Characterizations of $\alpha$-Quasi Uniformity and Theory of $\alpha$-P.Q. Metrics

Abstract

In Wu Fuzzy Systems and Mathematics 3:94–99, 2012, the author introduced concepts of $\alpha$-remote neighborhood mapping and $\alpha$-quasi uniform, and obtained many good results in $\alpha$-quasi uniform spaces. This chapter will further investigate properties of $\alpha$-remote neighborhood mapping, and give some characterizations of $\alpha$-quasi uniforms. Based on this, this chapter also introduces concept of $\alpha$-P.Q. metric, and establishes the relations between $\alpha$-quasi uniforms and $\alpha$-P.Q. metrics.

1 Introduction

Theory of quasi-uniformity in completely distributive lattices was firstly introduced by Erceg [1] and Hutton [2]. Then it was developed into various forms and was extended into different topological spaces [3–9]. In [10], the author introduced the concept of $\alpha$-quasi uniform in $\alpha$-layer order-preserving operator spaces, and revealed the relations between $\alpha$-layer topological spaces and $\alpha$-quasi uniform spaces. In this chapter, firstly, we further study properties of $\alpha$-remote neighborhood mappings. Then we discuss some characterizations of $\alpha$-quasi uniformities. Secondly, we introduce the concept of $\alpha$-P.Q. metrics, and establish the relations between $\alpha$-quasi uniforms and $\alpha$-P.Q. metrics.

2 Preliminaries

In this chapter, $X, Y$ will always denote nonempty crisp sets, A mapping $A:X\rightarrow L$ is called an $L$-fuzzy set. $L^X$ is the set of all $L$-fuzzy sets on $X$. An element $e\in L$ is called an irreducible element in $L$, if $p\vee q=e$ implies $p=e$ or $q=e$, where $p, q\in L$. The set of all nonzero irreducible elements in $L$ will be denoted by $M(L)$. If $x\in X$, $\alpha \in M(L)$, then $x_{\alpha }$ is called a molecule in $L^X$. The set of all molecules in $L^X$ is denoted by $M^*(L^X)$. If $A\in L^X$, $\alpha \in M(L)$, take $A_{[\alpha ]}=\{x\in X\mid A(x)\ge \alpha \}$ [3] and $A^{\alpha }=\vee \{x_{\alpha }\mid x_{\alpha }\not \le A\}$ [11]. It is easy to check $(A_{[\alpha ]})^{\prime }= A^{\alpha }_{[\alpha ]}$.

Let $(L^X, \delta )$ be an $L$-fuzzy topological space, $\alpha \in M(L)$. $\forall A\in L^X$, $D_{\alpha }(A)=\wedge \{G\in \delta ^{\prime }\mid G_{[\alpha ]}\supset A_{[\alpha ]}\}.$ Then the operator $D_{\alpha }$ is a $\alpha$-closure operator of some co-topology on $L^X$, denoted by $D_{\alpha }(\delta )$. We called $\alpha$-layer topology. The pair $(L^X,D_{\alpha }(\delta ))$ is called $\alpha$-layer co-topological space [11]. An $\alpha$-layer topological space$(L^X,D_{\alpha }(\delta ))$ is called an $\alpha$-$C_{II}$ space, if there is a countable base $\fancyscript{B}_{\alpha }$ of $D_{\alpha }(\delta )$.

A mapping $F_{\alpha }: L^X\rightarrow L^Y$ is called an $\alpha$-mapping, if $F_{\alpha }(A)_{[\alpha ]}=F_{\alpha }(B)_{[\alpha ]}$ whenever $A_{[\alpha ]}=B_{[\alpha ]}$, and $F_{\alpha }(A)=0_X$ whenever $A_{[\alpha ]}=\emptyset$. The mapping $F^{-1}_{\alpha }: L^Y\rightarrow L^X$ is called the reverse mapping of $F_{\alpha }$, if for each $B\in L^Y$, $F^{-1}_{\alpha }(B)=\vee \{A\in L^X\mid F_{\alpha }(A)_{[\alpha ]}\subset B_{[\alpha ]}\}.$ Clearly, $F^{-1}_{\alpha }$ is also an $\alpha$-mapping.

An $\alpha$-mapping $F_{\alpha }: L^X\rightarrow L^Y$ is called an $\alpha$-order-preserving homomorphism (briefly $\alpha$-$oph$), iff both $F_{\alpha }$ and $F^{-1}_{\alpha }$ are $\alpha$-union preserving mappings.

An $\alpha$-mapping $F_{\alpha }: L^X\rightarrow L^Y$ is called an $\alpha$-Symmetric mapping, if for every $A,B\in L^X$, we have

$$\begin{aligned} \exists C_{[\alpha ]}\not \subset A_{[\alpha ]}^{\alpha }, B_{[\alpha ]}\not \subset F_{\alpha }(C)_{[\alpha ]} \Leftrightarrow \exists D_{[\alpha ]}\not \subset B_{[\alpha ]}^{\alpha }, A_{[\alpha ]}\not \subset F_{\alpha }(D)_{[\alpha ]}. \end{aligned}$$

An $\alpha$-mapping $f_{\alpha }:L^X\rightarrow L^X$ is called an $\alpha$-remote neighborhood mapping, if for each $A\in L^X$ with $A_{[\alpha ]}\ne \emptyset$, we have $A_{[\alpha ]}\not \subset f_{\alpha }(A)_{[\alpha ]}$. The set of all $\alpha$-remote neighborhood mappings is denoted by $\fancyscript{F}_{\alpha }(L^X)$, (briefly by $\fancyscript{F}_{\alpha }$).

For $f_{\alpha },g_{\alpha }\in \fancyscript{F}_{\alpha }$, let’s define

• (1) $f_{\alpha }\le g_{\alpha } \Leftrightarrow \forall A\in L^X$, $f_{\alpha }(A)_{[\alpha ]} \subset g_{\alpha }(A)_{[\alpha ]}$.

• (2) $(f_{\alpha }\vee g_{\alpha })(A)=f_{\alpha }(A)\vee g_{\alpha }(A)$.

• (3) $(f_{\alpha }\odot g_{\alpha })(A)= \wedge \{f_{\alpha }(B)\mid \exists B\in L^X$, $B_{[\alpha ]}\not \subset g_{\alpha }(A)_{[\alpha ]}\}$.

An non-empty subfamily $\fancyscript{D}_{\alpha }\subset \fancyscript{F}_{\alpha }$ is called an $\alpha$-quasi-uniform, if $\fancyscript{D}_{\alpha }$ satisfies:

($\alpha$-U1) $\forall f_{\alpha }\in \fancyscript{D}_{\alpha },g_{\alpha }\in \fancyscript{F}_{\alpha }$ with $f_{\alpha }\le g_{\alpha }$, then $g_{\alpha }\in \fancyscript{D}_{\alpha }$.

($\alpha$-U2) $\forall f_{\alpha },g_{\alpha }\in \fancyscript{D}_{\alpha }$ implies $f_{\alpha }\vee g_{\alpha }\in \fancyscript{D}_{\alpha }$.

($\alpha$-U3) $\forall f_{\alpha }\in \fancyscript{D}_{\alpha }$, then $\exists g_{\alpha }\in \fancyscript{D}_{\alpha }$, such that $g_{\alpha }\odot g_{\alpha }\ge f_{\alpha }$.

$(L^X,\fancyscript{D}_{\alpha })$ is called an $\alpha$-quasi-uniform space. A subset $\fancyscript{B}_{\alpha }\subset \fancyscript{D}_{\alpha }$ is called a base of $\fancyscript{D}_{\alpha }$, if $\forall f_{\alpha }\in \fancyscript{D}_{\alpha }$, there is $g_{\alpha }\in \fancyscript{B}_{\alpha }$, such that $f_{\alpha }\le g_{\alpha }$. A subset $\fancyscript{A}_{\alpha }\subset \fancyscript{D}_{\alpha }$ is called a subbase of $\fancyscript{D}_{\alpha }$, if all of finite unions of the elements in $\fancyscript{A}_{\alpha }$ consist a base of $\fancyscript{D}_{\alpha }$. An $\alpha$-quasi-uniform $\fancyscript{D}_{\alpha }$ is called an $\alpha$-uniform, if $\fancyscript{D}_{\alpha }$ possesses a base whose elements are $\alpha$-symmetric. Usually, we call this base $\alpha$-symmetric base.

In [10], the author discussed the relation between an $\alpha$-quasi uniform space and an $\alpha$-layer co-topological space as following:

Let $\fancyscript{D}_{\alpha }$ be an $\alpha$-quasi uniform on $L^X$. $\forall A\in L^X$, Let’s take $c_{\alpha }(A)=\vee \{B\in L^X\mid \forall f_{\alpha }\in \fancyscript{D}_{\alpha }, A_{[\alpha ]}\not \subset f_{\alpha }(B)_{[\alpha ]}\}.$ Then $c_{\alpha }$ is an $\alpha$-closure operator of some $L$-fuzzy co-topology, which is denoted by $\eta _{\alpha }(\fancyscript{D}_{\alpha })$. Each $\alpha$-layer topological space $(L^X,D_{\alpha }(\delta ))$ can be $\alpha$-quasi uniformitale, i.e., there is an $\alpha$-quasi uniform $\fancyscript{D}_{\alpha }$, such that $D_{\alpha }(\delta )=\eta _{\alpha }(\fancyscript{D}_{\alpha })$.

Other definitions and notes not mentioned here can be seen in [12].

3 Properties of $\alpha$-Remote Neighborhood Mappings

Theorem 3.1

Let $f_{\alpha },g_{\alpha }\in \fancyscript{F}_{\alpha }$. Then

• (1) $f_{\alpha }\vee g_{\alpha }\in \fancyscript{F}_{\alpha }$, $f_{\alpha }\odot g_{\alpha }\in \fancyscript{F}_{\alpha }$.

• (2) $f_{\alpha }\odot g_{\alpha }\le f_{\alpha }, f_{\alpha }\odot g_{\alpha }\le g_{\alpha }$.

• (3) $(f_{\alpha }\odot g_{\alpha })\vee h_{\alpha }=(f_{\alpha }\vee h_{\alpha })\odot (g_{\alpha }\vee h_{\alpha })$,

$(f_{\alpha }\vee g_{\alpha })\odot h_{\alpha }=(f_{\alpha }\odot h_{\alpha })\vee (g_{\alpha }\odot h_{\alpha })$.

Theorem 3.2

Let $f_{\alpha }\in \fancyscript{F}_{\alpha }$. If for each $A\in L^X$,

$$\begin{aligned} f^{\nabla }_{\alpha }(A)=\wedge \{f_{\alpha }(B) \mid A_{[\alpha ]}\subset \bigcup \limits _{G_{[\alpha ]}\not \subset B^{\alpha }_{[\alpha ]}}f_{\alpha }(G)^{\alpha }_{[\alpha ]}\}, \end{aligned}$$

and

$$\begin{aligned} f^{\diamond }_{\alpha }(A)=\wedge \{f_{\alpha }(B) \mid \bigcup \limits _{G_{[\alpha ]}\not \subset B_{[\alpha ]}^{\alpha }} f_{\alpha }(G)_{[\alpha ]}^{\alpha }\not \subset f_{\alpha }(A)_{[\alpha ]}\}. \end{aligned}$$

Then

• (1) $f^{\nabla }_{\alpha },f^{\diamond }_{\alpha }\in \fancyscript{F}_{\alpha }$.

• (2) $f^{\diamond }_{\alpha }\le f^{\nabla }_{\alpha }\le f_{\alpha }$.

• (3) $f^{\diamond }_{\alpha }(A)=\wedge \{f^{\nabla }_{\alpha }(B)\mid B_{[\alpha ]}\not \subset f_{\alpha }(A)_{[\alpha ]}\} =(f^{\nabla }_{\alpha }\odot f_{\alpha })(A)$.

• (4) $f^{\nabla }_{\alpha }$ is $\alpha$-Symmetric.

Proof

(1) By $A_{[\alpha ]}\subset \bigcup \limits _{G_{[\alpha ]}\not \subset A_{[\alpha ]}^{\alpha }}f_{\alpha }(G)_{[\alpha ]}^{\alpha }$, we have $A_{[\alpha ]}\not \subset f^{\nabla }_{\alpha }(A)_{[\alpha ]}$. Thus $f^{\nabla }_{\alpha }\in \fancyscript{F}_{\alpha }$. Again by $A_{[\alpha ]}\not \subset f_{\alpha }(A)_{[\alpha ]}$, we get $\bigcup \limits _{G_{[\alpha ]}\not \subset A_{[\alpha ]}^{\alpha }}f_{\alpha }(G)_{[\alpha ]}^{\alpha }\not \subset f_{\alpha }(A)_{[\alpha ]}$. Therefore, $f^{\diamond }_{\alpha }\in \fancyscript{F}_{\alpha }$.

(2) The proof is obvious.

(3) For each $B\in L^X$,

$$\begin{aligned} \bigcup \limits _{G_{[\alpha ]}\not \subset B_{[\alpha ]}^{\alpha }} f_{\alpha }(G)_{[\alpha ]}^{\alpha }\not \subset f_{\alpha }(B)_{[\alpha ]}&\Leftrightarrow \exists D_{[\alpha ]}\subset \bigcup \limits _{G_{[\alpha ]}\not \subset A_{[\alpha ]}^{\alpha }} f_{\alpha }(G)_{[\alpha ]}^{\alpha }, D_{[\alpha ]}\not \subset f_{\alpha }(B)_{[\alpha ]} \\&\Leftrightarrow \exists D_{[\alpha ]}\subset \bigcup \limits _{G_{[\alpha ]}\not \subset B_{[\alpha ]}^{\alpha }} f_{\alpha }(G)_{[\alpha ]}^{\alpha },D_{[\alpha ]}\not \subset f_{\alpha }(B)_{[\alpha ]}. \end{aligned}$$

So

$$\begin{aligned} f^{\diamond }_{\alpha }(A)=\wedge \{f^{\nabla }_{\alpha }(D)\mid D_{[\alpha ]}\not \subset f_{\alpha }(D)_{[\alpha ]}\} =(f^{\nabla }_{\alpha }\odot f_{\alpha })(A). \end{aligned}$$

(4) $\forall D,E\in L^X$. If there is $A_{[\alpha ]}\not \subset D_{[\alpha ]}^{\alpha }$, such that

$$\begin{aligned} E_{[\alpha ]}\not \subset f^{\nabla }_{\alpha }(A)_{[\alpha ]}=\cap \{f_{\alpha }(B)_{[\alpha ]} \mid A_{[\alpha ]}\subset \bigcup \limits _{G_{[\alpha ]}\not \subset B_{[\alpha ]}^{\alpha }}f_{\alpha }(G)_{[\alpha ]}^{\alpha }\}. \end{aligned}$$

Then there is $B\in L^X$, satisfying $E_{[\alpha ]}\not \subset f_{\alpha }(B)_{[\alpha ]}$ and $A_{[\alpha ]}\subset \bigcup \limits _{G_{[\alpha ]}\not \subset (B_{[\alpha ]}^{\alpha }}f_{\alpha }(G)_{[\alpha ]}^{\alpha }$. Clearly, $\bigcup \limits _{G_{[\alpha ]}\not \subset B_{[\alpha ]}^{\alpha }}f_{\alpha }(G)_{[\alpha ]}^{\alpha }\not \subset D_{[\alpha ]}^{\alpha }$, i.e., $D_{[\alpha ]}\not \subset \bigcap \limits _{G_{[\alpha ]}\not \subset B_{[\alpha ]}^{\alpha }}f_{\alpha }(G)_{[\alpha ]}$. Thus, there is $C_{[\alpha ]}\not \subset B_{[\alpha ]}^{\alpha }$, such that $D_{[\alpha ]}\not \subset f_{\alpha }(C)_{[\alpha ]}$. Conclusively, we have

$$\begin{aligned} D_{[\alpha ]}\subset f_{\alpha }(B)_{[\alpha ]}^{\alpha }\subset \bigcup \limits _{G_{[\alpha ]}\not \subset B_{[\alpha ]}^{\alpha }}f_{\alpha }(G)_{[\alpha ]}^{\alpha }. \end{aligned}$$

and

$$\begin{aligned} D_{[\alpha ]}\not \subset \cap \{f_{\alpha }(C)_{[\alpha ]} \mid D_{[\alpha ]}\subset \bigcup \limits _{H_{[\alpha ]}\not \subset C_{[\alpha ]}^{\alpha }}f_{\alpha }(H)_{[\alpha ]}^{\alpha }\}= f^{\nabla }_{\alpha }(D)_{[\alpha ]}. \end{aligned}$$

Therefore, $f^{\nabla }_{\alpha }$ is $\alpha$-Symmetric.

Theorem 3.3

Let $f_{\alpha }$ be an $\alpha$-Symmetric remote neighborhood mapping, then

• (1) $C_{[\alpha ]}\not \subset f_{\alpha }(A)_{[\alpha ]}\Rightarrow A_{[\alpha ]}\subset \bigcup \limits _{D_{[\alpha ]}\not \subset C_{[\alpha ]}^{\alpha }}f_{\alpha }(D)_{[\alpha ]}^{\alpha }$.

• (2) $A_{[\alpha ]}\subset \bigcup \limits _{D_{[\alpha ]}\not \subset C_{[\alpha ]}^{\alpha }}f_{\alpha }(D)_{[\alpha ]}^{\alpha }\Rightarrow C_{[\alpha ]}\not \subset f_{\alpha }(A)_{[\alpha ]}$.

Proof

(1) Since $f_{\alpha }$ is an $\alpha$-Symmetric mapping. $\forall D_{[\alpha ]}\not \subset A_{[\alpha ]}^{\alpha }$, there is $B_{[\alpha ]}\not \subset C_{[\alpha ]}^{\alpha }$, satisfying $D_{[\alpha ]}\not \subset f_{\alpha }(B)_{[\alpha ]}$. Therefore,

$$\begin{aligned} A_{[\alpha ]}^{\alpha }\supset \bigcap \limits _{B_{[\alpha ]}\not \subset C_{[\alpha ]}^{\alpha }} f_{\alpha }(B)_{[\alpha ]}, ~~~i.e., A_{[\alpha ]}\subset \bigcup \limits _{D_{[\alpha ]}\not \subset C_{[\alpha ]}^{\alpha }}f_{\alpha }(D)_{[\alpha ]}^{\alpha }. \end{aligned}$$

(2) By $A_{[\alpha ]}\subset \bigcup \limits _{D_{[\alpha ]}\not \subset C_{[\alpha ]}^{\alpha }}f_{\alpha }(D)_{[\alpha ]}^{\alpha }$, we have $A_{[\alpha ]}^{\alpha }\supset \bigcap \limits _{D_{[\alpha ]}\not \subset C_{[\alpha ]}^{\alpha }}f_{\alpha }(D)_{[\alpha ]}$. So $\forall x\not \in A_{[\alpha ]}^{\alpha }$, there is $D_{[\alpha ]}\not \subset C_{[\alpha ]}^{\alpha }$, such that $x\not \in f_{\alpha }(D)_{[\alpha ]}$. Since $f_{\alpha }$ is $\alpha$-Symmetric. There is $B^x_{[\alpha ]}\not \subset x_{[\alpha ]}^{\alpha }$, such that $C_{[\alpha ]}\not \subset f_{\alpha }(B^x)_{[\alpha ]}$. Take $E=\vee \{B^x\mid x\not \in A_{[\alpha ]}^{\alpha }\}$, then $x\not \subset E_{[\alpha ]}^{\alpha }$. This implies $E_{[\alpha ]}^{\alpha }\subset A_{[\alpha ]}^{\alpha }$, i.e., $A_{[\alpha ]}\subset E_{[\alpha ]}$. Furthermore, we can conclude $C_{[\alpha ]}\not \subset f_{\alpha }(A)_{[\alpha ]}$. Otherwise, if $C_{[\alpha ]}\subset f_{\alpha }(A)_{[\alpha ]}\subset f_{\alpha }(E)_{[\alpha ]}$, then it contradicts with the statement: for each $B^x\le E , C_{[\alpha ]}\not \subset f_{\alpha }(B^x)_{[\alpha ]}$.

Theorem 3.4

Let $f_{\alpha }\in \fancyscript{F}_{\alpha }$. Then

$$\begin{aligned} (f^{\nabla }_{\alpha })_{\alpha }^{\nabla }\le f^{\nabla }_{\alpha }\odot f^{\nabla }_{\alpha } \le f^{\diamond }_{\alpha }\le f_{\alpha }\odot f_{\alpha }. \end{aligned}$$

Proof

$\forall A\in L^X$,

$$\begin{aligned} (f^{\nabla }_{\alpha })_{\alpha }^{\nabla }(A)&= \wedge \{f_{\alpha }^{\nabla }(B) \mid A_{[\alpha ]}\subset \bigcup \limits _{G_{[\alpha ]}\not \subset B_{[\alpha ]}^{\alpha }}f_{\alpha }^{\nabla }(G)_{[\alpha ]}^{\alpha }\} \\&\le \wedge \{f_{\alpha }^{\nabla }(C) \mid C_{[\alpha ]}\not \subset f_{\alpha }^{\nabla }(A)_{[\alpha ]}\}\\&= (f_{\alpha }^{\nabla }\odot f_{\alpha }^{\nabla })(A). \end{aligned}$$

By Theorem 2 (2), $f_{\alpha }^{\nabla }\le f_{\alpha }$, we have

$$\begin{aligned} (f_{\alpha }^{\nabla }\odot f_{\alpha }^{\nabla })(A)&= \wedge \{f_{\alpha }^{\nabla }(C)\mid C_{[\alpha ]}\not \subset f_{\alpha }^{\nabla }(A)_{[\alpha ]}\}\\&\le \wedge \{f_{\alpha }^{\nabla }(C) \mid C_{[\alpha ]}\not \subset f_{\alpha }(A)_{[\alpha ]}\}\\&= f^{\diamond }_{\alpha }(A). \end{aligned}$$

Therefore

$$\begin{aligned} f^{\diamond }_{\alpha }(A)=\wedge \{f_{\alpha }^{\nabla }(C) \mid C_{[\alpha ]}\not \subset f_{\alpha }(A)_{[\alpha ]}\} \le \wedge \{f_{\alpha }(C) \mid C_{[\alpha ]}\not \subset f_{\alpha }(A)_{[\alpha ]}\}=(f_{\alpha }\odot f_{\alpha })(A). \end{aligned}$$

Theorem 3.5

Let $f_{\alpha }\in \fancyscript{F}_{\alpha }$. Then

• (1) $f^{\nabla }_{\alpha }\le f_{\alpha }\odot f_{\alpha }$.

• (2) $f_{\alpha }\odot f_{\alpha }\odot f_{\alpha }=f^{\diamond }_{\alpha }$.

Proof

(1) By Theorem 2, we have

$$\begin{aligned} f^{\nabla }_{\alpha }(A)&= \wedge \{f_{\alpha }(B) \mid A_{[\alpha ]}\subset \bigcup \limits _{G_{[\alpha ]}\not \subset B_{[\alpha ]}^{\alpha }}f_{\alpha }(G)_{[\alpha ]}^{\alpha }\}\\&\le \wedge \{f_{\alpha }(B) \mid A_{[\alpha ]}\not \subset f_{\alpha }(A)_{[\alpha ]}\} \\&= f_{\alpha }\odot f_{\alpha }(A). \end{aligned}$$

(2) By Theorem 3 (2), we have

$$\begin{aligned} \bigcup \limits _{D_{[\alpha ]}\not \subset C_{[\alpha ]}^{\alpha }}f_{\alpha }(D)_{[\alpha ]}^{\alpha }\not \subset f_{\alpha }(A)_{[\alpha ]}&\Rightarrow \exists B_{[\alpha ]}\subset \bigcup \limits _{D_{[\alpha ]}\not \subset C_{[\alpha ]}^{\alpha }}f_{\alpha }(D)_{[\alpha ]}^{\alpha }, B_{[\alpha ]}\not \subset f_{\alpha }(A)_{[\alpha ]}\\&\Rightarrow \exists B_{[\alpha ]}\not \subset f_{\alpha }(A)_{[\alpha ]}, C_{[\alpha ]}\not \subset f_{\alpha }(B)_{[\alpha ]}\\&\Rightarrow C_{[\alpha ]}\not \subset f_{\alpha }\odot f_{\alpha }(A)_{[\alpha ]}. \end{aligned}$$

This shows $f_{\alpha }\odot f_{\alpha }\odot f_{\alpha }\le f^{\diamond }_{\alpha }$. On the other hand, by Theorem 2 (2) and (1) above, it is easy to find $f^{\diamond }_{\alpha }\le f^{\nabla }_{\alpha }\odot f_{\alpha }\le f_{\alpha }\odot f_{\alpha }\odot f_{\alpha }$. Therefore, (2) holds.

4 Characterizations of $\alpha$-Quasi Uniformities

Theorem 4.1

An non-empty subfamily $\fancyscript{D}_{\alpha }\subset \fancyscript{F}_{\alpha }$ is an $\alpha$-uniform, iff $\fancyscript{D}_{\alpha }$ satisfies:

• (1) $f_{\alpha }\in \fancyscript{D}_{\alpha },g_{\alpha }\in \fancyscript{F}_{\alpha }$ with $f_{\alpha }\le g_{\alpha }$, then $g_{\alpha }\in \fancyscript{D}_{\alpha }$.

• (2) $f_{\alpha },g_{\alpha }\in \fancyscript{D}_{\alpha }$ implies $f_{\alpha }\vee g_{\alpha }\in \fancyscript{D}_{\alpha }$.

• (3) $f_{\alpha }\in \fancyscript{D}_{\alpha }$, then $\exists g_{\alpha }\in \fancyscript{D}_{\alpha }$, such that $g^{\diamond }_{\alpha }\ge f_{\alpha }$.

Proof

Necessity. Since $\fancyscript{D}_{\alpha }$ is an $\alpha$-uniform, (1) and (2) hold. Furthermore, if $\fancyscript{B}_{\alpha }\subset \fancyscript{D}_{\alpha }$ is an $\alpha$-symmetric base. So for every $f_{\alpha }\in \fancyscript{D}_{\alpha }$, there is $g_{\alpha }\in \fancyscript{B}_{\alpha }$, such that $g_{\alpha }\odot g_{\alpha }\odot g_{\alpha }\odot g_{\alpha }\ge f_{\alpha }$. By Theorem 5 (2), $g^{\diamond }_{\alpha }=g_{\alpha }\odot g_{\alpha }\odot g_{\alpha } \ge g_{\alpha }\odot g_{\alpha }\odot g_{\alpha }\odot g_{\alpha }\ge f_{\alpha }$.

Sufficiency. If $\fancyscript{B}_{\alpha }=\{g^{\nabla }_{\alpha }\mid g_{\alpha }\in \fancyscript{D}_{\alpha }\}$. For every $f_{\alpha }\in \fancyscript{D}_{\alpha }$, there is $g_{\alpha }\in \fancyscript{D}_{\alpha }$, such that $g^{\diamond }_{\alpha }\ge f_{\alpha }$. By Theorem 4, $g_{\alpha }\odot g_{\alpha }\ge g^{\diamond }_{\alpha }\ge f_{\alpha }$. Then $\fancyscript{D}_{\alpha }$ is an $\alpha$-quasi-uniform. By Theorem 2 (2)and (4), we know $g^{\nabla }_{\alpha }\ge g^{\diamond }_{\alpha }\ge f_{\alpha }$. This shows $\fancyscript{B}_{\alpha }$ is an $\alpha$-symmetric base of $\fancyscript{D}_{\alpha }$. Therefore $\fancyscript{D}_{\alpha }$ is an $\alpha$-uniform.

Theorem 4.2

An non-empty subfamily $\fancyscript{D}_{\alpha }\subset \fancyscript{F}_{\alpha }$ is an $\alpha$-uniform, iff $\fancyscript{D}_{\alpha }$ satisfies:

• (1) $f_{\alpha }\in \fancyscript{D}_{\alpha },g_{\alpha }\in \fancyscript{F}_{\alpha }$ with $f_{\alpha }\le g_{\alpha }$, then $g_{\alpha }\in \fancyscript{D}_{\alpha }$.

• (2) $f_{\alpha },g_{\alpha }\in \fancyscript{D}_{\alpha }$ implies $f_{\alpha }\vee g_{\alpha }\in \fancyscript{D}_{\alpha }$.

• (3) $f_{\alpha }\in \fancyscript{D}_{\alpha }$, then $\exists g_{\alpha }\in \fancyscript{D}_{\alpha }$, such that $g^{\nabla }_{\alpha }\ge f_{\alpha }$.

Proof

Necessity. Since $\fancyscript{D}_{\alpha }$ is an $\alpha$-uniform, (1) and (2) hold. By Theorem 6, For every $f_{\alpha }\in \fancyscript{D}_{\alpha }$, there is $g_{\alpha }\in \fancyscript{D}_{\alpha }$, such that $g^{\diamond }_{\alpha }\ge f_{\alpha }$. So according to Theorem 2 (2), $g^{\nabla }_{\alpha }\ge g^{\diamond }_{\alpha }\ge f_{\alpha }$. Thus (3) holds.

Sufficiency. If $\fancyscript{B}_{\alpha }=\{g^{\nabla }_{\alpha }\mid g_{\alpha }\in \fancyscript{D}_{\alpha }\}$. By (3), For every $f_{\alpha }\in \fancyscript{D}_{\alpha }$, there is $g_{\alpha }\in \fancyscript{D}_{\alpha }$, such that $g^{\nabla }_{\alpha }\ge f_{\alpha }$. Again, there is $h_{\alpha }\in \fancyscript{D}_{\alpha }$, such that $h^{\nabla }_{\alpha }\ge g_{\alpha }$. By Theorem 4, we get

$$\begin{aligned} h_{\alpha }\odot h_{\alpha }\ge h^{\nabla }_{\alpha }\odot h^{\nabla }_{\alpha }\ge (h^{\nabla }_{\alpha })^{\nabla }_{\alpha }\ge g^{\nabla }_{\alpha }\ge f_{\alpha }. \end{aligned}$$

So $\fancyscript{B}_{\alpha }$ is an $\alpha$-symmetric base of $\fancyscript{D}_{\alpha }$. Therefore, $\fancyscript{D}_{\alpha }$ is an $\alpha$-uniform.

5 $\alpha$-P.Q. Metric and its Properties

A binary mapping $d_{\alpha }: L^X\times L^X\rightarrow [0,+\infty )$ is called an $\alpha$-mapping, if $\forall (A,B),(C,D)\in L^X\times L^X$ satisfying $A_{[\alpha ]}=C_{[\alpha ]}$ and $B_{[\alpha ]}=D_{[\alpha ]}$, then $d_{\alpha }(A,B)=d_{\alpha }(A,B)$.

Definition 5.1

An $\alpha$-mapping $d_{\alpha }: L^X\times L^X\rightarrow [0,+\infty )$ is called an $\alpha$-P.Q metric on $L^X$, if

($\alpha$-M1) $d_{\alpha }(A,A)=0$.

($\alpha$-M2) $d_{\alpha }(A,C)\le d_{\alpha }(A,B)+d_{\alpha }(B,C)$.

($\alpha$-M3) $d_{\alpha }(A,B)=\bigwedge \limits _{C_{[\alpha ]} \subset B_{[\alpha ]}}d_{\alpha }(A,C)$.

Theorem 5.1

Let $d_{\alpha }$ be an $\alpha$-P.Q. metrics on $L^X$. $\forall r\in (0,+\infty )$, a mapping $P_r: L^X\rightarrow L^X$ is defined by $\forall A\in L^X$,

$$\begin{aligned} P_{\alpha }^r(A)=\vee \{B\in L^X\mid d_{\alpha }(A,B)\ge r\}. \end{aligned}$$

Then

• (1) $P_{\alpha }^r$ is $\alpha$-symmetric mapping.

• (2) $\forall A,B\in L^X$, $B_{[\alpha ]}\subset P_{\alpha }^r(A)_{[\alpha ]} \Leftrightarrow d_{\alpha }(A,B)\ge r$.

• (3) $\forall A\in L^X, r>0$, $A_{[\alpha ]}\not \subset P_{\alpha }^r(A)_{[\alpha ]}$.

• (4) $\forall r,s\in (0,\infty )$, $P^r_{\alpha }\odot P^s_{\alpha }\ge P^{r+s}_{\alpha }$.

• (5) $\forall A\in L^X, r>0$, $P_{\alpha }^r(A)_{[\alpha ]}=\bigcap \limits _{s<r}P_{\alpha }^s(A)_{[\alpha ]}$.

• (6) $\forall A\in L^X$, $\bigcap \limits _{r>0}P_{\alpha }^s(A)_{[\alpha ]}=\emptyset$.

Proof

Since $d_{\alpha }$ is an $\alpha$-mapping, (1), (5) and (6) are easy.

(2) Clearly, $d_{\alpha }(A,B)\ge r\Rightarrow B_{[\alpha ]}\subset P_{\alpha }^r(A)_{[\alpha ]}$. Conversely. $\forall x\in B_{[\alpha ]}$, $\exists x\in D_x\in L^X$, such that $d_{\alpha }(A,D_x)\ge r$. So $d_{\alpha }(A,\{x_{\alpha }\})\ge d_{\alpha }(A,D_x)\ge r$. Thereby $d_{\alpha }(A,B)=\bigwedge \limits _{x\in B_{[\alpha ]}} d_{\alpha }(A,\{x_{\alpha }\})\ge r$.

(3) Suppose $A_{[\alpha ]}\subset P_{\alpha }^r(A)_{[\alpha ]}$. By (2), we get $d_{\alpha }(A,A)\ge r$. This is a contradiction with ($\alpha$-M1).

(4) $\forall A,B\in L^X$, if $B_{[\alpha ]}\not \subset P_{\alpha }^r\odot P_{\alpha }^s(A)_{[\alpha ]}$, then there is $D\in L^X$, such that $D_{[\alpha ]}\not \subset P^s_{\alpha }(A)_{[\alpha ]}$ and $B_{[\alpha ]}\not \subset P^r_{\alpha }(D)_{[\alpha ]}$. By (2), we have $d_{\alpha }(A,D)<s,d_{\alpha }(D,B)<r$. Then by ($\alpha$-M2), we gain $d_{\alpha }(A,B)<r+s$. Therefore $B_{[\alpha ]}\not \subset P_{\alpha }^{r+s}(A)_{[\alpha ]}$. This means $P^r_{\alpha }\odot P^s_{\alpha }\ge P^{r+s}_{\alpha }$.

Theorem 5.2

If a family of $\alpha$-mappings $\{P^r_{\alpha }\mid P_r: L^X\rightarrow L^X,r>0\}$ satisfies the conditions (2)–(5) in Theorem 8. For each $A,B\in L^X$, let’s denote

$$\begin{aligned} d_{\alpha }(A,B)=\wedge \{r\mid B_{[\alpha ]}\not \subset P_{\alpha }^r(A)_{[\alpha ]}\}. \end{aligned}$$

Then

• (1) $d_{\alpha }(A,B)<r\Leftrightarrow B_{[\alpha ]}\not \subset P_{\alpha }^r(A)_{[\alpha ]}$.

• (2) $d_{\alpha }$ is $\alpha$-P.Q. metric on $L^X$.

Proof

(1) By Theorem 8 (2),(5), we have

$$\begin{aligned} d_{\alpha }(A,B)<r\Leftrightarrow \exists s<r, B_{[\alpha ]}\not \subset P^s_{\alpha }(A)_{[\alpha ]} \Leftrightarrow B_{[\alpha ]}\not \subset P^r_{\alpha }(A)_{[\alpha ]}. \end{aligned}$$

(2) $\forall A,B\in L^X, r,s>0$, if $d_{\alpha }(A,B)>r+s$, then

$$\begin{aligned} B_{[\alpha ]}\subset P^{r+s}_{\alpha }(A)_{[\alpha ]} \subset (P^r_{\alpha }\odot P^s_{\alpha }(A))_{[\alpha ]}. \end{aligned}$$

Thus $\forall C\in L^X$, we have $C_{[\alpha ]}\not \subset P^s_{\alpha }(A)_{[\alpha ]}$ and $B_{[\alpha ]}\not \subset P^r_{\alpha }(C)_{[\alpha ]}$. this implies $d_{\alpha }(A,C)>s$, and $d_{\alpha }(A,B)>r$. Hence $d_{\alpha }(A,B)+d_{\alpha }(A,C)>r+s$. Consequently, we obtain $d_{\alpha }(A,B)+d_{\alpha }(A,C)\ge d_{\alpha }(A,B)$.

Theorem 5.3

An $\alpha$-mapping $d_{\alpha }: L^X\times L^X\rightarrow [0,+\infty )$ satisfies ($\alpha -M1$),($\alpha -M2$) and ($\alpha -M3^{*}$), then for each $C_{[\alpha ]}\subset B_{[\alpha ]}$, $d_{\alpha }(C,B)=0$.

($\alpha -M3^{*}) \forall A\in L^X , r>0, A_{[\alpha ]}\not \subset P_{\alpha }^r(A)_{[\alpha ]}$.

Proof

By ($\alpha$-M2), for each $B,C\in L^X$, satisfying $C_{[\alpha ]}\subset B_{[\alpha ]}$, we have $d_{\alpha }(A,B)<d_{\alpha }(A,C)+d_{\alpha }(C,B)$. Here we can conclude $d_{\alpha }(C,B)=0$. Otherwise, if $d_{\alpha }(C,B)=s>0$, then $B_{[\alpha ]}\subset P^s_{\alpha }(C)_{[\alpha ]}$. it contracts with $C_{[\alpha ]}\not \subset P^s_{\alpha }(C)_{[\alpha ]}$ according to ($\alpha$-M3$^{*}$). Therefore $d_{\alpha }(C,B)=0$.

Theorem 5.4

An $\alpha$-mapping $d_{\alpha }: L^X\times L^X\rightarrow [0,+\infty )$ is an $\alpha$-P.Q metric on $L^X$, iff  $d_{\alpha }$ satisfies ($\alpha$-M1),($\alpha$-M2) and ($\alpha$-M3$^*$).

Proof

We only need to prove ($\alpha$-M3)$\Leftrightarrow$($\alpha$-M3$^*$). By Theorem 8 (2), it is easy to check ($\alpha$-M3)$\Rightarrow$($\alpha$-M3$^*$). If the converse result is not true, then there are $r,s>0$, such that

$$\begin{aligned} d_{\alpha }(A,B)<s<r\le \bigcap \limits _{C_{[\alpha ]}\subset B_{[\alpha ]}}d_{\alpha }(A,C). \end{aligned}$$

so for each $C_{[\alpha ]}\subset B_{[\alpha ]}$,

$$\begin{aligned} r\le d_{\alpha }(A,C)\le d_{\alpha }(A,B)+d_{\alpha }(B,C)<s+d_{\alpha }(B,C). \end{aligned}$$

Thus, $0<r-s<d_{\alpha }(B,C)$, which implies $C_{[\alpha ]}\subset P^{r-s}_{\alpha }(B)_{[\alpha ]}$. As a result

$$\begin{aligned} B_{[\alpha ]}=(\vee \{C\mid C_{[\alpha ]}\subset B_{[\alpha ]}\})_{[\alpha ]}\subset P^{r-s}_{\alpha }(B)_{[\alpha ]}. \end{aligned}$$

However, it contradicts with ($\alpha$-M3$^*$). Therefore ($\alpha$-M3) holds.

Theorem 5.5

$d_{\alpha }$ is $\alpha$-P.Q. metric on $L^X$. Then $\{P_{\alpha }^r\mid r>0\}$ satisfying (3)–(5) in Theorem 8 is an $\alpha$-base of some $\alpha$-uniform, which is called the $\alpha$-uniform induced by $d_{\alpha }$.

Given an $\alpha$-quasi-uniform $\fancyscript{D}_{\alpha }$, if we say $\fancyscript{D}_{\alpha }$ is metricable, we mean there is an $\alpha$-P.Q. metric $d_{\alpha }$, such that $\fancyscript{D}_{\alpha }$ is induced by $d_{\alpha }$.

Theorem 5.6

An $\alpha$-quasi-uniform spaces $(L^X,\fancyscript{D}_{\alpha })$ is $\alpha$-P.Q. metricable iff it has a countable $\alpha$-base.

Proof

By Theorem 12, the necessity is obvious. Let’s prove the sufficiency.

Let $\fancyscript{D}_{\alpha }$ be an $\alpha$-uniformity on $L^X$, which has a countable $\alpha$-base $\fancyscript{B}_{\alpha }=\{P^n_{\alpha }\mid n\in N\}$. Let’s take $g^1_{\alpha }=P^1_{\alpha }$, then there is $g^2_{\alpha }\in \fancyscript{B}_{\alpha }$, such that $g^2_{\alpha }\odot g^2_{\alpha }\odot g^2_{\alpha }\ge g^1_{\alpha }\vee P^2_{\alpha }$. In addition, there is $g^3_{\alpha }\in \fancyscript{B}_{\alpha }$, such that $g^3_{\alpha }\odot g^3_{\alpha }\odot g^3_{\alpha }\ge g^2_{\alpha }\vee P^3_{\alpha }$. The process can be repeated again and again, then $\{g^n_{\alpha }\in n\in N\}$ is also an $\alpha$-base. Obviously, $g^{n+1}_{\alpha }\odot g^{n+1}_{\alpha }\odot g^{n+1}_{\alpha } \ge g^n_{\alpha }$. Let’s take $\varphi _{\alpha }:L^X\rightarrow L^X$, defined by: $\forall A\in L^X$,

$$\begin{aligned} \varphi ^r_{\alpha }(A)=\left\{ \begin{array} {c@{ \quad }c} g^n_{\alpha }(A), &{}\frac{1}{2^n}<r\le \frac{1}{2^{n-1}},\\ 0_X, &{}r>1. \end{array}\right. \end{aligned}$$

Clearly, $\forall r>0$, $A_{[\alpha ]}\not \subset \varphi ^r_{\alpha }(A)_{[\alpha ]}$. And $\forall \frac{1}{2^n}<r\le \frac{1}{2^{n-1}}$, we have

$$\begin{aligned} \varphi ^r_{\alpha }\odot \varphi ^r_{\alpha }\odot \varphi ^r_{\alpha } =g^n_{\alpha }\odot g^n_{\alpha }\odot g^n_{\alpha }\ge g^{n-1}_{\alpha }=\varphi ^{2r}_{\alpha }. \end{aligned}$$

Let’s define

$$\begin{aligned} f^r_{\alpha }(A)=\wedge \left\{ (\varphi ^{r_1}_{\alpha } \odot \varphi ^{r_2}_{\alpha }\odot \cdot \cdot \cdot \odot \varphi ^{r_k}_{\alpha }(A))\mid \sum \limits _{i=1}^k r_i=r\right\} . \end{aligned}$$

Then $\fancyscript{B}_{\alpha }=\{f^r_{\alpha }\mid r>0\}\subset \fancyscript{D}_{\alpha }$.

Finally, let’s prove $\fancyscript{B}_{\alpha }$ is an $\alpha$-base of $\fancyscript{D}_{\alpha }$ satisfying (3)-(6) in Theorem 8.

Step 1. For $f_{\alpha }\in \fancyscript{D}$, there is $n\in N$, such that $g^n_{\alpha }\ge f_{\alpha }$. So if $r\in (\frac{1}{2^{n+1}},\frac{1}{2^{n}}]$, then $\varphi ^{2r}_{\alpha }=g^n_{\alpha }$. Besides, it is easy to check, $\varphi ^{r_1}_{\alpha } \odot \varphi ^{r_2}_{\alpha }\odot \cdot \cdot \cdot \odot \varphi ^{r_k}_{\alpha }\ge \varphi ^{2r}_{\alpha }$ whenever $\sum \limits _{i=1}^k r_i=r$. Thus $f^r_{\alpha }\ge \varphi ^{2r}_{\alpha }=g^n_{\alpha }\ge f_{\alpha }$. This means $\fancyscript{B}_{\alpha }$ be an $\alpha$-base of $\fancyscript{D}_{\alpha }$.

Step 2. Obviously, $f_{\alpha }\in \fancyscript{B}$ satisfies (3) and (6) in Theorem 8. Furthermore, for $r,s>0, A\in L^X$, if there is $B_{[\alpha ]}\not \subset (f^r_{\alpha }\odot f^s_{\alpha }(A))_{[\alpha ]}$. Then there is $C\in L^X$, such that $B_{[\alpha ]}\not \subset f^r_{\alpha }(C)_{[\alpha ]}$ and $C_{[\alpha ]}\not \subset f^s_{\alpha }(A)_{[\alpha ]}$. So there are $\sum \limits _{i=1}^k r_i=r$ and $\sum \limits _{i=1}^m s_i=s$, such that

$$\begin{aligned} B_{[\alpha ]}\not \subset (\varphi ^{r_1}_{\alpha } \odot \varphi ^{r_2}_{\alpha }\odot \cdot \cdot \cdot \odot \varphi ^{r_k}_{\alpha }(C))_{[\alpha ]} \end{aligned}$$

and

$$\begin{aligned} C_{[\alpha ]}\not \subset (\varphi ^{s_1}_{\alpha } \odot \varphi ^{s_2}_{\alpha }\odot \cdot \cdot \cdot \odot \varphi ^{s_m}_{\alpha }(A))_{[\alpha ]}. \end{aligned}$$

Thus

$$\begin{aligned} B_{[\alpha ]}\not \subset (\varphi ^{r_1}_{\alpha } \odot \varphi ^{r_2}_{\alpha }\odot \cdot \cdot \cdot \odot \varphi ^{r_k}_{\alpha })\odot (\varphi ^{s_1}_{\alpha } \odot \varphi ^{s_2}_{\alpha }\odot \cdot \cdot \cdot \odot \varphi ^{s_m}_{\alpha })(A)_{[\alpha ]}. \end{aligned}$$

As a result, $B_{[\alpha ]}\not \subset f^{r+s}_{\alpha }(A)_{[\alpha ]}$. Consequently, $f^{r+s}_{\alpha }\le f^r_{\alpha }\odot f^s_{\alpha }$. This is the proof of (4) in Theorem 8.

Step 3. for $r>s>0$, $A\in L^X$,

$$\begin{aligned} f^r_{\alpha }(A)&= \wedge \left\{ (\varphi ^{r_1}_{\alpha } \odot \varphi ^{r_2}_{\alpha }\odot \cdot \cdot \cdot \odot \varphi ^{r_k}_{\alpha }(A)\mid \sum \limits _{i=1}^k r_i=r\right\} \\&\le \wedge \left\{ (\varphi ^{s_1}_{\alpha } \odot \varphi ^{s_2}_{\alpha }\odot \cdot \cdot \cdot \odot \varphi ^{s_m}_{\alpha }\odot \varphi ^{r-s}_{\alpha }(A))\mid \sum \limits _{i=1}^m s_i=m\right\} \\&\le \wedge \left\{ (\varphi ^{s_1}_{\alpha } \odot \varphi ^{s_2}_{\alpha }\odot \cdot \cdot \cdot \odot \varphi ^{s_m}_{\alpha }(A))\mid \sum \limits _{i=1}^m s_i=m\right\} =f^s_{\alpha }(A). \end{aligned}$$

Hence $f^r_{\alpha }\le \bigwedge \limits _{s<r}f^s_{\alpha }$.

Conversely. Let’s prove the reverse result.

If $B\in L^X$, $B_{[\alpha ]}\not \subset f^r_{\alpha }(A)_{[\alpha ]}$, then there is $\sum \limits _{i=1}^k r_i=r$, such that $B_{[\alpha ]}\not \subset \varphi ^{r_1}_{\alpha } \odot \varphi ^{r_2}_{\alpha }\odot \cdot \cdot \cdot \odot \varphi ^{r_k}_{\alpha }(A)_{[\alpha ]}$. So there is $C\in L^X$, such that $C_{[\alpha ]}\not \subset (\varphi ^{r_2}_{\alpha } \odot \varphi ^{r_2}_{\alpha }\odot \cdot \cdot \cdot \odot \varphi ^{r_k}_{\alpha }(A)_{[\alpha ]}$ and $B_{[\alpha ]}\not \subset \varphi ^{r_1}_{\alpha } (C)_{[\alpha ]}$. By $\varphi ^{r_1}_{\alpha }=\bigwedge \limits _{t<r_1}\varphi ^t_{\alpha }$, there is $t<r_1$, such that $B_{[\alpha ]}\not \subset \varphi ^{t}_{\alpha } (C)_{[\alpha ]}$, i.e., $B_{[\alpha ]}\not \subset \varphi ^{t}_{\alpha }\odot \varphi ^{r_2}_{\alpha }\odot \cdot \cdot \cdot \odot \varphi ^{r_k}_{\alpha }(A)_{[\alpha ]}$. Let’s take $s=t+\sum \limits _{i=2}^k r_i$, we have $s<r$ and $B_{[\alpha ]}\not \subset f^{s}_{\alpha }(A)_{[\alpha ]}$. Therefore $f^r_{\alpha }\ge \bigwedge \limits _{s<r}f^s_{\alpha }$. Therefore (5) in Theorem 8 holds.

Theorem 5.7

Each $\alpha$-$C_{II}$ $\alpha$-layer topological space is P.Q.-metriclizable.

Proof

Let $(L^X,D_{\alpha }(\delta ))$ be an $\alpha$-$C_{II}$ space, $\{P_n\mid n\in N\}$ be an $\alpha$-base. $\forall n\in N$, $f^{P_n}_{\alpha }:L^X\rightarrow L^X$ is defined as: $\forall A\in L^X$,

$$\begin{aligned} f_{\alpha }^{P_n}(A)=\left\{ \begin{array}{ll} 0_X, &{}A_{[\alpha ]}\subset P_{n[\alpha ]}.\\ P_n, &{}A_{[\alpha ]}\not \subset P_{n[\alpha ]}. \end{array}\right. \end{aligned}$$

Let’s take $\fancyscript{D}^*=\{f_{\alpha }^{P_n}\mid n\in N\}$ and

$$\begin{aligned} \fancyscript{B}_{\alpha }=\{f_{\alpha }\mid \exists f_{\alpha }^{P_{n_i}}\in \fancyscript{D}^*, i=1,2,\cdot \cdot \cdot ,m, f_{\alpha }=\bigvee \limits _{i=1}^m f_{\alpha }^{P_{n_i}}\}. \end{aligned}$$

Then $\fancyscript{B}_{\alpha }$ is an $\alpha$-base of some uniform denoted by $\fancyscript{D}_{\alpha }$ and clearly, $\eta _{\alpha }=\eta _{\alpha }(\fancyscript{D}_{\alpha })$. Furthermore, since $\fancyscript{B}_{\alpha }$ is countable, we know $(L^X,D_{\alpha }(\delta ))$ is P.Q.-metriclizable.

Theorem 5.8

An $\alpha$-layer co-topology $D_{\alpha }(\delta )$ on $L^X$ can be $\alpha$-P.Q. metriclizable iff there is a Sequence of $\alpha$-remote neighborhood mappings $\{f^n_{\alpha }\}_{n\in N}$ satisfying

• (1) $\forall n\in N, f^n_{\alpha }\le f^{n+1}_{\alpha }\odot f^{n+1}_{\alpha }\odot f^{n+1}_{\alpha }$,

• (2) $\forall a\in M^*(L^X)$, $\{f^{n}_{\alpha }(a)\}_{n\in N}$ is the $\alpha$-remote neighborhood family of $a$.

Proof

Necessary. If $D_{\alpha }(\delta )$ can be $\alpha$-P.Q. metriclizable, there is an $\alpha$-P.Q. metric, say $d_{\alpha }$. Let’s take $f^n_{\alpha }=P^{\frac{1}{3^n}}_{\alpha }$. By Theorem 8, it is clear that (1) and (2) hold.

Sufficiency. If $\{f^n_{\alpha }\}_{n\in N}$ satisfies (1) and (2). Clearly, $\{f^n_{\alpha }\}_{n\in N}$ is countable. So by Theorem 12, it is an $\alpha$-base of some $\alpha$-quasi uniform $\fancyscript{D}_{\alpha }$. Therefore $D_{\alpha }(\delta )=\eta _{\alpha }(\fancyscript{D}_{\alpha })$. This means $D_{\alpha }(\delta )$ is $\alpha$-P.Q. metriclizable.

Theorem 5.9

An $\alpha$-layer co-topology $D_{\alpha }(\delta )$ on $L^X$ can be $\alpha$-P.Q. metriclizable iff there is a Sequence of $\alpha$-symmetric remote neighborhood mappings $\{f^n_{\alpha }\}_{n\in N}$ satisfying

• (1) $\forall n\in N, f^n_{\alpha }\le f^{n+1}_{\alpha }\odot f^{n+1}_{\alpha }\odot f^{n+1}_{\alpha }$,

• (2) $\forall a\in M^*(L^X)$, $\{f^{n}_{\alpha }(a)\}_{n\in N}$ is the $\alpha$-remote neighborhood family of $a$.

Theorem 5.10

An $\alpha$-layer co-topology $D_{\alpha }(\delta )$ on $L^X$ can be $\alpha$-P.Q. metriclizable iff there is a Sequence of $\alpha$-symmetric remote neighborhood mappings $\{f^n_{\alpha }\}_{n\in N}$ satisfying

• (1) $\forall n\in N, f^n_{\alpha }\le (f^{n+1}_{\alpha })^{\diamond }\le f^{n+1}_{\alpha }$,

• (2) $\forall a\in M^*(L^X)$, $\{f^{n}_{\alpha }(a)\}_{n\in N}$ is the $\alpha$-remote neighborhood family of $a$.

Proof

Necessity. It is similar to that of Theorem 14.

Sufficiency. If $\{f^n_{\alpha }\}_{n\in N}$ satisfies (1) and (2). By Theorem 2, $\forall n\in N$, $f^n_{\alpha }\le (f^{n+1}_{\alpha })^{\diamond }\le (f^{n+1}_{\alpha })^{\nabla }\le f^{n+1}_{\alpha }$. By Theorem 4, $\forall n\in N$, $f^n_{\alpha }\le (f^{n+1}_{\alpha })^{\diamond }\le f^{n+1}_{\alpha }\odot f^{n+1}_{\alpha }$.

Therefore $\forall n\in N$,

$$\begin{aligned} f^n_{\alpha }\le f^{n+1}_{\alpha }\odot f^{n+1}_{\alpha }\le (f^{n+2}_{\alpha })^{\nabla }\odot (f^{n+2}_{\alpha })^{\nabla }\le (f^{n+2}_{\alpha })^{\nabla }. \end{aligned}$$

Again, by Theorem 4,

$$\begin{aligned} f^n_{\alpha }&\le (f^{n+2}_{\alpha })^{\nabla }\odot (f^{n+2}_{\alpha })^{\nabla } \le ((f^{n+4}_{\alpha })^{\nabla })^{\nabla }\odot ((f^{n+4}_{\alpha })^{\nabla })^{\nabla } \\&\le (f^{n+4}_{\alpha })^{\nabla }\odot (f^{n+4}_{\alpha })^{\nabla }\odot (f^{n+4}_{\alpha })^{\nabla } \odot (f^{n+4}_{\alpha })^{\nabla } \\&\le (f^{n+4}_{\alpha })^{\nabla }\odot (f^{n+4}_{\alpha })^{\nabla }\odot (f^{n+4}_{\alpha })^{\nabla }. \end{aligned}$$

$\forall n\in N$, $g_{\alpha }^n=(f^{4n-3}_{\alpha })^{\nabla }$. Then $\{g_{\alpha }^n\}_{n\in N}$ is a family of $\alpha$-symmetric remote neighborhood mappings. It is clear that $\{g^{n}_{\alpha }(a)\}_{n\in N}$ is the $\alpha$-remote neighborhood family of $a$.

Theorem 5.11

An $\alpha$-layer co-topology $D_{\alpha }(\delta )$ on $L^X$ can be $\alpha$-P.Q. metriclizable iff there is a Sequence of $\alpha$-remote neighborhood mappings $\{f^n_{\alpha }\}_{n\in N}$ satisfying

• (1) $\forall n\in N, f^n_{\alpha }\le (f^{n+1}_{\alpha })^{\nabla }\le f^{n+1}_{\alpha }$,

• (2) $\forall a\in M^*(L^X)$, $\{f^{n}_{\alpha }(a)\}_{n\in N}$ is the $\alpha$-remote neighborhood family of $a$.