhood mappings of p is exactly $\left\{B_r|r\in [0,+\infty )\right\}$.

Proof. 

We first prove the following result:

$$c\le B_r\left(b\right)\iff p(b,c)\le r.$$

In fact, $c\le B_r\left(b\right)\Rightarrow p(b,c)\le r$ is trivial. Conversely, let $p(b,c)\le r$. For each $s>r$, by the definition of p, it is obvious that $c\le B_s\left(b\right)$. From (R4), we have $c\le {\displaystyle \underset{s>r}{\bigwedge }}{B}_s\left(b\right)=B_r\left(b\right)$, as desired.

(A1) can be obtained from (R1).

(A2) Suppose that $p(b,a)=r$ and $p(c,b)=s$. Then, $a\le B_r\left(b\right)$ and $b\le B_s\left(c\right)$. Hence, $a\le B_r\circ B_s\left(c\right)$. From (R2), we know $a\le B_{r+s}\left(c\right)$, which implies $p(c,a)\le r+s$. Therefore, $p(c,b)+p(b,a)\ge p(c,a)$.

(B1) can be obtained from the following implications:

$$p(b,a)\le r\iff a\le B_r\left(b\right)\iff \forall c\ll a,c\le B_r\left(b\right)\iff {\displaystyle \underset{c\ll b}{\bigvee }}p(b,c)\le r.$$

(A3) is obvious from (R4), as desired. □

Theorem 12.

Suppose that p is the first pseudo-metric on $L^X$. Then, the topological space induced by this metric is $T_1$ if and only if p satisfies (A4).

Proof. 

$(\Leftarrow )$. Let $b\in M\left(L^X\right)$. Then, according to (1) in Theorem 1, we can obtain $\overline{b}={\displaystyle \underset{r>0}{\bigwedge }}{D}_r\left(b\right)$. Let $h\ll \overline{b}$. Then, for each $r>0$, we have $h\ll D_r\left(b\right)\le B_r\left(b\right)$, which implies $p(b,h)=0$. Therefore, by (A4) we know $h\le b$. Consequently, $b=\overline{b}$.

$(\Rightarrow )$. If $p(b,a)=0$, then for any $r>0$ we can obtain $a\le D_r\left(b\right)$. Thus, $a\le {\displaystyle \underset{r>0}{\bigwedge }}{D}_r\left(b\right)=\overline{b}=b$, as desired. □

4. The Relationships Among the Four Basic Spheres

This section analyzes the relationships that exist among four fundamental types of spheres in the first metric space on $L^X$. Subsequently, we demonstrate several of their distinctive and interrelated properties.

Theorem 13.

If p is the first pseudo-metric, then

(1) 

$D_r\left(a\right)={\displaystyle \underset{s<r}{\bigvee }}{D}_s\left(a\right)$;

(2) 

${\displaystyle \underset{c\ll a}{\bigvee }}{D}_r\left(c\right)=D_r\left(a\right)$.

Proof. 

(1). Obviously, $D_r\left(a\right)\ge {\displaystyle \underset{s<r}{\bigvee }}{D}_s\left(a\right)$. Conversely, for each $s<r$, there exists t with $s<t<r$ such that $B_s\left(b\right)\le D_t\left(b\right)$, and then by Theorem 9, we have $D_r\left(a\right)={\displaystyle \underset{s<r}{\bigvee }}{B}_s\left(a\right)\le {\displaystyle \underset{s<r}{\bigvee }}{D}_s\left(a\right)$.

(2). It can be obtained from Theorem 3, as desired. □

Theorem 14.

Suppose that p is the first pseudo-metric. Then,

(1) 

$B_r\left(b\right)={\displaystyle \underset{r<s}{\bigwedge }}{B}_s\left(b\right)$;

(2) 

${\displaystyle \underset{r<s}{\bigwedge }}{\displaystyle \underset{c\ll b}{\bigvee }}{B}_s\left(c\right)=B_r\left(b\right)$.

Proof. 

(1). It can be obtained from (2) in Theorem 1.

(2). If $c\ll b$, then for each $s>0$, we can obtain $B_s\left(c\right)\ll B_s\left(b\right)$. Therefore, ${\displaystyle \underset{c\ll b}{\bigvee }}{B}_s\left(c\right)\le B_s\left(b\right)$. According to (2) in Theorem 1, we have

$${\displaystyle \underset{r<s}{\bigwedge }}{\displaystyle \underset{c\ll b}{\bigvee }}{B}_s\left(c\right)\le {\displaystyle \underset{r<s}{\bigwedge }}{B}_s\left(b\right)=B_r\left(b\right).$$

Furthermore, ${\displaystyle \underset{c\ll b}{\bigvee }}{D}_s\left(c\right)\le {\displaystyle \underset{c\ll b}{\bigvee }}{B}_s\left(c\right)$. So, by (1) in Theorem 1 and (2) in Theorem 13, we can deduce the following: $B_r\left(b\right)={\displaystyle \underset{r<s}{\bigwedge }}{D}_s\left(b\right)={\displaystyle \underset{r<s}{\bigwedge }}{\displaystyle \underset{c\ll b}{\bigvee }}{D}_s\left(c\right)\le {\displaystyle \underset{r<s}{\bigwedge }}{\displaystyle \underset{c\ll b}{\bigvee }}{B}_s\left(c\right)\le {\displaystyle \underset{r<s}{\bigwedge }}{B}_s\left(b\right)=B_r\left(b\right),$ as desired. □

Theorem 15.

Suppose that p is the first pseudo-matric. Then,

(1) $Q_r\left(a\right)={\displaystyle \underset{r<s}{\bigvee }}{Q}_s\left(a\right)$;

(2) ${\displaystyle \underset{c\ll a}{\bigvee }}{Q}_r\left(c\right)=Q_r\left(a\right)$.

Proof. 

(1). Obviously, ${\displaystyle \underset{r<s}{\bigvee }}{Q}_s\left(a\right)\le Q_r\left(a\right)$. Conversely, let $c\ll Q_r\left(a\right)$. Then, there exists $e\in M$ such that $c\le e$ and $p(e,a)>r$, and then $p(c,a)\ge p(e,a)>r$. Choose s such that $p(c,a)>s>r$. Then, we have $c\le Q_s\left(a\right)\le {\displaystyle \underset{r<s}{\bigvee }}{Q}_s\left(a\right)$. Therefore, $Q_r\left(a\right)\le {\displaystyle \underset{r<s}{\bigvee }}{Q}_s\left(a\right)$.

(2). Obviously, ${\displaystyle \underset{c\ll a}{\bigvee }}{Q}_r\left(c\right)\le Q_r\left(a\right)$. Conversely, let $e\ll Q_r\left(a\right)$. Then, we can obtain $p(e,a)>r$. If $e\nleqq {\displaystyle \underset{c\ll a}{\bigvee }}{Q}_r\left(c\right)$, then for each $c\ll a$, it holds that $e\nleqq Q_r\left(c\right)$, which means $p(e,c)\le r$. Therefore, by (B1) we can deduce $r<p(e,a)={\displaystyle \underset{c\ll a}{\bigvee }}p(e,c)\le r$. But this is a contradiction. Hence, $e\le {\displaystyle \underset{c\ll a}{\bigvee }}{Q}_r\left(c\right)$. As a result, ${\displaystyle \underset{c\ll a}{\bigvee }}{Q}_r\left(c\right)\ge Q_r\left(a\right)$, as desired. □

Theorem 16.

Suppose that p is the first pseudo metric. Then,

(1) 

$P_r\left(a\right)={\displaystyle \underset{s<r}{\bigwedge }}{P}_s\left(a\right)$;

(2) 

${\displaystyle \underset{s<r}{\bigwedge }}{\displaystyle \underset{e\ll a}{\bigvee }}{P}_s\left(e\right)=P_r\left(a\right).$

Proof. 

(1). It can be obtained from Theorem 2.

(2). It is obvious that ${\displaystyle \underset{e\ll a}{\bigvee }}{P}_s\left(e\right)\le P_s\left(a\right)$. Conversely, if $s<r$, then $P_r\left(a\right)\le Q_s\left(a\right)$. Therefore, by (2) in Theorem 15 and (1), we can obtain the following equations: $P_r\left(a\right)\le {\displaystyle \underset{s<r}{\bigwedge }}{Q}_s\left(a\right)={\displaystyle \underset{s<r}{\bigwedge }}{\displaystyle \underset{e\ll a}{\bigvee }}{Q}_s\left(e\right)\le {\displaystyle \underset{s<r}{\bigwedge }}{\displaystyle \underset{e\ll a}{\bigvee }}{P}_s\left(e\right)\le {\displaystyle \underset{s<r}{\bigwedge }}{P}_s\left(a\right)=P_r\left(a\right),$ as desired. □

Theorem 17.

Let p be the first pseudo-metric. Then, for each $a\in M\left(L^X\right)$ and $r\in [0,+\infty )$, there are

(1) 

$Q_r\left(a\right)={\displaystyle \underset{r<s}{\bigvee }}{Q}_s\left(a\right)$;

(2) 

$Q_r\left(a\right)={\displaystyle \underset{z\nleqq a^{\prime }}{\bigvee }}{B}_r{\left(z\right)}^{\prime }$. Thus, $Q_r\left(a\right)$ is an open set.

Proof. 

(1). Clearly, ${\displaystyle \underset{r<s}{\bigvee }}{Q}_s\left(a\right)\le Q_r\left(a\right)$. In order to prove that ${\displaystyle \underset{r<s}{\bigvee }}{Q}_s\left(a\right)\ge Q_r\left(a\right)$, we take $c\ll Q_r\left(a\right)$. Thus, we know that there exists an $e\in M\left(L^X\right)$ such that $c\le e$ and $p(e,a)>r$. By (B1), we have $p(c,a)\ge p(e,a)>r$. Take s such that $p(c,a)>s>r$. Then, $c\le Q_s\left(a\right)\le {\displaystyle \underset{r<s}{\bigvee }}{W}_s\left(a\right)$, which shows that $Q_r\left(a\right)\le {\displaystyle \underset{r<s}{\bigvee }}{Q}_s\left(a\right)$.

(2). Let $b\ll Q_r\left(a\right)$. Then, from (1), there exists an $s>r$ such that $b\ll Q_s\left(a\right)$. From the proof process of (1), we know $p(b,a)>s$. If $b\nleqq {\displaystyle \underset{z\nleqq a^{\prime }}{\bigvee }}{B}_r{\left(z\right)}^{\prime }$, then for each $z\nleqq a^{\prime }$, we have $b\nleqq B_r{\left(z\right)}^{\prime }$, i.e., $B_r\left(z\right)\nleqq b^{\prime }$. Therefore, there exists a point $x\in M\left(L^X\right)$ such that $x\ll B_r\left(z\right)$ and $x\nleqq b^{\prime }$. Hence, $p(z,x)\le r<s$. By (A3), there exists a $y=y\left(z\right)$ such that $y\nleqq z^{\prime }$ and $p(b,y)<s$. Let

$$q=\bigvee \{y=y\left(z\right)|z\nleqq a^{\prime }\}.$$

Then, $q\nleqq z^{\prime }$, i.e., $z\nleqq q^{\prime }$. Since $z\nleqq a^{\prime }$ implies $z\nleqq q^{\prime }$, we can get $q^{\prime }\le a^{\prime }$, i.e., $a\le q$. So, for each $c\ll a\le q$, there exists a $y=y\left(z\right)$ such that $c\le y$, which implies $p(b,c)\le p(b,y)<s$. Again by (B1), we have $p(b,a)\le s$, which contradicts $p(b,a)>s$. Therefore,

$${\displaystyle \underset{z\nleqq a^{\prime }}{\bigvee }}{B}_r{\left(z\right)}^{\prime }\ge Q_r\left(a\right).$$

Conversely, let $b\nleqq Q_r\left(a\right)$. Then, there exists $c\ll b$ such that $c\nleqq W_r\left(a\right)$. So, for each $s>r$, $p(c,a)\le r<s$. Thus, for each $z\nleqq a^{\prime }$ (i.e., $a\nleqq z^{\prime }$), there exists $x\nleqq c^{\prime }$ such that $p(z,x)<s$ from (A3). Hence, we have $x\le D_s\left(z\right)$. Again, since $x\nleqq c^{\prime }$, we can get $D_s\left(z\right)\nleqq c^{\prime }$ (i.e., $c\nleqq D_s{\left(z\right)}^{\prime }$), which implies $b\nleqq {\displaystyle \underset{r<s}{\bigvee }}{D}_s{\left(z\right)}^{\prime }$. According to (1) in Theorem 1, we have $b\nleqq B_r{\left(z\right)}^{\prime }$. Therefore, $b\nleqq W_r\left(a\right)$ implies $b\nleqq B_r{\left(z\right)}^{\prime }$. As a result $B_r{\left(z\right)}^{\prime }\le Q_r\left(a\right)$. So, ${\displaystyle \underset{z\nleqq a^{\prime }}{\bigvee }}{B}_r{\left(z\right)}^{\prime }\le Q_r\left(a\right)$, as desired. □

Theorem 18.

If p is the first pseudo-metric on $L^X$, then ${\displaystyle \underset{s>r}{\bigwedge }}{\displaystyle \underset{z\nleqq b^{\prime }}{\bigvee }}{Q}_s{\left(z\right)}^{\prime }=B_r\left(b\right).$

Proof. 

First, we prove ${\displaystyle \underset{z\nleqq b^{\prime }}{\bigvee }}{Q}_s{\left(z\right)}^{\prime }\le B_s\left(b\right)$. Let $a\ll {\displaystyle \underset{z\nleqq b^{\prime }}{\bigvee }}{Q}_s{\left(z\right)}^{\prime }$. Then, there exists an $e\in M\left(L^X\right)$ such that $a\ll e$ and $e\le {\displaystyle \underset{z\nleqq b^{\prime }}{\bigvee }}{Q}_s{\left(z\right)}^{\prime }$, i.e., $e^{\prime }\ge {\displaystyle \underset{z\nleqq b^{\prime }}{\bigwedge }}{Q}_s\left(z\right)$. From this, we know that for each $x\nleqq e^{\prime }$, there exists a $z\nleqq b^{\prime }$ such that $x\nleqq W_s\left(z\right)$, which implies $p(x,z)\le s$. Take $u>s$. According to (A3), there exists a $y=y\left(x\right)$ such that $y\nleqq x^{\prime }$ and $p(b,y)<u$. Let $q=\bigvee \{y=y\left(x\right)|x\nleqq e^{\prime }\}$. Then, $q\nleqq x^{\prime }$, i.e., $x\nleqq q^{\prime }$. Thus, if $x\nleqq e^{\prime }$, then we have $x\nleqq q^{\prime }$. So, we get $q^{\prime }\le e^{\prime }$ (i.e., $e\le q$). By $a\ll e\le q$, we know that there exists a $y=y\left(x\right)$ such that $a\le y$ and $x\nleqq e^{\prime }$. From (A1) and (A2), we have $p(b,a)\le p(b,y)<u$. Hence, $p(b,a)\le s$, and then $a\le B_s\left(b\right)$, which shows ${\displaystyle \underset{z\nleqq b^{\prime }}{\bigvee }}{Q}_s{\left(z\right)}^{\prime }\le B_s\left(b\right).$

Secondly, we prove $B_r\left(b\right)\le {\displaystyle \underset{z\nleqq b^{\prime }}{\bigvee }}{Q}_s{\left(z\right)}^{\prime }$$(s>r)$, i.e., $B_r{\left(b\right)}^{\prime }\ge {\displaystyle \underset{z\nleqq b^{\prime }}{\bigwedge }}{Q}_s\left(z\right)$. Let $x\ll {\displaystyle \underset{z\nleqq b^{\prime }}{\bigwedge }}{Q}_s\left(z\right).$ Then, $x\ll W_s\left(z\right)$ for any $z\nleqq b^{\prime }$. Thus, there exists $e\in M\left(L^X\right)$ such that $x\le e$ and $p(e,z)>s$. Hence, $p(x,z)\ge p(e,z)>s$. Now, we come to prove $x\le B_r{\left(b\right)}^{\prime }$. Let $x\nleqq B_r{\left(b\right)}^{\prime }$. Then, $B_r\left(b\right)\nleqq x^{\prime }$. Furthermore, take $a\le B_r\left(b\right)$ such that $a\nleqq x^{\prime }$. Then, $p(b,a)\le r<s$. By (A3), we know that there exists a $z\nleqq b^{\prime }$ such that $p(x,z)<s$. But this contradicts $p(x,z)>s$. Therefore, $B_r\left(b\right)\le {\displaystyle \underset{z\nleqq b^{\prime }}{\bigvee }}{Q}_s{\left(z\right)}^{\prime }$$(s>r)$.

According to $B_r\left(b\right)\le {\displaystyle \underset{z\nleqq b^{\prime }}{\bigvee }}{Q}_s{\left(z\right)}^{\prime }$$(s>r)$, ${\displaystyle \underset{z\nleqq b^{\prime }}{\bigvee }}{Q}_s{\left(z\right)}^{\prime }\le B_s\left(b\right)$, and (R3), we can obtain $B_r\left(b\right)\le {\displaystyle \underset{s>r}{\bigwedge }}{\displaystyle \underset{z\nleqq b^{\prime }}{\bigvee }}{Q}_s{\left(z\right)}^{\prime }\le {\displaystyle \underset{s>r}{\bigwedge }}{B}_s\left(b\right)=B_r\left(b\right)$, as desired. □

Theorem 19.

If p is the first pseudo-metric, then

(1) 

$D_r\left(b\right)={\displaystyle \underset{s<r}{\bigvee }}{B}_s\left(b\right)$;

(2) 

$B_r\left(b\right)={\displaystyle \underset{r<s}{\bigwedge }}{D}_s\left(b\right)$.

Proof. 

$\left(1\right)$ and $\left(2\right)$ can be obtained from Theorem 9 and (1) in Theorem 1, respectively, as desired. □

Theorem 20.

If p is the first pseudo metric, then

(1) 

$D_r\left(b\right)\ge {\displaystyle \underset{z\nleqq b^{\prime }}{\bigvee }}{P}_r{\left(z\right)}^{\prime }$;

(2) 

$P_r\left(a\right)={\displaystyle \underset{s<r}{\bigwedge }}{\displaystyle \underset{z\nleqq a^{\prime }}{\bigvee }}{D}_s{\left(z\right)}^{\prime }$.

Proof. 

(1) can be obtained directly from (2) in Theorem 2.

(2). First, we prove ${\displaystyle \underset{z\nleqq a^{\prime }}{\bigvee }}{D}_s{\left(z\right)}^{\prime }\le P_s\left(a\right)$. Let $b\nleqq P_s\left(a\right);$ then, there exits $c\ll b$ such that $c\nleqq P_s\left(a\right)$, and then $p(c,a)<s$. Thus, for each $z\nleqq a^{\prime }$, i.e., $a\nleqq z^{\prime }$, according to (A3), there exits $y\nleqq c^{\prime }$ such that $p(z,y)<s$, and then $y\le D_s\left(z\right)$, which means $c\nleqq D_s{\left(z\right)}^{\prime }$. So, from $c\ll b$, we have $b\nleqq {\displaystyle \underset{z\nleqq a^{\prime }}{\bigvee }}{D}_s{\left(z\right)}^{\prime }$. Therefore, ${\displaystyle \underset{z\nleqq a^{\prime }}{\bigvee }}{D}_s{\left(z\right)}^{\prime }\le P_s\left(a\right)$. Secondly, by (2) in Theorem 16, we can obtain

$$Q_s\left(a\right)={\displaystyle \underset{z\nleqq a^{\prime }}{\bigvee }}{B}_s{\left(z\right)}^{\prime }\le {\displaystyle \underset{z\nleqq a^{\prime }}{\bigvee }}{D}_s{\left(z\right)}^{\prime }\le P_s\left(a\right).$$

In addition, when $s<r$, we have $P_r\left(a\right)\le Q_s\left(a\right)$. So, from (1) in Theorem 15, we can get the following $P_r\left(a\right)\le {\displaystyle \underset{s<r}{\bigwedge }}{Q}_s\left(a\right)={\displaystyle \underset{s<r}{\bigwedge }}{\displaystyle \underset{z\nleqq a^{\prime }}{\bigvee }}{B}_s{\left(z\right)}^{\prime }\le {\displaystyle \underset{s<r}{\bigwedge }}{\displaystyle \underset{z\nleqq a^{\prime }}{\bigvee }}{D}_s{\left(z\right)}^{\prime }\le {\displaystyle \underset{s<r}{\bigwedge }}{P}_s\left(a\right)=P_r\left(a\right)$, as desired. □

Theorem 21.

If p is the first pseudo-metric, then

(1) 

$B_r\left(b\right)\ge {\displaystyle \underset{z\nleqq b^{\prime }}{\bigvee }}{P}_r{\left(z\right)}^{\prime }$;

(2) 

$P_r\left(a\right)={\displaystyle \underset{s<r}{\bigwedge }}{\displaystyle \underset{z\nleqq a^{\prime }}{\bigvee }}{B}_s{\left(z\right)}^{\prime }$.

Proof. 

(1). It can be obtained from (1) of Theorem 19 and $B_r\left(b\right)\ge D_r\left(b\right)$. Generally speaking, the inequality can not be equal. Otherwise, if $B_r\left(b\right)={\displaystyle \underset{z\nleqq b^{\prime }}{\bigvee }}{P}_r{\left(z\right)}^{\prime }$, then from (1) of Theorem 19, we can get $D_r\left(b\right)\ge B_r\left(b\right)$. Obviously, this is a contradiction.

(2). According to (2) in Theorem 16 and (2) in Theorem 19, we can obtain the following equations: $P_r\left(a\right)\le {\displaystyle \underset{s<r}{\bigwedge }}{Q}_s\left(a\right)={\displaystyle \underset{s<r}{\bigwedge }}{\displaystyle \underset{z\nleqq a^{\prime }}{\bigvee }}{B}_s{\left(z\right)}^{\prime }\le {\displaystyle \underset{s<r}{\bigwedge }}{\displaystyle \underset{z\nleqq a^{\prime }}{\bigvee }}{D}_s{\left(z\right)}^{\prime }=P_r\left(a\right)$, as desired. □

Theorem 22.

If p is the first pseudo-metric, then

(1) 

$P_r\left(a\right)={\displaystyle \underset{s<r}{\bigwedge }}{Q}_s\left(a\right)$;

(2) 

$Q_r\left(a\right)={\displaystyle \underset{r<s}{\bigvee }}{P}_s\left(a\right)$.

Proof. 

(1) is straightforward from the proof process of (2) in Theorem 20.

(2). By the definitions of $Q_r\left(a\right)$ and $P_s\left(a\right)$, it is obvious that $Q_r\left(a\right)\ge {\displaystyle \underset{r<s}{\bigvee }}{P}_s\left(a\right)$. On the contrary, for each $e\ll Q_r\left(a\right)$, there exits $b\ge e$ such that $p(b,a)>r$. And furthermore, from (A1) and (A2), we have $p(e,a)\ge p(b,a)>r$. Let $p(e,a)=t>r$. Then, by $e\le P_t\left(a\right)$, we can obtain $e\le {\displaystyle \underset{r<s}{\bigvee }}{P}_s\left(a\right)$. As a result, ${\displaystyle \underset{r<s}{\bigvee }}{P}_s\left(a\right)\ge Q_r\left(a\right)$, as desired. □

Theorem 23.

If p is the first pseudo-metric, then

(1) 

$Q_s\left(a\right)\le {\displaystyle \underset{z\nleqq a^{\prime }}{\bigvee }}{D}_s{\left(z\right)}^{\prime }$;

(2) 

$D_r\left(b\right)\le {\displaystyle \underset{s>r}{\bigwedge }}{\displaystyle \underset{z\nleqq b^{\prime }}{\bigvee }}{Q}_s{\left(z\right)}^{\prime }$.

Proof. 

(1). The proof is straightforward from the proof process of (2) in Theorem 19.

(2). By Theorem 18, we have ${\displaystyle \underset{s>r}{\bigwedge }}{\displaystyle \underset{z\nleqq b^{\prime }}{\bigvee }}{Q}_s{\left(z\right)}^{\prime }=B_r\left(b\right)\ge D_r\left(b\right)$. Generally speaking, the inequality can not be equal. Otherwise, $D_r\left(b\right)=B_r\left(b\right)$, which is a contradiction, as desired. □

5. Symmetry Properties of Two Types of Basic Spheres

This section examines the two fundamental types of spheres, $B_r\left(a\right)$ and $Q_r\left(a\right)$, within the first metric space. Two key results have been established:

$$Q_r\left(a\right)={\displaystyle \underset{z\nleqq a^{\prime }}{\bigvee }}{B}_r{\left(z\right)}^{\prime }(\mathrm{Theorem}16),$$

$${\displaystyle \underset{s>r}{\bigwedge }}{\displaystyle \underset{z\nleqq b^{\prime }}{\bigvee }}{Q}_s{\left(z\right)}^{\prime }=B_r\left(b\right)(\mathrm{Theorem}17).$$

However, these relationships lack symmetry. To address this issue and achieve the symmetric form $B_r\left(b\right)={\displaystyle \underset{z\nleqq b^{\prime }}{\bigvee }}{Q}_r{\left(z\right)}^{\prime }$, we introduce a new fuzzy p-metric on $L^X$. Furthermore, we demonstrate that the L-fuzzy real line satisfies this new metric.

Definition 5.

A $fuzzy$ p-metric on $L^X$ is a mapping $p:M\times M\to [0,+\infty )$ satisfying conditions (A1), (A2), (B1), and the following:

${\left(\mathrm{A}3\right)}^{*}$$\forall a,b\in M$,$\exists y\nleqq b^{\prime }$ such that $p(a,y)\le r\iff \exists x\nleqq a^{\prime }$ such that $p(b,x)\le r.$

Theorem 24.

If p is a $fuzzy$ p-metric, then p is the first pseudo-metric.

Proof. 

It suffices to show that if p satisfies (A3)*, then it also satisfies (A3). This follows from the fact that (A3)* and (A3) are equivalent to the following conditions, respectively:

$${(A3-1)}^{*}\forall a,b\in M,\forall r>0B_r\left(a\right)\le b^{\prime }\iff B_r\left(b\right)\le a^{\prime };$$

$$(A3-1)\forall a,b\in M,\forall r>0D_r\left(a\right)\le b^{\prime }\iff D_r\left(b\right)\le a^{\prime }.$$

We now show that if p satisfies (A3-1)*, then it satisfies (A3-1). Suppose $D_r\left(a\right)\le b^{\prime }$ for $a,b\in M$ and $r>0$. By (1) in Theorem 18, we have the following: $B_s\left(a\right)\le b^{\prime }$ for each $s<r$, and then from (A3-1)*, it follows that $B_s\left(b\right)\le a^{\prime }$. Thus, we obtain $D_r\left(b\right)={\displaystyle \underset{s<r}{\bigvee }}{B}_s\left(b\right)\le a^{\prime }$. Therefore, p satisfies (A3-1), as desired. □

Indeed, while every fuzzy p-metric is a first pseudo-metric, the converse does not hold in general. We, thus, present the following result:

Theorem 25.

Let p be the first pseudo-metric and define $s_{ab}=inf\left\{r\right|D_r\left(a\right)\nleqq b^{\prime }\}$. Then,

(w1) 

$s_{ab}=s_{ba}$;

(w2) 

If $s_{ab}=0$, then for all $t>0$, $B_t\left(a\right)\nleqq b^{\prime }\iff B_t\left(b\right)\nleqq a^{\prime }$;

(w3) 

If $s_{ab}>0$, then the following hold:

(a) 

If $t<s_{ab}$, then $D_t\left(a\right)\le b^{\prime }$ and $B_t\left(a\right)\le b^{\prime }$;

(b) 

If $t=s_{ab}$, then $D_t\left(a\right)\le b^{\prime }$;

(c) 

If $t>s_{ab}$, then $D_t\left(a\right)\nleqq b^{\prime }$ and $B_t\left(a\right)\nleqq b^{\prime }$.

Proof. 

(w1) follows directly from the fact that pp satisfies (A3-1) and the definition of $s_{ab}$.

By the definition of $D_r$, we have ${\displaystyle \underset{r>0}{\bigvee }}{D}_r\left(a\right)={\displaystyle \underline{1}}$ for each $a\in M$. Now, suppose for contradiction that for some $b\in M$, ${\displaystyle \underset{r>0}{\bigvee }}{D}_r\left(a\right)={\displaystyle \underline{1}}\le b^{\prime }$. Then, $b^{\prime }={\displaystyle \underline{1}}$, which implies $b={\displaystyle \underline{0}}$. However, this contradicts $b\ne {\displaystyle \underline{0}}$, since ${\displaystyle \underline{0}}\notin M$. Therefore, for any $a,b\in M$, it must hold that $b\nleqq {\displaystyle \underset{r>0}{\bigwedge }}{D}_r{\left(a\right)}^{\prime }$. Consequently, there exists some $r>0$ such that $b\nleqq D_r{\left(a\right)}^{\prime }$, i.e., $D_r\left(a\right)\nleqq b^{\prime }$. By the definition $s_{ab}=infr|D_r\left(a\right)\nleqq b^{\prime }$, the set is non-empty and bounded below by 0, so $s_{ab}$ is well-defined and either $s_{ab}=0$ or $s_{ab}>0$.

(w2). Assume $s_{ab}=0$. Then, for every $ϵ>0$, there exists $r\in (0,ϵ)$ such that $D_r\left(a\right)\nleqq b^{\prime }$. Fix $t>0$ and choose $ϵ=t$. Then, there exists $r<t$ with $D_r\left(a\right)\nleqq b^{\prime }$. By (1) in Theorem 18, $D_r\left(a\right)={\bigvee }_{s<r}{B}_s\left(a\right)$, so there exists $s<r$ such that $B_s\left(a\right)\nleqq b^{\prime }$. Since $B_r\left(a\right)$ is increasing, $B_s\left(a\right)\le B_t\left(a\right)$. If $B_t\left(a\right)\le b^{\prime }$, then $B_s\left(a\right)\le b^{\prime }$, a contradiction. Hence, $B_t\left(a\right)\nleqq b^{\prime }$ for all $t>0$. By (A3), $D_r\left(a\right)\nleqq b^{\prime }\iff D_r\left(b\right)\nleqq a^{\prime }$. Thus, for every $r>0$, there exists $r^{\prime }<r$ such that $D_{r^{\prime }}\left(b\right)\nleqq a^{\prime }$, so $s_{ba}=0$. By symmetry, $B_t\left(b\right)\nleqq a^{\prime }$ for all $t>0$. Therefore, $B_t\left(a\right)\nleqq b^{\prime }\iff B_t\left(b\right)\nleqq a^{\prime }$ for all $t>0$.

(w3). Assume $s_{ab}>0$.

(a) Let $t<s_{ab}$. By the definition of $s_{ab}$, for all $r<s_{ab}$, we have $D_r\left(a\right)\le b^{\prime }$. In particular, $D_t\left(a\right)\le b^{\prime }$. Now choose r such that $t<r<s_{ab}$. Then, $D_r\left(a\right)\le b^{\prime }$. By (1) in Theorem 18, $D_r\left(a\right)={\bigvee }_{s<r}{B}_s\left(a\right)$, so $B_t\left(a\right)\le D_r\left(a\right)\le b^{\prime }$. Hence, $B_t\left(a\right)\le b^{\prime }$.

(b) Let $t=s_{ab}$. For every $r<s_{ab}$, from part (a), we have $B_r\left(a\right)\le b^{\prime }$. By (1) in Theorem 18, $D_{s_{ab}}\left(a\right)={\bigvee }_{r<s_{ab}}{B}_r\left(a\right)$. Since each $B_r\left(a\right)\le b^{\prime }$, it follows that $D_{s_{ab}}\left(a\right)\le b^{\prime }$.

(c) Let $t>s_{ab}$. Since $s_{ab}=inf\{r\mid D_r\left(a\right)\nleqq b^{\prime }\}$, there exists rr such that $s_{ab}<r<t$ and $D_r\left(a\right)\nleqq b^{\prime }$.

Since $r<t$ and D is increasing, $D_r\left(a\right)\le D_t\left(a\right)$, so $D_t\left(a\right)\nleqq b^{\prime }$. By (1) in Theorem 18, $D_r\left(a\right)={\bigvee }_{s<r}{B}_s\left(a\right)\le B_r\left(a\right)$ (because $B_s\left(a\right)\le B_r\left(a\right)$ for all $s<r$). Thus, $D_r\left(a\right)\le B_r\left(a\right)$. Since $D_r\left(a\right)\nleqq b^{\prime }$, it follows that $B_r\left(a\right)\nleqq b^{\prime }$. As $r<t$ and B is increasing, $B_r\left(a\right)\le B_t\left(a\right)$, so $B_t\left(a\right)\nleqq b^{\prime }$.

This completes the proof of (w3). □

For the subcase (b) in (w3), whether $B_{s_{ab}}\left(a\right)\le b^{\prime }$ or $B_{s_{ab}}\left(a\right)\nleqq b^{\prime }$, it is not enough to confirm when p is the first pseudo metric. In other words, both cases of $B_{s_{ab}}\left(a\right)\le b^{\prime }$ and $B_{s_{ab}}\left(a\right)\nleqq b^{\prime }$ are possible. The following two examples respectively provide the two situations.

Example 1.

Let $L=[0,1]$ and define a mapping $p:(0,1]\times (0,1]\to [0,+\infty )$ by the equation

$$p(a,b)=\left\{\begin{array}{cc}0,\hfill & if a\ge b;\hfill \\ 1,\hfill & if a<b.\hfill \end{array}\right.$$

Then, p is a first pseudo-metric on $[0,1]$.

For this, we need verify that p satisfies (A1), (A2), (B1), and (A3-1).

(A1) and (A2) are trivial.

(B1). If $a,b\in (0,1]$ and $a\ge b$, then $p(a,b)=0$. From (A1), we can obtain ${\displaystyle \underset{x\ll b}{\bigvee }}p(a,x)=0$. Thus, we have $p(a,b)={\displaystyle \underset{x<b}{\bigvee }}p(a,x)$. Similarly, when $a<b$, it is true that $p(a,b)={\displaystyle \underset{x<b}{\bigvee }}p(a,x)=1$. So, p satisfies (B1).

(A3-1). By definition, we can prove $D_r\left(a\right)=a$ for each $a\in (0,1]$. It is obvious that $D_r\left(a\right)\le b^{\prime }\iff D_r\left(b\right)\le a^{\prime }$ for any $a,b\in M$.

Let $a,b\in (0,1]$, satisfying $b\ne 1$ and $a<b^{\prime }$. Then, we can obtain $s_{ab}=1$ and $B_{s_{ab}}\left(a\right)\nleqq b^{\prime }$.

Example 2.

Let $L=[0,1]$ and define mapping $p:(0,1]\times (0,1]\to [0,+\infty )$ as follows:

$$p(a,b)=max\{b-a,0\}.$$

Then, it is easy to verify that p is a first pseudo-metric. Furthermore, let $a,b\in (0,1]$ and $a<b^{\prime }$. Then, we can obtain $s_{ab}=b^{\prime }-a$ and $B_{s_{ab}}\left(a\right)\le b^{\prime }$.

Definition 6.

A lattice $L^X$ is termed a strong molecular lattice if for every $a\in M$ and any non-empty family ${\left\{g_i\right\}}_{i\in \Gamma }\subseteq M$, it follows that ${\displaystyle \underset{g\in {\Gamma }_a}{\bigwedge }}g\nleqq a^{\prime }$.

Theorem 26.

If p is a first pseudo-metric on a strong molecular lattice $L^X$, then p is a $fuzzy$ p-metric.

Proof. 

We aim to show that if p is a first pseudo-metric on a strong molecular lattice $L^X$, then it is a fuzzy p-metric. That is, for all $a,b\in M$ and $r>0$, we must prove

$B_r\left(a\right)\nleqq b^{\prime }⟺B_r\left(b\right)\nleqq a^{\prime }$. We proceed by assuming $B_r\left(a\right)\nleqq b^{\prime }$ and showing $B_r\left(b\right)\nleqq a^{\prime }$. The converse follows by symmetry.

Assume $B_r\left(a\right)\nleqq b^{\prime }$. Consider two cases:

Case I: $D_r\left(a\right)\nleqq b^{\prime }$. By the property (A3-1) of the first pseudo-metric (which is equivalent to (A3)), we have, $D_r\left(a\right)\nleqq b^{\prime }\iff D_r\left(b\right)\nleqq a^{\prime }$. Since $D_r\left(b\right)\le B_r\left(b\right)$, it follows that $B_r\left(b\right)\nleqq a^{\prime }$.

Case II: $D_r\left(a\right)\le b^{\prime }$. From $B_r\left(a\right)\nleqq b^{\prime }$, there exists a molecule $c\in M$ such that $c\le B_r\left(a\right),c\nleqq b^{\prime },\mathrm{and}p(a,c)=r$. We now show that there exists $e\in M$ with $e\nleqq a^{\prime }$ and $p(b,e)=r$, which implies $B_r\left(b\right)\nleqq a^{\prime }$. For every $s>r$, since $p(a,c)=r<s$, by (A3) there exists $g_s\in M$ such that $g_s\nleqq a^{\prime }\mathrm{and}p(b,g_s)<s$. If $p(b,g_s)<r$, then since $D_r\left(b\right)\le a^{\prime }$ (which follows from $D_r\left(a\right)\le b^{\prime }$ and (A3-1)), we would have $g_s\le a^{\prime }$, contradicting $g_s\nleqq a^{\prime }$. Hence, $r\le p(b,g_s)<s$. Define: $g={\bigwedge }_{s>r}{g}_s$. Since $L^X$ is a strong molecular lattice and $g_s\nleqq a^{\prime }$ for all $s>r$, we have $g\nleqq a^{\prime }$. By the way-below relation ≪, there exists $e\in M$ such that $e\ll g\mathrm{and}e\nleqq a^{\prime }$. If $p(b,e)<r$, then $e\le D_r\left(b\right)\le a^{\prime }$, contradicting $e\nleqq a^{\prime }$. Therefore, $r\le p(b,e)$. Moreover, since $e\ll g\le g_s$ for all $s>r$, and $p(b,g_s)<s$, we have $p(b,e)\le p(b,g_s)<s\mathrm{for}\mathrm{all}s>r$. Taking the infimum over $s>r$, we conclude that $p(b,e)=r$. Thus, $e\le B_r\left(b\right)$ and $e\nleqq a^{\prime }$, so $B_r\left(b\right)\nleqq a^{\prime }$. In both cases, $B_r\left(a\right)\nleqq b^{\prime }$ implies $B_r\left(b\right)\nleqq a^{\prime }$. By symmetry, the converse holds. Therefore, p satisfies $B_r\left(a\right)\nleqq b^{\prime }\iff B_r\left(b\right)\nleqq a^{\prime }$, which is the defining condition for a fuzzy p-metric, as desired. □

Theorem 27.

If p is a $fuzzy$ p-metric, then ${\displaystyle \underset{z\nleqq b^{\prime }}{\bigvee }}{Q}_r{\left(z\right)}^{\prime }=B_r\left(b\right)$.

Proof. 

Let $p(b,a)\le r$. Noticing (A3-1)*, for every $x\nleqq a^{\prime }$, i.e., $a\nleqq x^{\prime }$, there exists $z\nleqq b^{\prime }$ such that $p(x,z)\le r$. If $x\ll Q_r\left(z\right)$, then there exists $e\in M$ such that $x\le e$ and $p(e,z)>r$. From (A1) and (A2), we have $p(x,z)\ge p(e,z)>r$, which is a contradiction. Therefore, $x\nleqq {\displaystyle \underset{z\nleqq b^{\prime }}{\bigwedge }}{Q}_r\left(z\right)$. In summary, as long as $x\nleqq a^{\prime }$, there is $x\nleqq {\displaystyle \underset{z\nleqq b^{\prime }}{\bigwedge }}{Q}_r\left(z\right)$. Hence, ${\displaystyle \underset{z\nleqq b^{\prime }}{\bigwedge }}{Q}_r\left(z\right)\le a^{\prime }$, i.e., $a\le {\displaystyle \underset{z\nleqq b^{\prime }}{\bigvee }}{Q}_r{\left(z\right)}^{\prime }$. So, ${\displaystyle \underset{z\nleqq b^{\prime }}{\bigvee }}{Q}_r{\left(z\right)}^{\prime }\ge B_r\left(b\right).$ On the contrary, let $a\ll {\displaystyle \underset{z\nleqq b^{\prime }}{\bigvee }}{Q}_r{\left(z\right)}^{\prime }$. Then, by a below ${\displaystyle \underset{z\nleqq b^{\prime }}{\bigvee }}{Q}_r{\left(z\right)}^{\prime }$ (see Preliminaries), there exists $e\in M$ such that $a\ll e$ and $e\le {\displaystyle \underset{z\nleqq b^{\prime }}{\bigvee }}{Q}_r{\left(z\right)}^{\prime }$, i.e., $e^{\prime }\ge {\displaystyle \underset{z\nleqq b^{\prime }}{\bigwedge }}{Q}_r\left(z\right)$. Therefore, for every $x\nleqq e^{\prime }$, there exists $z\nleqq b^{\prime }$ such that $x\nleqq Q_r\left(z\right)$, and then $p(x,z)\le r$. From (A3-1)*, for every x, there exists $y=y\left(x\right)$ such that $y\nleqq x^{\prime }$ and $p(b,y)\le r$. Let $q=\bigvee \{y=y\left(x\right)|x\nleqq e^{\prime }\}$. Then, we can obtain $q\nleqq x^{\prime }$, i.e., $x\nleqq q^{\prime }$. In short, as long as $x\nleqq e^{\prime }$, there is $x\nleqq q^{\prime }$. Consequently, $q^{\prime }\le e^{\prime }$ (i.e., $e\le q$). Based on $a\ll e\le q$, there exists $y=y\left(x\right)$ such that $a\le y$ and $x\nleqq e^{\prime }$. From (A1) and (A2), we have $p(b,a)\le p(b,y)\le r$. So, $a\le B_r\left(b\right)$. Therefore, ${\displaystyle \underset{z\nleqq b^{\prime }}{\bigvee }}{Q}_r{\left(z\right)}^{\prime }\le B_r\left(b\right)$, as desired. □

Theorem 28.

Suppose that p satisfies (A1), (A2), and (B1). Then, p is a fuzzy p-metric if and only if p satisfies ${\displaystyle \underset{z\nleqq b^{\prime }}{\bigvee }}{Q}_r{\left(z\right)}^{\prime }=B_r\left(b\right)$.

Proof. 

(⇒). It is straightforward from Theorem 26.

(⇐). We only need prove that p satisfies (B3-1)*: $\forall a,b\in M$, $\exists y\nleqq b^{\prime }$ such that $p(a,y)\le r\iff \exists z\nleqq a^{\prime }$ such that $p(b,z)\le r$. This proof is as follows: because $\exists y\nleqq b^{\prime }$ such that $p(a,y)\le r\iff B_r\left(a\right)\nleqq b^{\prime }$, from ${\displaystyle \underset{z\nleqq a^{\prime }}{\bigvee }}{Q}_r{\left(z\right)}^{\prime }=B_r\left(a\right)$, we have ${\displaystyle \underset{z\nleqq a^{\prime }}{\bigvee }}{Q}_r{\left(z\right)}^{\prime }\nleqq b^{\prime }$, i.e., $b\nleqq {\displaystyle \underset{z\nleqq a^{\prime }}{\bigwedge }}{Q}_r\left(z\right)$. Thus, there exists $z\nleqq a^{\prime }$ such that $b\nleqq Q_r\left(z\right)$, and then $p(b,z)\le r$. On the contrary, the proof is similar, as desired. □

Corollary 1.

If p is a first pseudo metric, then it is a $fuzzy$ p-metric if and only if ${\displaystyle \underset{z\nleqq b^{\prime }}{\bigvee }}{Q}_r{\left(z\right)}^{\prime }=B_r\left(b\right)$.

Next we prove that an L-real line is a $fuzzy$ p-metric. First, we show a lemma as follows:

Lemma 1.

Let $\mathbf{R}\left(L\right)$ be an L-real line. Define two mappings $\epsilon \left(e\right)=sup\left\{t\right|e\le {\displaystyle l_t^{\prime }}\}$ and $\sigma \left(e\right)=inf\left\{t\right|e\le {\displaystyle r_t^{\prime }}\}$ for every $e\in M$. Then, there exists $a\nleqq {\lambda }^{\prime }$ such that $\epsilon \left(\mu \right)\le \epsilon \left(a\right)+s$ if and only if there exists $b\nleqq {\mu }^{\prime }$ such that $\sigma \left(\lambda \right)\ge \sigma \left(b\right)-s$ for all $\lambda ,\mu \in M$.

Proof. 

Suppose that $\epsilon \left(\mu \right)\le \epsilon \left(a\right)+s$. Then, for each $t>0$, $\epsilon \left(\mu \right)<\epsilon \left(a\right)+s+t$. This means $\mu \nleqq {\displaystyle l_{ϵ\left(a\right)+s+t}^{\prime }}$ or $l_{\epsilon \left(a\right)+s+t}\nleqq {\mu }^{\prime }.$ Thus, there exists a molecule $b\le l_{\epsilon \left(a\right)+s+t}$ such that $b\nleqq {\mu }^{\prime }$. By Theorem 6, we have $l_{\epsilon \left(a\right)+s+t}\le {\displaystyle r_{\epsilon \left(a\right)+s+t}^{\prime }}$, and then $\sigma \left(b\right)\le \epsilon \left(a\right)+s+t$. Thereby, $\sigma \left(b\right)-s-2t<\epsilon \left(a\right)$. Furthermore, from $a\le {\displaystyle l_{\epsilon \left(a\right)}^{\prime }}$ and Theorem 6, we have $\lambda \nleqq a^{\prime }\ge l_{\epsilon \left(a\right)}\ge {\displaystyle r_{\sigma \left(b\right)-s-2t}^{\prime }}.$ Therefore, $\sigma \left(\lambda \right)>\sigma \left(b\right)-s-2t$. Because t is arbitrary, we can obtain $\sigma \left(\lambda \right)\ge \sigma \left(b\right)-s$, as desired. □

Theorem 29.

Suppose that $\mathbf{R}\left(L\right)$ is an L-real line. For any $a,b\in M$, define $p(b,a)=max\left\{\epsilon \right(b)-\epsilon (a),\sigma (a)-\sigma (b),0\}$. Then, p is a $fuzzy$ p-metric.

Proof. 

According to Theorems 5 and 6, we can obtain that p satisfies (A1), (A2), and (B1). Thus, we only need prove (B3)*.

For any $\lambda ,\mu \in M$, if there exists $a\nleqq {\lambda }^{\prime }$ such that $p(\mu ,a)=max\left\{\epsilon \right(\mu )-\epsilon (a),\sigma (a)-\sigma (\mu ),0\}\le r$, i.e., both $\epsilon \left(\mu \right)-\epsilon \left(a\right)\le r$ and $\sigma \left(a\right)-\sigma \left(\mu \right)\le r$ are valid. So, we can obtain $\epsilon \left(\mu \right)\le \epsilon \left(a\right)+r$ and $\sigma \left(\mu \right)\ge \sigma \left(a\right)-r$. By Lemma 1, there exists $b\nleqq {\mu }^{\prime }$ and $c\nleqq {\mu }^{\prime }$ such that $\sigma \left(\lambda \right)\ge \sigma \left(b\right)-r$ and $\epsilon \left(\lambda \right)\le \epsilon \left(c\right)+r$, respectively. Therefore, $b\wedge c\nleqq {\mu }^{\prime }$. Furthermore, because $\mu$ is an irreducible element, we can choose $d\le b\wedge c$ such that $d\nleqq {\mu }^{\prime }$. Hence, we have $\sigma \left(\lambda \right)\ge \sigma \left(b\right)-r\ge \sigma \left(d\right)-r$ and $\epsilon \left(\lambda \right)\le \epsilon \left(c\right)+r\le \epsilon \left(d\right)+r$. Consequently, $p(\lambda ,d)=max\left\{\epsilon \right(\lambda )-\epsilon (d),\sigma (d)-\sigma (\lambda ),0\}\le r$, as desired. □

Remark 1.

Let $S=\{p\mid p\mathit{is}a\mathit{Shi}\mathit{pseudo-metric}\}$ and $F=\{p\mid p\mathit{is}\mathit{a}\mathit{fuzzy}\mathit{p-metric}\}$. Then, according to Theorems 5 and 26, we have $F\subset S$.

6. Conclusions

Recently, in [10,11,12], we provided a series of interesting conclusions about fuzzy metrics. Especially in [12], we demonstrated the conclusion that has been achieved so far:

Metrization Theorem. If a $[0,1]$-topological space $(X,\delta )$ is $T_1$ and regular, and δ has a σ-locally finite base, then it is Deng, Erceg, Chen, and Yang-Shi metrizable.

In this paper, our aim is to continue studying a type of metric in L-topology. Firstly, we provide a distinctive construction theorem for this type of metric. Moreover, we show its equivalent characterization by a $C-nbd$ cluster, which is composed of one of the four types of basic spheres defined in this paper. Incidentally, we carefully investigate the topological properties of four types of basic spheres and obtain a series of meaningful results related to them, such as their opening and closing properties, symmetrical properties, and so on. Finally, to solve the problem of related symmetry relationship between two types of basic spheres $B_r\left(a\right)$ and $Q_r\left(a\right)$, we propose a new fuzzy p-metric and prove that an L-real line $\mathbf{R}\left(L\right)$ satisfies the kind of fuzzy p-metric.

In the future, we will consider whether or not these results will be able to be generalized to Chen’s metric and further investigate the properties of Chen’s metric in L-topology (see [10]). Additionally, we will continue to study the lattice-valued topological spaces, each of whose topologies has a $\sigma$-locally finite base. In addition, we also intend to inquire into some questions on the fuzzifying metric topology (see [18,29,36,38,44], etc.) and other metrics, such as $S_b$ metrics [59], b-metrics [60,61], strong b-metric [62], ⊕-sb-metric, which was studied recently [64], and so on.