# Analyzing Some Structural Properties of TopologicalB-Algebras

Authors: Narciso C. Gonzaga
Journal: International Journal of Mathematics and Mathematical Sciences
 (2019)

---

## Abstract

In this study, we investigate the topology onB-algebras: an algebraic system of propositional logic. We define here the notion of topological B-algebras (briefly, T
B-algebras) and some properties are investigated. A characterization of T
B-algebras based on neighborhoods is provided. We also provide a filterbase that generates a unique B-topology, making a T
B-algebra in which the filterbase is a neighborhood base of the constant element, provided that the given B-algebra is commutative. Finally, we investigate subalgebras of T
B-algebras and introduce the notion of quotient T
B-algebras of the given B-algebra.


---

## Body

## 1. Introduction

J. Neggers and H. S. Kim [1] introduced the concept of B-algebras in 2002. It is an algebraic system which is related to B
C
K/B
C
I-algebras but seems to have more profound properties without being excessively complicated. The authors in [2, 3] proved that there is a direct correspondence between B-algebras and groups. Thus, certain properties and results on B-algebras were established via the properties of groups. Readers may refer to [4–8] for some of these “group-alike” properties and results on B-algebras.The theory of topological groups is one of the already well-defined concepts in both algebra and topology. Thus, in connection to the researches previously mentioned, this study initiates the notion of topologicalB-algebras. Some fundamental properties of a topological B-algebra that anchors on the basic topological and algebraic concepts will be investigated. This will provide the foundation of future investigations regarding the overall structure of a topological B-algebra.


## 2. Preliminaries

In this section, we recall some elementary concepts inB-algebras that are necessary in this paper.AB-algebra [1] is defined as a triple (
X
;
∗
,
0
) where X is a nonempty set with “0” being its constant element and “∗” being a binary operation that satisfies the following conditions for all x
,
y
,
z
∈
X:(
B
1
)
x
∗
x
=
0,(
B
2
)
x
∗
0
=
x,(
B
3
)
(
x
∗
y
)
∗
z
=
x
∗
(
z
∗
(
0
∗
y
)
).In addition,(
X
;
∗
,
0
) is said to be acommutative
B-algebra if the following condition holds for all x
,
y
∈
X:(
B
4
)
x
∗
(
0
∗
y
)
=
y
∗
(
0
∗
x
).Throughout the remainder of this study,X shall conveniently denote the B-algebra (
X
;
∗
,
0
) unless otherwise specified.LetX be a B-algebra. A nonempty subset A is said to be asubalgebra of X [9] if x
∗
y
∈
A for all x
,
y
∈
A. In addition, A is said to be anormal subset of X [10] if (
x
∗
a
)
∗
(
y
∗
b
)
∈
A whenever x
∗
y
,
a
∗
b
∈
A. Accordingly, every normal subset of X is a subalgebra of X. Thus, it is more precise to say that A is anormal subalgebra of X if the condition of normality is satisfied.Some basic notations and properties of (commutative)B-algebras are the following.Definition 1 (see [7]).
LetX be a B-algebra. For each x
∈
X, x
-
1 denotes the expression 0
∗
x.Lemma 2 (see [1]).
LetX be a B-algebra. Then, for all x
,
y
,
z
∈
X,(i)
(
x
∗
y
)
∗
y
-
1
=
x;
(ii)
x
∗
(
y
∗
z
)
=
x
∗
z
-
1
∗
y;
(iii)
x
∗
y
=
0 if and only if x
=
y.Lemma 3 (see [11]).
IfX is a B-algebra, then x
-
1
-
1
=
x for all x
∈
X.Lemma 4 (see [10]).
LetX be a B-algebra. Then x
∗
y
=
(
y
∗
x
)
-
1 for all x
,
y
∈
X.Lemma 5 (see [1]).
LetX be a commutative B-algebra. Then, for every x
,
y
∈
X, x
-
1
∗
y
-
1
=
y
∗
x.Theorem 6 (see [12]).
LetX be a B-algebra. Then X is commutative if and only if x
∗
(
x
∗
y
)
=
y for all x
,
y
∈
X.On the other hand, the following notions appear in [6]. Let X be a B-algebra and N a normal subalgebra of X. For x
∈
X, theleft coset and theright coset of N in X by x are, respectively, the sets(1)
x
∗
N
=
x
∗
n
:
n
∈
N
,
N
∗
x
=
n
∗
x
:
n
∈
N
.Denote the set of all distinct left cosets of N in X by X
/
N and define a binary operation “⊛” on X
/
N as follows: for all x
∗
N
,
y
∗
N
∈
X
/
N, (
a
∗
N
)
⊛
(
b
∗
N
)
=
(
a
∗
b
)
∗
N. Then (
X
/
N
;
⊛
,
N
) is a B-algebra and is called thequotient
B
-algebra of X by N. For simplicity of this study, we shall denote the quotient B-algebra (
X
/
N
;
⊛
,
N
) by X
/
N only, unless otherwise specified.We finish this section by providing some new classes of sets inB-algebras and their properties that will be needed in this study.Definition 7.
LetA and B be nonempty subsets of a B-algebra X:(i)
For eachx
∈
X, the sets x
∗
A and A
∗
x are called theleft andright translates of A by x, respectively.
(ii)
Theproduct of A and B, denoted by A
∗
B, is given by(2)
A
∗
B
=
⋃
a
∈
A
a
∗
B
=
⋃
b
∈
B
A
∗
b
=
a
∗
b
:
a
∈
A
,
b
∈
B
.We also recall the following notation which has appeared first in [6].Definition 8.
LetA be a nonempty subset of a B-algebra X. Theinverse of A, denoted by A
-
1, is the set {
a
-
1
:
a
∈
A
}.Note that ifA is a subalgebra of a B-algebra X, then, as explained earlier, the sets x
∗
A and A
∗
x are left and right cosets of A in X, respectively.Some obvious results on the notations presented above are the following.Remark 9.
LetA and B be nonempty subsets of a B-algebra X:(i)
A
∗
0
=
A and 0
∗
A
=
A
-
1.
(ii)
If0
∈
B, then A
⊆
A
∗
B.
(iii)
A
∗
B
⊆
C
∗
D if and only if A
⊆
C and B
⊆
D.The following result, which appears in [6], will be considered in this study.Corollary 10.
LetN be a subalgebra of a B-algebra X. Then X is a normal subalgebra of X if and only if x
∗
(
x
∗
N
)
⊆
N for all x
∈
X.Proposition 11.
LetA be a nonempty subset of a B-algebra X. Then the following statements hold:(i)
A
-
1
-
1
=
A.
(ii)
For allx
,
y
∈
X, A
∗
(
x
∗
y
)
=
A
∗
y
-
1
∗
x.
(iii)
IfA is a subalgebra of X, then A
∗
A
=
A.
(iv)
IfA is a subalgebra of X, then (
x
∗
a
)
∗
A
=
x
∗
A, for all x
∈
X and for all a
∈
A.Proof.
LetA be a nonempty subset of X and x
,
y
∈
X:(i)
It follows from Lemma3.
(ii)
It follows from Lemma2
(
i
i
).
(iii)
It follows from Remark9
(
i
i
) and from the fact that A is a subalgebra of X.
(iv)
SupposeA is a subalgebra of X. Let x
∈
X and a
∈
A. Then for p
∈
(
x
∗
a
)
∗
A, there is an a
′
∈
A such that p
=
(
x
∗
a
)
∗
a
′. By (
B
3
), p
=
x
∗
a
′
∗
a
-
1, with a
′
∗
a
-
1
∈
A. Hence, p
∈
x
∗
A and so (
x
∗
a
)
∗
A
⊆
x
∗
A. Further, if q
∈
x
∗
A, then there is an a
1
∈
A such that q
=
x
∗
a
1. By Lemma 3, Definition 1,and Lemma 4, (
B
1
), (
B
3
), and (
B
2
), we have(3)
q
=
x
∗
a
1
=
x
∗
a
1
-
1
-
1
=
0
∗
a
1
∗
x
=
x
∗
a
∗
x
∗
a
∗
a
1
∗
x
=
x
∗
a
∗
a
1
∗
x
∗
x
∗
a
-
1
=
x
∗
a
∗
a
1
∗
x
∗
a
∗
x
=
x
∗
a
∗
a
1
∗
a
∗
x
∗
x
-
1
=
x
∗
a
∗
a
1
∗
a
∗
x
-
1
∗
x
-
1
=
x
∗
a
∗
a
1
∗
a
∗
0
=
x
∗
a
∗
a
1
∗
a
∈
x
∗
a
∗
A
.
Therefore,x
∗
A
⊆
(
x
∗
a
)
∗
A. Consequently, (
x
∗
a
)
∗
A
=
x
∗
A.Meanwhile, we consider the following properties that are present in a commutativeB-algebra.Proposition 12.
LetA and B be nonempty subsets of a commutative B-algebra X. Then the following hold:(i)
For allx
,
y
∈
X, (
A
∗
x
)
∗
(
B
∗
y
)
=
(
A
∗
B
)
∗
(
x
∗
y
).
(ii)
If(
x
∗
A
)
∗
x
⊆
B, for all x
∈
X, then A
⊆
B
-
1.Proof.
(i) It suffices to show that(
a
∗
x
)
∗
(
b
∗
y
)
=
(
a
∗
b
)
∗
(
x
∗
y
) for all x
,
y
,
a
,
b
∈
X. Now, by (
B
3
), Lemma 5, (
B
2
), Lemma 4, and Theorem 6,(4)
a
∗
x
∗
b
∗
y
=
a
∗
b
∗
y
∗
x
-
1
=
a
∗
b
∗
x
-
1
∗
y
-
1
=
a
∗
b
∗
y
∗
x
∗
0
=
a
∗
b
∗
y
∗
x
∗
x
∗
y
∗
x
∗
y
=
a
∗
b
∗
y
∗
x
∗
x
∗
y
-
1
∗
x
∗
y
=
a
∗
b
∗
y
∗
x
∗
y
∗
x
∗
x
∗
y
=
a
∗
y
∗
x
∗
b
∗
y
∗
x
-
1
∗
x
∗
y
=
a
∗
y
∗
x
∗
y
∗
x
∗
b
∗
x
∗
y
=
a
∗
b
∗
x
∗
y
.(ii) Let a
∈
A. Then, for each x
∈
X, (
x
∗
a
)
∗
x
∈
(
x
∗
A
)
∗
x, and so, by the hypothesis, there is a b
∈
B such that b
=
(
x
∗
a
)
∗
x. Observe that by (
B
3
), (
B
1
), and (
B
2
),(5)
b
∗
x
-
1
=
x
∗
a
∗
x
∗
x
-
1
=
x
∗
a
∗
x
-
1
∗
x
-
1
=
x
∗
a
∗
0
=
x
∗
a
.
By Theorem6, Lemma 2
(
i
i
), Lemma 3, (
B
2
), (
B
1
), and Definition 1,(6)
a
=
x
∗
x
∗
a
=
x
∗
b
∗
x
-
1
=
x
∗
x
-
1
-
1
∗
b
=
x
∗
x
∗
b
=
0
∗
b
=
b
-
1
∈
B
-
1
.
Therefore,A
⊆
B
-
1.


## 3. Initial Properties of TopologicalB-Algebras

This section presents the definition of a topologicalB-algebra as well as some of its fundamental properties. Also, some characterizations of (commutative) topological B-algebras in terms of neighborhood and filters are investigated.Definition 13.
LetX be a B-algebra. A topology τ furnished on X is called a B-topology on X. In addition, (
X
,
τ
) is called atopological
B-algebra (or T
B-algebra) if τ is a B-topology on X and the binary operation ∗
:
X
×
X
→
X is continuous, where the Cartesian product topology on X
×
X is furnished by τ.Example 14.
Consider the groupR of real numbers under the usual addition operation “+”. Then it is well known that R together with “+” and the Euclidean topology E
1 is a topological group. Let “∗” be a binary operation on R defined by a
∗
b
=
a
-
b for all a
,
b
∈
R. Then it is easy to show that (
R
;
∗
,
0
) is a B-algebra. Now, one of the properties of the topological group (
R
,
E
1
) is that the map (
a
,
b
)
↦
a
-
b is continuous. This implies that *
:
R
×
R
→
R is a continuous map. Therefore, (
R
,
E
1
) is also a T
B-algebra.Example 15.
Consider theB-algebra X
=
{
0
,
a
,
b
,
c
} with the binary operation “∗” defined on the Cayley table provided in Table 1 (see [13]). Let τ
=
{
X
,
⌀
,
{
0
,
a
}
,
{
b
,
c
}
}. Then τ is a B-topology on X. By routine calculations, we see that (
X
,
τ
) is a T
B-algebra.
In view of Definition7
(
i
i
), the following remark is straightforward.Table 1
Cayley table for the binary operation “∗”.
∗
0
a
b
c
0
0
a
b
c
a
a
0
c
b
b
b
c
0
a
c
c
b
a
0Remark 16.
LetX be a B-algebra. Then ∗
(
A
×
B
)
=
A
∗
B for any nonempty subsets A and B of X.On the other hand, the following result provides a necessary and sufficient condition for aT
B-algebra.Theorem 17.
LetX be a B-algebra and τ a B-topology on the set X. Then (
X
,
τ
) is a T
B-algebra if and only if, for all x
,
y
∈
X and for every nbd W of x
∗
y, there are nbds U and V of x and y, respectively, such that U
∗
V
⊆
W.Proof.
Suppose that “∗” is a continuous map. Let x
,
y
∈
X and W a nbd of x
∗
y. Then ∗
-
1
(
W
) is a nbd of (
x
,
y
) in X
×
X. By the definition of Cartesian product topology, there exist nbds U and V of x and y, respectively, such that U
×
V
⊆
∗
-
1
(
W
). By Remark 16, U
∗
V
=
∗
(
U
×
V
). Hence, there are nbds U and V of x and y, respectively, such that U
∗
V
⊆
∗
(
∗
-
1
(
W
)
)
=
W.
Conversely, let(
a
,
b
)
∈
X
×
X and W a nbd of ∗
(
a
,
b
)
=
a
∗
b. Now, by the assumption, there exist nbds U and V of a and b, respectively, such that U
∗
V
⊆
W. By the definition of Cartesian product topology, it follows that U
×
V is a nbd of (
a
,
b
). Now, ∗
(
U
×
V
)
=
U
∗
V by Remark 16, and so there is a nbd U
×
V of (
a
,
b
) such that ∗
(
U
×
V
)
⊆
W. Therefore, “∗” is a continuous function, completing the proof of the theorem.Theorem 18.
Let(
X
,
τ
) be a T
B-algebra and a a fixed element of the B-algebra X. Then the following functions are homeomorphisms:(i)
R
a
:
(
X
,
τ
)
→
(
X
,
τ
) defined by R
a
(
x
)
=
x
∗
a.
(ii)
L
a
:
(
X
,
τ
)
→
(
X
,
τ
) defined by L
a
(
x
)
=
a
∗
x.
(iii)
ι
:
(
X
,
τ
)
→
(
X
,
τ
) defined by ι
(
x
)
=
x
-
1.Proof.
Leta be a fixed element of X. Consider the function φ
a
:
(
X
,
τ
)
→
(
X
,
τ
)
×
(
X
,
τ
) defined by φ
a
(
x
)
=
(
x
,
a
). Let x
∈
X and W a nbd of φ
a
(
x
). Then, by the definition of Cartesian product topology, there are τ-open sets U and V in X such that φ
a
(
x
)
=
(
x
,
a
)
∈
U
×
V
⊆
W. Thus, there is a nbd U of x such that φ
a
(
U
)
=
U
×
{
a
}
⊆
U
×
V
⊆
W. This shows that φ
a is continuous. Similarly, the function ϕ
a
:
(
X
,
τ
)
→
(
X
,
τ
)
×
(
X
,
τ
) defined by ϕ
a
(
x
)
=
(
a
,
x
) is continuous. Now, observe that R
a
=
∗
∘
φ
a and L
a
=
∗
∘
ϕ
a, with “∗” being a continuous map by Definition 13. Thus, both R
a and L
a are compositions of two continuous functions. Hence, R
a and L
a are continuous.
Next, considerR
a
-
1
:
(
X
,
τ
)
→
(
X
,
τ
) defined by R
a
-
1
(
x
)
=
x
∗
a
-
1. By the previous argument, R
a
-
1 is also continuous. By Definition 1 and Lemma 2
(
i
), we have(7)
R
a
-
1
∘
R
a
x
=
R
a
-
1
R
a
x
=
x
∗
a
∗
a
-
1
=
x
.Also, by Definition 1, (
B
3
), Lemma 3, (
B
1
), and (
B
2
), it follows that(8)
R
a
∘
R
a
-
1
x
=
R
a
R
a
-
1
x
=
x
∗
a
-
1
∗
a
=
x
∗
a
∗
a
-
1
-
1
=
x
.Hence, we have R
a
-
1
∘
R
a
=
1
X and R
a
∘
R
a
-
1
=
1
X. Consequently, R
a is a homeomorphism, proving (
i
).
On the other hand, forx
∈
X and W a nbd of ι
(
x
)
=
x
-
1
=
0
∗
x, there are nbds U and V of 0 and x, respectively, such that U
∗
V
⊆
W. Observe that, by Definition 8 and Remark 9, ι
(
V
)
=
V
-
1
=
0
∗
V
⊆
U
∗
V. Thus, there is a nbd V of x such that ι
(
V
)
⊆
W. Therefore, ι is continuous. Next, by Lemma 3, we have (
ι
∘
ι
)
(
x
)
=
ι
(
ι
(
x
)
)
=
x
-
1
-
1
=
x, and so ι
∘
ι
=
1
X. Hence, ι is a homeomorphism, proving (
i
i
i
).
Lastly, by Lemma4, L
a
(
x
)
=
a
∗
x
=
(
x
∗
a
)
-
1
=
ι
(
x
∗
a
)
=
ι
(
R
a
(
x
)
) for all x
∈
X. Thus, L
a
=
ι
∘
R
a; that is, L
a is a composition of two homeomorphisms. Consequently, L
a is a homeomorphism, proving (
i
i
).Corollary 19.
Let(
X
,
τ
) be a T
B-algebra and O
⊆
X. Then the following statements are equivalent:(i)
O is τ-open in X.
(ii)
O
-
1 is τ-open in X.
(iii)
x
∗
O is τ-open in X for all x
∈
X.
(iv)
O
∗
x is τ-open in X for all x
∈
X.Meanwhile, the following corollary is immediate from the fact that, in a homeomorphism of topological spaces, the closure of the image of a subset of the domain is equal to the image of the closure of the said subset.Corollary 20.
Let(
X
,
τ
) be a T
B-algebra. Then the following statements hold:(i)
U
-
1
¯
=
U
¯
-
1 for all U
⊆
X.
(ii)
x
∗
U
¯
=
x
∗
U
¯ for all U
⊆
X and for all x
∈
X.
(iii)
U
∗
x
¯
=
U
¯
∗
x for all U
⊆
X and for all x
∈
X.We shall now use Theorem18 in proving some results pertaining to neighborhoods in T
B-algebras. Let (
X
,
τ
) be a T
B-algebra and x
∈
X. Denote the neighborhood filter of x by η
x
(
X
,
τ
). Also, from Remark 9
(
i
i
i
), we have U
2
=
U
∗
U, where U
⊆
X.Theorem 21.
LetU be a neighborhood base of the constant element 0 of a T
B-algebra (
X
,
τ
). Then the following conditions hold:(i)
For everyU
∈
U, there is a V
∈
U such that V
2
⊆
U.
(ii)
For everyU
∈
U, there is a V
∈
U such that V
-
1
⊆
U.
(iii)
For everyU
∈
U and x
∈
X, there is a V
∈
U such that (
x
∗
V
)
∗
x
⊆
U.Proof.
LetU be an arbitrary element of U.(i)
We have0
∗
0
=
0. By the continuity of “∗” and by Theorem 17, there are members W
1 and W
2 of η
0
(
X
,
τ
) such that W
1
∗
W
2
⊆
U. Thus, there exists V
1
∈
η
0
(
X
,
τ
) such that V
1
⊆
W
1
∩
W
2. Since U is a neighborhood base of 0, there is a V
∈
U such that V
⊆
V
1. Therefore, by Remark 9
(
i
i
i
) and Remark 16,(9)
V
2
⊆
V
1
2
=
∗
V
1
×
V
1
⊆
∗
W
1
×
W
2
=
W
1
∗
W
2
⊆
U
.
(ii)
Similar to(
i
), but use the continuity of ι in Theorem 18
(
i
i
i
).
(iii)
Fixx
∈
X and let ν
x
:
X
→
X be defined by ν
x
(
y
)
=
(
x
∗
y
)
∗
x. Then ν
x
=
L
x
∘
R
x, and so, in view of Theorem 18, ν
x is continuous. Now, ν
x
(
0
)
=
(
x
∗
0
)
∗
x
=
0 by (
B
2
) and (
B
1
). This means that if P
∈
η
0
(
X
,
τ
), then ν
x
-
1
(
P
)
∈
η
0
(
X
,
τ
). Meanwhile, U
∈
U implies U
∈
η
0
(
X
,
τ
). Thus, we can choose V
∈
U such that V
⊆
ν
x
-
1
(
U
). Therefore, we have (
x
∗
V
)
∗
x
=
ν
x
(
V
)
⊆
ν
x
(
ν
x
-
1
(
U
)
)
⊆
U.On the other hand, we shall construct aB-topology on a B-algebra which is generated by a neighborhood base of the constant element, making the space a T
B-algebra provided that the B-algebra is commutative.Theorem 22.
LetX be a B-algebra and U a filterbase of X which satisfies the statements (i), (ii), and (iii) in Theorem 21. Then there is a unique B-topology τ such that U is a neighborhood base of the constant element 0 of X in the topology τ. Moreover, if X is commutative, then (
X
,
τ
) is a T
B-algebra.Proof.
Firstly, we claim that every member ofU contains the constant element 0. Let U
∈
U. Then, by the hypothesis, there exists V
∈
U such that V
2
⊆
U. Let v
∈
V. Then, by (
B
1
), 0
=
v
∗
v
∈
V
2. This proves the claim.
Next, let(10)
τ
=
O
⊆
G
:
for  every
x
∈
O
,
t
h
e
r
e
i
s
a
U
∈
U
s
u
c
h
t
h
a
t
U
∗
x
-
1
⊆
O
.We shall show that τ is a B-topology furnished from X. Clearly, X
,
⌀
∈
τ, and so (
T
1
) is satisfied. Let S
1
=
{
O
α
:
α
∈
A
} be a subclass of τ and set O
=
⋃
α
O
α. Let x
∈
O. Then x
∈
O
α for some α
∈
A. Thus, there is a U
∈
U such that U
∗
x
-
1
⊆
O
α
⊆
O. Hence, O
∈
τ. Further, let S
2
=
{
O
k
:
k
=
1,2
,
…
,
n
} be a finite subclass of τ and O
′
=
⋂
k
=
1
n
O
k. Let y
∈
O
′. Then y
∈
O
k for all k
=
1,2
,
…
,
n. Thus, for each k, there is a U
k
∈
U such that U
k
∗
y
-
1
⊆
O
k. Now, U a filterbase means that there is a U
∈
U such that U
⊆
U
k for all k. Therefore, there is a U
∈
U such that U
∗
y
-
1
⊆
⋂
k
=
1
n
U
k
∗
y
-
1
⊆
O
′. This shows that O
′
∈
τ. Consequently, τ is a B-topology on X.
We will now show thatU is a neighborhood base of 0. We claim first that every member of η
0
(
X
,
τ
) contains a member of U. Let N
∈
η
0
(
X
,
τ
). Then we have 0
∈
i
n
t
τ
N
⊆
N with i
n
t
τ
N
∈
τ. From the definition of τ, Remark 9
(
i
), and by the fact that 0
-
1
=
0, there is a U
∈
U such that U
=
U
∗
0
-
1
⊆
i
n
t
τ
N
⊆
N. This proves the claim. Next, we claim that U
⊆
η
0
(
X
,
τ
). It suffices to show that every member of U is a nbd of 0. Let U
∈
U. Then by the previous argument, 0
∈
U. Now, set(11)
B
=
x
∈
U
:
∃
V
∈
U
s
u
c
h
t
h
a
t
V
∗
x
-
1
⊆
U
.Then B
≠
⌀ since 0
∈
B. By construction, B
⊆
U. Let w
∈
B. Then there is a V
∈
U such that V
∗
w
-
1
⊆
U. Also, there is a P
∈
U such that P
2
⊆
V. Let y
∈
P
∗
w
-
1. Then y
=
p
∗
w
-
1 for some p
∈
P. Thus by (
B
2
) and by the fact that 0
∈
P, y
=
p
∗
w
-
1
=
(
p
∗
0
)
∗
w
-
1
∈
P
2
∗
w
-
1. Hence, by Remark 9
(
i
i
), P
∗
w
-
1
⊆
P
2
∗
w
-
1
⊆
V
∗
w
-
1
⊆
U. This means that y
∈
B, and so there is a P
∈
U such that P
∗
w
-
1
⊆
B. We have shown that B
∈
τ, and so B is τ-open. This proves the second claim; hence, U is a neighborhood base of 0. Consequently, since a neighborhood base generates one and only one topology on X, τ is unique.
Lastly, supposeX is commutative. We will then prove the continuity of  “∗”. First, we claim that if O
∈
τ and y
∈
X, then O
∗
y
∈
τ. Let x
∈
O
∗
y. Then x
=
u
∗
y for some u
∈
O. By (
B
4
) and Lemma 2
(
i
), y
∗
x
-
1
=
x
∗
y
-
1
=
(
u
∗
y
)
∗
y
-
1
=
u
∈
O, and so there is a U
∈
U such that U
∗
y
∗
x
-
1
-
1
=
U
∗
x
-
1
∗
y
⊆
O by the definition of τ. By Theorem 6, Proposition 11
(
i
i
), Lemma 3, and Remark 9
(
i
i
i
), we have(12)
U
∗
x
-
1
=
U
∗
y
∗
y
∗
x
-
1
=
U
∗
y
∗
x
-
1
-
1
∗
y
=
U
∗
x
-
1
∗
y
∗
y
⊆
O
∗
y
.This proves the claim. Next, let x
,
y
∈
X and W a nbd of x
∗
y. Then we have x
∗
y
∈
O
⊆
W for some O
∈
τ. Thus, by the definition of τ and Lemma 3, P
∗
(
x
∗
y
)
-
1
=
P
∗
(
y
∗
x
) for some P
∈
U. Also, P
∈
U implies U
2
⊆
P for some U
∈
U. So, we have U
2
∗
(
y
∗
x
)
⊆
P
∗
(
y
∗
x
)
⊆
O
⊆
W. Further, U
∈
U implies there is a V
′
∈
U such that (
x
∗
V
′
)
∗
x
⊆
U for all x
∈
X. But since X is commutative, it follows from Proposition 12
(
i
i
) that V
′
-
1
⊆
U. Also, V
′
∈
U implies there is a V
∈
U such that V
⊆
V
′
-
1. Now, x
∈
U
∗
x
-
1 since, by Lemma 3 and Definition 1, we have x
=
0
∗
(
0
∗
x
)
=
0
∗
x
-
1 and 0
∈
U. Similarly, y
∈
V
∗
y
-
1. Since U
,
V
∈
η
0
(
X
,
τ
), it follows that U
∗
x
-
1 and V
∗
y
-
1 are nbds of x and y, respectively. By Proposition 12
(
i
), Lemma 5, and Remark 9
(
i
i
i
)
(13)
U
∗
x
-
1
∗
V
∗
y
-
1
=
U
∗
V
∗
x
-
1
∗
y
-
1
=
U
∗
V
∗
y
∗
x
⊆
U
∗
V
′
-
1
∗
y
∗
x
⊆
U
2
∗
y
∗
x
⊆
W
.Therefore, by Theorem 17, (
X
,
τ
) is a T
B-algebra.


## 4. Subalgebras and Quotients ofT
B-Algebras

In this section, we provide some properties ofT
B-algebras in terms of its (normal) subalgebras. Also, we consider the construction of quotient T
B-algebras determined by a normal subalgebra of a B-algebra.Proposition 23.
Let(
X
,
τ
) be a T
B-algebra. Then (
X
,
τ
) is Hausdorff if and only if the trivial subalgebra 0 is a closed subset of X.Proof.
Suppose(
X
,
τ
) is Hausdorff. Let x
∈
X such that x
≠
0 and S
x the subclass of P
(
X
) containing all the open neighborhoods of x. Also, set(14)
S
=
⋃
x
∈
X
∖
0
S
x
.Then(15)
X
∖
0
=
⋃
S
∈
S
Sis an open subset of X which does not contain 0. Hence, {
0
}
=
X
∖
{
0
}
c is closed in X.
Conversely, assume that{
0
} is a closed subset of X. Observe that the binary operation “∗” is continuous. Thus, ∗
-
1
(
{
0
}
) is closed in X
×
X. But ∗
-
1
(
{
0
}
)
=
{
x
∗
x
:
x
∈
X
}
=
Δ by (
B
1
) and Lemma 2
(
i
i
i
). Therefore, (
X
,
τ
) is Hausdorff.Proposition 24.
Let(
X
,
τ
) be a T
B-algebra and Y a subalgebra of X.(i)
The closureY
¯ of Y is a subalgebra of X.
(ii)
IfY is a normal subalgebra of X, then so does Y
¯.
(iii)
If(
X
,
τ
) is Hausdorff and Y is commutative, then Y
¯ is commutative.Proof.
LetY be a subalgebra of X.(i)
We will first prove that, for every nonempty subsetsA and B of X, A
¯
∗
B
¯
⊆
A
∗
B
¯. Indeed, since by Remark 16 and by the continuity of “∗”, A
¯
∗
B
¯
=
∗
(
A
¯
×
B
¯
)
⊆
∗
(
A
×
B
)
¯
=
A
∗
B
¯. Now, suppose x
,
y
∈
Y
¯. Observe that by Proposition 11, Y
¯
∗
Y
¯
⊆
Y
∗
Y
¯
=
Y
2
¯
=
Y
¯. Hence, we see that x
∗
y
∈
Y
¯
∗
Y
¯
⊆
Y
¯. Therefore, Y
¯ is a subalgebra of X.
(ii)
SupposeY is normal in X. Then, by Corollary 10, x
∗
(
x
∗
Y
)
⊆
Y for all x
∈
X. By Corollary 20
(
i
i
), x
∗
x
∗
Y
¯
=
x
∗
(
x
∗
Y
)
¯
⊆
Y
¯ for all x
∈
X. Hence, by Corollary 10, Y
¯ is a normal subalgebra of X.
(iii)
Consider the functionα
:
X
×
X
→
X defined by α
(
x
,
y
)
=
x
∗
(
y
∗
(
y
∗
x
)
). Then α is continuous since “∗” is a continuous map. Now, (
X
,
τ
) being a Hausdorff space implies {
0
} is closed in X by Proposition 23. Hence, α
-
1
(
{
0
}
) is closed in X
×
X. Since Y is a commutative B-algebra, it follows that x
∗
(
y
∗
(
y
∗
x
)
)
=
x
∗
x
=
0 for all x
,
y
∈
Y by Theorem 6 and (
B
1
). Thus, Y
×
Y
⊆
α
-
1
(
{
0
}
); whence Y
¯
×
Y
¯
=
Y
×
Y
¯
⊆
α
-
1
(
{
0
}
). Note that (
0,0
)
∈
Y
¯
×
Y
¯. Therefore, {
0
}
⊆
α
Y
¯
×
Y
¯
⊆
α
(
α
-
1
(
{
0
}
)
)
⊆
{
0
}, and so α
Y
¯
×
Y
¯
=
{
0
}. This means that α
(
x
,
y
)
=
x
∗
(
y
∗
(
y
∗
x
)
)
=
0 for all x
,
y
∈
Y
¯. Now, by Lemma 2
(
i
i
) and Lemma 4,(16)
0
=
x
∗
y
∗
y
∗
x
=
x
∗
y
∗
x
-
1
∗
y
=
x
∗
x
∗
y
∗
y
.
Thus, by Lemma2
(
i
i
i
), x
∗
(
x
∗
y
)
=
y for all x
,
y
∈
Y
¯. Consequently, by Theorem 6, Y
¯ is commutative.We shall now consider the construction of quotientT
B-algebras by a normal subalgebra of the given B-algebra. Recall first that a function f
:
X
→
Y where X and X
′ are B-algebras is said to be a B
-homomorphism [9] if f
(
x
∗
X
y
)
=
f
(
x
)
∗
X
′
f
(
y
) for all x
,
y
∈
X, with “∗
X” and “∗
X
′” being the binary operations on X and Y, respectively. In addition, f is a B
-epimorphism if f is surjective.LetX be a B-algebra and N a subalgebra of X. Denote X
(
≡
N
) as the collection of all left cosets of X by N. Such members of X
(
≡
N
) are distinct, since these are the equivalence classes of an equivalence relation on X ([6], Theorem 3). Clearly, the mapping π
:
X
→
X
(
≡
N
) is surjective. If N is normal in X, then X
(
≡
N
) coincides with the quotient B-algebra X
/
N. Moreover, π is a B-epimorphism [9] and is called thecanonical projection of B-algebras.Proposition 25.
Let(
X
,
τ
) be a T
B-algebra, N a subalgebra of X, and A a nonempty subset of X. Then π
-
1
(
π
(
A
)
)
=
A
∗
N.Proof.
SupposeA is a nonempty subset of X. Let p
∈
π
-
1
(
π
(
A
)
). Then(17)
p
∗
N
=
π
p
∈
π
A
=
a
∗
N
:
a
∈
A
.This means that there is an a
1
∈
A such that p
∗
N
=
a
1
∗
N. Thus, there are n
1
,
n
2
∈
N such that p
∗
n
1
=
a
1
∗
n
2. Since N is a subalgebra of X, it follows that 0
∈
N and n
-
1
=
0
∗
n
∈
N for every n
∈
N. Now, by Lemma 2
(
i
) and (
B
3
),(18)
p
=
p
∗
n
1
∗
n
1
-
1
=
a
1
∗
n
2
∗
n
1
-
1
=
a
∗
n
1
-
1
∗
n
2
-
1
∈
a
∗
N
.This shows that π
-
1
(
π
(
A
)
)
⊆
A
∗
N. Next, suppose q
∈
A
∗
N. Then there exist a
∈
A and n
∈
N such that q
=
a
∗
n. By Proposition 11
(
i
v
), we have π
(
q
)
=
π
(
a
∗
n
)
=
(
a
∗
n
)
∗
N
=
a
∗
N. Hence, by Definition 7
(
i
i
),(19)
q
∈
π
-
1
a
∗
N
⊆
π
-
1
⋃
a
∈
A
a
∗
N
=
π
-
1
π
A
.Therefore, A
∗
N
⊆
π
-
1
(
π
(
A
)
). Consequently, π
-
1
(
π
(
A
)
)
=
A
∗
N.Theorem 26.
Let(
X
,
τ
) be a T
B-algebra and N a subalgebra of X. For the class X
(
≡
N
), define τ
X
(
≡
N
)
⊆
P
(
X
(
≡
N
)
) by(20)
τ
X
≡
N
=
O
⊆
X
≡
N
:
π
-
1
O
i
s
o
p
e
n
i
n
X
.Then τ
X
(
≡
N
) is a topology on X
(
≡
N
).Proof.
It is straightforward.Theorem 27.
Let(
X
,
τ
) be a T
B-algebra and N a subalgebra of X. Then, with respect to the topological space X
(
≡
N
)
,
τ
X
(
≡
N
), the surjective map π is continuous and open.Proof.
The continuity ofπ follows directly from the definition of τ
X
(
≡
N
). It remains to show that π is an open map. From the definition of τ
X
(
≡
N
), it suffices to show that π
-
1
(
π
(
U
)
)
∈
τ whenever U
∈
τ. So let U be τ-open in X. By Proposition 25 and Definition 7
(
i
i
),(21)
π
-
1
π
U
=
U
∗
N
=
⋃
n
∈
N
U
∗
n
.By Corollary 19
(
i
v
), for each n
∈
N, U
∗
n is τ-open in X, and so π
-
1
(
π
(
U
)
) is a union of some τ-open sets in X. Therefore, π
-
1
(
π
(
U
)
)
∈
τ.Finally, we introduce the notion ofquotient
T
B
-algebras of B-algebras which is stipulated in the result below.Theorem 28.
Let(
X
,
τ
) be a T
B-algebra. If N is a normal subalgebra of X, then(22)
X
≡
N
,
τ
X
≡
N
=
X
/
N
,
τ
X
/
Nis a T
B-algebra.Proof.
Leta
∗
N
,
b
∗
N
∈
X
/
N and W
∈
τ
X
/
N such that (
a
∗
N
)
⊛
(
b
∗
N
)
∈
W. Now, π a B-epimorphism implies there exist x
,
y
∈
X such that π
(
x
)
=
a
∗
N and π
(
y
)
=
b
∗
N. Moreover, π
(
x
∗
y
)
=
π
(
x
)
⊙
π
(
y
)
=
(
a
∗
N
)
⊛
(
b
∗
N
)
∈
W, and so x
∗
y
∈
π
-
1
(
W
). By Theorem 26, π
-
1
(
W
)
∈
τ. Since (
X
,
τ
) is a T
B-algebra, there are U
,
V
∈
τ such that x
∈
U, y
∈
V, and U
∗
V
⊆
π
-
1
(
W
) by virtue of Theorem 17. Also, we have π
(
U
∗
V
)
=
π
(
U
)
⊛
π
(
V
) since π is a B-homomorphism. Note that a
∗
N
∈
π
(
U
) and b
∗
N
∈
π
(
V
). Now,(23)
π
U
⊛
π
V
=
π
U
∗
V
⊆
π
π
-
1
W
⊆
W
.Since π is an open map by Theorem 27, π
(
U
)
,
π
(
V
)
∈
τ
X
/
N. Hence, there are τ
X
/
N-open sets S
=
π
(
U
) and T
=
π
(
V
) in X
/
N such that a
∗
N
∈
S, b
∗
N
∈
T, and S
⊛
T
⊆
W. Therefore, by Theorem 17, (
X
/
N
,
τ
X
/
N
) is a T
B-algebra.


## 5. Conclusion

We have presented the notion ofT
B-algebras and provided some of its initial properties. We showed that, in a B-algebra, a neighborhood base of the constant element generates a unique B-topology, making a T
B-algebra provided that the given B-algebra is commutative. Lastly, we provided some properties of (normal) subalgebras of T
B-algebras and investigated the notion of quotient T
B-algebras. As a consequence of this study, it is best recommended to explore the properties of T
B-algebras with respect to some known topological invariants (e.g., separability and compactness).



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*Source: 8683965-2019-07-15.xml*