EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS
Vol. 13, No. 4, 2020, 830-839
ISSN 1307-5543 – www.ejpam.com
Published by New York Business Global

On B-ideals in a Topological B-algebra and the Uniform
B-topological Space

Katrina E. Belleza1,∗, Jocelyn P. Vilela2

1 Department of Computer, Information Sciences and Mathematics, School of Arts and
Sciences, University of San Carlos, 6000 Cebu City, Philippines
2 Department of Mathematics and Statistics, College of Science and Mathematics, Center
of Graph Theory, Algebra and Analysis, Premier Research Institute of Science and
Mathematics, Mindanao State University-Iligan Institute of Technology, 9200 Iligan City,
Philippines

Abstract. This paper presents the characterizations and properties of B-ideals in a topological
B-algebra and introduces the uniform topology on a B-algebra in terms of its B-ideals. Moreover,
this paper shows that a uniform B-topological space is a topological B-algebra.

2020 Mathematics Subject Classifications: 54A05, 54F65, 54H99, 55M99

Key Words and Phrases: B-ideals, topological B-algebra, uniform B-topology

1. Introduction

Y.B. Jun et al. [4] in 1999 introduced topological BCI-algebras, provided some prop-
erties on this structure, and characterized a topological BCI-algebra in terms of neigh-
borhoods. In 2002, J.Neggers and H.S. Kim [9] introduced and investigated B-algebras.
In 2017, S. Mehrshad and J. Golzarpoor [7] provided some properties of uniform topology
and topological BE-algebras. A recent study on topological B-algebras was conducted by
N.C. Gonzaga, Jr. [6] in 2019, which characterized a topological B-algebra with respect
to neighborhoods.

This paper provides some properties of topological B-algebra, describes the B-ideals
in a topological B-algebra, and characterizes uniform B-topology in a B-algebra.

∗Corresponding author.
DOI: https://doi.org/10.29020/nybg.ejpam.v13i4.3807

Email addresses: kebelleza@usc.edu.ph (K. Belleza), jocelyn.vilela@g.msuiit.edu.ph (J. Vilela)

https://www.ejpam.com 830 c© 2020 EJPAM All rights reserved.



K. Belleza, J. Vilela / Eur. J. Pure Appl. Math, 13 (4) (2020), 830-839 831

2. Preliminaries

An algebra of type (2,0) is an algebra with a binary operation and a constant element.

Definition 1. [9] A B-algebra is a non-empty set X with a constant 0 and a binary
operation “ ∗ ” satisfying the following axioms for all x, y, z in X:

(B1) x ∗ x = 0 (B2) x ∗ 0 = x (B3) (x ∗ y) ∗ z = x ∗ [z ∗ (0 ∗ y)]

Example 1. [8] Let X = {0, a, b, c, d, e} be a set with the following Cayley table:

∗ 0 a b c d e
0 0 b a c d e
a a 0 b d e c
b b a 0 e c d
c c d e 0 b a
d d e c a 0 b
e e c d b a 0

Then (X, ∗, 0) is a B-algebra.

Definition 2. [6] Let A and B be nonempty subsets of a B-algebra X. The product of
A and B, denoted by A ∗B, is given by

A ∗B = {a ∗ b|a ∈ A, b ∈ B}.

Lemma 1. [9] Let (X, ∗, 0) be a B-algebra. Then for any x, y ∈ X,

(i) x ∗ y = 0 implies x = y; (ii) 0 ∗ x = 0 ∗ y implies x = y; (iii) 0 ∗ (0 ∗ x) = x.

Definition 3. [10] Let (X, ∗, 0) be a B-algebra. A nonempty subset N of X is called a
subalgebra of X if x ∗ y ∈ N for any x, y ∈ N .

Lemma 2. [3] Let X be a B-algebra. If {Nα : α ∈ A} is a nonempty collection of
subalgebras of X, then

⋂
α∈A

Nα is a subalgebra of X.

Definition 4. [8] Let (X, ∗, 0) be a B-algebra. A nonempty subset S of X is said to be
normal in X if for any x ∗ y, a ∗ b ∈ S, (x ∗ a) ∗ (y ∗ b) ∈ S.

Theorem 1. [10] Let N be a subalgebra of a B-algebra X. Then the following statements
are equivalent:

(i) N is a normal subalgebra; (ii) if x ∈ X and y ∈ N, then x ∗ (x ∗ y) ∈ N.

Suppose (X, ∗, 0) is a B-algebra and I a normal subalgebra of X. The relation “ ∼=I ”
defined by x ∼=I y if and only if x ∗ y, y ∗ x ∈ I is a congruence relation on X for any
x, y ∈ X. That is, ∼=I is an equivalence relation and for each a, b, x, y ∈ X, if x ∼=I y and
a ∼=I b, then a ∗ x ∼=I b ∗ y. Let Ix = {y : y ∼=I x} denote the equivalence class of x and
X/I = {Ix : x ∈ X}. Then X/I is a B-algebra called the quotient B-algebra under the
binary operation given by Ix ∗ Iy = Ix∗y [3].



K. Belleza, J. Vilela / Eur. J. Pure Appl. Math, 13 (4) (2020), 830-839 832

Definition 5. [1] Let (X, ∗, 0) be a B-algebra and I a nonempty subset of X. Then I is
called a B-ideal of X if it satisfies the following: for any x, y in X,

(i) 0 ∈ I; (ii) if x ∗ y ∈ I and y ∈ I, then x ∈ I.
Remark 1. Not every B-ideal is a normal subset of a B-algebra X. Consider Example
1. Then {0, c} is a B-ideal but is not normal since c ∗ a = c, b ∗ d = c ∈ {0, c} with
(c ∗ b) ∗ (a ∗ d) = b /∈ {0, c}.
Remark 2. Any nonempty normal subset of X is a subalgebra. Hence, if I is a normal
B-ideal, then X/I is a B-algebra.

Let X be a set. A topology (or topological structure) in X is a family τ of subsets of
X that satisfies the following:

(i) Each union of members of τ is also a member of τ ;

(ii) Each finite intersection of members of τ is also a member of τ ; and

(iii) ∅ and X are members of τ .

A couple (X, τ) consisting of a set X and a topology τ in X is called a topological space.
We also say “τ is the topology of the space X”. The members of τ are called open sets
of (X, τ). Let (X, τX) and (Y, τY ) be topological spaces. A map f : X → Y is called
continuous if the inverse image of each open set in Y is open in X (that is, if f−1 maps
τY into τX) [2].

Definition 6. [2] Let {Yα|α ∈ A} be any family of topological spaces. For each α ∈ A,
let τα be the topology for Yα. The cartersian product topology in

∏
α Yα is that having for

subbasis all sets < Uβ >= ρ−1β (Uβ), where ρ :
∏
α Yα → Yα, Uβ ranges over all members

of τβ and β over all elements of A.

Definition 7. [6] Let X be a B-algebra. A topology τ furnished on X is called a B-
topology on X. A B-topological space (X, τ) is called a topological B-algebra if τ is a
B-topology on X and the binary operation ∗ : X ×X → X is continuous, where X ×X
is furnished by the Cartesian product topology.

Let (X, τ) be a topological space and A ⊂ X. By a neighborhood of an element x in X
(denoted as U(x)) is meant any open set (that is, member of τ) containing x. The interior
Int(A) of A is the largest open set contained in A, that is, Int(A)=

⋃
{U |U ∈ τ, U ⊂ A}.

A point a is an interior point of A if a ∈ Int(A), that is, there exists U(a) ∈ τ such that
U(a) ⊂ A. A is open if and only if Int (A)= A. A set Y ⊂ X is a closed set in X if its
complement is open. A point x ∈ X is adherent to Y if each neighborhood of x contains
at least one point of Y . The set Y = {x ∈ X|∀U(x), U(x) ∩ Y 6= ∅} of all points in X
adherent to Y is called the closure of Y [2].

Theorem 2. [6] Let X = (X, ∗, 0) be a B-algebra and τ a B-topology on the set X. Then
(X, τ) is a topological B-algebra if and only if for all x, y in X and for every neighborhood
W of x∗y, there are neighborhoods U and V of x and y, respectively, such that U ∗V ⊆W .

Throughout this article we will denote a B-topological space (X, τ), topological B-
algebra (X, ∗, τ), or a B-algebra (X, ∗, 0) as simply, X.



K. Belleza, J. Vilela / Eur. J. Pure Appl. Math, 13 (4) (2020), 830-839 833

3. B-ideals in Topological B-algebras

Example 2. Consider the B-algebra X = {0, a, b, c} with the binary operation “∗” defined
on the Cayley table provided. Let τ =

{
X,∅, {0, b}, {a, c}

}
. Then τ is a B-topology on

X and (X, ∗, τ) is a topological B-algebra.

∗ 0 a b c
0 0 a b c
a a 0 c b
b b c 0 a
c c b a 0

Remark 3. Not every B-ideal of a B-algebra X is either an open or closed set in a
topological B-algebra. This remark is illustrated in the next example.

Example 3. Consider the topological B-algebra in Example 2. Let I = {0, c}. Then I
is a B-ideal of X. Observe that I /∈ τ implying that I is not an open set in X. Also,
X\I = {a, b} /∈ τ implying that I is not a closed set in X.

However if a B-ideal A in a topological B-algebra X is open, it is also a closed B-ideal
in X. This is formally stated in the next theorem.

Theorem 3. If A is an open B-ideal of a topological B-algebra X, then A is also closed.
Proof. Suppose A is an open B-ideal of a topological B-algebra X. Let x ∈ X\A.

Since A is a B-ideal of X, x ∗ x = 0 ∈ A by (B1). By Theorem 2, there exists U(x)
such that U(x) ∗ U(x) ⊆ A. We claim that U(x) ⊆ X\A. Assume on the contrary that
U(x) * X\A, that is, U(x) ∩ A 6= ∅. Then there exists y ∈ U(x) ∩ A. Note that for all
z ∈ U(x), z ∗ y ∈ U(x) ∗U(x) ⊂ A. Since y ∈ A and A is a B-ideal, z ∈ A. So, U(x) ⊆ A
which implies that x ∈ A, a contradiction. Hence, X\A is open. Therefore, A is closed in
X.

The next theorem is a characterization of an open set (containing 0) in a topological
B-algebra.

Theorem 4. Let X be a topological B-algebra and A ⊂ X such that 0 ∈ A. Then A is
open if and only if 0 is an interior point of A.

Proof. Suppose A is open. Since 0 ∈ A, 0 is an interior point of A. Conversely,
suppose 0 is an interior point of A. Then there exists U(0) such that U(0) ⊆ A. Let
y ∈ A. By (B1), y ∗ y = 0 ∈ U(0). Since X is a topological B-algebra, by Theorem 2,
there exist U(y) such that U(y) ∗ U(y) ⊆ U(0). It remains to show that U(y) ⊆ A. Let
x ∈ U(y). By (B1), x∗x = 0 ∈ A. If x ∈ A, we are done. Suppose x /∈ A. Then x /∈ U(0).
This implies that x /∈ U(y) ∗ U(y). By (B2), x ∗ 0 = x /∈ U(y) ∗ U(y). This implies that
x /∈ U(y) which is a contradiction. Therefore, U(y) ⊆ A and A is open.

The following corollary follows from Theorem 4.



K. Belleza, J. Vilela / Eur. J. Pure Appl. Math, 13 (4) (2020), 830-839 834

Corollary 1. Let X be a topological B-algebra. Then 0 is an interior point of a B-ideal
I if and only if I is open.

The next example illustrates that an open subset of a topological B-algebra may not
be a B-ideal in which the observation is formally stated as a remark.

Example 4. Consider the topological B-algebra in Example 2. Examine the open set
{a, c}. Note that {a, c} is not a B-ideal since 0 /∈ {a, c} and b ∗ c = a and c ∈ {a, c} but
b /∈ {a, c}.

Remark 4. Not every open subset of a topological B-algebra X is a B-ideal of X.

However, if all open sets are neighborhoods of 0, every open subset of a topological
B-algebra X is a B-ideal of X. This is formally stated in the next theorem which is a
characterization of a B-ideal in a topological B-algebra.

Theorem 5. Let X be a topological B-algebra and I an open subset of X. If 0 ∈
⋂
U∈τ

U ,

then I is a B-ideal of X.
Proof. Let x ∗ y ∈ I where y ∈ I. Since I is open, by Theorem 2, there exist

V (x) and V (y) such that V (x)∗V (y) ⊆ U(x∗y) ⊆ I. By (B2), x = x∗0 ∈ V (x)∗V (y) ⊆ I.
Therefore, I is a B-ideal of X.

Lemma 3. Let X be a topological B-algebra, I0 ⊆ X where I0 contains 0 is such that if
0 ∈ U , then I0 ⊆ U for all U ∈ τ . Then for any x ∈ I0 and U(x) ∈ τ, I0 ⊆ U(x).

Proof. Suppose x ∈ I0. By (B2), x ∗ 0 = x ∈ U(x). By Theorem 2, there exist
V (x) and V (0) such that V (x) ∗ V (0) ⊆ U(x). By (B1) and the hypothesis, 0 = x ∗ x ∈
V (x) ∗ I0 ⊆ V (x) ∗ V (0) ⊆ U(x). This implies that U(x) is an open set containing 0.
Therefore, I0 ⊆ U(x).

The next theorem gives another characterization of a B-ideal.

Theorem 6. Let X be a topological B-algebra and I an open subset of X containing 0
such that if 0 ∈ U , then I ⊆ U for all U ∈ τ . Then I is a B-ideal of X.

Proof. Suppose x∗y, y ∈ I for any x, y ∈ X. By Theorem 2, there exist U(x) and U(y)
such that U(x) ∗U(y) ⊂ I. By (B2), the hypothesis, and Lemma 3, x = x ∗ 0 ∈ U(x) ∗ I ⊂
U(x) ∗ U(y) ⊂ I. Hence, I is a B-ideal of X.

Note that the converse of Theorem 6 is not always true as shown in the next example.

Example 5. Let X be the topological B-algebra in Example 2. Then the trivial B-ideal
X is not the smallest open set containing 0.

However, if the B-ideal I is closed, I is also open but may not be the smallest open
set containing 0. This is stated in the next theorem which is the converse of Theorem 3



K. Belleza, J. Vilela / Eur. J. Pure Appl. Math, 13 (4) (2020), 830-839 835

Theorem 7. Let X be a topological B-algebra and I a closed B-ideal of X. Then I is
also open.

Proof. Suppose I is a closed B-ideal of X. Assume on the contrary that I is not an
open set in X. By Theorem 4, 0 is not an interior point of I. This implies that for all
U(0) ∈ τ , U(0) * I. Let I0 be open with property defined in Lemma 3. Then I0 * I.
Hence, (X\I)∩I0 6= ∅ and so there exists z ∈ (X\I)∩I0. Note that (X\I)∩I0 is an
open set containing z. By Lemma 3, I0 ⊂ (X\I). This implies that 0 ∈ X\I which is a
contradiction. Therefore, I is open.

The next corollary follows directly from Theorems 3 and 7.

Corollary 2. Suppose X is a topological B-algebra and I a B-ideal of X. Then I is an
open subset of X if and only if I is a closed subset of X.

Theorem 8. Let I be a family of normal B-ideals in a B-algebra X. Then there is a
topology τ = {U ⊆ X|∀x ∈ U,∃I ∈ I such that Ix ⊆ U} such that (X, ∗, τ) is a topological
B-algebra.

Proof. Note that for all x ∈ X, there exists I ∈ I such that Ix ⊆ X. This implies
that X ∈ τ . Suppose ∅ /∈ τ . Then there exists x ∈ ∅ such that for all I ∈ I, Ix * U
which is a contradiction. Hence, ∅ ∈ τ . Let y ∈ U1 ∩ U2, where U1, U2 ∈ τ . Then y ∈ U1

and y ∈ U2 which imply that there exist I1, I2 ∈ I such that I1y ⊆ U1 and I2y ⊆ U2. Let
I = I1 ∩ I2 ∈ I.
Claim 1: Iy ⊆ I1y, Iy ⊆ I2y.
Suppose x ∈ Iy. Then y ∼=I x which implies that y ∗ x ∈ I ⊆ I1. Hence, y ∼=I1 x implying
that x ∈ I1y so that Iy ⊆ I1y . Similarly, Iy ⊆ I2y. This proves claim 1.
Since I1y ⊆ U1 and I2y ⊆ U2, it follows that Iy ⊆ (U1 ∩ U2). Hence, U1 ∩ U2 ∈ τ. Let
y ∈

⋃
α∈A

Uα where Uα ∈ τ for all α ∈ A. Then y ∈ Uβ for some β ∈ A. This implies that

there exists Iβ ∈ I such that Iβy ⊆ Uβ ⊆
⋃
α∈A

Uα. Hence, Uα
α∈A
∈ τ. This implies that τ is a

B-topology.
Claim 2: For any I ∈ I and x ∈ X, Ix ∈ τ
Let y ∈ Ix. Then y ∼=I x. We will show that Iy ⊆ Ix. Let z ∈ Iy. Then z ∼=I y. By
transitivity, z ∼=I x. Hence, z ∈ Ix so that Iy ⊆ Ix. This proves claim 2.
Suppose x ∗ y ∈ U ∈ τ . Then there exists I ∈ I such that Ix∗y ⊆ U . Note that Ix and Iy
are open sets containing x and y, respectively. Then Ix ∗ Iy = Ix∗y ⊆ U . This implies that
∗ is continuous. Therefore, (X, ∗, τ) is a topological B-algebra by Theorem 2.

4. Uniform Topology on B-Algebras

Throughout this section, all B-ideals of a B-algebra X are normal B-ideals of X.
The following definitions are parallel to that of [5], page 340-341.

Suppose X is a B-algebra and U, V ⊆ X ×X, consider the following notations:



K. Belleza, J. Vilela / Eur. J. Pure Appl. Math, 13 (4) (2020), 830-839 836

(i) U−1 = {(y, x)|(x, y) ∈ U}; (iii) U [[x]] = {y|(x, y) ∈ U};

(ii) U ◦ V = {(x, z)|∃y ∈ X, (x, y) ∈ V, (y, z) ∈ U}; (iv) ∆ = {(x, x)|x ∈ X}.

Suppose Ω is an arbitrary family of B-ideals in a B-algebra X and A ⊆ X. Consider
the following notations:

(i) UI = {(x, y) ∈ X ×X|x ∼=I y}; (iii) K = {U ⊆ X ×X|UI ⊆ U,∃UI ∈ K?};

(ii) K? = {UI : I ∈ Ω}; (iv) UI [[A]] =
⋃
a∈A

UI [[a]].

Remark 5. K? ⊆ K.

Definition 8. By a uniformity on a B-algebra X, we shall mean a nonempty collection
K of subsets of X ×X which satisfies the following conditions for any U, V ∈ K:

(i) ∆ ⊆ U ; (iv) U ∩ V ∈ K; and

(ii) U−1 ∈ K; (v) If U ⊆W ⊆ X ×X then W ∈ K.

(iii) W ◦W ⊆ U, for some W ∈ K;

The pair (X,K) is called a uniform B-structure.

Example 6. Consider the B-algebra X = {0, a, b, c, d, e} in Example 1. The normal B-
ideals of X are {X, I} where I = {0, a, b}. By routine calculations, (X,K) is a uniform
B-structure where K? = {UX , UI}, UX [[0]] = UX [[a]] = UX [[b]] = UX [[c]] = UX [[d]] =
UX [[e]] = {0, a, b, c, d, e} = X, UI [[0]] = UI [[a]] = UI [[b]] = {0, a, b}, UI [[c]] = UI [[d]],=
UI [[e]] = {c, d, e}.

Remark 6. (X,K∗) is not a uniform B-structure as shown in the next example.

Example 7. Consider the B-algebra X = {0, a, b, c, d, e} in Example 1 and the B-ideal
I = {0, a, b} in Example 6. By Theorem 1, X and I are normal B-ideals of X. Hence,
K? = {UX , UI} where UX = {(x, y) ∈ X × X|x ∗ y, y ∗ x ∈ X} = X × X and UI =
{(0, 0), (a, a), (b, b), (c, c), (d, d), (e, e),
(0, b), (b, 0), (a, 0), (0, a), (b, a), (a, b), (c, e), (e, c), (d, c), (c, d), (e, d), (d, e)}. Let M = I ∪
{0, e} = {0, a, b, e}. Then UM = UI ∪ {(0, e), (e, 0), (a, d), (d, a), (b, c),
(c, b)}. Note that UI ⊆ UM ⊆ X×X. Moreover, M /∈ Ω since d∗a = e, a ∈M but d /∈M .
Hence, UM /∈ K?. This implies that K? does not satisfy condition (v) of Definition 8.

However, in view of Remark 5, the next theorem states that the pair (X,K) is a uniform
B-structure.

Theorem 9. Let Ω be an arbitrary family of B-ideals in a B-algebra X. Then (X,K) is
a uniform B-structure.



K. Belleza, J. Vilela / Eur. J. Pure Appl. Math, 13 (4) (2020), 830-839 837

Proof. Suppose Ω is an arbitrary family of B-ideals in a B-algebra X and U, V ∈ K.
Then there exist UI , UJ ∈ K? such that UI ⊆ U and UJ ⊆ V , respectively. (i) Let (x, x)
∈ ∆. Since x ∼=I x, it follows that (x, x) ∈ UI . Hence, (x, x) ∈ U so that ∆ ⊆ U . (ii) Let
(x, y) ∈ UI . Then x ∼=I y and y ∼=I x. This implies that (y, x) ∈ UI . Hence, (y, x) ∈ U .
It follows that (x, y) ∈ U−1 with UI ⊆ U−1 so that U−1 ∈ K. (iii) Consider UI ∈ K and
(x, z) ∈ UI ◦ UI . Then there exists y ∈ X such that (x, y), (y, z) ∈ UI . This implies that
x ∼=I y and y ∼=I z. Hence, x ∼=I z. It follows that (x, z) ∈ UI ⊆ U so that UI ◦ UI ⊆ U .
(iv) Let UI , UJ ∈ K?. Note that a B-ideal is a subalgebra of X. By Lemma 2, Ω is closed
under finite intersections.

Claim: UI ∩ UJ = UI∩J ∈ K?.
Let (x, y) ∈ UI ∩UJ . Then x ∼=I y and x ∼=J y. These imply that x∗y, y ∗x ∈ I, J . Hence,
x∗y, y∗x ∈ I∩J implying that x ∼=I∩J y and so (x, y) ∈ UI∩J . Therefore, UI∩UJ ⊆ UI∩J .
The converse is similar. This proves the claim.
Let (x, y) ∈ UI∩J = UI ∩ UJ . Then x ∼=I y and x ∼=J y. Thus, (x, y) ∈ UI and (x, y)
∈ UJ . Hence, (x, y) ∈ U and (x, y) ∈ V . Therefore, (x, y) ∈ U ∩ V so that UI∩J ⊆ U ∩ V .
Consequently, U ∩ V ∈ K. It remains to show that K satisfies condition (v). Let U ∈ K
such that U ⊆ V ⊆ X × X. Then there exists UI ∈ K? such that UI ⊆ U ⊆ V . Hence,
V ∈ K. Therefore, K satisfies condition (v) and (X,K) is a uniform B-structure.

Definition 9. Let (X,K) be a uniform B-structure. If τ is a topology on X, then τ is
called a uniform B-topology and the pair (X, τ) is called a uniform B-topological space.

Example 8. Consider the uniform structure (X,K) in Example 6. Then the family
τ = {X,∅, {0, a, b}, {c, d, e}} is a uniform B-topology on X. Thus, (X, τ) is a uniform
B-topological space.

Theorem 10. Suppose (X,K) is a uniform B-structure. Then τ = {G ⊆ X|∀x ∈ G,∃U ∈
K, U [[x]] ⊆ G} is a uniform B-topology on X.

Proof. Suppose (X,K) is a uniform B-structure. Note that for all x ∈ X and U ∈ K,
U [[x]] ⊆ X. Hence, X ∈ τ . Also, ∅ ∈ τ by definition. Let x ∈

⋃
Gi∈τ,i∈A

Gi. Then there

exists j ∈ A such that x ∈ Gj . Since Gj ∈ τ , there exists Uj ∈ K such that Uj [[x]] ⊆ Gj .
This implies that Uj [[x]] ⊆

⋃
Gi∈τ,i∈A

Gi. Hence,
⋃

Gi∈τ,i∈A
Gi ∈ τ. Suppose G,H ∈ τ such

that x ∈ G ∩ H. Then there exist U, V ∈ K such that U [[x]] ⊆ G and V [[x]] ⊆ H. Let
W = U ∩ V . By Definition 8(iv), W ∈ K.
Claim: W [[x]] ⊆ U [[x]] ∩ V [[x]].
Let y ∈ W [[x]]. Then (x, y) ∈ U and (x, y) ∈ V . This implies that y ∈ U [[x]] and
y ∈ V [[x]]. Hence, W [[x]] ⊆ U [[x]] ∩ V [[x]]. This proves the claim.
By the claim, W [[x]] ⊆ U [[x]] ⊆ G and W [[x]] ⊆ V [[x]] ⊆ H. Hence, W [[x]] ⊆ G∩H. This
implies that G ∩H ∈ τ . Therefore, τ is a B-topology on X.

The next remark follows from Definition 9 and Theorem 10.

Remark 7. Suppose X is a B-topological space.



REFERENCES 838

(i) Then (X, τ) in Theorem 10 is a uniform B-topological space.

(ii) For any UI ∈ K∗ and x ∈ X, x ∈ UI [[x]] and UI [[x]] ∈ τ , that is, UI [[x]] is a
neighborhood of x.

Lemma 4. Suppose X is a B-algebra such that U ⊆ V for any U, V ∈ K. Then U [[x]] ⊆
V [[x]] for all x ∈ X.

Proof. Let U ⊆ V for any U, V ∈ K and x ∈ X. Suppose a ∈ U [[x]]. Then (x, a)
∈ U ⊆ V . This implies that (x, a) ∈ V . Therefore, a ∈ V [[x]].

Theorem 11. Suppose X is a uniform B-topological space. Then X is a topological
B-algebra.

Proof. Let (X,K) be a uniform structure. By Theorem 10 and Remark 7(i), there is a
uniform B-topology τ = {G ⊆ X|∀x ∈ G,∃U ∈ K, U [[x]] ⊆ G}. Suppose x ∗ y ∈ U(x ∗ y)
where x, y ∈ X. By Theorem 10, there exists G ∈ K such that G[[x ∗ y]] ⊆ U(x ∗ y).
Then there exists GI ∈ K? such that GI ⊆ G for some B-ideal I of X. By Lemma 4,
GI [[x ∗ y]] ⊆ G[[x ∗ y]]. Note that GI [[x]] and GI [[y]] are open neighborhoods of x and y,
respectively by Remark 7(ii).
Claim: GI [[x]] ∗GI [[y]] ⊆ GI [[x ∗ y]].
Suppose a ∗ b ∈ GI [[x]] ∗ GI [[y]]. Then (x, a), (y, b) ∈ GI . This implies that x ∼=I a and
y ∼=I b. Hence, x ∗ y ∼=I a ∗ b. It follows that a ∗ b ∈ GI [[x ∗ y]]. This proves the claim.
Hence, GI [[x]] ∗GI [[y]] ⊆ U(x ∗ y). By Theorem 2, X is a topological B-algebra.

The converse of Theorem 11 follows directly from Definition 9 provided that the B-
topology is a uniform B-topology. This is formally stated in the next corollary.

Corollary 3. Suppose X is a topological B-algebra. If τ is a uniform B-topology, then X
is a uniform B-topological space.

Acknowledgements

This research is funded by the Commission on Higher Education (CHED) and Min-
danao State University-Iligan Institute of Technology, Philippines.

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