Uniform BL-algebras

Abstract

In this paper, we define the notion of uniform BL-algebras and derive some conditions under which the operations of BL-algebras are uniformly continuous. Also, some properties of uniform topology are discussed. Finally, we use some types of congruence relations to construct some uniformities and analyze the relationship between these uniformities.

1 Introduction

The notion of uniform space was first introduced by Weil (1938) as a tool which in contrast with the metrics was used for studying topological spaces with no countability assumptions. It has also been investigated by other mathematicians, yet none was able to develop a desirable theory. However, the systematic theory of uniform spaces was first presented by Bourbaki (1948). Bourbaki insisted that uniformity is an independent theory which is linked to the theory of topological spaces. Thus, the concept of a uniform space can be considered either as axiomatizations of some geometric notions which are close to, but quite independent of the concept of a topological space or as the tools which are selected for studying a topological space.

The theory of uniform spaces shows striking analogies with the theory of metric spaces contrast to the realm of its applicability which is much broader. For example, on every topological group, there are three natural uniformities which are useful in application of topological groups.

BL-algebras were also proposed by Hájek (1998). In order to study the basic logic framework of fuzzy set theory, he carried out several experiments on BL-algebras which were based on both logical and algebraic standpoints. Likewise, the study of BL-algebras endowed with a topology has experienced a tremendous growth over the recent years. Moreover Borzooei et al. (2011) defined the (semi) topological BL-algebras and introduced the least conditions in which all operation were continuous. Borzooei et al. (2012a) also studied metrizability on (semi) topological BL-algebras.

In the recent years uniformity has been studied on lattices and algebraic structures associated with logic. See, for example, Haveshki et al. (2007) and Weber (1991, 1993).

The main purpose of the present paper is to introduce uniform BL-algebras and study the uniform continuity of operations in BL-algebras. This topic is introduced and discussed thoroughly in Sect. 3. A topology induced by a uniformity is called uniform topology. The relationships between the uniform topologies on BL-algebras will be discussed in Sect. 4. Finally, in the last Section, with the help of ideals and filters, some uniformities are introduced and their properties are studied.

2 Preliminary

BL-algebras

A BL-algebra is an algebra $(A,\wedge ,\vee ,\odot ,\rightarrow ,0,1)$ of type (2, 2, 2, 2, 0, 0) such that $(A,\wedge ,\vee ,0,1)$ is a bounded lattice, $(A,\odot ,1)$ is a commutative monoid and for any $a,b,c\in A:$

$(BL_1)$ :

$c \le a \rightarrow b \Leftrightarrow a \odot c \le b$,

$(BL_2)$ :

$a \wedge b = a \odot (a \rightarrow b)$,

$(BL_3)$ :

$(a \rightarrow b) \vee (b \rightarrow a) = 1$.

Let A be a BL-algebra. We define $a' = a \rightarrow 0$ and denote $(a')'$ by ${a''}$. The map $c:A \rightarrow A$ defined by $c(a) = a'$, for any $a \in A$, is called the negation map. Also, we assume ${a^0} = 1$ and ${a^n} = {a^{n - 1}} \odot a$ for all natural numbers n. The following properties hold in A:

$(\mathcal {BL}_1)$ :

$x \odot y \le x,y$ and $x \odot 0 = 0,$

$(\mathcal {BL}_2)$ :

$x \le y$ implies $x \odot z \le y \odot z,$

$(\mathcal {BL}_3)$ :

$x \le y$ iff $x \rightarrow y = 1,$

$(\mathcal {BL}_4)$ :

$1 \rightarrow x = x$, $x \rightarrow x = 1,$$x \rightarrow 1 = 1$ and $1 \odot x = x,$

$(\mathcal {BL}_5)$ :

$y \le x \rightarrow y,$

$(\mathcal {BL}_6)$ :

$x \rightarrow (y \rightarrow z) = (x \odot y) \rightarrow z = y \rightarrow (x \rightarrow z),$

$(\mathcal {BL}_7)$ :

$1' = 0$ and $0' = 1,$

$(\mathcal {BL}_8)$ :

$x' = 1 \Leftrightarrow x = 0,$

$(\mathcal {BL}_9)$ :

if $x \le y,$ then $y \rightarrow z\le x \rightarrow z$ and $z \rightarrow x \le z \rightarrow y,$

$(\mathcal {BL}_{10})$ :

$(x \wedge y)'' = x'' \wedge y''$ and $(x \vee y)'' = x'' \vee y'',$

$(\mathcal {BL}_{11})$ :

$(x \odot y)'' = x'' \odot y''$ and $(x \rightarrow y)'' = x'' \rightarrow y'',$

$(\mathcal {BL}_{12})$ :

$x''' = x',$

$(\mathcal {BL}_{13})$ :

$x \odot x' = 0,$$x \odot y = 0$ iff $x \le y',$

$(\mathcal {BL}_{14})$ :

$x \rightarrow y' = y \rightarrow x' = (x \odot y)' = x'' \rightarrow y',$

$(\mathcal {BL}_{15})$ :

$(x \rightarrow y) \odot (y \rightarrow z) \le (x \rightarrow z),$

$(\mathcal {BL}_{16})$ :

$(x \rightarrow y) \odot (a \rightarrow b) \le (x * a) \rightarrow (y * b),$ where $* \in \left\{ { \wedge , \vee , \odot } \right\} ,$

$(\mathcal {BL}_{17})$ :

$(x \rightarrow a) \odot (b \rightarrow y) \le (a \rightarrow b) \rightarrow (x \rightarrow y),$

$(\mathcal {BL}_{18})$ :

$x \rightarrow y \le (a \rightarrow x) \rightarrow (a \rightarrow y),$

$(\mathcal {BL}_{19})$ :

$a\vee b=((a\rightarrow b)\rightarrow b)\wedge ((b\rightarrow a)\rightarrow a).$

Let A be a BL-algebra. Then:

• (i) a filter of A is a nonempty subset F of A such that for all $a, b \in A$,

  $({F_1})$ :

  $a,b \in F$ implies $a \odot b \in F,$

  $({F_2})$ :

  $a \in F$ and $a \le b$ imply $b \in F$ (Hájek 1998).

• (ii) an ideal of A is a nonempty subset I of A such that for all $a, b \in A$,

  $({I_1})$ :

  if $a, b \in I,$ then $a\oslash b \in I;$ where $a \oslash b:={a'} \rightarrow b,$

  $({I_2})$ :

  if $a \le b$ and $b \in I,$ then $a \in I$ (Lele and Nganou 2013).

Proposition 2.1

Hájek (1998) and Lele and Nganou (2013) Let A be a BL-algebra. Then:

• (i) a subset F of A is a filter iff $1\in F$ and

  $$\begin{aligned} (x\in F,\ x\rightarrow y\in F)\Rightarrow y\in F, \end{aligned}$$

• (ii) a set I containing 0 of A is an ideal iff for every $x,y \in A,$$x' \odot y \in I$ and $x \in I$ imply $y \in I.$

Definition 2.2

Lele and Nganou (2013) Let $(A, \wedge , \vee , \odot , \rightarrow , 0,1)$ be a BL-algebra and X any subset of A. The set of complement elements (with respect to X) is denoted by N(X) and is defined by $N(X) = \{ x \in A: x' \in X\}$.

Proposition 2.3

Lele and Nganou (2013) Let F be a filter, I an ideal of A and $I'=\{x':x\in I\}.$ Then:

$({\mathcal {N}}_1)$ :

N(I) is a filter and $I' \subseteq N(I).$

$({\mathcal {N}}_2)$ :

N(F) is the ideal generated by ${F'}.$

$({\mathcal {N}}_3)$ :

$I = N(N(I)).$

$({\mathcal {N}}_4)$ :

$F \subseteq N(N(F)).$

$({\mathcal {N}}_5)$ :

$N(F) = N(N(N(F))).$

A congruence relation on BL-algebra A is an equivalence relation $\equiv$ on A such that if $x\equiv y$ and $u\equiv v$ and $*\in \{\wedge ,\vee ,\odot ,\rightarrow \},$ then $x * u\equiv y* v.$

Proposition 2.4

Let F be a filter and I an ideal of a BL-algebra A. Then the following relations are congruence relations on A.

$$\begin{aligned} x\mathop \equiv \limits ^ {F}y \iff x \rightarrow y \in F \,and \,\, y \rightarrow x \in F,\\ x \mathop \equiv \limits ^{I}y\Longleftrightarrow x' \odot y \in I\ and\ \ y' \odot x \in I. \end{aligned}$$

Let A be a BL-algebra, F a filter and I an ideal in it. Then we assume that:

• (i) $x/F=\{y\in A: x\mathop \equiv \limits ^{F} y\}$ and $A/F=\{x/F:x\in A\},$

• (ii) $x/I=\{y\in A: x\mathop \equiv \limits ^{I} y\}$ and $A/I=\{x/I:x\in A\},$

• (iii) A / F and A / I,  both, are quotient BL-algebras with the following operations:

  $$\begin{aligned} x/F*y/F=(x*y)/F\ \mathrm{and}\ x/I*y/I=(x*y)/I, \end{aligned}$$

where $*\in \{\wedge ,\vee ,\odot ,\rightarrow \}.$

Definition 2.5

Borzooei et al. (2011) Let $\tau$ be a topology on BL-algebra A and $* \in \left\{ { \wedge , \vee , \odot , \rightarrow } \right\} .$ Then $*$ is:

• (i) continuous in second (first) variable if for each $a \in A,$ the map ${l_a}:A \rightarrow A$ (${r_a}:A \rightarrow A$) is defined by $x \rightarrow a * x$ ($x \rightarrow x * a$), is continuous, or equivalently, for any $x \in A,$ and any open neighborhood U of $a * x$ ($x * a$), there exists an open neighborhood V of x such that $a * V \subseteq U$ ($V * a \subseteq U$),

• (ii) continuous in each variable separately if it is continuous in first and second variable. Moreover, $(A, \tau )$ is a semitopological BL-algebra if the operations $\wedge , \vee , \odot , \rightarrow$ are continuous in each variable separately,

• (iii) continuous if for any $x,y \in A$ and any open neighborhood W of $x*y,$ there exist two open neighborhoods U and V of x and y,  respectively, such that $U*V \subseteq W.$ The pair $(A, \tau )$ is a topological BL-algebra if the operations $\wedge , \vee , \odot , \rightarrow$ are continuous.

Proposition 2.6

Borzooei et al. (2011) Let A be a BL-algebra and $\tau$ a topology on A. Then $(A,\tau )$ is a topological BL-algebra if and only if $\odot$ and $\rightarrow$ are continuous.

Definition 2.7

Borzooei et al. (2012a) Let $\tau$ be a topology on BL-algebra A,  $W \subseteq A$ and $W^n=\{x^n:x\in W\}.$ Then:

• (i) W is an n-quasifilter if $x \in {W^n}$ and $x \le y$ imply $y \in {W^n},$

• (ii) W is a quasifilter if, for each $n\ge 1,$W is an n-quasifilter,

• (iii) W is a quasifilter neighborhood of 1 if W is a neighborhood of 1 which is also a quasifilter,

• (iv) ${\mathcal {W}}$ is a QN-fundamental system if ${\mathcal {W}}$ is a fundamental system of open neighborhoods of 1 which all are quasifilters.

Definition 2.8

Let A be a BL-algebra and ${\tau }$ be a topology on it. Then $(A, {\tau })$ satisfies the open condition if for any filter F,  the canonical epimorphism ${\pi _F}:A \rightarrow \frac{A}{F}$ is open.

Definition 2.9

Borzooei et al. (2012b) Let A be a BL-algebra. We call a filter satisfies the maximum condition if for each $x \in A, \, \, \frac{x}{F}$ has a maximum. Also we call A satisfies the maximum condition if for any filter F,  F satisfies the maximum condition.

Topological Spaces

A topological space $(X, \tau )$ is said to be:

• (i) a ${T_0}$-space if for each $x \ne y \in X$, there is at least one in an open neighborhood excluding the other,

• (ii) a ${T_1}$-space if for each $x \ne y \in X$, each has an open neighborhood not containing the other,

• (iii) a ${T_2}$-space (or a Hausdorff space) if for each $x \ne y \in X,$ there two disjoint open neighborhoods U, V of x and y,  respectively,

• (iv) a regular space if for each $x \in U \in \tau ,$ there exists an open set H such that $x \in H \subseteq {\overline{H}} \subseteq U.$ A regular ${T_1}$-space is called a ${T_3}$-space,

• (v) a normal space if for each closed set S and each open set U contains S,  there is an open set H such that $S \subseteq H \subseteq {\overline{H}} \subseteq U,$

• (vi) a completely regular space if for every closed set F in X and for each $x \in X$ not belonging to F,  there exists a continuous function $f:X \rightarrow [0,1],$ such that $f(x)=0$ and $f(F)=1,$

• (vii) a Tychonoff space is a ${T_1}$-completely regular space Husain (1966).

Theorem 2.10

Engelking (1989) (i) Every $T_1$ second countable regular space is a normal space. (ii) Every $T_1$ normal space is a Tychonoff space.

Uniform Spaces

Let X be a nonempty set. A uniformity on X is a nonempty family ${\mathcal {U}}$ of subsets of $X\times X$ such that:

$(U_1)$ :

$\bigtriangleup =\{(x,x): x\in X\}\subseteq U$ for each $U\in {\mathcal {U}},$

$(U_2)$ :

if $U\in {\mathcal {U}},$ then $U^{-1}=\{(x,y)\in X\times X: (y,x)\in U\}$ belongs to ${\mathcal {U}},$

$(U_3)$ :

if $U\in {\mathcal {U}},$ then $V\circ V\subseteq U ,$ for some $V\in {\mathcal {U}},$ where

$$\begin{aligned} \, \, V\circ V=\{(x,y)\in X\times X:\exists z\in X\ \textit{s.t.}\ (x,z), (z,y)\in V\}, \end{aligned}$$

$(U_4)$ :

$U , V\in {\mathcal {U}},$ then $U \cap V\in {\mathcal {U}},$

$(U_5)$ :

$U\in {\mathcal {U}}, U \subseteq V,$ then $V\in {\mathcal {U}}.$

The pair $(X,{\mathcal {U}})$ is called a uniform space.

Let $(X,{\mathcal {U}})$ be a uniform space. $U\in {\mathcal {U}}$ is said to be symmetric if $U=U^{-1}.$ A subfamily ${\mathcal {B}}$ of ${\mathcal {U}}$ is called a base for ${\mathcal {U}}$ if each member of ${\mathcal {U}}$ contains a member of ${\mathcal {B}} .$ A subfamily ${\mathcal {S}}$ of ${\mathcal {U}}$ is called a subbase for ${\mathcal {U}}$ if the collection of all finite intersections of members of ${\mathcal {S}}$ is a base for ${\mathcal {U}}$ (Engelking 1989).

Lemma 2.11

Engelking (1989) A nonempty family ${\mathcal {B}}$ of subsets $X\times X$ is a base for the uniformity

$$\begin{aligned} {\mathcal {U}}=\{U\subseteq X\times X: \exists B \in {\mathcal {B}}\ \textit{s.t.}\ B\subseteq U \} \end{aligned}$$

if and only if

(B1) :

$\bigtriangleup =\{(x,x): x\in X\}\subseteq U$ for each $U\in {\mathcal {B}},$

(B2) :

if U belongs to ${\mathcal {B}} ,$ then $U^{-1}$ contains a member of ${\mathcal {B}},$

(B3) :

if U belongs to ${\mathcal {B}},$ then $V\circ V\subseteq U,$ for some V belonging to ${\mathcal {B}},$

(B4) :

if U and V are in ${\mathcal {B}},$ then $W\subseteq U \cap V,$ for some W belonging to ${\mathcal {B}}.$

Lemma 2.12

Engelking (1989) Let X be a set and ${\mathcal {S}} \subset {\mathcal {P}}(X \times X)$ be a family such that for every $U \in {\mathcal {S}}$ the following conditions hold.

$({S_1})$ :

$\Delta \subseteq U.$

$({S_2})$ :

${U^{ - 1}}$ contains a member of ${\mathcal {S}}.$

$({S_3})$ :

There exists $V \in {\mathcal {S}}$ such that $V\circ V \subseteq U.$

Then there exists a unique uniformity ${\mathcal {U}}$ for which ${\mathcal {S}}$ is a subbase.

Suppose that $(X,{\mathcal {U}})$ and $(Y,\mathcal {V)}$ are uniform spaces. Then the product of $(X,{\mathcal {U}})$ and $(Y,\mathcal {V)}$ is a uniform space $(Z,{\mathcal {W}})$ with the underlying set $Z=X\times Y$ and the uniformity ${\mathcal {W}}$ on Z whose base consists of the sets

$$\begin{aligned} W_{U,V}= & {} \{((x,y),(x',y'))\in Z\times Z: \\&\quad (x,x')\in U, (y,y')\in V\}, \end{aligned}$$

where $U\in {\mathcal {U}}$ and $V\in {\mathcal {V}}.$ The uniformity ${\mathcal {W}}$ is called the product of ${\mathcal {U}}$ and ${\mathcal {V}}$ and is written as ${\mathcal {W}}={\mathcal {U}}\times {\mathcal {V}} .$

Definition 2.13

Engelking (1989) Let $(A, {\mathcal {U}})$ be a uniform space. The set

$$\begin{aligned} \tau = \left\{ {G \subseteq A\left| { \, \forall x \in G \, \, \, \exists U \in {\mathcal {U}} \,\,\textit{s.t.}\ \, \,U[x] \subseteq G} \right. } \right\} , \end{aligned}$$

where $U[x]: = \left\{ {y \in A\left| {(x,y) \in U} \right. } \right\} ,$ is a topology on A which is called the uniform topology onAinduced by${\mathcal {U}}.$

Definition 2.14

Engelking (1989) Let $f:(X,{\mathcal {U}})\rightarrow (Y,{\mathcal {V}})$ be a map between uniform spaces. The map f is uniformly continuous if for each $V\in {\mathcal {V}}$ there exists $U\in {\mathcal {U}}$ such that $(f(x),f(y))\in V$ for all $(x,y)\in U,$ that is, $(f\times f)(U)\subseteq V.$ We denote $f\times f$ by $f^{(2)}.$

Theorem 2.15

Engelking (1989) Let $(X,\tau )$ be a $T_1$ topological space. Then $\tau$ is uniform topology iff $(X,\tau )$ is Tychonoff space.

3 Uniform BL-algebras

Definition 3.1

Let A be a BL-algebra and ${\mathcal {U}}$ be a uniformity on A and $\ * \in \{ \wedge , \vee ,\odot , \rightarrow \}.$ Then $* :(A \times A,{\mathcal {U}}\times {\mathcal {U}}) \rightarrow (A,{\mathcal {U}})$ is uniformly continuous if for any $W \in {\mathcal {U}},$ there exist $U,V \in {\mathcal {U}}$ such that $U * V \subseteq W$ or equivalently, for any $(x,x') \in U$ and $(y,y') \in V$, $(x * y, x' * y') \in W$. The pair $(A,{\mathcal {U}})$ is called uniform BL-algebra if each $\ * \in \{ \wedge , \vee ,\odot , \rightarrow \}$ is uniformly continuous.

Example 3.2

Let $\wedge ,\vee ,\odot$ and $\rightarrow$ on the real unit interval $A=[0,1]$ be defined as follows:

$$\begin{aligned} x\wedge y= & {} x \odot y = \min \{ x,y\},\ \ x\vee y=\max \{x,y\}, \\x \rightarrow y= & {} \left\{ \begin{array}{l} 1 \quad x \le y, \\ y \quad \mathrm{otherwise}. \\ \end{array} \right. \end{aligned}$$

Then A is a BL-algebra. Let $\varepsilon > 0$ and

$$\begin{aligned} {U_\varepsilon } = \{ (x,y) \in A \times A:\left| {x - y} \right| < \varepsilon \}. \end{aligned}$$

Then the set $\{U_\varepsilon :\varepsilon >0\}$ is a base for a uniformity ${\mathcal {U}}$ on A. It is easy to see that $\odot$ is uniformly continuous and $\rightarrow$ is not continuous in (x, 0),  for every $x\in A.$

Proposition 3.3

Let ${\mathcal {U}}$ be a uniformity on a B-algebra A. If $\odot$ and $\rightarrow$ are uniformly continuous, then $\wedge$ is uniformly continuous, too.

Proof

Suppose that $U \in {\mathcal {U}}$. Since $\odot$ and $\rightarrow$ are uniformly continuous, there exist ${V_1},V_2,{W_1}$ and $W_2$ in ${\mathcal {U}}$ such that ${V_1} \odot {W_1} \subseteq U$ and ${V_2} \rightarrow {W_2} \subseteq {W_1}.$ Hence we have

$$\begin{aligned} {V_1} \odot ({V_2} \rightarrow {W_2}) \subseteq {V_1} \odot {W_1} \subseteq U. \end{aligned}$$

If $\eta = {V_1} \cap {V_2} \in {\mathcal {U}},$ then

$$\begin{aligned} \eta \wedge {W_2} \subseteq \eta \odot (\eta \rightarrow {W_2}) \subseteq {V_1} \odot ({V_2} \rightarrow {W_2}) \subseteq U. \end{aligned}$$

$\square$

Proposition 3.4

If ${\mathcal {U}}$ is a uniformity on a B-algebra A and $\wedge , \rightarrow$ are uniformly continuous, then $\vee$ is uniformly continuous.

Proof

Suppose $W \in {\mathcal {U}}.$ Since $\wedge$ is uniformly continuous, there exist ${V_1},{K_1} \in {\mathcal {U}}$ such that ${V_1} \wedge {K_1} \subseteq W$. The operation $\rightarrow$ is uniformly continuous, hence there exist H, M, L and N in ${\mathcal {U}}$ such that $H \rightarrow M \subseteq {V_1}$ and $L \rightarrow N \subseteq H.$ So

$$\begin{aligned} (L \rightarrow N) \rightarrow M \subseteq H \rightarrow M \subseteq {V_1}. \end{aligned}$$

Let $\eta = M \cap N \in {\mathcal {U}}$. Then $(L \rightarrow \eta ) \rightarrow \eta \subseteq (L \rightarrow N) \rightarrow M \subseteq H \rightarrow M \subseteq {V_1}$. Similarly there are ${L_1},{\eta _1} \in {\mathcal {U}}$ such that

$$\begin{aligned} ({L_1} \rightarrow {\eta _1}) \rightarrow {\eta _1} \subseteq {K_1}. \end{aligned}$$

Consider $K = L \cap {\eta _1} \in {\mathcal {U}}$ and $V = {L_1} \cap \eta \in {\mathcal {U}}.$ Then

$$\begin{aligned}&((K \rightarrow V) \rightarrow V) \wedge ((V \rightarrow K) \rightarrow K) \subseteq ((L \rightarrow \eta ) \rightarrow \eta ) \\&\quad \wedge (({L_1} \rightarrow {\eta _1}) \rightarrow {\eta _1}). \end{aligned}$$

We claim that

$$\begin{aligned} V \vee K \subseteq ((V \rightarrow K) \rightarrow K) \wedge ((K \rightarrow V) \rightarrow V). \end{aligned}$$

For every $((x,y) \vee ({x_1},{y_1})) \in V \vee K$ we have

$$\begin{aligned} (x \vee {x_1},y \vee {y_1})= & {} (((x \rightarrow {x_1}) \rightarrow {x_1}) \wedge (({x_1} \rightarrow x) \rightarrow x),\\&\quad ((y \rightarrow {y_1}) \rightarrow {y_1}) \wedge (({y_1} \rightarrow y) \rightarrow y))\\= & {} ((x \rightarrow {x_1}) \rightarrow {x_1},(y \rightarrow {y_1}) \rightarrow {y_1}) \\&\quad \wedge (({x_1} \rightarrow x) \rightarrow x,({y_1} \rightarrow y) \rightarrow y))\\= & {} ((x \rightarrow {x_1},y \rightarrow {y_1}) \rightarrow ({x_1},{y_1})) \\&\quad \wedge (({x_1} \rightarrow x,{y_1} \rightarrow y) \rightarrow (x,y))\quad \quad \\= & {} (((x,y) \rightarrow ({x_1},{y_1})) \rightarrow ({x_1},{y_1})) \\&\quad \wedge ((({x_1},{y_1}) \rightarrow (x,y)) \rightarrow (x,y)), \end{aligned}$$

thus

$$\begin{aligned} (x \vee {x_1},y \vee {y_1}) \in ((V \rightarrow K) \rightarrow K) \wedge ((K \rightarrow V) \rightarrow V). \end{aligned}$$

Therefore

$$\begin{aligned} V \vee K\subseteq & {} ((V \rightarrow K) \rightarrow K) \wedge ((K \rightarrow V) \rightarrow V)\\ {}\subseteq & {} ((L \rightarrow \eta ) \rightarrow \eta ) \wedge (({L_1} \rightarrow {\eta _1}) \rightarrow {\eta _1})\\ {}\subseteq & {} {V_1} \wedge {K_1} \subseteq W. \end{aligned}$$

$\square$

Corollary 3.5

Let ${\mathcal {U}}$ be a uniformity on a B-algebra A. If $\odot$ and $\rightarrow$ are uniformly continuous, then $(A,{\mathcal {U}})$ is a uniform BL-algebra.

Proof

By Proposition 3.3, $\wedge$ is uniformly continuous and by Proposition 3.4, $\vee$ is uniformly continuous. $\square$

Proposition 3.6

Let $c:A \rightarrow A$ and ${c^{(2)}} :A \times A \rightarrow A \times A$ be the negation maps. Then:

• (i) the negation map c is one-to-one iff c is onto. Moreover, the inverse of c is c,  i.e. $c^{-1}=c,$

• (ii) if c is one-to-one and uniformly continuous, then ${c^{(2)}}({\mathcal {U}}) = {\mathcal {U}},$ where ${\mathcal {U}}$ is a uniformity on A.

Proof

(i) Let c be onto. Then for any $y \in X$ there exists $x \in X$ such that $y = c(x).$ By $(\mathcal {BL}_{12}),$ we have

$$\begin{aligned} cc(y) = ccc(x) = c(x) \Rightarrow cc(y) = y. \end{aligned}$$

So $c = {c^{ - 1}}$ i.e c is one-to-one.

Conversely, let c be one-to-one. Then for any $y \in X,$${c^{ - 1}}(c(y)) = y$. By $(\mathcal {BL}_{12}),$

$$\begin{aligned} c(c(c(y)))= & {} c(y) \Rightarrow {c^{ - 1}}(c(c(c(y)))) \\= & {} {c^{ -1}}(c(y)) \Rightarrow c(c(y)) = y. \end{aligned}$$

So c is onto.

(ii) Let $V\in {\mathcal {U}}.$ Since the negation map c is uniformly continuous, there exists $U \in {\mathcal {U}}$ s.t ${c^{(2)}}(U) \subseteq V$. By (i), ${c^{(2)}}$ is bijection, so $U \subseteq {c^{(2)}}(V).$ Hence ${c^{(2)}}(V) \in {\mathcal {U}}$ . As ${c^{(2)}}({c^{(2)}}(V)) = V$ we get that $V \in {c^{(2)}}({\mathcal {U}}).$ Hence ${\mathcal {U}} \subseteq {c^{(2)}}({\mathcal {U}})$. This also concludes that ${c^{(2)}}({\mathcal {U}})\subseteq {\mathcal {U}}.$$\square$

A BL-algebra A is an MV-algebra if, for each $x\in A,$$x''=x.$ Thus by Proposition 3.6(1), if the negation map c in BL-algebra A is one-to-one or onto, then A is an MV-algebra. Uniform MV-algebras are already studied see for instance (Barbieri 1998; Graziano 2000). By Proposition 3.6(2), if ${\mathcal {U}}$ is a uniformity on an MV-algebra which makes c uniformly continuous, then ${c^{(2)}}({\mathcal {U}})={\mathcal {U}}.$

Proposition 3.7

If $(A,{\mathcal {U}})$ is a uniform BL-algebra, then for any $W \in {\mathcal {U}}$, there exists $U \in {\mathcal {U}}$ such that $U \subseteq W,$$U \odot U \subseteq W$ and $U\rightarrow U \subseteq W.$

Proof

Since $\odot$ and $\rightarrow$ are uniformly continuous, there exist $V_1,V_2,U_1,U_2 \in {\mathcal {U}}$ such that ${V_1}\rightarrow {U_1} \subseteq W$ and ${V_2} \odot {U_2} \subseteq W$. Now suppose that $U = U_1 \cap U_2\cap {V_1} \cap {V_2}$ then $U \rightarrow U \subseteq W$ and $U \odot U \subseteq W$. For any $(x,y) \in U$, since $(1,1) \in U,$ by $(\mathcal {BL}_{4}),$$(x,y) = (1\rightarrow x,1\rightarrow y) =(1,1)\rightarrow (x,y) \in W$. Then $U \subseteq W$. $\square$

Proposition 3.8

Let ${\mathcal {U}}$ be a uniformity on BL-algebra A and $\rightarrow$ be uniformly continuous. Then $\oslash$ is uniformly continuous.

Proof

Assume $W \in {\mathcal {U}}.$ Since $\rightarrow$ is uniformly continuous, there exist ${V_1},{V_2} \in {\mathcal {U}}$ such that ${V_1} \rightarrow {V_2} \subseteq W$ and for ${V_1} \in {\mathcal {U}}$ there exists ${V_3} \in {\mathcal {U}}$ such that $c^{(2)}(V_3) \subseteq V_1.$ Thus

$$\begin{aligned} {V_3} \oslash {V_2} = {c^{(2)}}({V_3}) \rightarrow {V_2} \subseteq {V_1} \rightarrow {V_2} \subseteq W. \end{aligned}$$

$\square$

Theorem 3.9

Let ${\mathcal {U}}$ be a uniformity on BL-algebra A. If A is an MV-algebra and the negation map c is uniformly continuous, then $\rightarrow$ is uniformly continuous if and only if $\odot$ is uniformly continuous.

Proof

Since A is an MV-algebra, for all $x\in A,$$x''=x.$ This concludes that the negation map c is onto. Now let $\rightarrow$ be uniformly continuous and $U\in {\mathcal {U}}.$ By Proposition 3.6, the negation map c is injective and $c^{(2)}(U)\in {\mathcal {U}}.$ For some ${U_1},{U_2} \in {\mathcal {U}},$${U_1} \rightarrow {U_2} \subseteq c^{(2)}(U)$. The negation map c is onto and uniformly continuous, so by using of Proposition 3.6, there exists ${U_3} \in {\mathcal {U}}$ s.t ${c^{(2)}}({U_3}) = {U_2}$. By $(\mathcal {BL}_{14}),$ we have

$$\begin{aligned} {c^{(2)}}({U_1} \odot {U_3}) = {U_1} \rightarrow {c^{(2)}}({U_3}) = {U_1} \rightarrow {U_2} \subseteq {c^{(2)}}(U). \end{aligned}$$

The map c is injective, so ${U_1} \odot {U_3} \subseteq U$.

Let $\odot$ be uniformly continuous and $U\in {\mathcal {U}}.$ From Proposition 3.6, we conclude there is a $W \in {\mathcal {U}}$ such that ${c^{(2)}}(W) = U$. Because $\odot$ is uniformly continuous, there are $M,N \in {\mathcal {U}}$ such that $M \odot N \subseteq W$. By Proposition 3.6, $c^{(2)}(N)\in {\mathcal {U}}.$ For each $(x,y)\in M$ and every $(a,b)\in N,$ by $(\mathcal {BL}_{14}),$ we have

$$\begin{aligned} (x,y)\rightarrow c^{(2)}(a,b)= & {} (x,y)\rightarrow (a',b')=(x\rightarrow a',y\rightarrow b')\\= & {} ((x\odot a)',(y\odot b)')\\ {}= & {} c^{(2)}(x\odot a,y\odot b). \end{aligned}$$

Hence

$$\begin{aligned} M\rightarrow c^{(2)}(N)=c^{(2)}(M\odot N)\subseteq c^{(2)}(W)\subseteq U. \end{aligned}$$

This implies that $\rightarrow$ is uniformly continuous. $\square$

Proposition 3.10

Let A be a BL-algebra such that for any $x, y \in A, x \wedge y =0=x \odot y$ iff $x=0$ or $y=0.$ Then there is a uniformity ${\mathcal {U}}$ on A such that $\wedge , \vee$ and $\odot$ are uniformly continuous.

Proof

Let ${\mathcal {F}}$ be a family of filters of A which is closed under intersections and for each $x \ne y,$ there are $F, J \in {\mathcal {F}}$ such that $F[x] \cap J[y] = \phi ,$ where $F[x] = \big \{ y \in A:y \ne 0, \, y \rightarrow x \in F \big \}.$

If $B = \left\{ {F[x]: \, F \in {\mathcal {F}}, \,x \in A} \right\} \cup \left\{ 0 \right\} ,$ then B is a base for a topology on A. Clearly, $A \subseteq \cup B.$ Let ${F_1}[{x_1}]$ and ${F_2}[{x_2}]$ be two members of B and $a \in {F_1}[{x_1}] \cap {F_2}[{x_2}].$ Suppose $F = {F_1} \cap {F_2} \in {\mathcal {F}}.$ Then $F[a] \subseteq {F_1}[{x_1}] \cap {F_2}[{x_2}]$ because if $z \in F[a],$ then $z \rightarrow a$ and $a \rightarrow {x_i}$ are in ${F_i}.$ By $(\mathcal {BL}_{15}),$

$$\begin{aligned} (z \rightarrow a) \odot (a \rightarrow {x_i}) \le z \rightarrow {x_i}, \end{aligned}$$

hence $z \rightarrow {x_i} \in {F_i},$ and so $z \in {F_i}[{x_i}].$ Thus B is a base for a topology $\tau$ on A. We prove that $(A,\tau )$ is a compact space. Let $\left\{ {{U_i}: \, i \in I} \right\}$ be an open covering of A. Then for some $i\in I,$$1\in U_i.$ From that B is a base for $\tau$ we get a $F\in {\mathcal {F}}$ and a $x\in A,$ such that $1 \in F[x]\subseteq U_i.$ Consequently, $x \in F.$ If $y \ne 0$ is an arbitrary element of A,  since $x \le y \rightarrow x,$ we conclude that $y \rightarrow x \in F$ and so $y \in F[x].$ This implies that $A{\setminus }\{0\}\subseteq U_i.$ On the other hand, for some $j\in I,$${0 \in {U_j}}.$ Therefore, $A \subseteq U_i\cup {U_j}$ which implies that A is compact. It is easy to prove that this space is Hausdorff. Since $(A,\tau )$ is a compact Hausdorff space, it is a Tychonoff space and so by Engelking (1989, Theorem 8.1.20), there is a uniformity ${\mathcal {U}}$ on A such that $\tau$ is the uniform topology induced by ${\mathcal {U}}.$ Now we prove that $(A, \wedge , \vee , \odot ,\tau )$ is a topological BL-algebra. For do this, let $* \in \left\{ { \wedge , \vee , \odot } \right\}$ and x and y be in A. In the following cases we prove that $*$ is continuous.

Case 1: Let $x * y = 0.$ If $* \in \left\{ { \wedge , \odot } \right\}$, then by hypothesis $x=0$ or $y=0.$ Without loss of generality, we assume that $x=0.$ Then $\left\{ 0 \right\}$ and A are two open neighborhoods of x and y,  respectively, such that $\left\{ 0 \right\} * A \subseteq \left\{ 0 \right\} .$ If $* = \vee ,$ then since $x,y \le x \vee y = 0,$ we get that $x=y=0.$ Now $\left\{ 0 \right\}$ is an open neighborhood of x and y such that $\left\{ 0 \right\} \vee \left\{ 0 \right\} \subseteq \left\{ 0 \right\} .$

Case 2: Let $x * y \ne 0.$ Then there is a $F \in {\mathcal {F}}$ such that $x * y \in F[x * y].$ Suppose $a \in F[x]$ and $b \in F[y],$ then $a \rightarrow x$ and $b \rightarrow y,$ both, belong to F. By $(\mathcal {BL}_{16}),$ we have

$$\begin{aligned} (a \rightarrow x) \odot (b \rightarrow y) \le (a * b \rightarrow x * y), \end{aligned}$$

hence $F[x] * F[y] \subseteq F[x * y].$

From cases 1 and 2, we conclude that $*$ is continuous. Since A is compact, the operations $\wedge ,\vee$ and $\odot$ are uniformly continuous. $\square$

Theorem 3.11

Let F be a compact filter with the maximum condition in a Hausdorff topological BL-algebra $(A,\tau ).$ If the set $\frac{A}{F}$ is finite, then there is a uniformity ${\mathcal {U}}$ on A such that $(A,{\mathcal {U}})$ is a uniform BL-algebra.

Proof

First we prove that for each $x,y \in A,$$F \odot x = F \odot y$ iff $x=y.$ If $x=y,$ clearly $F \odot x = F \odot y.$ Let $F \odot x = F \odot y.$ Since $x \in F \odot x,$ for some $f \in F, \, x = f \odot y.$ By $(\mathcal {BL}_{1})$$x \le y.$ Similarly, $y\le x.$ Hence $x=y.$ The set $\frac{A}{F}$ is finite, hence by Borzooei (2012a, Proposition 3.5), there exist ${x_1},{x_2},\ldots ,{x_n} \in A$ such that $A = (F \odot {x_1}) \cup \cdots \cup (F \odot {x_n}).$ Because the operation $\odot$ is continuous, for each $1 \le i \le n, \, F \odot {x_i}$ is compact and so A is also compact . Since A is a compact and Hausdorff space ,we conclude that A is normal. Therefore, by Engelking (1989, Theorem 8.1.20), there exists a uniformity ${\mathcal {U}}$ on A such that $\tau$ is the topology induced by it. Since A is compact, $*\in \{\wedge ,\vee ,\odot ,\rightarrow \}$ is uniformly continuous. $\square$

Recall, BL-algebra A is prime if for every $x,y,z \in A,x \odot y \le z$ implies $x\le z$ or $y \le z$ (Borzooei et al. 2012b).

Proposition 3.12

Let W be a 1-quasifilter in a prime BL-algebra A. Then the relation

$$\begin{aligned} x\mathop \equiv \limits ^W y \Leftrightarrow x \rightarrow y,y \rightarrow x \in W \end{aligned}$$

is a congruence relation on A.

Proof

Clearly, for every $x,y \in A, \, x\mathop \equiv \limits ^W x$ and if $x\mathop \equiv \limits ^W y,$ then $y\mathop \equiv \limits ^W x.$ Now let $x\mathop \equiv \limits ^W y$ and $y\mathop \equiv \limits ^W z.$ Then by $(\mathcal {BL}_{15})$, $(x \rightarrow y )\odot (y \rightarrow z) \le x \rightarrow z.$ Since A is a prime BL-algebra, $x \rightarrow y \le x \rightarrow z$ or $y \rightarrow z \le x \rightarrow z.$ Since W is a 1-quasifilter, $x \rightarrow z \in W.$ In a similar way one can show that, $z \rightarrow x \in W$ and so $x\mathop \equiv \limits ^W z.$ This shows that $\mathop \equiv \limits ^W$ is an equivalence relation on A. Let $x\mathop \equiv \limits ^W y$ and $a \mathop \equiv \limits ^W b.$ If $* \in \left\{ { \wedge , \vee , \odot } \right\} ,$ then by $(\mathcal {BL}_{16}),$$(x \rightarrow y) \odot (a \rightarrow b) \le (a * x) \rightarrow (b * y).$ Since A is prime and W is a 1-quasifilter, we get that $(a * x) \rightarrow (b * y) \in W.$ With a similar argument $(b * y) \rightarrow (a * x)$ is in W. Hence $a *x\mathop \equiv \limits ^W b *y.$ By $(\mathcal {BL}_{17}),$$(x \rightarrow y) \odot (b \rightarrow a) \le (y \rightarrow b) \rightarrow (x \rightarrow a),$ so $(y \rightarrow b) \rightarrow (x \rightarrow a)$ belongs to W. Similarly, we can get that $(x \rightarrow a) \rightarrow (y \rightarrow b) \in W.$ Hence $x \rightarrow a \mathop \equiv \limits ^W y \rightarrow b.$$\square$

Theorem 3.13

Let ${\mathcal {W}}$ be the family of all 1-quasifilters in prime BL-algebra A. For every $W\in {\mathcal {W}},$ there is a uniformity ${\mathcal {U}}$ on A such that $(A,{\mathcal {U}} )$ is a uniform BL-algebra.

Proof

Let $W\in {\mathcal {W}}$ and ${W_0} = \left\{ {(x,y):x\mathop {\equiv y}\limits ^W } \right\} .$ Assume that ${\mathcal {B}}= \{ W_0 \}.$ Obviously, $\Delta \subseteq W_0$ and $W_0^{-1}=W_0\in {\mathcal {B}}.$ Thus conditions (i) and (ii) of Lemma 2.11 hold. If $(x,y)\in W_0\circ W_0,$ then for some $z\in A,$$(x,z),(z,y)\in W_0.$ Consequently, $x\rightarrow z,z\rightarrow x, z\rightarrow y$ and $y\rightarrow z$ are in W. By $(\mathcal {BL}_{15}),$$(x\rightarrow z)\odot (z\rightarrow y)\le x\rightarrow y.$ Since A is prime BL-algebra, $x\rightarrow z\le x\rightarrow y$ or $z\rightarrow y\le x\rightarrow y.$ This implies that $x\rightarrow y\in W.$ With a similar argument $y\rightarrow x\in W.$ Hence $(x,y)\in W_0.$ Therefore $W_0\circ W_0\subseteq W_0.$ By Lemma 2.11, ${\mathcal {B}}$ is a subbase for a uniformity ${\mathcal {U}}$ on A. By Proposition 3.12, ${W_0} * {W_0} \subseteq {W_0},$ for each $* \in \left\{ { \wedge , \vee , \odot , \rightarrow } \right\} .$ Hence $(A,{\mathcal {U}})$ is a uniform BL-algebra. $\square$

4 Filters and uniform topology

Proposition 4.1

Let $\tau$ be a topology on a BL-algebra A such that $\rightarrow$ is continuous in each variable separately. Let ${\mathcal {F}}$ be a family of open filters which is closed under intersections and for each $x \ne 1,$ there is a $F \in {\mathcal {F}}$ such that $1 \notin x/F.$ Then there is a nontrivial uniform topology $\nu$ on A coarser than $\tau$ such that $(A,\nu )$ is a topological BL-algebra.

Proof

It is easy to prove that the set $B = \big \{ x/F: \, F \in {\mathcal {F}}, x \in A \big \}$ is a base for the topology $\nu =\big \{ V \subseteq A: \,\forall x \in V \exists F \in {\mathcal {F}} \, \, s.t \, \, x/F \subseteq V \big \}$ on A. We prove that $\nu$ is coarser than $\tau .$ For this, let $x \in V \in \nu .$ Then for some $F \in {\mathcal {F}},$$x \in x/F \subseteq V.$ Since $x \rightarrow x=1 \in F$ and the operation $\rightarrow$ is continuous in each variable separately, there exists $U \in \tau$ such that $x \in U \in \tau$ and $U \rightarrow x \subseteq F$ and $x \rightarrow U \subseteq F.$ But $U \subseteq V$ because if $z \in U,$ then $z \rightarrow x$ and $x \rightarrow z,$ both, belong to F and so $z \in x/F \subseteq V.$ This proves that $\nu$ is coarser than $\tau .$ For each $x,y\in A,$ if $y\in \overline{x/F},$ then there is a $z\in y/F\cap x/F.$ This implies that $y\in x/F$ and so x / F is closed in $(A,\nu ).$ Thus for any $F \in {\mathcal {F}}$ and $x \in A,$x / F is an open and closed set in $(A,\nu ),$ hence $(A,\nu )$ is a regular space. $(A,\nu )$ is a ${T_1}$-space because if $x\not = y,$ then

$$\begin{aligned} x\rightarrow y\not =1\ or\ y\rightarrow x\not =1. \end{aligned}$$

Without loss of generality we assume that $x\rightarrow y\not =1.$ By hypothesis $1\not \in (x\rightarrow y)/F$ for some $F\in {\mathcal {F}}.$ Consequently, $x\not \in y/F$ and $y\not \in x/F.$ So $(A,\nu )$ is a $T_1$ space. Since $(A,\nu )$ is a $T_1$ space and $\nu$ has a base of open and closed sets, the pair $(A,\nu )$ is a Tychonoff space. By Engelking (1989, Theorem 8.1.20), $\nu$ is a uniform topology. Finally, if $* \in \left\{ { \odot , \rightarrow } \right\} ,$ for any $F \in {\mathcal {F}}$ and $x,y \in A, x/F * y/F = x *y/F$ concludes that $*$ is continuous. By Borzooei (2011, Theorem 3.5), $(A,\nu )$ is a topological BL-algebra. $\square$

Proposition 4.2

Let $(A,\tau )$ be a second countable topological BL-algebra and let for each $a \in A$ the mapping ${t_a}(x) = a \odot x$ be an open and closed map from A into A. Then $\tau$ is a uniform topology iff for each $x \ne 1$ and each open neighborhood U of 1,  there exists an open set V such that $1 \in V \subseteq {\overline{V}} \subseteq U$ and $x \notin V.$

Proof

Let for each $x \ne 1$ and each open neighborhood U of 1,  there exists an open set V such that $1 \in V \subseteq {\overline{V}}\subseteq U$ and $x \notin V.$ Then by Borzooei (2011, Proposition 4.4), $(A,\tau )$ is a ${T_1}$-space. We prove that it is regular. To do this, let $x \in U \in \tau .$ If $x=1,$ then clearly there is a $V \in \tau$ such that $1 \in V \subseteq {\overline{V}} \subseteq U.$ Hence we assume that $x \ne 1.$ Since the operation $\odot$ is continuous, there is an open neighborhood W of 1 such that $x \odot W \subseteq U.$ Let V be an open set such that $1 \in V \subseteq {\overline{V}} \subseteq W.$ Since ${t_x}$ is an open and closed map, the set $x \odot V$ is an open neighborhood of x and $x \odot {\overline{V}}$ is a closed set. Thus

$$\begin{aligned} x \in x \odot V \subseteq \overline{x \odot V }\subseteq \overline{ x \odot {\overline{V}} }= x \odot {\overline{V}} \subseteq x \odot W \subseteq U. \end{aligned}$$

Hence $(A,\tau )$ is a regular space. Because $(A,\tau )$ is second countable, by Engelking (1989, Theorem 1.5.16), it is a Tychonoff space. By Engelking (1989, Theorem 8.1.20), $\tau$ is a uniform topology. Conversely, if $\tau$ is a uniform topology, then $(A,\tau )$ is a regular space. Obviously, for each $x \ne 1$ and each open neighborhood U of 1,  there exists an open neighborhood V of 1 such that $1 \in V \subseteq {\overline{V}} \subseteq U$ and $x \notin V.$

$\square$

Proposition 4.3

Let $(A, \tau )$ be a second countable topological BL-algebra and let for each $a \in A$ the mapping ${r_a}(x) = x \rightarrow a$ be an open map from A into A. Then $\tau$ is a uniform topology if for each $x \ne 1$ and each open neighborhood U of 1,  there exists an open set V such that $1 \in V \subseteq {\overline{V}} \subseteq U,$${\overline{V}}$ is compact and $x \notin V.$

Proof

Suppose for each $x \ne 1$ and each open neighborhood U of 1, there exists an open set V such that $1 \in V \subseteq {\overline{V}} \subseteq U,$${\overline{V}}$ is compact and $x \notin V.$ By Borzooei (2011, Proposition 4.5), $(A, \tau )$ is a Hausdorff space. We prove that it is regular. To do this, let $x \in U \in \tau .$ If $x=1,$ by hypothesis, there is a $V \in \tau$ such that $1 \in V \subseteq {\overline{V}} \subseteq U.$ Hence we suppose that $x \ne 1.$ Since the operation $\rightarrow$ is continuous, there is an open neighborhood W of 1 such that $W \rightarrow x \subseteq U.$ Let V be an open set such that $1 \in V \subseteq {\overline{V}} \subseteq W$ and ${\overline{V}}$ is compact. Since ${r_x}$ is an open and continuous map, the set $V \rightarrow x$ is an open neighborhood of x and ${\overline{V}} \rightarrow x$ is a compact set. But $(A, \tau )$ is Hausdorff, hence the set ${\overline{V}} \rightarrow x$ is closed. We have

$$\begin{aligned} V \rightarrow x \subseteq \overline{V \rightarrow x} \subseteq \overline{{\overline{V}} \rightarrow x} = {\overline{V}} \rightarrow x \subseteq W \rightarrow x \subseteq U \end{aligned}$$

which proves that $(A, \tau )$ is a regular space. Since $(A, \tau )$ is second countable, by Engelking (1989, Theorem 1.5.16), it is a Tychonoff space. Therefore, by Engelking (1989, Theorem 8.1.20), $\tau$ is a uniform topology. $\square$

Theorem 4.4

Let ${\mathcal {F}}$ be a family of filters in a BL-algebra A which is closed under intersections. If for each $x \ne y$

• (i) there is a $F \in {\mathcal {F}}$ such that $y \notin F \odot x,$

• (ii) for each $F, J \in {\mathcal {F}},$ if $F \odot x \cap J \odot y$ is a nonempty set, then either $x \in J \odot y$ or $y \in F \odot x,$

  then there exists a uniform topology $\tau$ on A such that $\odot$ is continuous and $(A, \tau )$ is a ${T_1}$ space.

Proof

Let $\tau = \left\{ {G \subset A:\forall x \in G \, \, \exists F \in {\mathcal {F}} \, \, s.t \, \,F \odot x \subseteq G} \right\} .$ Then $\tau$ is closed under arbitrary unions and intersections, hence it is a topology on A. If $x \ne y,$ then by (i), for some $F, J \in {\mathcal {F}}, x \notin J \odot y$ and $y \notin F \odot x.$ Hence $x \notin K \odot y$ and $y \notin K \odot x,$ where $K = F \cap J.$ This implies that $(A,\tau )$ is a ${T_1}$-space. Since F is a filter, continuity of $\odot$ is clear. We show that for each $x \in A$ and $F \in {\mathcal {F}},$$F \odot x$ is closed. To do this, let $x \in A, F \in {\mathcal {F}}$ and $y\in \overline{F\odot x}.$

If $y=x,$ clearly, $y \in F \odot x.$ Let $y \ne x.$ Then by (i), there exists a $J \in {\mathcal {F}}$ such that $x \notin J \odot y.$ Since y is in closure of $F \odot x, J \odot y \cap F \odot x$ is a nonempty set. By (ii), $y \in F \odot x.$ Hence $F \odot x$ is closed in A. Let us show that $(A,\tau )$ is a normal space. Let S be a closed subset of A and G an open neighborhood of it. Since for any $x \in S$ there is a ${F_x} \in {\mathcal {F}}$ such that ${F_x} \odot x \subseteq G,$ hence $S \subseteq \bigcup \nolimits _{x \in S} {{F_x}} \odot x \subseteq G.$ On the other hand, since for any $x \in A,$${F_x} \odot x$ is closed and $\tau$ is closed under arbitrary intersections, we deduce the set $\bigcup \nolimits _{x \in S} {{F_x}} \odot x$ is closed in A. This implies that $(A,\tau )$ is a normal space. Now by Engelking (1989, Theorem 8.1.20), $\tau$ is a uniform topology. $\square$

Recall, a topological space $(X, \tau )$ is locally compact if for each $x \in X$ there is an open neighborhood U of x such that ${\overline{U}}$ is compact. If $(X, \tau )$ is Hausdorff, then X is locally compact iff for each $x \in U \in \tau$ there is an open set V such that $x \in V \subseteq {\overline{V}} \subseteq U$ and ${\overline{V}}$ is compact. It is easy to see that any compact space is locally compact (Engelking 1989).

Theorem 4.5

Let $(A, \tau )$ be a Hausdorff BL-algebra and for each $a \in A,$${t_a}(x) = x \odot a$ be an open map from A into A. Then $\tau$ is a uniform topology if there is a locally compact open neighborhood of 1.

Proof

Let U be a locally compact open neighborhood of 1. Let $a \in A.$ Since ${t_a}(x) = x \odot a$ is an open map of U onto $U \odot a,$ by Engelking (1989, Theorem 3.3.15), the set $U \odot a$ is an open locally compact subset of A. As $A = \bigcup \nolimits _{x \in A} {U \odot x}$ is a union of open locally compact subspaces of A, then by Engelking (1989, Exercise 3.3.B(a)), it is a locally compact space. Since $(A, \tau )$ is Hausdorff and locally compact, by Engelking (1989, Theorem 3.3.1), it is a Tychonoff space. By Engelking (1989, Theorem 8.1.20), $\tau$ is a uniform topology.

$\square$

Theorem 4.6

Let $(A, \tau )$ be a Hausdorff semitopological BL-algebra. Suppose for every $a\in A{\setminus }\{1\},$$r_a(x)=x\rightarrow a$ and $l_a(x)=a\rightarrow x$ are two maps from A into A such that if S is compact in A,  the sets $r_a^{-1}(S)$ and $l_a^{-1}(S)$ are compact sets. If F is a locally compact open filter in A,  then $\tau$ is a uniform topology.

Proof

First we show that for each $a\in A{\setminus }\{1\}$ the set ${r_a}^{ - 1}(F)$ is locally compact. To do this, suppose $1\not =a\in A$ and $x\in {r_a}^{ - 1}(F).$ Since F is locally compact and ${r_a}(x) \in F,$ there is an open set V in A such that ${r_a}(x) \in V \subseteq {\overline{V}} \subseteq F$ and ${\overline{V}}$ is compact in A. From the fact that $r_a$ is continuous, we infer that $r_a^{-1}(V)$ is open in A. By hypothesis, $r_a^{-1}({\overline{V}})$ is compact. The set $\overline{r_a^{-1}(V)}$ is a closed subset of $r_a^{-1}({\overline{V}}),$ so it is compact in A. Now we have

$$\begin{aligned} x \in {r_a}^{ - 1}(V ) \subseteq \overline{{r_a}^{ - 1}(V )} \subseteq {r_a}^{ - 1}({\overline{V}}) \subseteq {r_a}^{ - 1}(F) \end{aligned}$$

which implies that ${r_a}^{ - 1}(F)$ is locally compact. Similarity ${l_a}^{ - 1}(F)$ is also locally compact. Hence $\frac{a}{F} = {r_a}^{ - 1}(F) \cap {l_a}^{ - 1}(F)$ is a locally compact open subset of A. Since $A = \bigcup \nolimits _{a \in A} {\frac{a}{F}}$ is a union of locally compact open subspaces of A,  by Engelking (1989, Exercise 3.3.B(a)), $(A, \tau )$ is locally compact. By (Engelking 1989, Theorem 3.3.1), it is a Tychonoff space. By (Engelking 1989, Theorem 8.1.20), $\tau$ is a uniform topology. $\square$

Theorem 4.7

Let ${\mathcal {W}}$ be a QN-fundamental system in a Hausdorff topological BL-algebra $(A, \tau ).$ Suppose for every $a\in A{\setminus }\{1\},$$r_a(x)=x\rightarrow a$ and $l_a(x)=a\rightarrow x$ are two maps from A into A such that if S is compact in A,  the sets $r_a^{-1}(S)$ and $l_a^{-1}(S)$ are compact sets. If for each $a \in A, \, {t_a}(x) = a \odot x$ is an open map from A into A,  then $\tau$ is a uniform topology provided there exists a compact open neighborhood of 1.

Proof

Let U be a compact open neighborhood of 1 and $x \in U.$ Since the operation $\odot$ is continuous, there exist two open neighborhoods ${V_x}$ and ${H_x}$ of 1 such that $x \odot {V_x} \subseteq U$ and ${H_x} \odot {H_x} \subseteq {V_x}.$ Clearly, $\left\{ {x \odot {H_x}:x \in U} \right\}$ is an open covering of U. Since U is compact, the union of a finite number of sets $\left\{ {{x_i} \odot {H_{{x_i}}}:1 \le i \le n} \right\}$ covers U. Let $V = \bigcap \nolimits _{i = 1}^n {{H_{{x_i}}}} .$ Then V is an open neighborhood of 1 such that

$$\begin{aligned} U \odot V\subseteq & {} \left( \bigcup \limits _{i = 1}^n {{x_i}} \odot {H_{{x_i}}}\right) \odot V \subseteq \bigcup \limits _{i = 1}^n {{x_i}} \odot {H_{{x_i}}} \odot {H_{{x_i}}}\\\subseteq & {} \bigcup \limits _{i = 1}^n {{x_i}} \odot {V_{{x_i}}} \subseteq U. \end{aligned}$$

Since $U \cap V$ is an open neighborhood of 1,  there is a $W \in {\mathcal {W}}$ such that $W \subseteq U \cap V.$ Thus ${W^2} \subseteq U \odot V \subseteq U.$ But then by induction, ${W^n} = {W^{n - 1}} \odot W \subseteq U \odot V \subseteq U.$ Let $F = \bigcup \nolimits _{n \ge 1} {{W^n}} .$ By Borzooei (2012a, Proposition 3.13), F is a filter. Since for each $a \in A, {t_a}$ is an open map, F is an open set. If $y \in {\overline{F}},$ then since $F \odot y$ is an open neighborhood of y,  there is a $f \in F$ such that $f \odot y \in F.$ Since F is a filter, $y \in F.$ Thus F is a closed set in the compact space U and so F is compact. Now by Theorem  4.6, $\tau$ is a uniform topology. $\square$

5 Some uniformities on BL-algebras and their properties

Theorem 5.1

Let $\varrho$ and $\eta$ be two families of filters and ideals, respectively, of BL-algebra A which are closed under intersection. If for each $F \in \varrho$ and $I\in \eta$

$$\begin{aligned} {U_F}= & {} \left\{ (x,y) \in A \times A: x\mathop \equiv \limits ^{F} y \right\} \ and\\ {U_I}= & {} \left\{ (x,y) \in A \times A: x\mathop \equiv \limits ^{I}y\right\} , \end{aligned}$$

then $K = \{ {U_F}\left| {F \in \varrho }\right. \}$ and $H = \{ {O_I}\left| {I \in \eta } \right. \}$ are bases for two uniformities ${\mathcal {K}}$ and ${\mathcal {H}}$ on A. Moreover, $(A,{\mathcal {K}})$ and $(A,{\mathcal {H}})$ are uniform BL-algebras.

Proof

It is easy to prove that K, H satisfy $(B_1), (B_2), (B_3)$ and $(B_4)$ of Lemma 2.11. Hence they are bases for two uniformities

$$\begin{aligned} {\mathcal {K}}=\{U \subseteq A \times A: {{U_F} \subseteq U,\ for \, some \,\, U_F} \in K \} \end{aligned}$$

and

$$\begin{aligned} {\mathcal {H}}=\{U \subseteq A \times A: {{U_I} \subseteq U,\ for \, some \,\, U_I} \in H\}. \end{aligned}$$

Let $*\in \{\odot ,\rightarrow \}$ and $I\in \eta .$ If (x, y) and (a, b) are in $U_I,$ then $x \mathop \equiv \limits ^{I}y$ and $a \mathop \equiv \limits ^{I}b.$ The relation $\mathop \equiv \limits ^{I}$ is a congruence relation, so $U_I*U_I\subseteq U_I.$ By Corollary  3.5, $(A,{\mathcal {H}})$ is a uniform BL-algebra. In a similar way, $(A,{\mathcal {K}})$ is also a uniform BL-algebra. $\square$

Theorem 5.2

In BL-algebra A assume that $O = \big \{ (x,y)\left| x' = y' \right. \big \}.$ Then $\{ O\}$ is a subbase for a uniformity ${\mathcal {L}}$ on A,  such that $(A,{\mathcal {L}})$ is a uniform BL-algebra.

Proof

It is easy to see $\{O\}$ satisfies $(S_1),(S_2)$ and $(S_3)$ of Lemma 2.12. Let ${\mathcal {L}}$ be uniformity generated by $\{O\}.$ We show that $(A, {\mathcal {L}})$ is a uniform BL-algebra. To do this, we prove that $O * O \subseteq O,$ where $* \in \{ \odot , \rightarrow \}.$ For any $(z,k), \,(x,y) \in O,$ by $(\mathcal {BL}_{11}),$ we have

$$\begin{aligned} (x * z)'= & {} (x * z)''' = (x * z)'' \rightarrow 0 = (x'' * z'') \rightarrow 0 \\= & {} (y'' * k'') \rightarrow 0 = (y * k)'' \rightarrow 0 = (y * k)', \end{aligned}$$

therefore $(x * z, y * k) \in O.$ So $(A, {\mathcal {L}})$ is a uniform BL-algebra. $\square$

Notation From now on, ${\mathcal {K}}$ and ${\mathcal {H}}$ are the uniformities introduced in Theorem 5.1 and $\varrho$ and $\eta$ are as in that theorem. Also ${\mathcal {L}}=\{U\subseteq A\times A: O\subseteq U\}$ is the uniformity introduced in Theorem 5.2.

Proposition 5.3

In a BL-algebra A suppose $F\in \varrho$ and $I\in \eta .$ Then:

• (1) $U_I[x]=I$ iff $x\in I,$

• (2) $U_F[x]=F$ iff $x\in F,$

• (3) $U_I=U_{N(I)},$

• (4) if $\varrho =\{N(I):I\in \eta \}$, then $\mathcal {K=H}.$

Proof

By Proposition 2.4, $U_I[x]=x/I$ and $U_F[x]=x/F,$ hence (1) and (2) are obvious. We prove (3). If $(x,y) \in {U_{N(I)}},$ then $x\mathop \equiv \limits ^{N(I)} y.$ By $(\mathcal {BL}_{13}),$

$$\begin{aligned} 0=x' \odot x\mathop \equiv \limits ^{N(I)} x' \odot y\ \mathrm{and}\ 0=y' \odot y\mathop \equiv \limits ^{N(I)} y' \odot x. \end{aligned}$$

Hence $(x' \odot y)' \in N(I), (x \odot y')' \in N(I).$ By $({\mathcal {N}}_3)$ of Proposition 2.3, $x \odot y', \, x' \odot y \in N(N(I)) = I.$ Therefore $(x,y) \in {U_{I}}.$

Consider $(x,y) \in {U_{I}},$ then by the compatibility with $\rightarrow$, we have $x \rightarrow y\mathop \equiv \limits ^I 1$ and $y \rightarrow x\mathop \equiv \limits ^I 1.$ By definition of congruence $(x \rightarrow y)' \in I, (y \rightarrow x)' \in I .$ Hence $x \rightarrow y \in N(I)$ and $y \rightarrow x \in N(I).$ So $(x,y) \in {U_{N(I)}} .$

By (3), the proof of (4) is clear. $\square$

Proposition 5.4

Let F be a filter in BL-algebra A. Then

$$\begin{aligned} U_{N(F)}= & {} \left\{ (x,y)\in A\times A: x'\mathop \equiv \limits ^F y'\right\} \\= & {} \left\{ (x,y)\in A\times A: x''\mathop \equiv \limits ^F y''\right\} . \end{aligned}$$

Moreover, if the negation map c is onto and $\eta =\{N(F):F\in \varrho \},$ then $\mathcal {K=H}.$

Proof

By $(\mathcal {BL}_{14}),$ we have

$$\begin{aligned} U_{N(F)}= & {} \left\{ (x,y)\in A\times A:x\mathop \equiv \limits ^{N(F)} y\right\} \\= & {} \left\{ (x,y):A\times A: x'\odot y, y'\odot x\in N(F)\right\} \\= & {} \left\{ (x,y)\in A\times A:(x'\odot y)',(y'\odot x)'\in F\right\} \\= & {} \left\{ (x,y)\in A\times A: x'\rightarrow y',y'\rightarrow x'\in F\right\} \\= & {} \left\{ (x,y)\in A\times A:x'\mathop \equiv \limits ^F y'\right\} . \end{aligned}$$

By $(\mathcal {BL}_{12}),$ it is easy to show that $U_{N(F)}=\{(x,y)\in A\times A: x''\mathop \equiv \limits ^F y''\}.$

Now let the negation map c be onto and $\eta =\{N(F):F\in \varrho \}.$ Then by Proposition 3.6(i), for every $x\in A,$$x''=x.$ If $F\in \varrho ,$

$$\begin{aligned} U_{N(F)}= & {} \left\{ (x,y)\in A\times A: x''\mathop \equiv \limits ^F y''\right\} \\= & {} \left\{ (x,y)\in A\times A: x\mathop \equiv \limits ^F y\right\} =U_F. \end{aligned}$$

Hence $\mathcal {K=H}.$$\square$

Theorem 5.5

If F is a filter in BL-algebra A,  then $\{U_F, U_{N(F)}\}$ is a base for a uniformity ${\mathcal {U}}$ such that $(A,{\mathcal {U}})$ is a uniform BL-algebra.

Proof

By $(\mathcal {BL}_{11})$ and Proposition 5.4, we have

$$\begin{aligned} U_{NN(F)}= & {} \left\{ (x,y)\in A\times A:x\mathop \equiv \limits ^{NN(F)} y\right\} \\= & {} \left\{ (x,y):A\times A: x\rightarrow y, y\rightarrow x\in NN(F)\right\} \\= & {} \left\{ (x,y)\in A\times A:(x\rightarrow y)'',(y\rightarrow x)''\in F\right\} \\= & {} \left\{ (x,y)\in A\times A: x''\rightarrow y'',y''\rightarrow x''\in F\right\} \\= & {} \left\{ (x,y)\in A\times A:x''\mathop \equiv \limits ^F y''\right\} =U_{N(F)}. \end{aligned}$$

By $({\mathcal {N}}_4)$ of Proposition 2.3, $F\subseteq NN(F),$ hence $U_F$ is a subset of $U_{N(F)}.$ It is easy to prove that $\{U_F, U_{N(F)}\}$ satisfies $(B_1),(B_2),(B_3)$ and $(B_4)$ of Lemma 2.11, so the set $\{U_F, U_{N(F)}\}$ is a base of a uniformity ${\mathcal {U}}$ on A. If $*\in \{\odot ,\rightarrow \},$ then $U_F*U_F\subseteq U_F$ and $U_{N(F)}*U_{N(F)}\subseteq U_{N(F)}.$ This implies that $\odot$ and $\rightarrow$ are uniformly continuous maps. By Corollary 3.5, $(A,{\mathcal {U}})$ is a uniform BL-algebra. $\square$

Proposition 5.6

In a BL-algebra A the following properties hold.

• (i) O[1] is a filter.

• (ii) $O = \left\{ {(x,y) \in A \times A\left| {x'' = y''} \right. } \right\} .$

• (iii) $O= {U_{O[1]}},$

• (iv) if $O[1]\in \varrho ,$ then $\mathcal {L\subseteq K}.$

Proof

• (i) Let $x,y \in {O}[1].$ Then $(1,x),(1,y) \in O.$ By $(\mathcal {BL}_{11})$ and $(\mathcal {BL}_{12})$, we have

  $$\begin{aligned} (x \odot y)'= & {} (x \odot y)'''= (x \odot y)'' \rightarrow 0 \\= & {} (x'' \odot y'') \rightarrow 0 = 1 \rightarrow 0 = 0. \end{aligned}$$

  So $x \odot y \in {O}[1].$

  If $x \le y$ and $x \in O[1],$ then $x''=1.$ By $(\mathcal {BL}_{11})$ and $(\mathcal {BL}_{12}),$

  $$\begin{aligned} (x \rightarrow y)'= & {} (x \rightarrow y)''' = (x \rightarrow y)'' \rightarrow 0 \\= & {} (x'' \rightarrow y'' ) \rightarrow 0 = (1 \rightarrow y'' ) \rightarrow 0 = y'. \end{aligned}$$

  Hence $y \in O[1].$

• (ii) By $(\mathcal {BL}_{12})$ the proof is clear.

• (iii) By $(\mathcal {BL}_{11}),$

  $$\begin{aligned} {U_{O[1]}}= & {} \left\{ {(x,y)\left| {x \rightarrow y,y \rightarrow x \in O[1]} \right. } \right\} \\= & {} \left\{ {(x,y)\left| {(x \rightarrow y)'=1'=(y \rightarrow x)' } \right. } \right\} \\= & {} \left\{ {(x,y)\left| {(x \rightarrow y)''=1=(y \rightarrow x)'' } \right. } \right\} \\= & {} \left\{ {(x,y)\left| {x'' = y''} \right. } \right\} = O. \end{aligned}$$

• (iv) The proof is clear. $\square$

Example 5.7

Let $A=\{0,a,b,c,d,e,f,1\}.$ Define $\odot$ and $\rightarrow$ as follows:

$\odot$	0	a	b	c	d	e	f	1
0	0	0	0	0	0	0	0	0
a	0	a	a	a	0	a	a	a
b	0	a	a	b	0	a	a	b
c	0	a	b	c	0	a	b	c
d	0	0	0	0	d	d	d	d
e	0	a	a	a	d	e	e	e
f	0	a	a	b	d	e	e	f
1	0	a	b	c	d	e	f	1

$\rightarrow$	0	a	b	c	d	e	f	1
0	1	1	1	1	1	1	1	1
a	d	1	1	1	d	1	1	1
b	d	f	1	1	d	f	1	1
c	d	e	f	1	d	e	f	1
d	c	c	c	c	1	1	1	1
e	0	c	c	c	d	1	1	1
f	0	b	c	c	d	f	1	1
1	0	a	b	c	d	e	f	1

also assume that

$$\begin{aligned} 0< a< b< c< 1,\ \ 0<d<e<f<1,\ \ a<e,\ \ b<f. \end{aligned}$$

It is easy to see $(A, \wedge , \vee , \odot , \rightarrow ,0,1)$ is a BL-algebra. Consider the ideal $I = \left\{ {0,d} \right\} ,$ then $N(I)=\left\{ {c,1} \right\}$ is a filter. Let $\varrho = \{ N(I)\}$ and $\eta = \{ I\}$. We have

$$\begin{aligned} U_{N(I)}= & {} \Delta \cup \{ (a,e),(e,a),(b,f),(f,b),(c,1),\\&\quad (1,c),(0,d),(d,0) \}= {U_I} . \end{aligned}$$

Then by Proposition 5.3(4), ${\mathcal {K}}={\mathcal {H}}$ and $(A,{\mathcal {K}})$ is a uniform BL-algebra. Also,

$$\begin{aligned} O= & {} \Delta \cup \{(a,b),(b,a),(a,c),(c,a),(b,c),(c,b),\\&\quad (e,f),(f,e),(e,1),(1,e),(f,1),(1,f) \} . \end{aligned}$$

The set $O[1] = \{ e,f,1\}$ is a filter and $N(O[1])=\{0\}.$ By Proposition 5.6, $U_{O[1]}=O.$ It is easy to check that

$$\begin{aligned} U_{\{0\}}= & {} \Delta \cup \{(1,0),(0,1),(a,b),(b,a),(a,c),(c,a), (b,c),\\&(c,b),(e,f),(f,e),(e,1),(1,e),(f,1),(1,f)\}. \end{aligned}$$

By Theorem 5.5, $\{U_{O[1]},U_{\{0\}}\}$ is a base for a uniformity ${\mathcal {U}}$ such that $(A,{\mathcal {U}})$ is a uniform BL-algebra.

Proposition 5.8

In a BL-algebra A,  let $\tau _{\varrho }$ and $\tau _{\eta }$ be uniform topologies induced by ${\mathcal {K}}$ and ${\mathcal {H}}$ on A,  respectively. Then:

• (i) for every $F\in \varrho$ and $I\in \eta ,$F is in $\tau _{\varrho }$ and $I\in \tau _{\eta },$

• (ii) for each $x\in A$ and $F\in \varrho$ and $I\in \eta ,$$U_F[x]\in \tau _{\varrho }$ and $U_I[x]\in \tau _{\eta },$

• (iii) $(A,\tau _{\varrho })$ and $(A,\tau _{\eta })$ are topological BL-algebras.

Proof

(i) By Definition 2.13,

$$\begin{aligned}&\tau _{\varrho }=\{G\subseteq A: \forall x\in G\ \exists F\in \varrho \ s.t\ U_F[x]\subseteq G\},\\&\tau _{\eta }=\{G\subseteq A: \forall x\in G\ \exists I\in \eta \ s.t\ U_I[x]\subseteq G\}. \end{aligned}$$

By Proposition 5.3, it is clear that $F\in \tau _{\varrho }$ and $I\in \tau _{\eta }$ for any $F\in \varrho$ and $I\in \eta .$

(ii) Let $x\in A,$$F\in \varrho$ and $y\in U_F[x].$ By $(\mathcal {BL}_{15}),$ it is easy to see that $U_F[y]\subseteq U_F[x].$ Hence $U_F[x]\in \tau _{\varrho }.$ In a similar way, $U_I[x]\in \tau _{\eta },$ when $I\in \eta .$

(iii) By Theorem 5.1, the pairs $(A,{\mathcal {K}})$ and $(A,{\mathcal {H}})$ are uniform BL-algebras, so $(A,\tau _{\varrho })$ and $(A,\tau _{\eta })$ are topological BL-algebras. $\square$

Theorem 5.9

If $\varrho$ is a subbase for a topology $\tau$ on A such that $(A,\rightarrow ,\tau )$ is a topological BL-algebra, then there is a uniformity ${\mathcal {U}}$ on A such that $(A,{\mathcal {U}})$ is uniform BL-algebra and $\tau$ is the uniform topology induced by ${\mathcal {U}}.$

Proof

Let $F\in \varrho$ and $(x,y)\in U_F.$ Then $x\rightarrow y$ and $y\rightarrow x,$ both are in F. Since $F\in \tau ,$ there exist $F_0, F_1\in \tau$ such that $x\in F_0,$$y\in F_1$ and

$$\begin{aligned} F_0\rightarrow F_1\subseteq F\ and\ F_1\rightarrow F_0\subseteq F. \end{aligned}$$

Hence $(x,y)\in F_0\times F_1\subseteq U_F.$ This says that $U_F$ is open in $(A\times A,\tau \times \tau ).$ Let $x\in G\in \tau .$ Then for some $F\in \varrho ,$$x\in F\subseteq G.$ By Proposition 5.3(2), $x\in U_F[x]=F\subseteq G.$ This implies that $G\in \tau _\varrho .$ Hence $\tau \subseteq \tau _\varrho .$ By Proposition 5.8(ii), $\tau _\varrho \subseteq \tau .$ Thus $\tau =\tau _{\varrho }.$ Therefore, $\tau$ is the uniform topology induced by ${\mathcal {K}}.$ By Theorem 5.1, $(A,{\mathcal {K}})$ is uniform BL-algebra. $\square$

Theorem 5.10

If $\eta$ is a subbase for a topology $\tau$ on A such that $(A,\tau )$ is a topological BL-algebra, then there is a uniformity ${\mathcal {U}}$ on A such that $(A,{\mathcal {U}})$ is uniform BL-algebra and $\tau$ is the uniform topology induced by ${\mathcal {U}}.$

Proof

Let $I\in \eta$ and $(x,y)\in U_I.$ Then $x'\odot y$ and $y'\odot x,$ both are in I. Since $\odot$ and $\rightarrow$ are continuous, there are $I_0$ and $I_1$ in $\eta$ such that $x\in I_0,$$y\in I_1$ and

$$\begin{aligned} I_0'\odot I_1\subseteq I\ \mathrm{and}\ I_1'\odot I_0\subseteq I. \end{aligned}$$

This implies that $I_0\times I_1\subseteq U_I.$ Hence $U_I\in \tau \times \tau .$

Now we prove that $\tau =\tau _{\eta }.$ Let $I\in \eta ,$$x\in A$ and $y\in U_I[x].$ By the previous paragraph for some $I_0$ and $I_1$ in $\eta$ we have $I_0\times I_1\subseteq U_I.$ Thus $y\in I_1\subseteq U_I[x].$ Therefore, $U_I[x]\in \tau .$ Since $\{U_I{[x]}:I\in \eta , x\in A\}$ is a base for $\tau _{\eta },$$\tau _{\eta }\subseteq \tau .$ Proposition 5.3(i) easily implies that $\tau \subseteq \tau _{\eta }.$ By Theorem 5.1, ${\mathcal {H}}$ is a uniformity on A such that $(A,{\mathcal {H}})$ is uniform BL-algebra and $\tau$ is the uniform topology induced by ${\mathcal {H}}.$$\square$

Proposition 5.11

Suppose A is a BL-algebra, I an ideal and q from A onto A / I is quotient map. Then:

• (i) the set $S\subseteq A/I$ is an ideal in A / I iff $S=J/I$ for some ideal J of A containing I, 

• (ii) if J is an ideal of A containing I, then $U_{J/I}=q^{(2)}(U_J),$

• (iii) if $\eta _I=\{J: J\ is\ an\ ideal\ in\ A, I\subseteq J\},$ then $\eta _{A/I}=\{U_{J/I}: J\in \eta _I\}$ is a base for a uniformity on A / I such that q is uniformly continuous. Moreover, A / I is uniform BL-algebra.

Proof

• (i) Suppose S is an ideal of A / I,  we define $J = \{ x \in A: \frac{x}{I} \in S \}.$ Since S is an ideal, $\frac{0}{I} \in S,$ so $0 \in J.$ Let $x, x' \odot y \in J,$ then $\frac{x}{I}, \frac{{x'}}{I} \odot \frac{y}{I} \in S.$ Because S is an ideal, by (ii) of Proposition 2.1, $\frac{y}{I} \in S$ and so $y \in J.$ Thus J is an ideal of A such that $I \subseteq J.$ Let $H=J/I=\{ \frac{x}{I}: x \in J \}.$ If $\frac{z}{I}\in H,$ then for some $x\in J,$$\frac{z}{I}=\frac{x}{I}.$ Consequently, $x'\odot z\in I\subseteq J.$ Since $x\in J,$ by (ii) of Proposition 2.1, z is in J. This implies that $H\subseteq S.$ It is easy to see that $S\subseteq H.$ Hence $H = S.$

  Conversely, suppose J is an ideal of A including I and $S = \left\{ {\frac{x}{I}\left| {x \in J} \right. } \right\} ,$ then $0 \in J$ implies that $\frac{0}{I} \in S.$ If $\frac{x}{I}, \frac{{x' \odot y}}{I} \in S,$ then $x,x' \odot y \in J.$ By (ii) of Proposition 2.1, $y \in J.$ Hence $\frac{y}{I} \in S.$ Consequently S is an ideal of $\frac{A}{I}.$

• (ii)
  $$\begin{aligned} \left( \frac{x}{I},\frac{y}{I}\right) \in U_{J/I}\Leftrightarrow & {} \frac{x'\odot y}{I},\frac{y'\odot x}{I}\in J/I\\\Leftrightarrow & {} x'\odot y, y'\odot x\in J\Leftrightarrow (x,y)\in U_J. \end{aligned}$$

  Hence $U_{J/I}=q^{(2)}(U_J).$

• (iii) Since $\eta _I$ is closed under intersections, it is easy to show that $\eta _{A/I}=\{U_{J/I}: J\in \eta _I\}$ is a base for a uniformity on A. By (ii), q is uniformly continuous. Let $J\in \eta _I$ and $*\in \{\odot ,\rightarrow \}.$ Since $\mathop \equiv \limits ^{J}$ is congruence relation, $U_J*U_J\subseteq U_J.$ Hence

  $$\begin{aligned} U_{J/I}*U_{J/I}= & {} q^{(2)}(U_J)*q^{(2)}(U_J)=q^{(2)}(U_J*U_J)\\\subseteq & {} q^{(2)}(U_J)=U_{J/I}. \end{aligned}$$

  By Corollary 3.5, A / I is uniform BL-algebra. $\square$

Proposition 5.12

Suppose F is a filter in BL-algebra A and q from A onto A / F is quotient map. Then:

• (i) the set $S\subseteq A/F$ is a filter in A / F iff $S=K/F,$ for some filter K of A containing F, 

• (ii) if K is a filter of A containing F, then $U_{K/F}=q^{(2)}(U_K),$

• (iii) if $\varrho _F=\{K: K\ is\ a\ filter\ in\ A, F\subseteq K\},$ then ${\mathcal {K}}_{A/F}=\{U_{K/F}: K\in \varrho _F\}$ is a base for a uniformity on A / F such that q is uniformly continuous. Moreover, A / F is uniform BL-algebra.

Proof

(i) Let $S\subseteq A/F$ be a filter and $K=\{x\in A:\frac{x}{F}\in S\}.$ From that $\frac{1}{F}\in S,$ we get that $1\in K.$ If x and $x\rightarrow y$ are two elements of K,  then $\frac{x}{F},\frac{x}{F}\rightarrow \frac{y}{F}$ belong to S. By Proposition 2.1(ii), $\frac{y}{F}\in S,$ hence $y\in K.$ Proposition 2.1(ii), implies that K is a filter. Clearly $F\subseteq K.$ Put $H=K/F=\{\frac{x}{F}:x\in K\}.$ If $\frac{z}{F}\in H,$ then there is a $x\in K$ such that $\frac{z}{F}=\frac{x}{F}.$ Since $x\rightarrow z$ and x are in $F\subseteq K,$ by Proposition 2.1(ii), $z\in K.$ Hence $\frac{z}{F}\in S.$ It is easy to prove that $S\subseteq H,$ so $S=H.$

Conversely, let $S=K/F,$ where F and K are filters of A and $F\subseteq K.$ Clearly $\frac{1}{F}\in S.$ If $\frac{x}{F}$ and $\frac{x}{F}\rightarrow \frac{y}{F}$ are in K / F,  then x and $x\rightarrow y,$ both, are in K. Since K is a filter, by Proposition 2.1(ii), $y\in K.$ So $\frac{y}{F}\in K/F,$ which implies that K / F is a filter. (ii)

$$\begin{aligned} \left( \frac{x}{I},\frac{y}{I}\right) \in U_{K/F}\Leftrightarrow & {} \frac{x\rightarrow y}{F},\frac{y\rightarrow x}{F}\in K/F\\\Leftrightarrow & {} x\rightarrow y, y\rightarrow x\in K\Leftrightarrow (x,y)\in U_K. \end{aligned}$$

Hence $U_{K/F}=q^{(2)}(U_K).$

(iii) The proof is similar to Proposition 5.11. $\square$

Example 5.13

Let A BL-algebra of Example 3.2. Then for any $a\in [0,1],$$F_a=(a,1]$ is a filter and $I_a=[0,a)$ is an ideal in [0, 1]. Clearly, $N(F_a)=\{0\},$$N(I_a)=A{\setminus }\{0\},$$U_{I_a}=A\times A=U_{N(I_a)}$ and

$$\begin{aligned} U_{F_a}= & {} \{(x,y)\in A\times A: a<x\le y\ or\ a<y\le x\}, \\ U_{N(F_a)}= & {} A\times A, \end{aligned}$$

Thus $K=\{U_{F_a}: a\in A\}$ is a base for a uniformity ${\mathcal {K}}$ and $H=\{U_{I_a}:a\in A\}=\{A\times A\}$ is a base for uniformity ${\mathcal {H}}=\{A\times A\}.$

But $O=\{(x,y): x' = y' \} =(A{\setminus }\{0\}\times A{\setminus }\{0\})\cup \{(0,0)\}.$ Hence ${\mathcal {L}}=\{ A\times A, O\}.$ Also, by Propositions 5.11 and 5.12, if $F=F_{1/2}$ and $I=I_{1/2},$ then for $a\le 1/2,$

$$\begin{aligned} U_{{F_a}/F}= & {} q^{(2)}(U_{F_a})=\{(x/F,y/F):a<x\\&\le y\ or\ a<y\le x\} \end{aligned}$$

and for $a>1/2,$$U_{{I_a}/I}=q^{(2)}(U_{I_a})=A/I\times A/I.$

Proposition 5.14

Let A and B be two BL-algebras. Then:

• (i) if $\varrho _{A}$ and $\varrho _B$ are two families of filters of A and B,  respectively, which are closed under intersections, then $\{U_{F_1\times F_2}: F_1\in \varrho _A, F_2\in \varrho _B\}$ is a base for product uniformity such that $A\times B$ is uniform BL-algebra.

• (ii) if $\eta _{A}$ and $\eta _B$ are two families of ideals of A and B,  respectively, which are closed under intersections, then $\{U_{I_1\times I_2}: I_1\in \varrho _A, I_2\in \varrho _B\}$ is a base for product uniformity such that $A\times B$ is uniform BL-algebra.

Proof

(i) The set $\{F_1\times F_2: F_1\in \varrho _A, F_2\in \varrho _B\}$ is a family of filters in $Z=A\times B$ that is closed under intersections. By Theorem 5.1, $\{U_{F_1\times F_2}: F_1\in \varrho _A, F_2\in \varrho _B\}$ is a base for a uniformity ${\mathcal {K}}_{A\times B}$ such that $(A\times B,{\mathcal {K}}_{A\times B})$ is a uniform BL-algebra. For any $F_1\in \varrho _A$ and $F_2\in \varrho _B,$ we have

$$\begin{aligned} U_{F_1 \times F_2}= & {} \left\{ {((x,y),(x_1,y_1)) \in Z \times Z: (x,y)\mathop \equiv \limits ^{{F_1} \times {F_2} }(x_1,y_1) } \right\} \\= & {} \left\{ {((x,y),(x_1,y_1)) \in Z \times Z: x\mathop \equiv \limits ^{F_1} x_1,y\mathop \equiv \limits ^{F_2} y_1 } \right\} \\= & {} \left\{ ((x,y),(x_1,y_1)) \in Z \times Z: (x,x_1) \in U_{F_1},\right. \\&\quad \left. (y,y_1) \in U_{F_2} \right\} = W_{U_{F_1},U_{F_2}} \end{aligned}$$

Hence ${\mathcal {K}}_{A\times B}$ is product uniformity on $A\times B.$

(ii) The proof is similar to that of (i). $\square$