EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS
Vol. 15, No. 4, 2022, 1948-1956
ISSN 1307-5543 – ejpam.com
Published by New York Business Global

On Dual B-Topological Spaces Determined by
Filterbase and Some Sets in a Dual B-algebra

Katrina E. Belleza1,∗, Jimboy R. Albaracin2

1 Department of Computer, Information Science, and Mathematics, School of Arts and
Sciences, University of San Carlos, Talamban, Cebu City, Philippines
2 Mathematics and Statistics Programs, College of Science, University of the Philippines
Cebu, Cebu City, Philippines

Abstract. This paper presents dual B-topologies that are determined by filterbase and some
sets in a dual B-algebra. Also, some properties of a filterbase in a dual B-topological space are
provided. In particular, a commutative dual B-topological space and a symmetric B-topological
space are topological dual B-algebras.

2020 Mathematics Subject Classifications: 46H10, 54A05, 54F65, 54H99, 55M99

Key Words and Phrases: Topological algebra, dual B-algebra, topological dual B-algebra, dual
B-topological space, tdB-algebra

1. Introduction

In 1998, D.S. Lee and D.N. Ryu [5] introduced the notion of a topological BCK-
algebra. Moreover, they derived a filter base generating a BCK-algebra topology. On
the following year, Y.B. Jun et al. [4] gave a filterbase generating a BCI-topology and
making a BCI-algebra into a topological BCI-algebra for which the filterbase is a fun-
damental system of neighborhoods. In 2019, K.E. Belleza and J.P. Vilela introduces and
characterized the notion of a dual B-algebra [2]. Moreover on the following year, K.E.
Belleza introduces the dual B-topological space and a tdB-algebra involving dual B-ideals
and dual B-subalgebras.

2. Preliminaries

Definition 1. [2] A dual B-algebra XD is a triple (XD, ◦, 1) where XD is a non-empty
set with a binary operation “ ◦ ” and a constant 1 satisfying the following axioms for all
x, y, z in XD:

∗Corresponding author.
DOI: https://doi.org/10.29020/nybg.ejpam.v15i4.4594

Email addresses: kebelleza@usc.edu.ph (K. Belleza), jralbaracin@up.edu.ph (J. Albaracin)

https://www.ejpam.com 1948 © 2022 EJPAM All rights reserved.



K. Belleza, J. Albaracin / Eur. J. Pure Appl. Math, 15 (4) (2022), 1948-1956 1949

(DB1) x ◦ x = 1; (DB2) 1 ◦ x = x; (DB3) x ◦ (y ◦ z) = ((y ◦ 1) ◦ x) ◦ z.

Lemma 1. [2] Let XD be a dual B-algebra. For any x, y in XD, x ◦ y = 1 implies x = y.

Theorem 1. [2] Let X = (X, ◦, 1) be any algebra of type (2, 0). Then X is a dual B-
algebra if and only if for any x, y, z ∈ X,

(i) x ◦ x = 1; (ii) x = (x ◦ 1) ◦ 1; (iii) (x ◦ y) ◦ (x ◦ z) = y ◦ z.

Definition 2. [2] Let XD be a dual B-algebra. Define a binary operation “ + ” on X as
follows: x+ y = (x ◦ 1) ◦ y for all x, y in XD. A dual B-algebra is said to be commutative
if x+ y = y + x, that is, (x ◦ 1) ◦ y = (y ◦ 1) ◦ x for all x, y in XD.

Proposition 1. [2] Suppose XD is a commutative B-algebra. Then for all x, y in XD,
x ◦ (y ◦ z) = y ◦ (x ◦ z).

Let XD be a dual B-algebra such that x ◦ y = y ◦ x for all x, y ∈ XD. Then we say
that XD satisfies a symmetric condition [2].

Lemma 2. [2] Let XD be a dual B-algebra satisfying a symmetric condition. Then for
all x, y, z ∈ XD, (x ◦ y) ◦ (z ◦ y) = x ◦ z.

Definition 3. [1] Let XD be a dual B-algebra and S a nonempty subset of XD. Then S
is called a dual B-subalgebra of XD if S itself is a dual B-algebra with binary operation
of XD on S.

Definition 4. [1] Let XD be a dual B-algebra. A subset F of XD is called a dual B-filter
if it satisfies the following axioms: for all x, y in XD,

(dF1) 1 ∈ F ; (dF2) x ◦ y ∈ F and x ∈ F imply y ∈ F.

Definition 5. [3] 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). A family B ⊂ τ is called a basis for τ if each open set is the union of
members of B. 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). [3]

Theorem 2. [3] Let B ⊂ τ . The following two properties of B are equivalent:
(i) B is a basis for τ ;
(ii) for each G ∈ τ and each x ∈ G, there is a U ∈ B with x ∈ U ⊂ G.



K. Belleza, J. Albaracin / Eur. J. Pure Appl. Math, 15 (4) (2022), 1948-1956 1950

Definition 6. [3] Let (X, τ) be a topological space. 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.

Definition 7. [3] Let {Yα | α ∈ A} be any family of topological spaces. For each α ∈ A,
let τα be the topology for Yα. The Cartesian product topology in

∏
α Yα is that having for

subbasis all sets ⟨Uβ⟩ = ρ−1
β (Uβ), where ρ :

∏
α Yα → Yα, Uβ ranges over all members of

τβ and β over all elements of A.

Definition 8. [1] Let XD be a dual B-algebra. A topology τ on XD is called a dual
B-topology and the couple (XD, τ) is called a dual B-topological space.

Remark 1. Let XD be a dual B-algebra and nonempty A,B ∈ XD. Then

A ◦B = {a ◦ b | a ∈ A, b ∈ B}.

Definition 9. [1] The triple (XD, ◦, τ) is called a topological dual B-algebra (or tdB-
algebra) if τ is a dual B-topology and the binary operation ◦ : XD × XD → XD is
continuous where the topology on XD ×XD is the Cartesian product topology.

Theorem 3. [1] Let XD be a dual B-algebra and τ a dual B-topology. Then (XD, ◦, τ)
is a tdB-algebra if and only if for all x, y ∈ XD and U(x ◦ y), there exists U(x) and U(y)
such that U(x) ◦ U(y) ⊆ U(x ◦ y).

3. Dual B-topological Space Determined by Some Sets

Suppose XD is a dual B-algebra. For each V ⊆ XD and x ∈ XD, let us denote the
following notations:

(i) V [x] = {y ∈ XD | x ◦ y ∈ V }; (ii) V ′[x] = {y ∈ XD | y ◦ x, x ◦ y ∈ V }.

Remark 2. Let XD be a dual B-algebra and V ⊆ XD. Then V ′[x] ⊆ V [x] for any
x ∈ XD.

Example 1. Consider the set XD = {1, a, b, c, d, e} and binary operation ◦ as defined in
the table below.

◦ 1 a b c d e
1 1 a b c d e
a b 1 a d e c
b a b 1 e c d
c c d e 1 a b
d d e c b 1 a
e e c d a b 1

Then XD is a dual B-algebra [1]. Let V = {a, d, e}. Then V [b] = {1, c, e} and V ′[b] =
{c, e}.



K. Belleza, J. Albaracin / Eur. J. Pure Appl. Math, 15 (4) (2022), 1948-1956 1951

Proposition 2. Suppose XD is a dual B-algebra and V ⊆ XD such that 1 ∈ V . Then
x ∈ V ′[x]. In particular, V = {1} if and only if V [x] = {x} = V ′[x] for any x ∈ X.

Proof. Suppose XD is a dual B-algebra and V ⊆ XD such that 1 ∈ V . By (DB1),
x ◦ x = 1 ∈ V for all x ∈ XD. This implies that x ∈ V ′[x].

Suppose V = {1}. Then V [x] = {x} = V ′[x] for any x ∈ X. Let V [x] = {x} = V ′[x].
Then x ◦ x ∈ V . Thus, 1 ∈ V . Suppose a ∈ V such that a ̸= 1. Then there exists y ∈ X
such that x ◦ y = a with x ̸= y. Hence, y ∈ V [x], a contradiction. Therefore, V = {1}.

Proposition 3. Let XD be a dual B-algebra and U, V ⊂ XD. If U ⊆ V , then U [x] ⊆ V [x]
and U ′[x] ⊆ V ′[x] for any x ∈ XD.

Proof. Suppose XD is a dual B-algebra and U, V ⊂ XD. Let y ∈ U [x]. Then
x ◦ y ∈ U ⊆ V . Hence, y ∈ V [x] which implies that U [x] ⊆ V [x]. Similarly, U ′[x] ⊆ V ′[x].

Proposition 4. Let XD be a dual B-algebra satisfying the symmetric condition and V ⊆
XD such that for all p, q ∈ V and x ∈ XD, p ◦ (x ◦ q) = 1 implies x ∈ V . Then
V ′[x] ◦ V ′[y] ⊆ V ′[x ◦ y].

Proof. Let p◦q ∈ V ′[x]◦V ′[y] where p ∈ V ′[x] and q ∈ V ′[y]. Then p◦x, x◦p, q◦y, y◦q ∈
V . By (DB1), Lemma 2, (DB3), symmetric condition, and (DB2), 1 = (p ◦ x) ◦ (p ◦ x) =

(p ◦ x) ◦ [(p ◦ q) ◦ (x ◦ q)] = (p ◦ x) ◦
[
(p ◦ q) ◦ [(x ◦ y) ◦ (q ◦ y)]

]
= (p ◦ x) ◦

([
[(x ◦ y) ◦ 1] ◦

(p ◦ q)
]
◦ (q ◦ y)

)
= (p ◦ x) ◦

[
[(x ◦ y) ◦ (p ◦ q)] ◦ (q ◦ y)

]
. Since (p ◦ x), (q ◦ y) ∈ V and

(p◦x)◦
[
[(x◦y)◦(p◦q)]◦(q◦y)

]
= 1, this implies that (x◦y)◦(p◦q) ∈ V by the hypothesis.

Similarly, (p ◦ q) ◦ (x ◦ y) ∈ V . Hence, p ◦ q ∈ V ′[x ◦ y]. Therefore, V ′[x] ◦V ′[y] ⊆ V ′[x ◦ y].

Theorem 4. Let Ω be a family of nonempty subsets in a dual B-algebra XD that is closed
under finite intersections. Then the set τ = {U ⊆ XD | ∀x ∈ U,∃V ∈ Ω such that V [x] ⊆
U} is a dual B-topology on XD.

Proof. Let XD be a dual B-algebra and x ∈ XD. Note that V [x] ⊆ XD for any V ∈ Ω.
This implies that XD ∈ τ . Since ∅ does not contain any element, then it is vacuously true
that ∅ ∈ τ . Suppose U1, U2 ∈ τ and x ∈ U1 ∩ U2. Then there exist V1, V2 ∈ Ω such that
V1[x] ⊆ U1 and V2[x] ⊆ U2. Since V1∩V2 ⊆ V1, V2, it follows that (V1∩V2)[x] ⊆ V1[x] ⊆ U1

and (V1 ∩ V2)[x] ⊆ V2[x] ⊆ U2 by Proposition 3. Moreover by the hypothesis, V1 ∩ V2 ∈ Ω.
This implies that (V1∩V2)[x] ⊆ U1 and (V1∩V2)[x] ⊆ U2 or (V1∩V2)[x] ⊆ U1∩U2. Hence,
U1 ∩ U2 ∈ τ. Suppose x ∈

⋃
i∈A

Ui where Ui ∈ τ for all i ∈ A. Then there exists j ∈ A such

that x ∈ Uj . This implies that Vj [x] ⊆ Uj for some Vj ∈ Ω. Hence, Vj [x] ⊆
⋃
i∈A

Ui. It

follows that
⋃
i∈A

Ui ∈ τ. Therefore, τ is a dual B-topology on XD.

Henceforth, the dual B-topology τ in the following results is the dual B-topology in
Theorem 4.



K. Belleza, J. Albaracin / Eur. J. Pure Appl. Math, 15 (4) (2022), 1948-1956 1952

Theorem 5. Suppose XD is a dual B-topological space and Ω is a family of subsets in
a dual B-algebra XD that is closed under finite intersections. If ∅ ∈ Ω, then XD is a
tdB-algebra.

Proof. Suppose XD is a dual B-topological space and Ω is a family of subsets in a
dual B-algebra XD that is closed under finite intersections. Let V ⊆ XD. Then for all
x ∈ V , there exists ∅ ∈ Ω such that ∅[v] = ∅ ⊆ V . This implies that V ∈ τ . Hence
P(XD) ⊆ τ where P(XD) is the power set of XD. Since τ ⊆ P(XD), it follows that
τ = P(XD). Let x, y ∈ XD and U(x ◦ y) ∈ τ . Then there exist {x}, {y} ∈ τ such that
{x} ◦ {y} = {x ◦ y} ⊆ U(x ◦ y). Then XD is a tdB-algebra.

Theorem 6. Let Ω be a family of subsets in the dual B-algebra XD that is closed under
finite intersections. Suppose that for each U ∈ Ω, 1 ∈ U and for each x ∈ U , there exists
V ∈ Ω such that V [x] ⊆ U . Then the set B = {U [a] | U ∈ Ω, a ∈ XD} is a basis for the
dual B-topology.

Proof. First, we will show that B ⊆ τ . Suppose x ∈ U [a] ∈ B for any a ∈ XD and
U ∈ Ω. Then a◦x ∈ U . Moreover, there exists V ∈ Ω such that V [a◦x] ⊆ U . Let y ∈ V [x].
Then x◦y ∈ V . By Theorem 1(iii), (a◦x)◦(a◦y) = x◦y ∈ V . Hence, a◦y ∈ V [a◦x] ⊆ U .
This implies that y ∈ U [a]. Moreover, U [a] ∈ τ . That is, V [x] ⊆ U [a]. Consequently,
B ⊆ τ. Now let U ∈ τ and x ∈ U . Then there exists V ∈ Ω such that V [x] ⊆ U .
By Proposition 2 and Remark 2, x ∈ V [x]. Therefore, there exists V [x] ∈ B such that
x ∈ V [x] ⊆ U. By Theorem 2, B is a basis for the dual B-topology τ .

Theorem 7. Let Ω be a family of subsets in a commutative dual B-algebra XD that
is closed under finite intersections. Suppose that for each V ∈ Ω, 1 ∈ V and for each
x ∈ V ∈ Ω, there exists U ∈ Ω such that U [x] ⊆ V . Then XD is a tdB-algebra.

Proof. Let x, y ∈ XD and U ∈ τ such that x ◦ y ∈ U . Then there exists V ∈ Ω such
that V [x ◦ y] ⊆ U . By Remark 2 and Proposition 2 respectively, V ′[x ◦ y] ⊆ V [x ◦ y]
with x ∈ V ′[x] and y ∈ V ′[y]. We will show that V ′[x] ◦ V ′[y] ⊆ V ′[x ◦ y]. Suppose
a ∈ V ′[x]. Then a ◦ x, x ◦ a ∈ V . By (DB1), Theorem 1(iii), and Proposition 1, 1 =
(a ◦ y) ◦ (a ◦ y) = (a ◦ y) ◦ [(x ◦ a) ◦ (x ◦ y)] = (x ◦ a) ◦ [(a ◦ y) ◦ (x ◦ y)] and 1 =
(x ◦ y) ◦ (x ◦ y) = (x ◦ y) ◦ [(a ◦ x) ◦ (a ◦ y)] = (a ◦ x) ◦ [(x ◦ y) ◦ (a ◦ y)]. By Lemma 1, it
follows that (a ◦ y) ◦ (x ◦ y) = x ◦ a ∈ V and (x ◦ y) ◦ (a ◦ y) = a ◦ x ∈ V . This implies that
a ◦ y ∈ V ′[x ◦ y]. Hence, V ′[x] ◦ y ⊆ V ′[x ◦ y]. Suppose b ∈ V ′[y]. Then b ◦ y, y ◦ b ∈ V .
By Theorem 1(iii), (x ◦ b) ◦ (x ◦ y) = b ◦ y ∈ V and (x ◦ y) ◦ (x ◦ b) = y ◦ b ∈ V . This
implies that x ◦ b ∈ V ′[x ◦ y]. Hence, x ◦ V ′[y] ⊆ V ′[x ◦ y]. Assume on the contrary
that V ′[x] ◦ V ′[y] ⊈ V ′[x ◦ y]. By Proposition 2, V ′[x] ◦ y ∈ V ′[x] ◦ V ′[y] ⊈ V ′[x ◦ y]
and x ◦ V ′[y] ∈ V ′[x] ◦ V ′[y] ⊈ V ′[x ◦ y]. These are contradictions. Therefore, XD is a
tdB-algebra.

Lemma 3. Suppose Ω is an arbitrary family of dual B-filters in a dual B-algebra XD.
Then for all a ∈ V ∈ Ω, V [a] = V .



K. Belleza, J. Albaracin / Eur. J. Pure Appl. Math, 15 (4) (2022), 1948-1956 1953

Proof. Suppose Ω is an arbitrary family of dual B-filters in a dual B-algebra XD and
let a ∈ V ∈ Ω. Suppose x ∈ V [a]. Then a ◦x ∈ V . Since V is a dual B-filter and a ∈ V , it
follows that x ∈ V implying that V [a] ⊆ V . Conversely, suppose x ∈ V . Since V is a dual
B-filter, V is a dual B-subalgebra of XD. Then a ◦ x ∈ V . Hence, x ∈ V [a]. Therefore,
V [a] = V .

The next corollary follows from Lemma 3 and Theorem 7.

Corollary 1. Let Ω be a family of dual B-filters in a commutative dual B-algebra XD

closed under finite intersections such that 1 ∈ V for all V ∈ Ω. Then XD is a tdB-algebra.

4. Filterbase in a Dual B-algebra

Definition 10. Let XD be a dual B-topological space. A filterbase U in XD is a family
U = {Aα | α ∈ A} of subsets of XD having two properties:
(i) Aα ̸= ∅ for all α ∈ A;
(ii) for all α, β ∈ A, there exists γ ∈ A such that Aγ ⊆ Aα ∩Aβ.

Remark 3. Let XD be a dual B-topological space and x1 ∈ XD. The family {U(x1)} is
a filterbase called the neighborhood filterbase of x1.

Example 2. Suppose XD is a dual B-topological space. Any family W of subsets of XD

containing ∅ is not a filterbase. In particular, the dual B-topology τ on XD is not a
filterbase in XD.

Remark 4. The family of dual B-filters is not a subclass of a filterbase in a dual B-algebra
XD.

Example 3. Consider the dual B-algebra X = {1, a, b, c, d, e} in Example 1. Let F =
{{1, e}, {1, a, b}, {1, c}}. Then F is a family of dual B-filters in XD [1]. Note that
{1, e}, {1, a, b} ∈ F but {1, e} ∩ {1, a, b} = {1} /∈ F . This implies that F is not a fil-
terbase.

Remark 5. A filterbase is not a subclass of a family of dual B-filters in a dual B-algebra.

Example 4. Consider the dual B-algebra XD = {1, a, b, c, d, e} in Example 1 and Ω =
{{1, a, b}, {a}}. Then Ω is a filterbase of XD but {a} ∈ Ω is not a dual B-filter of XD

since 1 /∈ {a}.

The next results describes the dual B-topology τ determined by a filterbase Ω followed
by the relationship of Ω and τ if Ω is a family of dual B-filters.

Theorem 8. Let Ω be a filterbase in a dual B-algebra XD. Then the family τ = {O ⊆
XD | ∀a ∈ O,∃V ∈ Ω such that V ′[a] ⊆ O} is a dual B-topology on XD.



K. Belleza, J. Albaracin / Eur. J. Pure Appl. Math, 15 (4) (2022), 1948-1956 1954

Proof. Let Ω be a filterbase in a dual B-algebra XD. Since V ′[a] ⊆ XD for all V ∈ Ω
and a ∈ XD, it follows that XD ∈ τ . Since ∅ do not have any element, then vacuously
∅ ∈ τ . Suppose that Oα, Oβ ∈ τ and a ∈ Oα ∩ Oβ. Then there exist Vα, Vβ ∈ Ω such
that V ′

α[a] ⊆ Oα and V ′
β[a] ⊆ Oβ. Since Ω is a filterbase, there exists V ∈ Ω such that

V ⊆ Vα∩Vβ. By Proposition 3, V ′[a] ⊆ (Vα∩Vβ)
′[a] ⊆ V ′

α[a] ⊆ Oα. Similarly, V ′[a] ⊆ Oβ.
Hence, V ′[a] ⊆ Oα∩Oβ. This implies that Oα∩Oβ ∈ τ . Suppose Oα ∈ τ for all α ∈ A and
let a ∈

⋃
α∈A

Oα. Then a ∈ Oβ ∈ τ for some β ∈ A. This implies that there exists Vβ ∈ Ω

such that V ′
β[a] ⊆ Oβ. Hence, V ′

β[a] ⊆
⋃

α∈A
Oα. It follows that

⋃
α∈A

Oα ∈ τ. Therefore, τ is

a dual B-topology.

Theorem 9. Let XD be a dual B-topological space and Ω a filterbase in XD such that Ω
is a family of dual B-filters of XD. Then Ω is a proper subclass of τ .

Proof. Suppose XD is a dual B-topological space and Ω a filterbase in XD such that Ω
is a family of dual B-filters of XD. Note that ∅ /∈ Ω by Definition 10(i) but ∅ ∈ τ . This
implies that Ω ̸= τ . Let O ∈ Ω and x ∈ O. It remains to show that O′[x] ⊆ O. Suppose
a ∈ O′[x]. Then a ◦ x, x ◦ a ∈ O. Since O is a dual B-filter and x ∈ O, it follows that
a ∈ O. Hence, O′[x] ⊆ O. This implies that O ∈ τ . Therefore, Ω is a proper subclass of τ .

Theorem 10. Suppose XD is a dual B-topological space and let Ω be a filterbase in XD

such that for all V ∈ Ω and for all p, q ∈ V , (i) p ◦ 1 ∈ V ; and (ii) (p ◦ x) ◦ q = 1
implies x ∈ V . Then Ω is the neighborhood filterbase of 1 ∈ XD. That is, Ω is a family of
neighborhoods of 1 (∀V ∈ Ω, 1 ∈ V and V ∈ τ).

Proof. Suppose XD is a dual B-topological space and let Ω be a filterbase in XD and
p ∈ V . By (i), p ◦ 1 ∈ V . By (DB1) and (ii), (p ◦ 1) ◦ (p ◦ 1) = 1 implying that 1 ∈ V .
Claim: V ′[p] ⊆ V .
Let x ∈ V ′[p] . Then x ◦ p, p ◦ x ∈ V . This implies that p ◦ x = v for some v ∈ V . By
(DB1) and (ii), 1 = v ◦ v = (p ◦ x) ◦ v implying that x ∈ V . This proves the claim.
By the claim, V ∈ τ . Therefore, Ω is the neighborhood filterbase of 1 ∈ XD.

Lemma 4. Suppose XD is a dual B-topological space and let Ω be a filterbase in XD such
that for all V ∈ Ω and for all p, q ∈ V , (i) p◦1 ∈ V ; and (ii) (p◦x)◦ q = 1 implies x ∈ V .
Then V ′[a] is open in XD for all a ∈ XD.

Proof. Suppose XD is a dual B-topological space and let Ω be a filterbase in XD.
Suppose x ∈ V ′[a] for any a ∈ XD. Then a ◦ x, x ◦ a ∈ V . Note that by Theorem 10,
V ∈ τ . By Theorem 8, there exist Uα, Uβ ∈ Ω such that U ′

α[a ◦ x], U ′
β[x ◦ a] ⊆ V . Since

Ω is a filterbase in XD, there exist W ∈ Ω such that W ⊆ (Uα ∩ Uβ). This implies that
W ⊆ Uα and W ⊆ Uβ. By Proposition 3, it follows that W ′[a ◦ x] ⊆ U ′

α[a ◦ x] ⊆ V and
W ′[x ◦ a] ⊆ U ′

β[x ◦ a] ⊆ V .
Claim: W ′[x] ⊆ V ′[a].
Suppose y ∈ W ′[x]. Then x ◦ y, y ◦ x ∈ W . By (DB1) and Theorem 1 (iii), 1 = (x ◦ y) ◦



K. Belleza, J. Albaracin / Eur. J. Pure Appl. Math, 15 (4) (2022), 1948-1956 1955

(x◦ y) = [(a◦x)◦ (a◦ y)]◦ (x◦ y). Similarly, 1 = (y ◦x)◦ (y ◦x) = [(a◦ y)◦ (a◦x)]◦ (y ◦x).
Hence by (DB2),

(
1 ◦ [(a ◦x) ◦ (a ◦ y)]

)
◦ (x ◦ y) = 1 and

(
1 ◦ [(a ◦ y) ◦ (a ◦x)]

)
◦ (y ◦x) = 1.

By Theorem 10 and hypothesis (ii), (a ◦ x) ◦ (a ◦ y) ∈ W and (a ◦ y) ◦ (a ◦ x) ∈ W . This
implies that a ◦ y ∈ W ′[a ◦x] ⊆ U ′

α[a ◦x] ⊆ V . Similarly, y ◦a ∈ W ′[x ◦a] ⊆ U ′
β[x ◦a] ⊆ V .

It follows that y ∈ V ′[a]. This proves the claim. Therefore, V ′[a] ∈ τ . That is, V ′[a] is
open in XD for all a ∈ XD.

The next theorem identifies a dual B-topological space determined by a filterbase to
be a tdB-algebra provided some conditions.

Theorem 11. Suppose XD is a dual B-topological space satisfying the symmetric condi-
tion and Ω a filterbase in XD such that for all V ∈ Ω and for all p, q ∈ V , (i) p ◦ 1 ∈ V ;
and (ii) (p ◦ x) ◦ q = 1 implies x ∈ V . Then XD is a tdB-algebra.

Proof. Suppose XD is a dual B-topological space satisfying the symmetric condition
and Ω a filterbase in XD. Let x◦y ∈ O ∈ τ for any x, y ∈ XD. By Theorem 8, there exists
V ∈ Ω such that V ′[x ◦ y] ⊆ O. Note that by Lemma 4, Theorem 10, and Proposition
2, V ′[x], V ′[y] ∈ τ with x ∈ V ′[x] and y ∈ V ′[y]. By Proposition 4, V ′[x] ◦ V ′[y] ⊆ O.
Therefore by Theorem 3, XD is a tdB-algebra.

The last corollary follows from Theorem 11 and Definition 4 of a dual B-filter.

Corollary 2. Suppose XD is a dual B-topological space satisfying the symmetric condition
and Ω a filterbase in XD such that for all V ∈ Ω, V is a dual B-filter. Then XD is a
tdB-algebra.

5. Conclusion

Given a dual B-algebra XD and a family Ω of nonempty subsets of XD that is closed
under finite intersection, we can construct a dual B-topology on XD given by τ = {U ⊆
XD | ∀x ∈ U,∃V ∈ Ω such that V [x] ⊆ U}. If the empty set is a member of Ω, then XD is
a tdB-algbera. Furthermore, if Ω is a filterbase of XD, then τ = {O ⊆ XD | ∀a ∈ O,∃V ∈
Ω such that V ′[a] ⊆ O} is also a dual B-topology on XD. If the condition is imposed to
Ω such that for all V ∈ Ω and for all p, q ∈ V , (i) p ◦ 1 ∈ V ; and (ii) (p ◦ x) ◦ q = 1
implies x ∈ V , then XD is a tdB-algebra. Generally, in this paper we constructed two
dual B-topologies on XD and proved with some conditions that XD with these topologies
is a tdB-algebra.

Acknowledgement

This research is financially supported through an approved research load by the Re-
search, Development, Extension and Publications Office (RDEPO) of the University of
San Carlos during the 2nd term of A.Y. 2020-2021. The author would like to extend their
sincerest gratitude for this support.



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