In this section, we introduce the concept of diametric quantale, which extends the metric structure to quantale. The classical distance function is here being replaced by the notion of diameter satisfying certain properties. We give the respective characterizations in the terms of diametric quantales.

#### Definition 4.1

*The set* *C* *is called a conver of q quantale* *Q* *provided that* *C* ⊆ *Q* *and* ∨ *C* = 1.

#### Definition 4.2

*A diameter on a quantale* *Q* *is a mapping d* : *Q* ⟶ *R*^{+} ∪ {0} *(where* *R*^{+} ∪ {0} *is the set of nonegative reals) satisfying*:

*d*(0) = 0*;*

*a* ≤ *b* ⇒ *d*(*a*) ≤ *d*(*b*)*;*

*a*&*b* ≠ 0⇒ *d*(*a* ∨ *b*) ≤ *d*(*a*) + *d*(*b*)*;*

∀ *ϵ* > 0, *the set* $\begin{array}{}{U}_{\u03f5}^{d}\end{array}$ = {*x* ∈ *Q* ∣ *d*(*x*) < *ϵ*} *is a cover of* *Q*.

Sometimes we will drop condition (iv). In such a case, we will note that *d* is a prediameter.

#### Definition 4.3

(∗)

*A diameter is said to be a star diameter if for each* *x* ∈ *Q*, *and Y* ⊆ *Q* *such that* ∀ *y* ∈ *Y*, *x*&*y* ≠ 0, *then* *d*(*x* ∨ ⋁ *Y*) ≤ *d*(*x*) + *Sup*{*d*(*s*) + *d*(*t*) ∣ *s*, *t* ∈ *Y*, *s* ≠ *t*}.

(M)

*A diameter is said to be a metric diameter if for each* *x* ∈ *Q*, ∀ *ϵ* > 0, *there are* *y*, *z* ∈ *Q* *such that y*, *z* ≤ *x*, *and* *d*(*y*), *d*(*z*) < *ϵ*, *d*(*x*) < *d*(*y* ∨ *z*) + *ϵ*.

It is easy to check that (*M*) implies (∗). If *C* is a cover of a quantale *Q* and *x* ∈ *Q*, we denote *Cx* = ⋁{*c* ∈ *C* ∣ *c*&*x* ≠ 0}.

#### Definition 4.5

*Let* *Q* *be a quantale*, *d* *be a diameter of* *Q*, *μ*(*d*) = $\begin{array}{}\{{U}_{\u03f5}^{d}\mid \u03f5>0,\phantom{\rule{thinmathspace}{0ex}}\u03f5\in R\}\end{array}$ *is a family covers of* *Q*. *Let* $\begin{array}{}{\alpha}_{\u03f5}^{d}\end{array}$ *be the right adjoin of* $\begin{array}{}{U}_{\u03f5}^{d}\end{array}$. *If* ∀ *x* ∈ *Q*, *x* = $\begin{array}{}\underset{\u03f5>0}{\bigvee}{\alpha}_{\u03f5}^{d}x\end{array}$. *We say that the diameter* *d* *is compatible*.

If *x* ≥ $\begin{array}{}\underset{\u03f5>0}{\bigvee}{\alpha}_{\u03f5}^{d}x\end{array}$ *for* *x* ∈ *Q*, *we say that* *d* *is week compatile*.

*Let* *f* : *Q* ⟶ *P* *be a quantale epimorphism*, *d* *is prediameter of* *Q*. *We define* *d*(*y*) = *inf*{*d*(*x*) ∣ *y* ≤ *f*(*x*)}.

#### Theorem 4.6

*Let* *f* : *Q* ⟶ *P* *be a quantale epimorphism*, *d* *is a prediameter (diameter*, ∗-*diameter)*, *then* *d* *is a prediameter (diameter*, ∗-*diameter)*.

#### Proof

Firstly, we prove *d* is a prediameter of *P*.

By the definition of *d*, we have that *d*(0) = *inf*{*d*(*x*) ∣ 0 ≤ *f*(*x*)} = 0;

Let *a*, *b* ∈ *Q*, and *a* ≤ *b*, then *d*(*a*) = *inf*{*d*(*x*) ∣ *a* ≤ *f*(*x*)}, *d*(*b*) = *inf*{*d*(*y*) ∣ *b* ≤ *f*(*y*)}. Since *a* ≤ *b* ≤ *f*(*y*), then *d*(*a*) ≤ *d*(*b*).

Let *a*, *b* ∈ *P*, and *a*&*b* ≠ 0, then *d*(*a* ∨ *b*) = *inf*{*d*(*x*) ∣ *a* ∨ *b* ≤ *f*(*x*)}, *d*(*a*) = *inf*{*d*(*y*) ∣ *a* ≤ *f*(*y*)}, *d*(*b*) = *inf*{*d*(*z*) ∣ *b* ≤ *f*(*z*)}.

Since *inf*{*d*(*y*) ∣ *a* ≤ *f*(*y*)} + *inf*{*d*(*z*) ∣ *b* ≤ *f*(*z*)} = *inf*{*d*(*y*) + *d*(*z*) ∣ *a* ≤ *f*(*y*), *b* ≤ *f*(*z*)}, *d* is a prediameter of *P*, and *a* ≤ *f*(*y*), *b* ≤ *f*(*z*). 0 ≠ *a*&*b*, *a*&*b* ≤ *f*(*y*&*z*), then *f*(*y*&*z*) ≠ 0, and *y*&*z* ≠ 0. We have *d*(*y* ∨ *z*) ≤ *d*(*y*) + *d*(*z*), *a* ∨ *b* ≤ *f*(*y*) ∨ *f*(*z*) = *f*(*y* ∨ *z*). Thus *d*(*a* ∨ *b*) ≤ *d*(*a*) + *d*(*b*).

If *d* is a diameter of *Q*, we can prove that $\begin{array}{}{U}_{\u03f5}^{d}\end{array}$ = {*x* ∈ *Q* ∣ *d*(*x*) < *ϵ*} for *ϵ* > 0 is a cover of *Q*. Since *f*(1) = 1, then $\begin{array}{}f(\bigvee {U}_{\u03f5}^{d})=1,\end{array}$ and *d*(*y*) < *ϵ*, *d*(*f*(*y*)) < *ϵ*. Hence $\begin{array}{}1=f(\bigvee {U}_{\u03f5}^{d})=\bigvee f({U}_{\u03f5}^{d})\le \bigvee {U}_{\u03f5}^{\overline{d}},\end{array}$ that is $\begin{array}{}\bigvee {U}_{\u03f5}^{\overline{d}}=1.\end{array}$

Secondly, we can show that *d* is a ∗-diameter when *d* is a ∗-diameter of *P*.

Let *a* ∈ *P*, *B* ⊆ *P*, and ∀ *b* ∈ *B*, *a*&*b* ≠ 0, then *d*(*a*) = *inf*{*d*(*x*) ∣ *a* ≤ *f*(*x*)}, *d*(*b*) = *inf*{*d*(*y*) ∣ *b* ≤ *f*(*y*)}. By the definition of infimum, for all *ϵ* > 0, there exist *x*_{a},*x*_{b} ∈ *Q*, such that *d*(*x*_{a}) < *d*(*a*) + *ϵ*, *d*(*x*_{b}) < *d*(*b*) + *ϵ*, *a* ≤ *f*(*x*_{a}), *b* ≤ *f*(*x*_{b}), then 0 ≠ *a*&*b* ≤ $\begin{array}{}f({x}_{a}\vee (\underset{b\in B}{\bigvee}{x}_{b})),\end{array}$ Therefore *d*(*a* ∨ ⋁ *B*) ≤ $\begin{array}{}d({x}_{a}\vee (\underset{b\in B}{\bigvee}{x}_{b}))\end{array}$ ≤ *d*(*x*_{a})+ *Sup*{*d*(*x*_{s}) + *d*(*x*_{t}) ∣ *s* ≠ *t*, *s*, *t* ∈ *B*} < *d*(*a*) + *Sup*{*d*(*s*) + *d*(*t*) ∣ *s* ≠ *t*, *s*, *t* ∈ *B*} + 3*ϵ*. Thus *d* is a ∗-diameter of *P*. □

#### Theorem 4.7

*Let* *Q* *be a quantale*, *d* *is a diameter on* *Q*, *f* : *Q* ⟶ *P* *is a epimorphism*. *For all p* ∈ *P*, ∀ *ϵ* > 0, *we define* *Φ*_{ϵ}(*p*) = {*q* ∈ *Q* ∣ *q* ∈ $\begin{array}{}{U}_{\u03f5}^{d}\end{array}$, *f*(*q*)&*p* ≠ 0}, *ϕ*_{ϵ}(*p*) = ⋁ *Φ*_{ϵ}(*p*). *We have*:

*Let* *Q* *be a idempotent quantale*, *and* ∀ *q* ∈ $\begin{array}{}{U}_{\u03f5}^{d}\end{array}$, *f*(*q*) ≠ 0, *then* *ϕ*_{ϵ}(*f*(*q*)) ≥ *q;*

*ϕ*_{ϵ}(*A*) = $\begin{array}{}\underset{a\in A}{\bigvee}\end{array}$ *ϕ*_{ϵ}(*a*) *for all* *A* ⊆ *P;*

*If* *p* ≤ *f*(*q*), then *ϕ*_{ϵ}(*p*) ≤ $\begin{array}{}{U}_{\u03f5}^{d}\end{array}$ (*q*).

#### Proof

∀ *q* ∈ $\begin{array}{}{U}_{\u03f5}^{d}\end{array}$, *f*(*q*) ≠ 0, then *ϕ*_{ϵ}(*f*(*q*)) = ⋁ *Φ*_{ϵ}(*f*(*q*)) = ⋁ {*x* ∈ $\begin{array}{}{U}_{\u03f5}^{d}\end{array}$ ∣ *f*(*x*)&*f*(*q*) ≠ 0}. Since *Q* is a idempotent quantale, and *f*(*q*) ≠ 0, then *ϕ*_{ϵ}(*f*(*q*)) ≥ *q*.

∀ *A* ⊆ *P*, then

$$\begin{array}{}{\displaystyle {\varphi}_{\u03f5}(\bigvee A)=\bigvee {\mathit{\Phi}}_{\u03f5}(\vee A)=\bigvee \{q\in Q\mid f(q)\mathrm{\&}(\vee A)\ne 0\}}\\ {\displaystyle \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}=\bigvee \{q\in Q\mid (\underset{a\in A}{\bigvee}(f(q)\mathrm{\&}a))\ne 0\}}\\ {\displaystyle \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}=\bigvee \{q\in Q\mid \mathrm{\exists}\phantom{\rule{thinmathspace}{0ex}}a\in A,\phantom{\rule{thinmathspace}{0ex}}f(q)\mathrm{\&}a\ne 0\}}\\ {\displaystyle \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}=\bigvee \underset{a\in A}{\bigvee}\{q\in Q\mid f(q)\mathrm{\&}a\ne 0\}}\\ {\displaystyle \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}=\underset{a\in A}{\bigvee}\bigvee \{q\in Q\mid f(q)\mathrm{\&}a\ne 0\}}\\ {\displaystyle \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}=\underset{a\in A}{\bigvee}{\mathit{\Phi}}_{\u03f5}(a).}\end{array}$$

Since *ϕ*_{ϵ}(*p*) = ⋁ *Φ*_{ϵ}(*p*) = ⋁ {*q* ∈ $\begin{array}{}{U}_{\u03f5}^{d}\end{array}$ ∣ *f*(*q*)&*p* ≠ 0}, and $\begin{array}{}{U}_{\u03f5}^{d}\end{array}$ *q* = ⋁ {*x* ∈ *Q* ∣ *d*(*x*) < *ϵ*, *x*&*q* ≠ 0}. let *A* = {*q* ∈ *Q* ∣ *f*(*q*)&*p* ≠ 0}, *B* = {*x* ∈ *Q* ∣ *d*(*x*) < *ϵ*, *x*&*q* ≠ 0}. ∀ *q* ∈ *A*, and *q* ∈ $\begin{array}{}{U}_{\u03f5}^{d}\end{array}$, *f*(*q*)&*p* ≠ 0, then 0 ≠ *f*(*q*)&*p* ≤ *f*(*q*&*q*). We can see that *q*&*q* ≠ 0, thus *q* ∈ *B*, that is *ϕ*_{ϵ}(*p*) ≤ $\begin{array}{}{U}_{\u03f5}^{d}\end{array}$ *q*.

□

#### Theorem 4.8

*Let* *Q* *and* *P* *be quantales*, *f* : *Q* ⟶ *P* *be an epimorphism*. *d* *be induced by a compatible diameter* *d*, *then* *d* *is compatible*.

#### Proof

Let *d*(*u*) = *inf*{*d*(*a*) ∣ *u* ≤ *f*(*a*)} < *ϵ*, $\begin{array}{}u\mathrm{\&}f({\alpha}_{\u03f5}^{d})(x)\end{array}$ ≠ 0, then there exists *v* ∈ *Q*, such that *d*(*v*) < *ϵ*, *u* ≤ *f*(*v*). Since $\begin{array}{}0\ne u\mathrm{\&}f({\alpha}_{\u03f5}^{d})(x)\le f(v)\mathrm{\&}f({\alpha}_{\u03f5}^{d})(x)=f(v\mathrm{\&}{\alpha}_{\u03f5}^{d}(x)),\end{array}$ then $\begin{array}{}f(v\mathrm{\&}{\alpha}_{\u03f5}^{d}(x))\ne 0,\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\text{so that}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}v\mathrm{\&}{\alpha}_{\u03f5}^{d}(x)\ne 0.\end{array}$ Thus *v*&(⋁{*y* ∈ *Q* ∣ $\begin{array}{}{U}_{\u03f5}^{d}\end{array}$ *y* ≤ *x*}) ≠ 0, that is (⋁{*v*&*y* ∣ *y* ∈ *Q*, $\begin{array}{}{U}_{\u03f5}^{d}\end{array}$ *y* ≤ *x*}) ≠ 0, then ∃ *y*_{0} ∈ *Q*, $\begin{array}{}{U}_{\u03f5}^{d}\end{array}$ *y*_{0} ≤ *x*, *v*&*y*_{0} ≠ 0, therefore *v* ≤ *x*. Thus *u* ≤ *f*(*v*) ≤ *f*(*x*), $\begin{array}{}{U}_{\epsilon}^{\overline{d}}(f{\alpha}_{\u03f5}^{d}(x))\end{array}$ ≤ *f*(*x*), then $\begin{array}{}f{\alpha}_{\u03f5}^{d}(x)\le {\alpha}_{\u03f5}^{\overline{d}}(f(x)).\end{array}$ We have that ∀ *p* ∈ *P*, there exists *x* ∈ *Q*such that *p* = *f*(*x*), then *p* = *f*(*x*) ≤ $\begin{array}{}f(\underset{\u03f5>0}{\bigvee}{\alpha}_{\u03f5}^{d}x)=\underset{\u03f5>0}{\bigvee}f({\alpha}_{\u03f5}^{d}x)\le \underset{\u03f5>0}{\bigvee}{\alpha}_{\u03f5}^{\overline{d}}f(x)=\underset{\u03f5>0}{\bigvee}{\alpha}_{\u03f5}^{\overline{d}}p.\end{array}$ □

#### Definition 4.9

*Let* (*Q*, *d*), (*P*, *d*′) *be prediametric quantales*, *and* *f* : *Q* ⟶ *P* *be a quantale homomorphism*. *If* *d*′(*p*) < *ϵ* *for all* *p* ∈ *P*, *there exists an element* *q* ∈ *Q*, *such that* *d*(*q*) < *ϵ*, *p* ≤ *f*(*q*), *the quantale homomorphism* *f* *is called contractive*.

#### Theorem 4.10

*Let* (*Q*, *d*), (*P*, *d*′), (*K*, *d*″) *be prediametric quantales and* *f* *be a quantale homomorphism*. *Consider the following statements*:

*f* : (*Q*, *d*) ⟶ (*P*, *d*′) *is a contractive homomorphism;*

∀ *ϵ* > 0, *x* ∈ *Q*, *we have* $\begin{array}{}{U}_{\u03f5}^{{d}^{\prime}}\circ f(x)\le f({U}_{\u03f5}^{d}x);\end{array}$

∀ *ϵ* > 0, *x* ∈ *Q*, *we have* $\begin{array}{}f\circ {\alpha}_{\u03f5}^{d}(x)\le {\alpha}_{\u03f5}^{{d}^{\prime}}f(x).\end{array}$

That (1) ⟹ (2) ⟺ (3).

#### Proof

Firstly, we will prove the implication (1) ⟹ (2).

Let *p* ∈ $\begin{array}{}{U}_{\u03f5}^{{d}^{\prime}}\end{array}$ and *p*&*f*(*x*) ≠ 0. Since *f* is a contractive homomorphism, we have that there exists *q*′ ∈ *Q*, such that *d*(*q*′) < *ϵ*, *p* ≤ *f*(*q*′). But *p*&*f*(*x*) ≠ 0, then 0 ≠ *p*&*f*(*x*) ≤ *f*(*q*′)&*f*(*x*) = *f*(*q*′&*x*), which implies that *f*(*q*′&*x*) ≠ 0, *q*′&*x* ≠ 0. Thus

$$\begin{array}{}{\displaystyle {U}_{\u03f5}^{{d}^{\prime}}f(x)=\bigvee \{p\in P\mid {d}^{\prime}(p)<\u03f5\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}p\mathrm{\&}f(x)\ne 0\}}\\ {\displaystyle \phantom{\rule{2em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\le \{f(q)\in P\mid q\in Q,\phantom{\rule{thinmathspace}{0ex}}d(q)<\u03f5,\phantom{\rule{thinmathspace}{0ex}}q\mathrm{\&}x\ne 0\}}\\ {\displaystyle \phantom{\rule{2em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}=f(\bigvee \{q\in Q\mid d(q)<\u03f5,\phantom{\rule{thinmathspace}{0ex}}q\mathrm{\&}x\ne 0\})=f({U}_{\u03f5}^{d}(x)).}\end{array}$$

(2) ⟹ (3) Let *ε* > 0, *x* ∈ *Q*, then $\begin{array}{}f\circ {\alpha}_{\u03f5}^{d}(x)=f(\bigvee \{q\in Q\mid {U}_{\u03f5}^{d}q\le x\})=\bigvee \{f(q)\mid {U}_{\u03f5}^{d}q\le x,\phantom{\rule{thinmathspace}{0ex}}q\in Q\},\end{array}$ and $\begin{array}{}{\alpha}_{\u03f5}^{{d}^{\prime}}f(x)=\bigvee \{p\in P\mid {U}_{\u03f5}^{{d}^{\prime}}p\le f(x)\}.\end{array}$

Let *p* ∈ *P*, and *d*′(*p*) < *ϵ*, *p*&*f*(*q*) ≠ 0. B*y*(2), we have $\begin{array}{}{U}_{\u03f5}^{{d}^{\prime}}f(q)\le f({U}_{\u03f5}^{d}q),\end{array}$ then $\begin{array}{}p\le f({U}_{\u03f5}^{d}q)\end{array}$ = *f*(⋁{*y* ∈ *Q* ∣ *d*(*y*) < *ϵ*, *y*&*q* ≠ 0}) ≤ *f*(*x*). Thus ⋁{*p* ∈ *P* ∣ *d*′(*p*) < *ϵ*, *p*&*f*(*q*) ≠ 0} ≤ *f*(*x*), which implies $\begin{array}{}{U}_{\u03f5}^{{d}^{\prime}}\end{array}$ *f*(*q*) ≤ *f*(*x*), therefore*f*$\begin{array}{}({\alpha}_{\u03f5}^{d}(x))\le {\alpha}_{\u03f5}^{{d}^{\prime}}f(x).\end{array}$

(3) ⟹ (2) Since $\begin{array}{}{U}_{\u03f5}^{d}\u22a3{\alpha}_{\u03f5}^{d},\end{array}$ then $\begin{array}{}{U}_{\u03f5}^{{d}^{\prime}}f(x)\le f({U}_{\u03f5}^{d}x)\u27faf(x)\le {\alpha}_{\u03f5}^{d}f({U}_{\u03f5}^{{d}^{\prime}}(x)).\end{array}$ But *f*(*x*) ≤ $\begin{array}{}f({\alpha}_{\u03f5}^{d}{U}_{\u03f5}^{d}x)\end{array}$ ≤ $\begin{array}{}{\alpha}_{\u03f5}^{{d}^{\prime}}f({U}_{\u03f5}^{d}x),\end{array}$ which implies $\begin{array}{}{U}_{\u03f5}^{{d}^{\prime}}f(x)\le f({U}_{\u03f5}^{d}x).\end{array}$ □

#### Lemma 4.11

*Let* (*Q*, *d*) *be a diametric quantale*, *and* *d* *be a compatible diameter*, *β* : *Q* ⟶ 2 *is a quantale epimorphism*. *Let* *a*, *b* ∈ *Q*, *β*(*a*) = 1, *then there is a* *ϵ* > 0, *such that for all* *b* ∈ *Q* *with* *d*(*b*) < *ϵ* *either* *β* (*b*) = 0 *or* *b* ≤ *a*.

#### Proof

Since 1 = *β*(*a*) = $\begin{array}{}\beta (\underset{\u03f5>0}{\bigvee}{\alpha}_{\u03f5}^{d}(a))=\underset{\u03f5>0}{\bigvee}\beta ({\alpha}_{\u03f5}^{d}(a)),\end{array}$ then there exists *ϵ* > 0, such that $\begin{array}{}\beta ({\alpha}_{\u03f5}^{d}(a))=1.\end{array}$ Let *d*(*b*) < *ϵ*, and *b* ≰ *a*, then $\begin{array}{}b\mathrm{\&}{\alpha}_{\u03f5}^{d}\end{array}$(*a*) = 0. Otherwise, if $\begin{array}{}b\mathrm{\&}{\alpha}_{\u03f5}^{d}\end{array}$(*a*) ≠ 0, that is *b*&(⋁{*x* ∈ *Q* ∣ $\begin{array}{}{U}_{\u03f5}^{d}\end{array}$ *x* ≤ *a*}) = ⋁{*b*&*x* ∈ *Q* ∣ $\begin{array}{}{U}_{\u03f5}^{d}\end{array}$(*x*) ≤ *a*} ≠ 0, then there exists *x* ∈ *Q*, such that $\begin{array}{}{U}_{\u03f5}^{d}\end{array}$(*x*) ≤ *a*, *b*&*x* ≠ 0, then *b* ≤ *a*, but this contradicts the assumption that *b* ≰ *a*. Thus $\begin{array}{}b\mathrm{\&}{\alpha}_{\u03f5}^{d}\end{array}$(*a*) = 0, and 0 = $\begin{array}{}\beta (b\mathrm{\&}{\alpha}_{\u03f5}^{d}(a))=\beta (b)\mathrm{\&}\beta ({\alpha}_{\u03f5}^{d}(a)),\end{array}$ then *β*(*b*) = 0. Since *β*(*b*) ≠ 0, then *β*(*b*) = 1, that is $\begin{array}{}\beta (b)\mathrm{\&}\beta ({\alpha}_{\u03f5}^{d}(a))=\beta (b\mathrm{\&}{\alpha}_{\u03f5}^{d}(a))=1,\end{array}$ then $\begin{array}{}b\mathrm{\&}{\alpha}_{\u03f5}^{d}(a)\ne 0,\end{array}$ Therefore *b* ≤ *a*. □

Let (*Q*, *d*) be a diametric quantale, *d* be a compatible diameter. *Pt*(*Q*) denotes the collection of all points of *Q*.

Define: *ρ*_{d} : *Pt*(*Q*) × *Pt*(*Q*) ⟶ *R*^{+} ∪ {0}, *ρ*_{d}(*ξ*, *η*) = *inf*{*d*(*a*) ∣ *a* ∈ *Q*, *ξ*(*a*) = *η*(*a*) = 1}, for all *ξ*, *η* ∈ *Pt*(*Q*).

#### Theorem 4.12

*Let* (*Q*, *d*) *be a diameteric quantale*, *and* 1&1 ≠ 0, *d* *be a compatible diameter*, *then* *ρ*_{d} *is a metric on* *Pt*(*Q*) *and the induced topology coincides with the topology* *Σ*_{Q} = {*Σ*_{a} ∣ *a* ∈ *Q*} *of* *Pt*(*Q*), *Σ*_{a} = {*ξ* ∣ *ξ*(*a*) = 1}.

#### Proof

For every element *ξ*, *η* ∈ *Pt*(*Q*), it is easy to verify that *ρ*_{d}(*ξ*, *η*) ≥ 0.

Since ⋁ $\begin{array}{}{U}_{\u03f5}^{d}\end{array}$ = 1 for all *ϵ* > 0, then ∀ *ξ* ∈ *Pt*(*Q*), *ρ*_{d}(*ξ*,*ξ*) = *inf*{*d*(*a*) ∣ *ξ*(*a*) = 1}, hence *ρ*_{d}(*ξ*,*ξ*) = 0. For all *ξ*, *η* ∈ *Pt*(*Q*), *ϵ* > 0, we have *ρ*_{d}(*ξ*, *η*) = *inf*{*d*(*a*) ∣ *ξ*(*a*) = *η*(*a*) = 1} = 0. If *ξ*(*a*) = 1, there is an element *b* ∈ *Q*, such that *d*(*b*) < *ϵ*, and *ξ*(*b*) = *η*(*b*) = 1. Thus *b* ≤ *a*, therefore *η* (*a*) = 1. The symmetry is obvious. If *η*(*a*) = 1, then *ξ*(*a*) = 1. Thus *ξ* = *η*.

For all *ξ*, *η* ∈ *Pt*(*Q*), *ρ*_{d}(*ξ*, *η*) = *ρ*_{d}(*η*,*ξ*) is obvious.

For all *ξ*, *η*, *δ* ∈ *Pt*(*Q*), in the following, we will prove that *ρ*_{d}(*ξ*, *η*) ≤ *ρ*_{d}(*ξ*, *σ*) + *ρ*_{d}(*σ*, *η*).

Let *A* = {*a* ∈ *Q* ∣ *ξ*(*a*) = *η*(*a*) = 1}, *B* = {b ∈ *Q* ∣ *ξ*(*b*) = *δ*(*b*) = 1}, *C* = {*c* ∈ *Q* ∣ *δ*(*c*) = *η*(*c*) = 1},

∀ *b* ∈ *B*, ∀ *c* ∈ *C*, *ξ*(*b* ∨ *c*) = *ξ*(*b*) ∨ *ξ*(*c*) = 1 ∨ *ξ*(*c*) = 1, *η*(*b* ∨ *c*) = *η*(*b*) ∨ *η*(*c*) = 1 ∨ *η*(*b*) = 1, then *b* ∨ *c* ∈ *A*. Suppose *b*&*c* = 0, then 0 = *ξ*(*b*&*c*) = *ξ*(*b*)& *ξ*(*c*) = 1&1, but this contradicts the fact 1&1 ≠ 0. From Definition 4.2(iii), we have *d*(*b* ∨ *c*) ≤ *d*(*b*)+ *d*(*c*), then *inf*{*d*(*a*) ∣ *ξ*(*a*) = *η*(*a*) = 1} ≤ *inf*{*d*(*b*) ∣ *ξ*(*b*) = *δ*(*b*) = 1} +*inf*{*d*(*c*) ∣ *δ*(*c*) = *η*(*c*) = 1} = *inf*{*d*(*b*) + *d*(*c*) ∣ *ξ*(*b*) = *δ*(*b*) = 1, *δ*(*c*) = *η*(*c*) = 1}, that is *ρ*_{d}(*ξ*, *η*) ≤ *ρ*_{d}(*ξ*, *δ* ) + *ρ*_{d}( *δ*, *η*). Thus *ρ*_{d} is a metric on *Pt*(*Q*). Therefore (*Pt*(*Q*),*ρ*_{d}) is a metric space.

In the following, we will prove the induced topology (*Pt*(*Q*), *τ*_{ρd}) as *ρ*_{d} coincides with the topology *Σ*_{Q} = {*Σ*_{a} ∣ *a* ∈ *Q*} of *Pt*(*Q*), *Σ*_{a} = {*ξ* ∣ *ξ*(*a*) = 1}.

∀ *a* ∈ *A*, ∀ *ξ* ∈ *Σ*_{a}, then *ξ*(*a*) = 1. By Lemma 4.11, there exists an *ϵ* > 0. If *ρ*_{d}(*ξ*, *η*) < *ϵ*, we have an element *b* ∈ *Q*, such that *ξ*(*b*) = *η*(*b*) = 1, and *d*(*b*) ≤ *ϵ*, then *b* ≤ *a*. Thus *η*(*a*) = 1, that is *η* ∈ *Σ*_{a}, *ξ* ∈ {*η* ∈ *Pt*(*Q*) ∣ *ρ*_{d}(*ξ*, *η*) < *ϵ*} ⊆ *Σ*_{a}. Hence *Σ*_{Q} ⊆ *τ*_{ρd}.

Conversely, for all *ξ* ∈ *Pt*(*Q*), ∀ *ϵ* > 0, we put *a* = ∨ {*b* ∈ *Q* ∣ *d*(*b*) < *ϵ*, *ξ*(*b*) = 1}. For all *η* ∈ *Σ*_{a}, there exists *b* ∈ *Q*, such that *d*(*b*) < *ϵ*, and *ξ*(*b*) = *η*(*b*) = 1. Therefore *ρ*_{d}(*ξ*, *η*) ≤ *d*(*b*) < *ϵ*. If *ρ*_{d}(*ξ*, *η*) < *ϵ*, then there exists an element *b* ∈ *B*, such that *η*(*b*) = *ξ*(*b*) = 1, *d*(*b*) < *ϵ*. Thus *b* ≤ *a*, *η*(*a*) = 1. Hence *Σ*_{a} = {*η* ∣ *ρ*_{d}(*ξ*, *η*) < *ϵ*} $\begin{array}{}{B}_{\u03f5}^{{\rho}_{d}}(\xi ).\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\text{Since}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}U=\bigcup \{{B}_{\u03f5}^{{\rho}_{d}}(u)\mid u\in U\}=\underset{u\in U}{\bigvee}{\mathrm{\Sigma}}_{{a}_{u}}={\mathrm{\Sigma}}_{\underset{u\in U}{\bigvee}{a}_{u}}\end{array}$ for all *U* ∈ *τ*_{ρd}, then *U* ∈ *Σ*_{Q}, therefore *τ*_{ρd} ⊆ *Σ*_{Q}. □

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