Ratio Mathematica
Vol. 35, 2018, pp. 5-27

ISSN: 1592-7415
eISSN: 2282-8214

Quasi-Uniformity on BL-algebras

R.A. Borzooei ∗, N. Kouhestani †

Received: 05-06-2018 Accepted: 15-10-2018. Published: 18-12-2018

doi:10.23755/rm.v35i0.423

c©Borzooei, Kouhestani

Abstract

In this paper, by using the notation of filter in a BL-algebra A, we
introduce the quasi-uniformity Q and uniformity Q∗ on A. Then we
make the topologies T (Q) and T (Q∗) on A and show that
(A,∧,∨,�, T (Q)) is a compact connected topological BL-algebra and
(A, T (Q∗)) is a topological BL-algebra. Also we study Q∗-cauchy fil-
ters and minimal Q∗-filters on BL-algebra A and prove that the bi-
completion (Ã, Q̃) of quasi-uniform BL-algebra (A, Q) is a topological
BL-algebra.
2010 MSC: 06B10, 03G10.
Keywords : BL-algebra, (semi)topological BL-algebra, filter, Quasi-
uniforme space, Bicompletion

1 Introduction

BL-algebras have been introduced by Hájek [11] in order to investigate many-
valued logic by algebraic means. His motivations for introducing BL-algebras

∗Department of Mathematics, Shahid Beheshti University, Tehran, Iran; bor-
zooei@sbu.ac.ir
†Department of Mathematics, Sistan and Balouchestan University, Zahedan, Iran;

Kouhestani@math.usb.ac.ir

5



R. A. Borzooei, N. Kouhestan

were of two kinds. The first one was providing an algebraic counterpart of
a propositional logic, called Basic Logic, which embodies a fragment com-
mon to some of the most important many-valued logics, namely Lukasiewicz
Logic, Gödel Logic and Product Logic. This Basic Logic (BL for short) is
proposed as ”the most general” many-valued logic with truth values in [0,1]
and BL-algebras are the corresponding Lindenbaum-tarski algebras. The
second one was to provide an algebraic mean for the study of continuous
t-norms (or triangular norms) on [0,1]. In 1973, André Weil [24] introduced
the concept of a uniform space as a generalization of the concept of a metric
space in which many non-topological invariant can be defined. This concept
of uniformity fits naturally in the study of topological groups. The study of
quasi-uniformities started in 1948 with Nachbin’s investigations on uniform
preordered spaces. In 1960, Á. Csaszar introduced quasi-uniform spaces
and showed that every topological space is quasi-uniformizable. This result
established an interesting analogy between metrizable spaces and general
topological spaces. Just as a metrizable space can be studied with refer-
ence to particular compatible metric(s), a topological space can be studied
with reference to particular compatible quasi-uniformity(ies). In this and
some other respects, a quasi-uniformity is a more natural generalization of
a metric than is a uniformity. Quasi-uniform structures were also studied
in algebraic structures. In particular the study of paratopological groups
and asymmetrically normed linear spaces with the help of quasi-uniformities
is well known. See for example, [17], [18], [19], [20]. In the last ten years
many mathematicians have studied properties of BL-algebras endowed with
a topology. For example A. Di Nola and L. Leustean [9] studied compact
representations of BL-algebras, L. C. Ciungu [7] investigated some concepts
of convergence in the class of perfect BL-algebras, J. Mi Ko and Y. C. Kim
[21] studied relationships between closure operators and BL-algebras.
In [2] and [4] we study (semi)topological BL-algebras and metrizability on
BL-algebras. We showed that continuity the operations � and → imply
continuity ∧ and ∨. Also, we found some conditions under which a locally
compact topological BL-algebra become metrizable. But in there we can not
answer some questions, for example:
(i) Is there a topology U on BL-algebra A such that (A,U) be a (semi)topo-
logical BL-algebra?
(ii) Is there a topology U on a BL-algebra A such that (A,U) be a compact
connected topological BL-algebra?
(iii) Is there a topological BL-algebra (A,U) such that T0,T1 and T2 spaces
be equivalent?
(iv) If (A,→,U) is a semitopological BL-algebra, is there a topology V
coarsere than U or finer than U such that (A,V) be a (semi)topological

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Quasi-Uniformity on BL-algebras

BL-algebra?

Now in this paper, we answer to some above questions and get some
interesting results as mentioned in abstract.

2 Preliminary

Recall that a set X with a family U = {Uα}α∈I of its subsets is called a
topological space, denoted by (X,U), if X,∅ ∈ U, the intersection of any
finite numbers of members of U is in U and the arbitrary union of members
of U is in U. The members of U are called open sets of X and the complement
of X ∈U, that is X \U, is said to be a closed set. If B is a subset of X, the
smallest closed set containing B is called the closure of B and denoted by B
(or cluB). A subset P of X is said to be a neighborhood of x ∈ X, if there
exists an open set U such that x ∈ U ⊆ P . A subfamily {Uα : α ∈ J} of U is
said to be a base of U if for each x ∈ U ∈U there exists an α ∈ J such that
x ∈ Uα ⊆ U, or equivalently, each U in U is a union of members of {Uα}.
Let Ux denote the totality of all neighborhoods of x in X. Then a subfamily
Vx of Ux is said to form a fundamental system of neighborhoods of x, if for
each Ux in Ux, there exists a Vx in Vx such that Vx ⊆ Ux. (X,U) is said to
be compact, if each open covering of X is reducible to a finite open covering.
Also (X,U) is said to be disconnected if there are two nonempty, disjoint,
open subsets U,V ⊆ X such that X = U ∪V , and connected otherwise. The
maximal connected subset containing a point of X is called the component
of that point. Topological space (X,U) is said to be:
(i) T0 if for each x 6= y ∈ X, there is one in an open set excluding the other,
(ii) T1 if for each x 6= y ∈ X, each are in an open set not containing the
other,
(iii) T2 if for each x 6= y ∈ X, both are in two disjoint open set.(See [1])

Definition 2.1. [1] Let (A,∗) be an algebra of type 2 and U be a topology
on A. Then A = (A,∗,U) is called a
(i) left (right) topological algebra if for all a ∈ A, the map ∗a : A → A is
defined by x → a∗x ( x → x∗a) is continuous, or equivalently, for any x in
A and any open set U of a∗x (x∗a), there exists an open set V of x such
that a∗V ⊆ U (V ∗a ⊆ U).
(ii) semitopological algebra if A is a right and left topological algebra.
(iii) topological algebra if the operation ∗ is continuous, or equivalently, if
for any x,y in A and any open set (neighborhood) W of x ∗ y, there exist
two open sets (neighborhoods) U and V of x and y, respectively, such that
U ∗V ⊆ W.

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R. A. Borzooei, N. Kouhestan

Proposition 2.2. [1] Let (A,∗) be a commutative algebra of type 2 and U
be a topology on A. Then right and left topological algebras are equivalent.
Moreover, (A,∗,U) is a semitopological algebra if and only if it is right or
left topological algebra.

Definition 2.3. [1] Let A be a nonempty set and {∗i}i∈I be a family of
operations of type 2 on A and U be a topology on A. Then
(i) (A,{∗i}i∈I,U) is a right(left) topological algebra if for any i ∈ I, (A,∗i,U)
is a right (left) topological algebra.
(ii) (A,{∗i}i∈I,U) is a semitopological (topological) algebra if for all i ∈ I,
(A,∗i,U) is a semitopological (topological) algebra.

Definition 2.4. [11] A BL-algebra is an algebra A = (A,∧,∨,�,→, 0, 1) of
type (2, 2, 2, 2, 0, 0) such that (A,∧,∨, 0, 1) is a bounded lattice, (A,�, 1) is
a commutative monoid and for any a,b,c ∈ A,

c ≤ a → b ⇔ a� c ≤ b, a∧ b = a� (a → b), (a → b) ∨ (b → a) = 1.

Let A be a BL-algebra. We define a′ = a → 0 and denote (a′)′ by a′′. The
map c : A → A by c(a) = a′, for any a ∈ A, is called the negation map. Also,
we define a0 = 1 and an = an−1 �a, for all natural numbers n.

Example 2.5. [11] (i) Let “�” and “→” on the real unit interval I = [0, 1]
be defined as follows:

x�y = min{x,y} x → y =
{

1 , x ≤ y,
y , otherwise.

Then I = (I, min, max,�,→, 0, 1) is a BL-algebra.
(ii) Let � be the usual multiplication of real numbers on the unit in-

terval I = [0, 1] and x → y = 1 iff, x ≤ y and y/x otherwise. Then
I = (I, min, max,�,→, 0, 1) is a BL-algebra.

Proposition 2.6. [11] Let A be a BL-algebra. The following properties
hold.
(B1) x�y ≤ x,y and x� 0 = 0,
(B2) x ≤ y implies x�z ≤ y �z,
(B3) x ≤ y iff x → y = 1,
(B4) 1 → x = x, 1 �x = x,
(B5) y ≤ x → y,
(B6) x → (y → z) = (x�y) → z = y → (x → z),
(B7) x∨y = ((x → y) → y) ∧ ((y → x) → x),
(B8) x ≤ y ⇒ x → z ≥ y → z, z → x ≤ z → y,

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Quasi-Uniformity on BL-algebras

(B9) x → y ≤ (z → x) → (z → y),
(B10) x → y ≤ (y → z) → (x → z),
(B11) x → (y ∧z) = (x → y) ∧ (x → z),
(B12) (y ∧z) → x = (y → x) ∨ (z → x),
(B13) (y ∨z) → x = (y → x) ∧ (z → x),
(B14) x → y ≤ x�z → y �z,
(B15) (x → y) � (y → z) ≤ x → z,
(B16) (x → y) � (a → z) ≤ (x∨a) → (y ∨z),
(B17) (x → y) � (a → z) ≤ (x∧a) → (y ∧z),
(B18) (x → y) � (a → z) ≤ (x�a) → (y �z).

Definition 2.7. [11] A filter of a BL-algebra A is a nonempty set F ⊆ A
such that x,y ∈ F implies x�y ∈ F and if x ∈ F and x ≤ y imply y ∈ F ,
for any x,y ∈ A.

It is easy to prove that if F is a filter of a BL-algebra A, then for each
x,y ∈ F, x∧y, x∨y and x → y are in F

Proposition 2.8. [11] Let F be a subset of BL-algebra A such that 1 ∈ F .
Then the following conditions are equivalent.
(i) F is a filter.
(ii) x ∈ F and x → y ∈ F imply y ∈ F.
(iii) x → y ∈ F and y → z ∈ F imply x → z ∈ F .

Proposition 2.9. [11] Let F be a filter of a BL-algebra A. Define x ≡F
y ⇔ x → y,y → x ∈ F. Then ≡F is a congruence relation on A. Moreover,
if x/F = {y ∈ A : y ≡F x}, then
(i) x/F = y/F ⇔ y ≡F x,
(ii) x/F = 1/F ⇔ x ∈ F.

Definition 2.10. [2] (i) Let A be a BL-algebra and (A,{∗i},U) be a semi-
topological (topological) algebra, where {∗i}⊆{∧,∨,�,→}, then (A,{∗i},U)
is called a semitopological (topological) BL-algebra.

Remark 2.11. If {∗i} = {∧,∨,�,→}, we consider A = (A,U) instead of
(A,{∧,∨,�,→},U), for simplicity.

Proposition 2.12. [2] Let (A,{�,→},U) be a topological BL-algebra. Then
(A,U) is a topological BL-algebra.

Notation. From now on, in this paper, we use of BL-filter instead of
filter in BL-algebras.

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R. A. Borzooei, N. Kouhestan

Definition 2.13. [10] Let X be a non-empty set. A family F of nonempty
subsets of X is called a filter on X if (i) X ∈ F, (ii) for each F1,F2 of
elements of F, F1 ∩F2 ∈F and, (iii) if F ∈F and F ⊆ G, then G ∈F.

A subset B of a filter F on X is said to be a base of F if every set of F
contains a set of B.
If F is a family of nonempty subsets of X, then there exists the smallest
filter on X containing F, denoted with fil(F) and called generated filter by
F.

Definition 2.14. [10] A quasi-uniformity on a set X is a filter Q on X
such that
(i) 4 = {(x,x) ∈ X ×X : x ∈ A}⊆ q, for each q ∈ Q,
(ii) for each q ∈ Q, there is a p ∈ Q such that p◦p ⊆ q, where

p◦p = {(x,y) ∈ X ×X : ∃z ∈ A s.t (x,z), (z,y) ∈ p}.

The pair (X,Q) is called a quasi-uniform space.

If Q is a quasi-uniformity on a set X, q ∈ Q and q−1 = {(x,y) : (y,x) ∈
q}, then Q−1 = {q−1 : q ∈ Q} is also a quasi-uniformity on X called the
conjugate of Q. It is well-known that if Q satisfies condition: q ∈ Q implies
q−1 ∈ Q, then Q is a uniformity. Furthermore, Q∗ = Q∨Q−1 is a uniformity
on X. If Q and R are quasi-uniformities on X and Q ⊆ R, then Q is called
coarser than R. A subfamily B of quasi-uniformity Q is said to be a base for
Q if each q ∈ Q contains some member of B.(See [10])

Proposition 2.15. [22] Let B be a family of subsetes of X ×X such that
(i) 4⊆ q, for each q ∈B,
(ii) for q1,q2 ∈B, there exists a q3 ∈B such that q3 ⊆ q1 ∩ q2,
(iii) for each q ∈B, there is a p ∈B such that p◦p ⊆ q.
Then, there is the unique quasiuniformity Q = {q ⊆ X ×X : for some p ∈
B,p ⊆ q} on X for which B is a base.

The topology T(Q) = {G ⊆ X : ∀x ∈ G ∃q ∈ Q s.t q(x) ⊆ G} is called
the topology induced by the quasi-uniformity Q.

Definition 2.16. [10] (i) A filter G on quasi-uniform space (X,Q) is called
Q∗-cauchy filter if for each U ∈ Q, there is a G ∈G such that G×G ⊆ U.
(ii) A quasi-uniform space (X,Q) is called bicomplete if each Q∗-cauchy filter
converges with respect to the topology T(Q∗).
(iii) A bicompletion of a quasi-uniform space (X,Q) is a bicomplete quasi-
uniform space (Y,V) that has a T(V∗)-dense subspace quasi-unimorphic to

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Quasi-Uniformity on BL-algebras

(X,Q).
(iv) A Q∗-cauchy filter on a quasi-uniform space (X,Q) is minimal provided
that it contains no Q∗-cauchy filter other than itself.

Lemma 2.17. [10] Let G be a Q∗-cauchy filter on a quasi-uniform space
(X,Q). Then, there is exactly one minimal Q∗-cauchy filter coarser than G.
Furthermore, if B is a base for G, then {q(B) : B ∈ B and q is a symetric
member of Q∗} is a base for the minimal Q∗-cauchy filter coarser than G.

Lemma 2.18. [10] Let (X,Q) be a T0 quasi-uniform space and X̃ be the
family of all minimal Q∗-cauchy filters on (A,Q). For each q ∈ Q, let

q̃ = {(G,H) ∈ X̃ × X̃ : ∃G ∈G and H ∈H s.t G×H ⊆ q},

and Q̃ = fil{q̃ : q ∈ Q}. Then the following statements hold:
(i) (X̃,Q̃) is a T0 bicomplete quasi-uniform space and (X,Q) is a quasi-

uniformly embedded as a T((̃Q∗))-dense subspace of (X̃,Q̃) by the map i :

X → X̃ such that, for each x ∈ X, i(x) is the T(Q∗)-neighborhood filter at
x. Furthermore, the uniformities Q̃∗ and (̃Q∗) coincide.

Notation. From now on, in this paper we let A be a BL-algebra and
F be a family of BL-filters in A which is closed under intersection , unless
otherwise state.

3 Quasi-uniformity on BL-algebras

In this section, by using of BL-filters we introduce a quasi-uniformity Q
on BL-algebra A and stay some properties it. We show that (A,Q) is not
a T1 and T2 quasi-uniform space but it is a T0 quasi-uniform space. Also
we study Q∗-cauchy filters, minimal Q∗-cauchy filters and we make a quasi-
uniform space (Ã,Q̃) of minimal Q∗-cauchy filters of (A,Q) which admits
the structure of a BL-algebra.

Lemma 3.1. Let F be a BL-filter of BL-algebra A and F?(x) = {y : y →
x ∈ F}, for each x ∈ A. Then for each x,y ∈ A, the following properties
hold.
(i) x ≤ y implies F?(x) ⊆ F?(y),
(ii) F?(x) ∧F?(y) = F?(x∧y) = F?(x) ∩F?(y),
(iii) F?(x) ∨F?(y) ⊆ F?(x∨y),
(iv) F?(x) �F?(y) ⊆ F?(x�y),
(v) If for each a ∈ A, a�a = a, then F?(x) �F?(y) = F?(x�y),

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R. A. Borzooei, N. Kouhestan

(vi) x ∈ F ⇔ 1 ∈ F?(x) ⇔ F?(x) = A,
(vii) For a,b ∈ A, if a∨ b ∈ F?(x), then a,b ∈ F?(x),
(viii) If y ∈ F?(x), then F?(y) ⊆ F?(x).

Proof. (i) Let x,y ∈ A, such that x ≤ y and z ∈ F?(x). Then by (B8),
z → x ≤ z → y. Since F is a BL-filter and z → x ∈ F , z → y is in F and so
z ∈ F?(y).
(ii) Let x,y ∈ A, such that a ∈ F?(x) and b ∈ F?(y). Then a → x ∈ F
and b → y ∈ F and so (a → x) � (b → y) ∈ F . Since by (B17), (a →
x) � (b → y) ≤ (a ∧ b) → (x ∧ y), we get (a ∧ b) → (x ∧ y) ∈ F. Thus,
a ∧ b ∈ F?(x ∧ y). Now, if a ∈ F?(x ∧ y), since a → (x ∧ y) ∈ F and by
(B11), a → (x∧ y) = (a → x) ∧ (a → y), we conclude that a → x ∈ F and
a → y ∈ F. Hence a ∈ F?(x) ∩ F?(y). Finally, let a ∈ F?(x) ∩ F?(y). Since
a = a∧a, then a ∈ F?(x) ∧F?(y).
(iii), (iv) The proof is similar to the proof of (ii), by some modification.
(v) Let x,y ∈ A such that z ∈ F?(x�y). Then z → (x�y) ∈ F . By (B8),
z → (x�y) ≤ z → x and z → (x�y) ≤ z → y which imply that z → x,z →
y ∈ F. Hence z is in both F?(x) and F?(y) and so z = z�z ∈ F?(x)�F?(y).
(vi) The proof is clear.
(vii), (viii) The proof come from by (B13) and (B15).

Lemma 3.2. Let F be a BL-filter of BL-algebra A. Define F? = {(x,y) ∈
A×A : y ∈ F?(x)} and F∗? = F? ∩F−1? . Then
(i) F−1? = {(x,y) ∈ A×A : x → y ∈ F},
(ii) F∗? = {(x,y) ∈ A×A : x ≡F y} = F∗

−1
? ,

(iii) F∗? (x) = {y : x ≡F y},
(iv) F−1? (x) → y ⊆ F?(x → y),
(v) If •∈ {∧,∨,�,→}, then F∗? (x) •F∗? (y) ⊆ F∗? (x•y).

Proof. The proof of (i), (ii) and (iii) are clear.
(iv) Let a ∈ F−1? (x) → y. Then there exists a z ∈ F−1? (x) such that a = z → y
and x → z ∈ F. By (B10), (z → y) → (x → y) ≥ x → z. Since F is a filter,
(z → y) → (x → y) ∈ F. Hence a = z → y ∈ F?(x → y).
(v) Let a ∈ F∗? (x) and b ∈ F∗? (y). Then by (iii), a ≡F x and b ≡F y. By
Proposition 2.9, a• b ≡F x•y. Therefore, a• b ∈ F∗? (x•y).

Theorem 3.3. Let F be a family of BL-filters of BL-algebra A which is
closed under finite intersection. Then the set B = {F? : F ∈ F} is a base
for the unique quasi-uniformity Q = {q ⊆ A × A : ∃F ∈ F s.t F? ⊆ q}.
Moreover, Q∗ = {q ⊆ A×A : ∃F ∈F s.t F∗? ⊆ q}.

Proof. We prove that B satisfies in conditions (i), (ii) and (iii) of Proposition
2.15. For (i), it is easy to see that for each F ∈F, 4⊆ F?. Let F1,F2 ∈F

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Quasi-Uniformity on BL-algebras

and F = F1 ∩ F2. If (x,y) ∈ F?, then y → x ∈ F = F1 ∩ F2. Hence
(x,y) ∈ F1? ∩ F2?. This concludes that F? ⊆ F1? ∩ F2? and so (ii) is true.
Finally for (iii), let F ∈F and (x,y) ∈ F? ◦F?. Then there is a z ∈ A such
that (x,z) and (z,y) are both in F?. Hence z → x and y → z are in F. Since
F is a filter and by (B15), (y → z) � (z → x) ≤ y → x, we conclude that
y → x ∈ F. Hence F?◦F? ⊆ F? and so (iii) is true. Therefore, by Proposition
2.15, Q is a unique quasi-uniformity on A for which B is a base.

Now, we prove that

Q∗ = {q ⊆ A×A : ∃F ∈F s.t F∗? ⊆ q}.

First we prove that P = {q ⊆ A×A : ∃F ∈ F s.t F∗? ⊆ q} is a uniformity
on A. With a similar argument as above, we get {F∗? : F ∈F} is a base for
the quasi-uniformity P = {q ⊆ A×A : ∃F ∈ F s.t F∗? ⊆ q}. To prove that
P is a uniformity we have to show that for each q ∈P, q−1 is in P. Suppose
q ∈ P. Then there exists a F ∈ F, such that F∗? ⊆ q. By Lemma 3.2(ii),
F∗? = F

∗−1
? . Hence F

∗
? ⊆ q−1 and so q−1 ∈ P. Thus P is a uniformity on A

which contains Q. Since Q∗ = Q∨Q−1, then Q∗ ⊆P. On the other hand, if
q ∈P, then there is a F ∈F such that F∗? ⊆ q. Since F∗? = F? ∩F−1? ∈ Q∗,
we get that q ∈ Q∗. Therefore, Q∗ = P.

In Theorem 3.3, we call Q is quasi-uniformity induced by F, the pair
(A,Q) is quasi-uniform BL-algebra and the pair (A,Q∗) is uniform BL-
algebra.

Notation. From now on, F, Q and Q∗ are as in Theorem 3.3.

Example 3.4. Let I be the BL-algebra in Example 2.5 (i), and for each
a ∈ [0, 1), Fa = (a, 1]. Then Fa is a BL-filter in I and easily proved that
for each a,b ∈ [0, 1), Fa ∩ Fb = Fa∧b. Hence F = {Fa}a∈[0,1) is a family of
BL-filters which is closed under intersection. For each a ∈ [0, 1),

Fa? = (a, 1] × [0, 1], F−1a? = [0, 1] × (a, 1] and F
∗
a? = (a, 1] × (a, 1].

By Theorem 3.3, Q = {q : ∃a ∈ [0, 1) s.t (a, 1] × [0, 1] ⊆ q} and Q∗ = {q :
∃a ∈ [0, 1) s.t (a, 1] × (a, 1] ⊆ q}.

Recall that a map f from a (quasi)uniform space (X,Q) into a (quasi)uniform
space (Y,R) is (quasi) uniformly continuous, if for each V ∈ R, there exists
a U ∈ Q such that (x,y) ∈ U implies (f(x),f(y)) ∈ V. If f : (X,Q) ↪→ (Y,R)
is a quasi-uniform continuous map between quasi-uniform spaces, then f :
(X,Q∗) ↪→ (Y,R∗) is a uniform continuous map. (See [10])

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R. A. Borzooei, N. Kouhestan

Proposition 3.5. In BL-algebra A, for each a ∈ A, the mappings ta(x) =
a ∧ x, ra(x) = a ∨ x, la(x) = a � x and La(x) = a → x of quasi-uniform
BL-algebra (A,Q) into quasi-uniform BL-algebra (A,Q) are quasi-uniformly
continuous. Moreover, they are uniformly continuous mappings of uniform
BL-algebra (A,Q∗) into uniform BL-algebra (A,Q∗).

Proof. Let q ∈ Q. Then, there is a F ∈ F such that F? ⊆ q. If (x,y) ∈ F?,
then y → x ∈ F. By (B10) (a ∧ y) → (a ∧ x) ≥ y → x which implies that
(a∧y) → (a∧x) ∈ F ⊆ q. Hence ta is quasi-uniform continuous. Moreover,
ta : (A,Q

∗) ↪→ (A,Q∗) is uniform continuous. In a similar fashion and by use
of (B16), (B14) and (B9), we can prove that, respectively, ra, la and La are
quasi-uniform continuous of (A,Q) ↪→ (A,Q) and are uniform continuous of
(A,Q∗) ↪→ (A,Q∗).

Let (X,Q) be a (quasi)uniform space and B be a base for it. Recall (X,Q)
is
(i) T0 quasi-uniform if (x,y) and (y,x) are in

⋂
U∈B U, then x = y, for each

x,y ∈ X,
(ii) T1 quasi-uniform if 4 =

⋂
U∈B U,

(iii) T2 quasi-uniform if 4 =
⋂
U∈B U

−1 ◦U. (See [10])

Theorem 3.6. Quasi-uniform BL-algebra (A,Q) is not T1 and T2 quasi-
uniform. If {1} ∈ F, then (A,Q) is a T0 quasi-uniform space and uniform
BL-algebra (A,Q∗) is T0, T1 and T2 quasi-uniform space.

Proof. Let x,y ∈ A and F ∈ F. Since y → 1 = 1 ∈ F, we get that (1,y) ∈⋂
F∈F F?. Hence (A,Q) is not T0 quasi-uniform. Also since x → 1 = y → 1 ∈

F, we conclude that (1,x), (1,y) ∈ F?. Hence (x,y) ∈ F−1? ◦F? which implies
that 4 6=

⋂
F∈F F

−1
? ◦F?. So (A,Q) is not T2 quasi-unifom

Let {1} ∈ F and (x,y) and (y,x) be in
⋂
F∈F F?. Then for each F ∈ F,

x → y and y → x are in F. Hence x ≡{1} y, which implies that x = y.
Therefore, (A,Q) is T0 quasi-uniform. With a similar argument as above,
we can prove that (A,Q∗) is a T0 and T1 quasi-uniform space. To verify T2
quasi-uniformity, let (x,y) ∈

⋂
F∈F F

∗−1
? ◦F∗? . Then for each F ∈F there is

a z ∈ A such that (x,z) ∈ F∗−1? and (z,y) ∈ F∗? . By Lemma 3.2(ii), x ≡F y.
Since {1} ∈ F, we get that x = y. Therefore, (A,Q∗) is a T2 quasi-uniform
space.

Proposition 3.7. Let B be a base for a Q∗-cauchy filter G on quasi-uniform
BL-algebra (A,Q). Then the set {F∗? (B) : F ∈ F, B ∈ B} is a base for the
uniqe minimal Q∗-cauchy filter coarser than G.

14



Quasi-Uniformity on BL-algebras

Proof. By Lemma 2.17, the set {q(B) : B ∈ B, q−1 = q ∈ Q∗} is a base for
the unique minimal Q∗-cauchy filter G0 coarser than G. Let q−1 = q ∈ Q∗
and B ∈ B. Then for some F ∈ F, F∗? ⊆ q. So, F∗? (B) ⊆ q(B). Now, it is
easy to prove that the set {F∗? (B) : F ∈F, B ∈B} is a base for G0.

Proposition 3.8. F is a base for a minimal Q∗-cauchy filter on quasi-
uniform BL-algebra (A,Q).

Proof. Let C = {S ⊆ A : ∃F ∈F s.t F ⊆ S}. It is easy to prove that C is a
filter and F is a base for it. We prove that C is a Q∗-cauchy filter. For this,
let q ∈ Q. There is a F ∈ F such that F? ⊆ q. Since F is a filter, clearly
F × F ⊆ F? ⊆ q. Hence C is a Q∗-cauchy filter. Now, by Proposition 3.7,
the set {F∗? (F1) : F,F1 ∈ F} is a base for the unique minimal Q∗-cauchy
filter F0 coarser than C. To complete proof we show that for each F,F1 ∈F,
F∗? (F1) = F1. Let F,F1 ∈ F. If y ∈ F∗? (F1), then for some x ∈ F1, x ≡F y.
By Proposition 2.9, y ∈ F1. Hence F∗? (F1) ⊆ F1. Clearly, F1 ⊆ F∗? (F1).
Therefore, F1 = F

∗
? (F1). Thus proved that F is a base for F0.

Proposition 3.9. The set B = {F∗? (0) : F ∈ F} is a base for a minimal
Q∗-cauchy filter on quasi-uniform BL-algebra (A,Q).

Proof. Let C = {S ⊆ A : ∃F ∈ F s.t F∗? (0) ⊆ S}. It is easy to prove that C
is a filter and the set B = {F∗? (0) : F ∈ F} is a base for it. To prove that
C is a Q∗-cauchy filter, let q ∈ Q. There is a F ∈ F such that F? ⊆ q. If
x,y ∈ F∗? (0), then x ≡F y and so (x,y) ∈ F∗? ⊆ F? ⊆ q. This prove that
F∗? (0) × F∗? (0) ⊆ q. Hence C is a Q∗-cauchy filter. By Proposition 3.7, the
set {F∗? (F∗? (0)) : F ∈ F} is a base for the uniqe minimal Q∗-cauchy filter I
coarser than C. But it is easy to pove that fo each F ∈F, F∗? (F∗? (0)) = F∗? (0).
Therefore, B is a base for I.

Lemma 3.10. Let G and H be Q∗-cauchy filters on quasi-uniform BL-algebra
(A,Q). If • ∈ {∧,∨,�,→}, then G •H = {G • H : G ∈ G, H ∈ H} is a
Q∗-cauchy filter base on quasi-uniform BL-algebra (A,Q).

Proof. Let C = {S ⊆ A : ∃G, H s.t G ∈G, H ∈H, G•H ⊆ S}. It is easy to
prove that C is a filter and the set B = {G•H : G ∈G, H ∈H} is a base for
it. We prove that C is a Q∗-cauchy filter. For this, let q ∈ Q. Then for some a
F ∈F, F? ⊆ q. Since G,H are Q∗-cauchy filters, there are G ∈G and H ∈H
such that G×G ⊆ F? and H×H ⊆ F?. We show that G•H×G•H ⊆ F? ⊆ q.
Let g1,g2 ∈ G and h1,h2 ∈ H. Then (g1,g2), (g2,g1), (h1,h2), (h2,h1) are in
F?. So g1 ≡F g2 and h1 ≡F h2. By Proposition 2.9, g1 •h1 ≡F g2 •h2, which
implies that (g1 •h1,g2 •h2) ∈ F?.

15



R. A. Borzooei, N. Kouhestan

Theorem 3.11. There is a quasi-uniform space (Ã,Q̃) of minimal Q∗-cauchy
filters of quasi-uniform BL-algebra (A,Q) that admits a BL-algebra structure.

Proof. Let à be the family of all minimal Q∗-cauchy filters on (A,Q). Let
for each q ∈ Q,

q̃ = {(G,H) ∈ Ã× Ã : ∃G ∈G,H ∈H s.t G×H ⊆ q}.

If Q̃ = fil{q̃ : q ∈ Q}, then (Ã,Q̃) is a quasi-uniform space of minimal
Q∗-cauchy filters of (A,Q). Let G,H∈ Ã. Since G,H are minimal Q∗-cauchy
filters on A, then by Lemma 3.10, G∧H, G∨H, G�H and G →H are Q∗-
cauchy filter bases on A. Now, we define G fH, G gH, G }H and G ↪→H as
the minimal Q∗-cauchy filters contained G∧H, G∨H, G�H and G →H,
respectively. Thus, G fH, G gH, G }H and G ↪→H are in Ã. Now, we
will prove that (Ã,f,g,}, ↪→,I,F0) is a BL-algebra, where I is minimal
Q∗-cauchy filter in Proposition 3.9 and F0 is minimal Q∗-cauchy filter in
Proposition 3.8. For this, we consider the following steps:
(1) (Ã,f,g) is a bounded lattice.

Let G,H,K∈ Ã. We consider the following cases:
Case 1.1: G fG = G, G gG = G
By Proposition 3.7, S1 = {F∗? (G) : G ∈G,F ∈F} and S2 = {F∗? (G1 ∧G2) :
G1,G2 ∈G,F ∈F} are bases of the minimal Q∗-cauchy filters G and G fG,
respectively. First, we show that S2 ⊆ S1. Let F∗? (G1 ∧ G2) ∈ S2. Put
G = G1 ∩ G2, then G ∈ G. Let y ∈ F∗? (G). Then there is a x ∈ G such
that (x,y) ∈ F∗? . Since x ∧ x = x, it follows that (x ∧ x,y) ∈ F∗? and so
y ∈ F∗? (G1 ∧G2). Hence S2 ⊆ S1. Therefore, G fG ⊆G. By the minimality
of G, G fG = G. The proof of the other case is similar.
Case 1.2: G fH = H fG, G gH = H gG
By Proposition 3.7, S1 = {F∗? (G ∧ H) : G ∈ G,H ∈ H,F ∈ F} and S2 =
{F∗? (H ∧ G) : G ∈ G,H ∈ H,F ∈ F} are bases of G fH and H fG,
respectively. For each G ∈ G and H ∈ H, since G ∧ H = H ∧ G, for each
F ∈F, F∗? (G∧H) = F∗? (H ∧G). Hence G fH = H fG. The proof of the
other case is similar.
Case 1.3: G f (H fK) = (G fH) fK, G g (H gK) = (G gH) gK
By Proposition 3.7, the families

S1 = {F∗1?(F
∗
2?(G∧H) ∧K) : G ∈G,H ∈H,K ∈K,F1,F2 ∈F},

S2 = {F∗1?(G∧F
∗
2?(H ∧K) : G ∈G,H ∈H,K ∈K,F1,F2 ∈F}

are bases for the minimal Q∗-cauchy filters (G f H) f K and G f (H f K),
respectively. Let F∗1?(F

∗
2?G∧ (H ∧K) ∈ S2 and F = F1 ∩F2. Then F ∈ F.

16



Quasi-Uniformity on BL-algebras

Now, we show that F∗? (F
∗
? (G ∧ H) ∧ K) ⊆ F∗1?(G ∧ F∗2?(H ∧ K). Let y ∈

F∗? (F
∗
? (G ∧ H) ∧ K). Then there are x ∈ F∗? (G ∧ H), k ∈ K, g ∈ G and

h ∈ H such that y ≡F x ∧ k and x ≡F g ∧ h. Hence y ≡F (g ∧ h) ∧ k =
g∧(h∧k), which implies that y ∈ F∗? (G∧F∗? (H∧K) ⊆ F∗1?(G∧F∗2?(H∧K).
Therefore, G f (HfK) ⊆ (G fH) fK. By the minimality of (G fH) fK,
G f (H fK) = (G fH) fK. The proof of the other case is similar.
Case 1.4: G f (G gH) = G, G g (G fH) = G
It is enough to prove that G f (G g H) = G. The proof of the other case
is similar. By Proposition 3.7, the families S1 = {F∗? (G) : G ∈ G,F ∈ F}
and S2 = {F∗1?(G1 ∧ F∗2?(G2 ∨ H) : G1,G2 ∈ G,H ∈ H,F1,F2 ∈ F} are
bases for the minimal Q∗-cauchy filters G and Gf (GgH) , respectively. Let
F∗1?(G1 ∧F∗2?(G2 ∨H) ∈ S2. Put G = G1 ∩G2 and F = F1 ∩F2. We prove
that F∗? (G) ⊆ F∗1?(G1 ∧F∗2?(G2 ∨H). Let y ∈ F∗? (G). Then there is a g ∈ G
such that y ≡F g. If h ∈ H, since g = g∧ (g∨h), then y ≡F g∧ (g∨h) and
so y ∈ F∗1?(G1 ∧F∗2?(G2 ∨H). Hence G f (G gH) ⊆G. By the minimality of
G, we conclude that G f (G gH) = G.
Now the cases 1.1,1.2,1.3,1.4 imply that (Ã,f,g) is a lattice.
Case 1.5: The lattice (Ã,f,g) is bounded.
For this, for each G,H ∈ Ã, define G ≤H ⇔ G fH = G. It is clear that
(Ã,≤) is a partial ordered. Now, we prove that for each G ∈ Ã, I ≤G ≤F0.
First, we show that I ≤G. Let S ∈I. Then for some a F ∈F, F∗? (0) ⊆ S.
Since G is a minimal Q∗-cauchy filter, there is a G ∈G such that G×G ⊆ F?.
We show that F∗? (G ∧ F∗? (0)) ⊆ S. Let y ∈ F∗? (G ∧ F∗? (0)). Then there are
g ∈ G and x ∈ F∗? (0) such that y ≡F g∧x. On the other hand, since x ≡F 0,
we get g ∧x ≡F 0. Hence y ≡F 0 which implies that y ∈ F∗? (0) ⊆ S. Since
F∗? (G∧F∗? (0)) ∈GfI, then S ∈GfI. By the minimality of GfI, GfI = I.
Now, we prove that G ≤F0. By Proposition 3.7, the set S1 = {F∗? (G∧F1) :
G ∈ G, F,F1 ∈ F} is a base for G f F0. Let F∗? (G ∧ F1) ∈ S1. We prove
that F∗? (G) ⊆ F∗? (G∧F1). Let y ∈ F∗? (G). Then, there is a g ∈ G such that
y ≡F g = g∧1. Hence y ∈ F∗? (G∧F1). By the minimality of G, GfF0 = G.
(2) (Ã,}) is a commutative monoid
Case 2.1: (Ã,}) is a commutative semigroup.
We will prove that G } (H }K) = (G }H) }K. By Proposition 3.7, the sets

S1 = {F∗1?(G�F
∗
2?(H �K)) : G ∈G,H ∈H,K ∈K,F1,F2 ∈F},

S2 = {F∗1?(F
∗
2?(G�H) �K)) : G ∈G,H ∈H,K ∈K,F1,F2 ∈F}

are bases from G } (H }K) and (G }H) }K, respectively. Let F∗1?(F∗2?(G�
H)�K)) ∈ S2, F = F1∩F2 and y ∈ F∗? (G�F∗? (H�K). Then there are g ∈ G,
x ∈ F∗? (H�K), h ∈ H and k ∈ K such that y

F
≡ g�x and x

F
≡ h�k. Hence

17



R. A. Borzooei, N. Kouhestan

y
F
≡ g�(h�k) = (g�h)�k and so y ∈ F∗? (F∗? (G�H)�K) ⊆ F∗1?(F∗2?(G�H)�

K)). Therefore, S2 ⊆ S1 which implies that (G }H) }K ⊆ G } (H }K).
Now, by the minimality of G } (H }K), G } (H }K) = (G }H) }K. Fi-
nally, it is easy to prove that G }H = H }G.
Case 2.2: (Ã,}) is a monoid
We prove that G}F0 = G. By Proposition 3.7, the set S2 = {F∗? (G�F1) : G ∈
G,F,F1 ∈F} is a base for G }F0. It is clear that for each F∗? (G�F1) ∈ S2,
F∗? (G) ⊆ F∗? (G�F1) and this implies that G } F0 ⊆ G. By the minimality
of G, G }F0 = G.
(3) G } (G ↪→H) = G fH
By Proposition 3.7, the families

S1 = {F∗? (G∧H) : G ∈G,H ∈H,F ∈F},

S2 = {F∗1?(G1 �F
∗
2?(G2 → H)) : G1,G2 ∈G,H ∈H,F1,F2 ∈F}

are bases for G fH and G } (G ↪→H), respectively. Let F∗1?(G1 �F∗2?(G2 →
H)) ∈ S2, G = G1 ∩G2 and F = F1 ∩F2. We will prove that F∗? (G∧H) ⊆
F∗1?(G1 � F∗2?(G2 → H)). Let y ∈ F∗? (G ∧ H). Then there are g ∈ G and
h ∈ H such that y ≡F g ∧ h. It follows from g ∧ h = g � (g → h) which
y ∈ F∗1?(G1 �F∗2?(G2 → H)). Hence F∗? (G∧H) ⊆ F∗1?(G1 �F∗2?(G2 → H))
which implies that G}(G ↪→H) ⊆G fH. Now, by the minimality of G fH,
we get G } (G ↪→H) = G fH.
(4) G ≤H ↪→K⇔G }H≤K
First, we prove the following statements:
(a) G ≤H⇔G ↪→H = F0
(b) G ↪→ (H ↪→K) = G }H ↪→K.
(a) To prove it, let G ↪→H = F0. Then G} (G ↪→H) = G}F0 = G. By (3),
G fH = G and so G ≤H.
Conversely, let G ≤ H. By Proposition 3.7, the set S = {F∗? (G → H) : G ∈
G,H ∈ H,F ∈ F} is a base for G ↪→ H. Let F∗? (G → H) ∈ S. We prove
that 1 ∈ F∗? (G → H). Since by Lemma 3.10, G → H is a Q∗-cauchy filter
base, there are G1 ∈G and H1 ∈H such that (G1 → H1)×(G1 → H1) ⊆ F?.
Put G2 = G1 ∩ G and H2 = H1 ∩ H. It is easy to see that G2 ∧ H2 ⊆
F∗? (G2∧H2) ∈GfH. Since GfH = G, there is a G3 ∈G such that G3 ⊆ G1
and G3 ⊆ G2 ∧ H2. Since G3 6= φ, there are g3 ∈ G3, g ∈ G2 and h ∈ H2
such that g3 = g ∧ h. Since (g3 → h,g → h) and (g → h,g3 → h) both
are in (G1 → H1) × (G1 → H1) ⊆ F?, we get g → h ≡F g3 → h = 1
and so 1 ∈ F∗? (G → H). Hence F∗? (1) ⊆ F∗? (G → H). This implies that
G ↪→ H ⊆ F0. By the minimality of F0, G ↪→ H = F0. Therefore, we have
(a).

18



Quasi-Uniformity on BL-algebras

(b) By Proposition 3.7, the families

S1 = {F∗1?(G → F
∗
2?(H → K)) : G ∈G,H ∈H,K ∈K,F1,F2 ∈F},

S2 = {F∗1?(F
∗
2?(G�H) → K) : G ∈G,H ∈H,K ∈K,F1,F2 ∈F}

are bases of G ↪→ (H ↪→K) and (G}H) ↪→K, respectively. Let F∗1?(F∗2?(G�
H) → K) ∈ S2, F = F1 ∩F2 and y ∈ F∗? (G → F∗? (H → K)). Then there are
g ∈ G and x ∈ F∗? (H → K) such that y ≡F g → x. Also there are h ∈ H
and k ∈ K such that x ≡F h → k. Hence y ≡F g → x ≡F g → (h →
k) = (g �h) → k. Therefore, y ∈ F∗1?(F∗2?(G�H) → K). This implies that
(G } H) ↪→ K ⊆ G ↪→ (H ↪→ K). By the minimality of G ↪→ (H ↪→ K), we
get G ↪→ (H ↪→K) = G }H ↪→K. Hence we have (b).
Now, by (a) and (b), we have

G ≤H ↪→K⇔G ↪→ (H ↪→K) = F0 ⇔ (G }H) ↪→K = F0 ⇔G }H≤K.

So G ≤H ↪→K⇔G }H≤K.
(5) (G ↪→H) g (H ↪→G) = F0
By Proposition 3.7, the set

S = {F∗1?(F
∗
2?(G1 → H1)∨F

∗
3?(H2 → G2)) : G1,G2 ∈G,H1,H2 ∈H,F1,F2,F3 ∈F}

is a base for (G ↪→ H) g (H ↪→ G). Let F∗1?(F∗2?(G1 → H1) ∨ F∗3?(H2 →
G2)) ∈ S, G = G1 ∩G2, H = H1 ∩H2 and F = F1 ∩F2 ∩F3. We show that
1 ∈ F∗? (F∗? (G → H) ∨ F∗? (H → G)). Let g ∈ G and h ∈ H. Since A is a
BL-algebra, we have (g → h) ∨ (h → g) = 1. Since g → h ∈ F∗? (G → H)
and h → g ∈ F∗? (H → G), we have (g → h) ∨ (h → g) ∈ F∗? (F∗? (G →
H)∨F∗? (H → G)) and so 1 ∈ F∗? (F∗? (G → H)∨F∗? (H → G)). Hence F∗? (1) ⊆
F∗? (F

∗
? (G → H)∨F∗? (H → G)) which implies that (G ↪→H)g(H ↪→G) ⊆F0.

By the minimality of F0, (G ↪→H) g (H ↪→G) = F0.

4 Some topological properties on quasi-unifom

BL-algebra (A,Q)

Let T(Q) and T(Q∗) be topologies induced by Q and Q∗, respectively. Our
goal in this section is to study (semi)topological BL-algebras (A,T(Q)) and
(A,T(Q∗)). We prove that (A,∧,∨,�,T(Q)) is a compact connected topo-
logical BL-algebra and (A,T(Q∗)) is a regular topological BL-algebra. We
study separation axioms on (A,T(Q)) and (A,T(Q∗)). Also we stay condi-
tions under which (A,Q) becomes totally bounded. Finally, we show that if

19



R. A. Borzooei, N. Kouhestan

(A,Q) is a T0 quasi-uniform space, then the BL-algebra (Ã,Q̃) in Theorem
3.11 is the bicomplition topological BL-algebra of (A,Q).

Theorem 4.1. The set T(Q) = {G ⊆ A : ∀x ∈ G ∃F ∈ F s.t F?(x) ⊆
G} is the topology induced by Q on A such that (A,{∧,∨,�},T(Q)) is a
topological BL-algebras. Also (A,→,T(Q)) is a left topological BL-algebra.
Furthermore, if the negation map c(x) = x′ is one to one, then (A,T(Q)) is
a topological BL-algebra.

Proof. First we prove that T(Q) is a nonempty set. For this, we prove that
for each F ∈ F and each x ∈ A, F?(x) ∈ T(Q). Let F ∈ F, x ∈ A and
y ∈ F?(x). If z is an arbitrary element of F?(y), then z → y ∈ F. Since
y → x ∈ F, by (B15), we get z → x ∈ F. Hence F?(y) ⊆ F?(x) which implies
that F?(x) ∈ T(Q). Now we prove that T(Q) is a topology on A. Clearly,
φ,A ∈ T(Q). Also it is easy to prove that the arbitrary union of members
of T(Q) is in T(Q). Let G1, ...,Gn be in T(Q) and x ∈

⋂i=n
i=1 Gi. There are

F1, ...,Fn ∈ F such that Fi?(x) ⊆ Gi, for 1 ≤ i ≤ n. Let F = F1 ∩ ... ∩ Fn.
Then F ∈ F and F?(x) ⊆ F1?(x) ∩ ...∩Fn?(x) ⊆

⋂i=n
i=1 Gi. Hence T(Q) is a

topology. Since for each F ∈F, F? belongs to Q, then T(Q) is the topology
induced by Q. Now, by Lemmas 3.1, it is clear that (A,{∧,∨,�},T(Q))
is a topological BL-algebra. In continue, we prove that (A,→,T(Q)) is
a left topological BL-algebra. Let x,y,z ∈ A, and z ∈ F?(y). By (B9),
(x → z) → (x → y) ≥ z → y which implies that (x → z) → (x → y) ∈ F. So
x → z ∈ F?(x → y). Hence x → F?(y) ⊆ F?(x → y) and so (A,→,T(Q)) is
a left topological BL-algebra.
To complete the proof, suppose that the negation map c is one to one. Since
(A,→,T(Q)) is a topological BL-algebra, c is continuous. Now by [[2], The-
orem(3.15)], (A,T(Q)) is a topological BL-algebra.

Theorem 4.2. BL-algebra (A,T(Q)) is a connected and compact space and
each F ∈F, is a closed compact set in (A,T(Q)).
Proof. First we prove that if {Gi : i ∈ I} is an open cover of A in T(Q),
then for some i ∈ I, A = Gi. Let A =

⋃
i∈I Gi, where Gi ∈ T(Q). Then,

there are i ∈ I and F ∈F such that 1 ∈ Gi and F?(1) ⊆ Gi. By Lemma 3.1
(vi), A = F?(1). Hence A = Gi. Now, it is easy to show that (A,T(Q)) is
connected and compact. In continue we prove that each F ∈F, is a closed,
compact set in (A,T(Q)). For this, let F ∈ F and x ∈ F. Then, there is a
y ∈ F?(x) ∩ F. Since y ∈ F and y → x ∈ F , we get x ∈ F. Hence F = F.
Now, Since (A,T(Q)) is compact, F is compact.

Theorem 4.3. (i) BL-algebra (A,T(Q)) is not a T1 and T2 topological space.
(ii) BL-algebra (A,T(Q)) is a T0 topological space iff, for each 1 6= x ∈ A,
there is a F ∈F such that x 6∈ F.

20



Quasi-Uniformity on BL-algebras

Proof. (i) (A,T(Q)) is not a T1 and T2 topological space because for each
G ∈ T(Q), 1 ∈ G if and only if G = A.
(ii) Suppose for each 1 6= x ∈ A, there is a F ∈F such that x 6∈ F. We prove
that (A,T(Q)) is a T0 topological space. For this, let 1 6= x ∈ A. Then for
some F ∈F, x 6∈ F. Since 1 → x = x, then 1 6∈ F?(x). Moreover, since (A,→
,T(Q)) is a left topological BL-algebra, by [[2], Proposition(4.2)], (A,T(Q))
is a T0 topological space. Conversely, let (A,T(Q)) is a T0 topological space
and 1 6= x ∈ A. Then for some F ∈F, 1 6∈ F?(x). Hence x = 1 → x 6∈ F.

Theorem 4.4. The set T(Q∗) = {G ⊆ A : ∀x ∈ G ∃F ∈F s.t F∗? (x) ⊆ G}
is the topology induced by Q∗ on BL-algebra A such that (A,T(Q∗)) is a
topological BL-algebras.

Proof. By the similar argument as Theorem 4.1, we can prove that T(Q∗)
is the topology induced by Q∗ on A. By Lemma 3.2(v), (A,T(Q∗)) is a
topological BL-algebra.

Theorem 4.5. (i) BL-algebra (A,T(Q∗)) is connected iff, F = {A},
(ii) F has only a proper filter iff, each F ∈F is a component.

Proof. (i) Let F = {A}. Then it is easy to prove that T(Q∗) = {φ,A}. Hence
(A,T(Q∗)) is connected.
Conversely, let F 6= {A}. Then, there is a filter F ∈ F such that F 6= A.
Since for each x ∈ F, F∗? (x) ⊆ F, we conclude that F ∈ T(Q∗). Let y ∈ F.
Then there is a z ∈ F∗? (y) ∩ F. This proves that y ∈ F. Hence F is closed.
Now, since F is a closed and open subset of A, then A is not connected.
(ii) Let F has a proper filter F. By the similar argument as (i), we get that
F is closed and open. We show that F is connected. Let G1 and G2 be in
T(Q∗) and F = (F ∩ G1) ∪ (F ∩ G2). Without loss of generality, Suppose
that 1 ∈ F ∩ G1, then F ⊆ F∗? (1) ⊆ G1. Hence F ∩ G1 = F, which implies
that F is connected. Therefore, F is a component.
Conversely, suppose each F ∈F is a component. If F1 and F2 are in F, then
F1 ∩F2 is in F and is component. Hence F1 = F1 ∩F2 = F2.

Recall that a topological space (X,U) is regular if for each x ∈ G ∈ U
there is a U ∈U such that x ∈ U ⊆ U ⊆ G.

Theorem 4.6. BL-algebra (A,T(Q∗)) is a regular space.

Proof. First we prove that for each F ∈ F and x ∈ A, F∗? (x) = F∗? (x). Let
y ∈ F∗? (x). Then there is a z ∈ F∗? (y) ∩ F∗? (x). Hence y ≡F z ≡F x which
implies that y ∈ F∗? (x). Therefore, F∗? (x) = F∗? (x). Now if x ∈ G ∈ T(Q∗),
then for some a F ∈ F, x ∈ F∗? (x) = F∗? (x) ⊆ G. Hence (A,T(Q∗)) is a
regular space.

21



R. A. Borzooei, N. Kouhestan

Theorem 4.7. On BL-algebra (A,T(Q∗)) the follwing statements are equiv-
alent.
(i) (A,T(Q∗)) is a T0 space,
(ii)

⋂
F∈F F

∗
? (1) = {1},

(iii) (A,T(Q∗)) is a T1 space,
(iv) (A,T(Q∗)) is a T2 space.

Proof. (i ⇒ ii) Let (A,T(Q∗)) be a T0 space and 1 6= x ∈ A. By [[2], Propo-
sition(4.2)], there is a F ∈F such that 1 6∈ F∗? (x). Hence x 6∈ F. This implies
that x 6∈ F∗? (1). Therefore, x 6∈

⋂
F∈F F

∗
? (1).

(ii ⇒ i) Let
⋂
F∈F F

∗
? (1) = {1} and 1 6= x ∈ A. Then for some a F ∈ F,

x 6∈ F. Hence 1 6∈ F∗? (x). Now by [[2], Proposition(4.2)], (A,T(Q∗)) is a T0
space.
By Theorems 4.4 and 4.6, (A,T(Q∗)) is a regular topological BL-algebra.
Hence by [[2], Theorem(4.7)], the statements (ii), (iii) and (iv) are equiva-
lent.

Example 4.8. In Example 3.4, For each a ∈ [0, 1) and x ∈ [0, 1]

Fa∗(x) =

{
[0,x] , x ≤ a,
[0,1] , x > a.

F−1a∗ (x) =

{
[x,1] , x ≤ a,
(a,1] , x > a.

F∗a∗(x) =




x , x < a,
a , x = a
(a,1] , x > a.

If T(Q) is the induced topology by Q and G ∈ T(Q), then for each x ∈ G,
there is a a ∈ [0, 1) such that F∗a?(x) ⊆ G. Hence [0,x] ⊆ G or G = [0, 1]. If
G ∈ T(Q) and G 6= [0, 1], then for each x ∈ G, [0,x] ⊆ G. If g = supG, then
G = [0,g] or [0,g). Therefore T(Q) = {[0,x] : x ∈ [0, 1]}∪{[0,x) : x ∈ [0, 1]}.
Also if T(Q∗) is topology induced by Q∗ and G ∈ T(Q∗), then for each x ∈ G,
there is a a ∈ [0, 1) such that F∗a?(x) ⊆ G. Hence if G ∈ T(Q∗), then for some
a ∈ [0, 1), a ∈ G or (a, 1] ⊆ G.
Now since for each a ∈ [0, 1), F∗a?(1) = (a, 1], we get that

⋂
a∈[0,1) F

∗
a?(1) =

{1}. Hence by Theorems 4.4, 4.6 and 4.7, (A,T(Q∗)) is a Ti regular topolog-
ical BL-algebra, when 0 ≤ i ≤ 2.

Theorem 4.9. Let (A,→,U) be a semitopological BL-algebra and F0 be an
open proper BL-filter in A. Then, there exists a nontrivial topology V on A
such that V ⊆U and (A,V) is a topological BL-algebra.

Proof. Let F be a collection of BL-open filters in A which closed under finite
intersection and F0 ∈F. Let Q be the quasi-uniformity induced by F. Since

22



Quasi-Uniformity on BL-algebras

F0 6= A, by Lemma 3.1(vi), there is a x ∈ A such that F∗0?(x) 6= A. So T(Q∗)
is a nontrivial topology. We prove that T(Q∗) ⊆ U. Let x ∈ G ∈ T(Q∗).
Then, there is a F ∈ F such that F∗? (x) ⊆ G. Since x → x = 1 ∈ F ∈ U,
there is a U ∈ U such that x ∈ U and U → x ⊆ F and x → U ⊆ F . If
z ∈ U, then z → x,x → z ∈ F and so z ∈ F∗? (x). Hence x ∈ U ⊆ G.
Therefore, T(Q∗) is a nontrivial topology coaser than U and so by Theorem
4.4, (A,T(Q∗)) is a topological BL-algebra.

Example 4.10. Let I be the BL-algebra in Example 2.5(ii), and U be a
topology on I with the base S = {(a,b] ∩ I : a,b ∈ R}. We prove that
(I,→,U) is a semitopological BL-algebra. Let x,y ∈ I, and x → y ∈ (a,b].
If x ≤ y, then [0,x] and (ax,y] are two open neighborhoods of x and y,
respectively, such that (0,x] → y ⊆ (a, 1] and x → (ax,y] ⊆ (a, 1]. If x > y
and y = 0, then (0,x] and {0} are two open neighborhoods of x and 0,
respectively, such that (0,x] → 0 ⊆ [0,b] and x → {0} ⊆ [0,b]. If x > y
and y 6= 0, then (y/b,y/a] and (ax,bx] are two open sets of x,y, respectively,
such that (y/b,y/a] → y ⊆ (a,b] and x → (ax,bx] ⊆ (a,b]. It is easy to
prove that F = {(0, 1],A} is a collection of BL-filters which is closed under
intersection. Now since for each x ∈ A, A∗?(x) = A and (0, 1]∗?(x) = (0, 1], we
conclude T(Q∗) = {φ, (0, 1],A}. By Theorem 4.9, (A,T(Q∗)) is a topological
BL-algebra.

Recall a quasi-uniform space (X,Q) is totally-bounded if for each q ∈ Q,
there exist sets A1, ...,An such that X =

⋃i=n
i=1 Ai and for each 1 ≤ i ≤ n,

Ai ×Ai ⊆ q.( See [10])

Theorem 4.11. The following conditions on BL-algebra (A,T(Q∗)) are equiv-
alent.
(i) For each F ∈F, A/F is finite,
(ii) (A,Q) is totally bounded,
(iii) (A,T(Q∗)) is compact.

Proof. (i ⇒ ii) Let for each F ∈F, A/F be finite. We prove that (A,Q) is
totally bounded. For this it is enough to prove that, for each F ∈ F, there
are a1, ...,an ∈ A, such that for each 1 ≤ i ≤ n, ai/F × ai/F ⊆ F?. Let
F ∈F. Since A/F is finite, there are a1, ...,an ∈ A, such that A = ∪ni=1ai/F.
For each 1 ≤ i ≤ n, ai/F ×ai/F ⊆ F? because if (x,y) ∈ ai/F ×ai/F, then
x ≡F ai ≡F y and so (x,y) ∈ F?. This proves that (A,Q) is totally bounded.
(ii ⇒ iii) Let (A,Q) be totally bounded and F ∈ F. There exist sets
A1, ...,An, such that

⋃i=n
i=1 Ai = A and for each 1 ≤ i ≤ n, Ai ×Ai ⊆ F?. Let

1 ≤ i ≤ n and x,y ∈ Ai. Since (x,y) and (y,x) are in F?, we get x ≡F y.
This proves that Ai = ai/F, for some ai ∈ Ai.

23



R. A. Borzooei, N. Kouhestan

Now to prove that (A,T(Q∗)) is compact let A =
⋃
i∈I Gi, where each Gi is

in T(Q∗). Then there are H1, ...,Hn ∈ {Gi : i ∈ I}, such that ai ∈ Hi, for
each 1 ≤ i ≤ n. Now suppose x ∈ A, then x ∈ ai/F, for some 1 ≤ i ≤ n, and
so x ∈ F∗? (ai) ⊆ Hi. Therefore, A ⊆

⋃n
i=1 Hi, which shows that (A,T(Q

∗)) is
compact.
(iii ⇒ i) Let F ∈F. Since {F∗? (x) : x ∈ A} is an open cover of A in T(Q∗),
then there are a1, ...,an ∈ A, such that A ⊆

⋃n
i=1 F

∗
? (ai). Now, it is easy to

see that A/F = {a1/F,...,an/F}.

In the end, we prove that the quasi-uniform Bl-algeba (Ã,Q̃) in Theorem
3.11, is T0 bicomplition quasi-uniform of BL-algebra (A,Q).

Theorem 4.12. If quasi-uniform BL-algebra (A,Q) is T0, then

(i) (Ã,Q̃) is the bicompletion of (A,Q).

(ii) (Ã,T(Q̃)) is a topological BL-algebra.

(iii) A is a sub BL-algebra of Ã.

(iv) (Ã,T(Q̃∗)) is a topological BL-algebra.

Proof. (i) By Theorem 3.11 and Lemma 2.18, (Ã,Q̃) is an unique T0-bicompletion

quasi-uniform of (A,Q) and the mapping i : A → Ã by i(x) = {W ⊆
A : W is a T(Q∗) − neighborhood of x} is a quasi-uniform embedded and
clT(Q∗)i(A) = Ã.
(ii) It is clear that

T(Q̃) = {S ⊆ Ã : ∀G ∈ S ∃F ∈F s.t F̃?(G) ⊆ S}.

Let • ∈ {∧,∨,�} and •̃ ∈ {f,g,}}. We have to prove that for each
G,H∈ Ã, F̃?(G)•̃F̃?(H) ⊆ F̃?(G•̃H). Let G1 ∈ F̃?(G) and H1 ∈ F̃?(H). Then,
there are G ∈ G, G1 ∈ G1, H ∈ H and H1 ∈ H1 such that G × G1 ⊆ F?,
H×H1 ⊆ F?. By Proposition 3.7, S1 = {F∗? (G•H) : G ∈G,H ∈H,F ∈F}
and S2 = {F∗? (G1 • H1) : G1 ∈ G1,H1 ∈ H1,F ∈ F} are bases of G•̃H
and G1•̃H1, respectively. We show that G1•̃H1 ∈ F̃?(G•̃H). For this, it is
enough to show that F∗? (G • H) × F∗? (G1 • H1) ⊆ F?. Let (y,y1) ∈ F∗? (G •
H) × F∗? (G1 • H1) ⊆ F?. Then, there are g ∈ G, g1 ∈ G1, h ∈ H and
h1 ∈ H1 such that y ≡F g • h and y1 ≡F g1 • h1. By (B17), (B18) and
(B19), we have (g1 → g) � (h1 → h) ≤ (g1 • h1) → (g • h). It follows from
(g,g1) ∈ G×G1 ⊆ F? and (h,h1) ∈ H×H1 ⊆ F? that g1 → g and h1 → h are
in F . Hence g1 •h1 → g •h ∈ F. Therefore, y1 → y ∈ F and so (y,y1) ∈ F?.
Thus we proved that F̃?(G)•̃F̃?(H) ⊆ F̃?(G•̃H).
(iii) Let • ∈ {∧,∨,�,→}, •̃ ∈ {f,g,}, ↪→} and a,b ∈ A. We shall prove

24



Quasi-Uniformity on BL-algebras

that i(a)•̃i(b) = i(a • b). By Proposition 3.7, the set S = {F∗? (Wa • Wb) :
F ∈ F, Wa,Wb are T(Q∗) −neighborhoods of a,b} is a base for i(a)•̃i(b).
Since F∗? (a•b) ⊆ F∗? (Wa •Wb) and F∗? (a•b) ∈ i(a•b), we deduce that filter
i(a)•̃i(b) is contained in the filter i(a•b). Since they are minimal Q∗-cauchy
filters, i(a)•̃i(b) = i(a• b). Hence A is a sub-BL-algebra of Ã.
(iv) By Lemma 2.18, Q̃∗ = (Q̃)∗. Hence

T(Q̃∗) = {S ⊆ Ã : ∀G ∈ S ∃F ∈F s.t F̃∗? (G) ⊆ S}.

We prove that (Ã,T(Q̃∗)) is a topological BL-algebra. Let •∈ {∧,∨,�,→}
and •̃ ∈ {f,g,}, ↪→} and let G•̃H∈ F̃∗? (G•̃H). We show that F̃∗? (G)•̃F̃∗? (H) ⊆
F̃∗? (G•̃H). Let G1 ∈ F̃∗? (G) and H1 ∈ F̃∗? (H). Then, there are G ∈ G,
G1 ∈ G1, H ∈ H and H1 ∈ H1 such that G × G1 ⊆ F∗? and H × H1 ⊆ F∗? .
By Proposition 3.7, F∗? (G1 • H1) ∈ G1•̃H1 and F∗? (G • H) ∈ G•̃H. We
have to prove that G1•̃H1 ∈ F̃∗? (G•̃H). For this, it is enough to show that
F∗? (G•H) ×F∗? (G1 •H1) ⊆ F∗? . Let y ∈ F∗? (G•H) and y1 ∈ F∗? (G1 •H1).
Then y ≡F g • h and y1 ≡F g1 • h1 for some g ∈ G, g1 ∈ G1, h ∈ H and
h1 ∈ H1. Since (g,g1), (h,h1) are in F∗? , we get g • h ≡F g1 • h1. Hence
(y,y1) ∈ F∗? .

5 Conclusions

The aim of this paper is to study In [2] and [4] we study (semi)topological
BL-algebras and metrizability on BL-algebras. We showed that continuity
the operations � and → imply continuity ∧ and ∨. Also, we found some
conditions under which a locally compact topological BL-algebra become
metrizable. But in there we can not answer some questions, for example:
(i) Is there a topology U on BL-algera A such that (A,U) be a (semi)topological
BL-algebra?
(ii) Is there a topology U on a BL-algebra A such that (A,U) be a compact
connected topological BL-algebra?
(iii) Is there a topological BL-algebra (A,U) such that T0,T1 and T2 spaces
be equivalent?
(iv) If (A,→,U) is a semitopological BL-algebra, is there a topology V
coarsere than U or finer than U such that (A,V) be a (semi)topological
BL-algebra?

Now in this paper, we answered to some above questions and got some
interesting results as mentioned in abstract.

25



R. A. Borzooei, N. Kouhestan

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