Properties of prime filters are studied in
BE
algebras as well as in commutative
BE
algebras. An equivalent condition is derived for a
BE
algebra to become a totally ordered set. A condition
L
is introduced in a commutative
BE
algebra in ordered to study some more properties of prime filters in commutative
BE
algebras. A set of equivalent conditions is derived for a commutative
BE
algebra to become a chain. Some topological properties of the space of all prime filters of
BE
algebras are studied.
AMS Mathematics Subject Classification : 06F35, 03G25, 08A30.
1. Introduction
The notion of BEalgebras was introduced and extensively studied by H.S. Kim and Y.H. Kim in
[5]
. These classes of
BE
algebras were introduced as a generalization of the class of
BCK
algebras of K. Iseki and S. Tanaka
[4]
. Some properties of filters of
BE
algebras were studied by S.S. Ahn and K.S. So in
[1]
. In
[6
,
7]
, the notion of normal filters is introduced in
BE
algebras. In
[2
,
3]
, S.S. Ahn and K.S. So introduced the notion of ideals in
BE
algebras and proved several characterizations of such ideals. Also they generalized the notion of upper sets in
BE
algebras, and discussed some properties of the characterizations of generalized upper sets related to the structure of ideals in transitive and selfdistributive
BE
algebras. Recently in 2012, S.S. Ahn, Y.H. Kim and J.M. Ko
[1]
introduced the notion of a terminal section of
BE
algebras and derived some characterizations of commutative
BE
algebras in terms of lattice ordered relations and terminal sections.
In this paper, the notion of prime filters is introduced in
BE
algebras. Some properties of prime filters and maximal filters are then studied. An equivalent condition is derived, in terms of prime filters, for the class of all filters of a
BE
algebra to become a totally ordered set. Properties of prime filters are also studied in commutative
BE
algebras. A condition
L
is introduced to study some properties of prime filters of
BE
algebras. Prime filters of a commutative
BE
algebra are characterized. A set of equivalent conditions is derived for a commutative
BE
algebra to become a chain. Some topological properties of the space of all prime filters of a
BE
algebra are studied. An equivalent condition is derived for every prime filter of a
BE
algebra to become a maximal filter.
2. Preliminaries
In this section, we present certain definitions and results which are taken mostly from the papers
[1]
,
[5]
and
[7]
for the ready reference of the reader.
Definition 2.1
(
[5]
). An algebra (
X
, ∗, 1) of type (2, 0) is called a
BE
algebra if it satisfies the following properties:
(1)
x
∗
x
= 1 (2)
x
∗ 1 = 1
3) 1 ∗
x
=
x
(4)
x
∗ (
y
∗
z
) =
y
∗ (
x
∗
z
) for all
x
,
y
,
z
∈
X
Theorem 2.2
(
[5]
). Let (
X
, ∗, 1) be a
BEalgebra. Then we have the following:

(1)x∗ (y∗x) = 1 (2)x∗ ((x∗y) ∗y)) = 1
We introduce a relation ≤ on a
BE
algebra
X
by
x
≤
y
if and only if
x
∗
y
= 1 for all
x
,
y
∈
X
. A
BE
algebra
X
is called selfdistributive if
x
∗ (
y
∗
z
) = (
x
∗
y
) ∗ (
x
∗
z
) for all
x
,
y
,
z
∈
X
. A
BE
algebra
X
is called commutative if (
x
∗
y
) ∗
y
= (
y
∗
x
) ∗
x
for all
x
,
y
∈
X
.
Definition 2.3
(
[1]
). A
BE
algebra (
X
, ∗, 1) is said to transitive if for all
x
,
y
,
z
∈
X
, it satisfies
y
∗
z
≤ (
x
∗
y
) ∗ (
x
∗
z
).
Definition 2.4
(
[1]
). Let (
X
, ∗, 1) be a
BE
algebra. A nonempty subset
F
of
X
is called a filter of
X
if, for all
x
,
y
∈
X
, it satisfies the following properties:
(a) 1 ∈
F
(b)
x
∈
F
and
x
∗
y
∈
F
imply that
y
∈
F
Definition 2.5
(
[5]
). Let (
X
_{1}
, ∗, 1) and (
X
_{2}
, ◦, 1′) be two
BE
algebras. Then a mapping
f
:
X
_{1}
→
X
_{2}
is called a homomorphism if
f
(
x
∗
y
) =
f
(
x
) ◦
f
(
y
) for all
x
,
y
∈
X
_{1}
.
It it clear that if
f
:
X
_{1}
→
X
_{2}
is a homomorphism, then
f
(1) = 1′. For any
x
,
y
∈
X
, A. Walendzaik
[8]
defined the operation ∨ as
x
∨
y
= (
y
∗
x
) ∗
x
. However, in a commutative
BE
algebra
X
, we can obtain for any
x
,
y
∈
X
, that
x
∨
y
= (
y
∗
x
) ∗
x
= (
x
∗
y
) ∗
y
=
y
∨
x
. For any nonempty subset A of a
BE
algebra
X
, ⟨
A
⟩ is the smallest filter containing
A
.
Theorem 2.6
(
[1]
).
If A is a nonempty subset of a selfdistributive BEalgebra X, then
Let
F
be a filter of a
BE
algebra
X
. Then ⟨
F
∪ {
a
}⟩ = {
x
∈
X

a^{n}
∗
x
∈
F
for some
n
∈ ℕ}. For
A
= {
a
}, we will denote ⟨{
a
}⟩, briefly by ⟨
a
⟩, we call it a principal filter of
X
. If
X
is selfdistributive, then ⟨
a
⟩ = {
x
∈
X

a
∗
x
= 1}.
3. Prime filters of BEalgebras
In this section, some properties of prime filters and maximal filters of
BE
algebra are studied. A necessary and sufficient condition is derived for a proper filter of a
BE
algebra to become a prime filter. Throughout this section,
X
stands for a
BE
algebra unless otherwise mentioned.
Definition 3.1.
A proper filter
P
of a
BE
algebra
X
is called prime if
F
∩
G
⊆
P
implies
F
⊆
P
or
G
⊆
P
for any two filters
F
and
G
of
X
.
Theorem 3.2.
A proper filter P of a BEalgebra is prime if and only if
⟨
x
⟩ ∩ ⟨
y
⟩ ⊆
P implies x
∈
P or y
∈
P for all x
,
y
∈
X
.
Proof
. Assume that
P
is a prime filter of
X
. Let
x
,
y
∈
X
be such that ⟨
x
⟩∩⟨
y
⟩ ⊆
P
. Since
P
is prime, it implies that
x
∈ ⟨
x
⟩ ⊆
P
or
y
∈ ⟨
y
⟩ ⊆
P
. Conversely, assume that the condition holds. Let
F
and
G
be two filters of
X
such that
F
∩
G
⊆
P
. Let
x
∈
F
and
y
∈
G
be the arbitrary elements. Then ⟨
x
⟩ ⊆
F
and ⟨
y
⟩ ⊆
G
. Hence ⟨
x
⟩ ∩ ⟨
y
⟩ ⊆
F
∩
G
⊆
P
. Then by the assumed condition, we get
x
∈
P
or
y
∈
P
. Thus
F
⊆
P
or
G
⊆
P
. Therefore
P
is prime. □
Theorem 3.3.
Let X be a BEalgebra and F a filter of X. Then for any a
,
b
∈
X
,
Proof
. Assume that ⟨
F
∪{
a
}⟩∩⟨
F
∪{
b
}=
F
for any
a
,
b
∈
X
. Since
a
∈ ⟨
F
∪{
a
}⟩ and
b
∈ ⟨
F
∪ {
b
}⟩, we get ⟨
a
⟩ ⊆ ⟨
F
∪ {
a
}⟩ and ⟨
b
⟩ ⊆ ⟨
F
∪ {
b
}⟩. Hence, it yields ⟨
a
⟩ ∩ ⟨
b
⟩ ⊆ ⟨
F
∪ {
a
}⟩∩⟨
F
∪ {
b
}⟩ =
F
. Therefore, it concludes that ⟨
a
⟩ ∩ ⟨
b
⟩ ⊆
F
.
Conversely, assume that ⟨
a
⟩ ∩ ⟨
b
⟩ ⊆
F
. Clearly
F
⊆ ⟨
F
∪ {
a
}⟩ ∩ ⟨
F
∪ {
b
}⟩. Let
x
∈ ⟨
F
∪ {
a
}⟩ ∩ ⟨
F
∪ {
b
}⟩. Since
F
is a filter, there exists
m
,
n
∈ ℕ such that
a^{m}
∗
x
∈
F
and
b^{n}
∗
x
∈
F
. Hence, there exists
m
_{1}
,
m
_{2}
∈
F
such that
a^{m}
∗
x
=
m
_{1}
and
b^{n}
∗
x
=
m
_{2}
. Hence
Hence
m
_{1}
∗
x
∈ ⟨
a
⟩. Similarly, we get
m
_{2}
∗
x
∈ ⟨
b
⟩. Since
we get that
m
_{1}
∗ (
m
_{2}
∗
x
) ∈ ⟨
a
⟩ and
m
_{1}
∗ (
m
_{2}
∗
x
) ∈ ⟨
b
⟩. Hence
Since
m
_{1}
,
m
_{2}
∈
F
and
F
is a filter, we get
x
∈
F
. Hence ⟨
F
∪{
a
}⟩∩⟨
F
∪{
b
}⊆
F
. Therefore, it concludes that ⟨
F
∪ {
a
}⟩ ∩ ⟨
F
∪ {
b
}=
F
. □
Definition 3.4.
A filter
F
of a
BE
algebra
X
is called proper if
F
≠
X
.
Definition 3.5.
A proper filter
M
of a
BE
algebra
X
is called a maximal filter if ⟨
M
∪ {
x
}⟩ =
X
for any
x
∈
X
−
M
.
Theorem 3.6.
Every maximal filter of a BEalgebra is a prime filter
.
Proof
. Let
M
be a maximal filter of a
BE
algebra
X
. Let ⟨
x
⟩ ∩ ⟨
y
⟩ ⊆
M
for some
x
,
y
∈
X
. Suppose
x
∉
M
and
y
∉
M
. Then ⟨
M
∪ {
x
}⟩ =
X
and ⟨
M
∪ {
y
}⟩ =
X
. Hence
Hence, by the Theorem 3.3, it yields that ⟨
x
⟩ ∩ ⟨
y
⟩ ⊈
M
, which is a contradiction. Hence
x
∈
M
or
y
∈
M
. Therefore
M
is a prime filter of
X
. □
Theorem 3.7.
Let X and Y be two BEalgebras and f
:
X
→
Y a homomorphism such that f
(
X
)
is a filter in Y. If F is a prime filter of Y and f
^{−1}
(
F
) ≠
X, then f
^{−1}
(
F
)
is a prime filter of X
.
Proof
. Since
f
(1) = 1 ∈
F
, we get 1 ∈
f
^{−1}
(
F
). Let
x
,
x
∗
y
∈
f
^{−1}
(
F
). Then
f
(
x
) ∈
F
and
f
(
x
) ∗
f
(
y
) =
f
(
x
∗
y
) ∈
F
. Since
F
is a filter in
Y
, it yields that
f
(
y
) ∈
F
. Hence
y
∈
f
^{−1}
(
F
). Therefore
f
^{−1}
(
F
) is a filter of
X
. Let
x
,
y
∈
X
be such that ⟨
x
⟩ ∩ ⟨
y
⟩ ⊆
f
^{−1}
(
F
). Let
u
∈ ⟨
f
(
x
)⟩ ∩ ⟨
f
(
y
)⟩. Then there exists
m
,
n
∈ ℕ such that
f
(
x
)
^{n}
∗
u
= 1 ∈
F
and
f
(
y
)
^{m}
∗
u
= 1 ∈
F
. Since
f
(
x
) ∈
f
(
X
) and
f
(
X
) is a filter, it implies that
u
∈
f
(
X
). Hence
u
=
f
(
a
) for some
a
∈
X
. Moreover,
f
(
x^{n}
∗
a
) =
f
(
y^{m}
∗
a
) = 1 ∈
F
because of
f
is a homomorphism. Hence
Hence
Since ⟨
x
⟩ ∩ ⟨
y
⟩ ⊆
f
^{−1}
(
F
), then by Theorem 3.3, we get
a
∈
f
^{−1}
(
F
). Hence
u
=
f
(
a
) ∈
F
. It concludes that ⟨
f
(
x
)⟩ ∩ ⟨
f
(
y
)⟩ ⊆
F
. Since
F
is a prime filter of
Y
, we get that ⟨
f
(
x
)⟩ ⊆
F
or ⟨
f
(
y
)⟩ ⊆
F
. Thus it yields that
f
(
x
) ∈
F
or
f
(
y
) ∈
F
. Therefore
x
∈
f
^{−1}
(
F
) or
y
∈
f
^{−1}
(
F
), which concludes that
f
^{−1}
(
F
) is a prime filter of
X
. □
Let us now denote that the class of all filters of a
BE
algebra
X
by
F
(
X
). Then in the following theorem, a necessary and sufficient condition is derived, in terms of primeness of filters, for the class
F
(
X
) to become a chain.
Theorem 3.8.
Let X be a BEalgebra. Then F
(
X
)
is a totally ordered set or a chain if and only if every proper filter of X is prime
.
Proof
. Assume that
F
(
X
) is a totally ordered set. Let
F
be a proper filter of
X
. Let
a
,
b
∈
X
be such that ⟨
a
⟩ ∩ ⟨
b
⟩ ⊆
F
. Since ⟨
a
⟩ and ⟨
b
⟩ are filters of
X
, we get that either ⟨
a
⟩ ⊆ ⟨
b
⟩ or ⟨
b
⟩ ⊆ ⟨
a
⟩. Hence, it concludes that
a
∈
F
or
b
∈
F
. Therefore
F
is prime.
Conversely assume that every proper filter of
X
is prime. Let
F
and
G
be two proper filters of
X
. Since
F
∩
G
is a proper filter of
X
, we get that
Hence
F
⊆
G
or
G
⊆
F
. Therefore
F
(
X
) is a totally ordered set. □
4. Prime filters of commutative BEalgebras
In this section, a condition
L
is introduced to study the properties of prime filters of commutative
BE
algebras. A set of equivalent conditions is derived for a commutative
BE
algebra to become a totally ordered set.
Proposition 4.1.
Let
(
X
, ∗, 1)
be a commutative BEalgebra and x
,
y
,
z
∈
X. Then the following conditions hold
.
(1)
x
∗ (
y
∨
z
) = (
z
∗
y
) ∗ (
x
∗
y
)
;
(2)
x
≤
y implies x
∨
y
=
y;
(3)
z
≤
x and x
∗
z
≤
y
∗
z imply y
≤
x
.
Proof
. (1). Let
x
,
y
,
z
∈
X
. Then
x
∗ (
y
∨
z
) =
x
∗ ((
z
∗
y
) ∗
y
) = (
z
∗
y
) ∗ (
x
∗
y
).
(2). Let
x
≤
y
. Then
x
∗
y
= 1. Hence
y
= 1 ∗
y
= (
x
∗
y
) ∗
y
= (
y
∗
x
) ∗
x
=
x
∨
y
.
(3). Let
z
≤
x
and
x
∗
z
≤
y
∗
z
. Then
z
∗
x
= 1 and (
x
∗
z
) ∗ (
y
∗
z
) = 1. Hence
Therefore, it concludes that
y
≤
x
. □
Definition 4.2.
A
BE
algebra
X
is said to satisfy the condition
L
if for all
x
,
y
∈
X
, there exists
u
∈
X
such that
u
≤
x
and
u
≤
y
.
Theorem 4.3.
Let X be a commutative BEalgebra. Then X satisfies the condition L if and only if for all x
,
y
∈
X, the greatest lower bound inf
{
x
,
y
} =
x
∧
y for brevity, is x
∧
y
= [(
x
∗
u
) ∨ (
y
∗
u
)] ∗
u where u
≤
x
,
y
.
Proof
. Assume that
X
satisfies the condition
L
. Let
u
≤
x
,
y
. Clearly
u
≤
x
∧
y
. Since
x
∗
u
≤ (
x
∗
u
) ∨ (
y
∗
u
), we get
Hence
x
∧
y
≤
x
. Similarly, we can obtain that
x
∧
y
≤
y
. Hence
x
∧
y
is a lower bound of
x
and
y
. Suppose
v
is another lower bound for
x
and
y
, i.e.
v
≤
x
,
y
. Hence
x
∗
u
≤
v
∗
u
and
y
∗
u
≤
v
∗
u
. Hence (
x
∗
u
) ∨ (
y
∗
u
) ≤
v
∗
u
. Therefore we get
Hence
x
∧
y
is the greatest lower bound of
x
and
y
. Converse is clear. □
In the following proposition, some properties of a commutative
BE
algebra with condition
L
are derived. Throughout this section,
X
stands for a commutative
BE
algebra which satisfies the condition
L
, unless otherwise mentioned.
Proposition 4.4.
Let
(
X
, ∗, 1)
be a commutative BEalgebra and x
,
y
,
z
∈
X
.
Then the following conditions hold
.
(1) (
x
∨
y
) ∗
z
= (
x
∗
z
) ∧ (
y
∗
z
)
(2)
x
∗ (
y
∧
z
) = (
x
∗
y
) ∧ (
x
∗
z
)
(3)
x
∗ (
x
∧
y
) =
x
∗
y
(4) (
x
∗
y
) ∨ (
y
∗
x
) = 1
(5) (
x
∧
y
) ∗
z
= (
x
∗
z
) ∨ (
y
∗
z
)
Proof
. (1). Since
x
,
y
≤
x
∨
y
, we get that (
x
∨
y
) ∗
z
≤
x
∗
z
and (
x
∨
y
) ∗
z
≤
y
∗
z
. Hence (
x
∨
y
) ∗
z
is a lower bound for
x
∗
z
and
y
∗
z
. Let
u
be a lower bound for
x
∗
z
and
y
∗
z
. Hence
u
≤
x
∗
z
and
u
≤
y
∗
z
and so
x
≤
u
∗
z
and
y
≤
u
∗
z
. Therefore
x
∨
y
≤
u
∗
z
and thus
u
≤ (
x
∨
y
) ∗
z
. Therefore (
x
∨
y
) ∗
z
is the greatest lower bound for
x
∗
z
and
y
∗
z
. Hence (
x
∨
y
) ∗
z
= (
x
∗
z
) ∧ (
y
∗
z
).
(2). Let
x
,
y
,
z
∈
X
. By the Theorem 4.3, we know that
y
∧
z
= ((
y
∗
u
)∨(
z
∗
u
))∗
u
where
u
≤
y
,
z
. Since
u
≤
y
, we get that (
y
∗
u
) ∗
u
= (
u
∗
y
) ∗
y
= 1 ∗
y
=
y
. Similarly, we get that (
z
∗
u
) ∗
u
=
z
. Hence we get that
(3). By replacing
y
by
x
and
z
by
y
in (2), we get
(4). Let
x
,
y
,
z
∈
X
. Then
(5). By using the dual argument, it can be followed by (1). □
Definition 4.5.
A filter
P
of a commutative
BE
algebra is called prime if
x
∨
y
∈
P
implies
x
∈
P
or
y
∈
P
for all
x
,
y
∈
F
.
Lemma 4.6.
Let X be a selfdistributive and commutative BEalgebra. Then for any a
,
b
∈
X, the following conditions hold:
(1)
a
≤
b implies
⟨
b
⟩ ⊆ ⟨
a
⟩
(2) ⟨
a
∨
b
⟩ = ⟨
a
⟩ ∩ ⟨
b
⟩.
Proof
. (1). Suppose
a
≤
b
. Let
x
∈ ⟨
b
⟩. Then
b
∗
x
= 1. Hence 1 =
b
∗
x
≤
a
∗
x
. Thus it yields that
x
∈ ⟨
a
⟩. Therefore ⟨
b
⟩ ⊆ ⟨
a
⟩.
(2). Since
a
,
b
≤
a
∨
b
, we get that ⟨
a
∨
b
⟩ ⊆ ⟨
a
⟩ and ⟨
a
∨
b
⟩ ⊆ ⟨
b
⟩. Hence ⟨
a
∨
b
⟩ ⊆ ⟨
a
⟩∩⟨
b
⟩. Conversely, let
x
∈ ⟨
a
⟩∩⟨
b
⟩. Then
a
∗
x
=
b
∗
x
= 1. Since
X
is commutative, by proposition 4.4(1), we get (
a
∨
b
)∗
x
= (
a
∗
x
)∧(
b
∗
x
) = 1∧1 = 1. Hence
x
∈ ⟨
a
∨
b
⟩. Thus ⟨
a
⟩ ∩ ⟨
b
⟩ ⊆ ⟨
a
∨
b
⟩. Therefore ⟨
a
∨
b
⟩ = ⟨
a
⟩ ∩ ⟨
b
⟩. □
In the following theorem, the class of all prime filters of a commutative
BE
algebra is characterized in terms of principal filters.
Theorem 4.7.
Let X be a selfdistributive and commutative BEalgebra and P a proper filter of X. Then the following conditions are equivalent
.
(1)
P is prime;
(2)
For any two filters F and G of X, F
∩
G
⊆
P implies F
⊆
P or G
⊆
P;
(3)
For any x
,
y
∈
X
, ⟨
x
⟩ ∩ ⟨
y
⟩ ⊆
P implies x
∈
P or y
∈
P
.
Proof
. The equivalency between (2) and (3) is proved in Theorem 3.2.
(1) ⇒ (2): Assume that
P
is a prime filter of
X
. Let
F
and
G
be two filters of
X
such that
F
∩
G
⊆
P
. Without loss of generality, assume that
F
⊈
P
. Then there exists
a
∈
X
such that
a
∈
F
and
a
∉
P
. Let
b
∈
G
be an arbitrary element. Clearly ⟨
a
⟩ ∩ ⟨
b
⟩ =
F
∩
G
⊆
P
. Hence ⟨
a
∨
b
⟩ ⊆
F
∩
G
⊆
P
. Thus
a
∨
b
∈
P
. Since
P
is prime and
a
∉
P
, we get that
b
∈
P
. Therefore
G
⊆
P
.
(2) ⇒ (1): Assume that the condition (2) holds. Let
x
,
y
∈
X
be such that
x
∨
y
∈
P
. Then we get that ⟨
x
⟩ ∩ ⟨
y
⟩ ⊆
P
. Hence, by condition (2), either ⟨
x
⟩ ⊆
P
or ⟨
y
⟩ ⊆
P
. Therefore
x
∈
P
or
y
∈
P
. □
The following theorem provides another characterization of prime filters in commutative
BE
algebras with condition
L
.
Theorem 4.8.
Let X be a commutative BEalgebra with condition L and F a filter of X. Then F is prime if and only if x
∗
y
∈
F or y
∗
x
∈
F for all x, y
∈
X
.
Proof
. Assume that
F
is a a prime filter in
X
. Since (
x
∗
y
)∨(
y
∗
x
) = 1 ∈
F
, we get either
x
∗
y
∈
F
or
y
∗
x
∈
F
. Conversely, assume that
x
∗
y
∈
F
or
y
∗
x
∈
F
for all
x
,
y
∈
X
. Let
x
∨
y
∈
F
. Suppose
x
∗
y
∈
F
. Then (
x
∗
y
) ∗
y
=
y
∨
x
∈
F
. Since
F
is a filter and
x
∗
y
∈
F
, we get that
y
∈
F
. Suppose
y
∗
x
∈
F
. Then (
y
∗
x
) ∗
x
=
x
∨
y
∈
F
. Since
F
is a filter and
y
∗
x
∈
F
, we get that
x
∈
F
. □
The following extension property of prime filters is a direct consequence of the above theorem.
Corollary 4.9.
Let X be a commutative BEalgebra with condition L and F a prime filter of X. If G is a filter of X such that F
⊆
G, then G is also prime
.
Theorem 4.10.
Let X be a commutative BEalgebra with condition L. Then the following conditions are equivalent
.
(1)
Every proper filter is a prime filter;
(2)
The filter
{1}
is a prime filter;
(3)
X is a totally ordered set with respect to BEordering
.
Proof
. (1) ⇒ (2): It is obvious.
(2) ⇒ (3): Assume that {1} is a prime filter. Let
x
,
y
∈
X
. Since {1} is prime, we get that either
x
∗
y
∈ {1} or
y
∗
x
∈ {1}. Hence
x
≤
y
or
y
≤
x
. Therefore
X
is totally ordered.
(3) ⇒ (1): Assume that
X
is a totally ordered set with respect to
BE
ordering ≤. Let
F
be a proper filter of
X
. Let
x
,
y
∈
X
. Hence
x
≤
y
or
y
≤
x
and thus
x
∗
y
= 1 ∈
F
or
y
∗
x
= 1 ∈
F
. Therefore
F
is prime. □
Theorem 4.11.
Let F be a filter of a commutative BEalgebra with condition L. For any x, y
∈
X, define a relation θ on X by
Then θ is a congruence on X
.
Proof
. Clearly
θ
is reflexive and symmetric. Let
x
,
y
,
z
∈
X
be such that (
x
,
y
) ∈
θ_{F}
and (
y
,
z
) ∈
θ
. Then
x
∗
y
∈
F
,
y
∗
x
∈
F
,
y
∗
z
∈
F
and
z
∗
y
∈
F
. Since
y
∗
z
∈
F
, we get
x
∗ (
y
∗
z
) ∈
F
. By a known property of filters, we get {[
x
∗ (
y
∗
z
)] ∗ [(
x
∗
y
) ∗ (
x
∗
z
)]} = 1 ∈
F
. Since
x
∗ (
y
∗
z
) ∈
F
and
x
∗
y
∈
F
, we get
x
∗
z
∈
F
. Similarly, we get
z
∗
x
∈
F
. Thus (
x
,
z
) ∈
θ
. Therefore
θ
is an equivalence relation on
X
. Let (
x
,
y
) ∈
θ
and (
u
,
v
) ∈
θ
. Then
x
∗
y
∈
F
,
y
∗
x
∈
F
,
u
∗
v
∈
F
and
v
∗
u
∈
F
. Since
x
∗
y
∈
F
, we get (
u
∗
x
)∗(
u
∗
y
) =
u
∗(
x
∗
y
) ∈
F
. Since
y
∗
x
∈
F
, we get (
u
∗
y
)∗(
u
∗
x
) =
u
∗(
y
∗
x
) ∈
F
. Hence (
u
∗
x
,
u
∗
y
) ∈
θ
. Again,
Hence
Since
u
∗
v
∈
F
, we get (
v
∗
y
) ∗ (
u
∗
y
) ∈
F
. Similarly (
u
∗
y
) ∗ (
v
∗
y
) ∈
F
. Hence (
u
∗
y
,
v
∗
y
) ∈
θ
. Thus (
u
∗
x
,
v
∗
y
) ∈
θ
. Hence
θ
is a congruence on
X
. □
For any commutative
BE
algebra
X
, let
C_{x}
be the congruence class generated by
x
∈
X
, i.e.
C_{x}
= {
y
∈
X

x
is congruent to
y
}. Define
X_{/F}
= {
C_{x}

x
∈
F
}.
Then clearly
X_{/F}
is a commutative
BE
algebra with respect to the operation ∗ defined on
X_{/F}
as follows:
It can also be observed that, for any
x
,
y
∈
X
,
C_{x}
≤
C_{y}
if and only if
C_{x}
∗
C_{y}
=
C
_{1}
is a
BE
ordering on
X_{/F}
.
Theorem 4.12.
Let X be a commutative BEalgebra with condition L and a proper filter of X. Then F is prime if and only if X_{/F} is a totally ordered set(chain)
.
Proof
. Assume that
F
is a prime filter in
X
. Then
x
∗
y
∈
F
or
y
∗
x
∈
F
for all
x
,
y
∈
X
. If
x
∗
y
∈
F
, then
C_{x}
∗
C_{y}
=
C
_{x∗y}
=
C
_{1}
. Hence
C_{x}
≤
C_{y}
. If
y
∗
x
∈
F
, then similar argument yields
C_{y}
≤
C_{x}
. Therefore
X/F
is a totally ordered set. Conversely, assume that
X_{/F}
is a totally ordered set. Let
x
,
y
∈
X
. then clearly
C_{x}
≤
C_{y}
or
C_{y}
≤
C_{x}
. Hence
C
_{x∗y}
=
C_{x}
∗
C_{y}
=
C
_{1}
or
C_{y∗x}
=
C_{y}
∗
C_{x}
=
C
_{1}
. Thus, it yields
x
∗
y
∈
F
or
y
∗
x
∈
F
. Therefore
F
is a prime filter in
X
. □
5. The space of prime filters ofBEalgebras
In this section, some topological properties of the space of all prime filters of
BE
algebras are studied. A necessary and sufficient condition is derived for a prime filter of a
BE
algebra to become maximal.
Theorem 5.1.
Let X ba a BEalgebra and a
∈
X. If F is a filter in X such that a
∉
F, then there exists a prime filter P such that a
∉
P and F
⊆
P
.
Proof
. Let
F
be a filter of
X
such that
a
∉
F
. Consider
ℑ
= {
G
∈
F
(
X
) 
a
∉
G
and
F
⊆
G
}. Clearly
F
∈
ℑ
. Then by the Zorn’s Lemma,
ℑ
has a maximal element, say
M
. Clearly
a
∉
M
. We now prove that
M
is prime. Let
x
,
y
∈
X
be such that ⟨
x
⟩ ∨ ⟨
y
⟩ ⊆
M
. Then by Theorem 3.3, we get
Since
a
∉
M
, we can obtain that
a
∉ ⟨
M
∪ {
x
}⟩ or
a
∉ ⟨
M
∪ {
y
}⟩. By the maximality of
M
, we get that ⟨
M
∪{
x
}⟩ =
M
or ⟨
M
∪{
y
}⟩ =
M
. Hence
x
∈
M
or
y
∈
M
. Therefore
M
is prime.
Corollary 5.2.
Let X be a commutative BEalgebra and
1 ≠
a
∈
X
.
Then there exists a prime filter P such that a
∉
P
.
Let
X
be a commutative
BE
algebra and
Spec_{F}
(
X
) denote the set of all prime filters of
X
. For any
A
⊆
X
, let
K
(
A
) = {
P
∈
Spec_{F}
(
X
) 
A
⊈
P
} and for any
x
∈
L
,
K
(
x
) =
K
({
x
}). Then we have the following observations:
Lemma 5.3.
Let X be a commutative BEalgebra with condition L. For any x
,
y
∈
L, the following holds:
(1)
K
(
x
) ∩
K
(
y
) =
K
(
x
∨
y
)
(2)
K
(
x
) ∪
K
(
y
) =
K
(
x
∧
y
)
(3)
K
(
x
) = ∅ ⇔
x
= 1
Proof
. (1). Let
P
∈
Spec_{F}
(
X
) be such that
P
∈
K
(
x
) ∩
K
(
y
). Then
x
∉
P
and
y
∉
P
. Since
P
is prime, we get
x
∨
y
∉
P
. Hence
P
∈
K
(
x
∨
y
). Therefore
K
(
x
) ∩
K
(
y
) ⊆
K
(
x
∨
y
). Conversely, assume that
P
∈
Spec_{F}
(
X
). Suppose
P
∈
K
(
x
∨
y
). Hence
x
∨
y
∉
P
. If
x
∈
P
, then
x
∨
y
∈
P
because of
x
≤
x
∨
y
. Thus it yields that
x
∉
P
. Therefore
P
∈
K
(
x
). Similarly, we get
P
∈
K
(
y
). Hence
P
∈
K
(
x
) ∩
K
(
y
). Therefore
K
(
x
∨
y
) ⊆
K
(
x
) ∩
K
(
y
).
(2). Let
P
∈
Spec_{F}
(
X
) be such that
P
∈
K
(
x
) ∪
K
(
y
). Then
P
∈
K
(
x
) or
P
∈
K
(
y
). Hence
x
∉
P
or
y
∉
P
. If
x
∧
y
∈
P
, then we get that both
x
and
y
must be in
P
. Hence
x
∧
y
∉
P
. Thus
P
∈
K
(
x
∧
y
). Therefore
K
(
x
) ∪
K
(
y
) ⊆
K
(
x
∧
y
). Conversely, let
P
∈
Spec_{F}
(
X
) be such that
P
∈
K
(
x
∧
y
). Then
x
∧
y
∉
P
. Since
x
∧
y
is the g.l.b of
x
and
y
, it concludes that
x
∉
P
and
y
∉
P
. Hence
P
∈
K
(
x
) ∪
K
(
y
). Therefore
K
(
x
∧
y
) ⊆
K
(
x
) ∪
K
(
y
).
(3). Since {1} ⊆
P
for all
P
∈
Spec_{F}
(
X
), it is obvious.
Proposition 5.4.
For any commutative BEalgebra X
,
=
Spec_{F}
(
X
).
Proof
. Let
P
∈
Spec_{F}
(
X
). Since
P
is a proper filter, there exists
a
∈
X
such that
a
∉
P
. Hence
P
∈
K
(
a
) ⊆
. Therefore
Spec_{F}
(
X
) ⊆
. Clearly
⊆
Spec_{F}
(
X
). Therefore
=
Spec_{F}
(
X
). □
Form the above proposition, it can be seen that {
K
(
x
) 
x
∈
X
} forms a covering of
Spec_{F}
(
X
). Hence {
K
(
x
) 
x
∈
X
} is an open base for a topology on
Spec_{F}
(
X
) which is called a hullkernel technology . In the following, we will discuss the properties of this topology.
Lemma 5.5.
Let X be a commutative BEalgebra. Then the following hold
.
(1)
For any x
∈
X
,
K
(⟨
x
⟩) =
K
(
x
)
;
(2)
For any two filters F, G of X, K
(
F
) ∩
K
(G) =
K
(
F
∩
G
).
Proof
. (1) Let
P
∈
Spec_{F}
(
X
) be such that
P
∈
K
(⟨
x
⟩). Then ⟨
x
⟩ ⊈
P
. Hence
x
∉
P
. Therefore
P
∈
K
(
x
). Thus
K
(⟨
x
⟩) ⊆
K
(
x
). Conversely, let
P
∈
K
(
x
). Then
x
∉
P
. Hence ⟨
x
⟩ ⊈
P
. Therefore
P
∈
K
(⟨
x
⟩). Hence
K
(
x
) ⊆
K
(⟨
x
⟩). Therefore
K
(⟨
x
⟩) =
K
(
x
).
(2). Let
P
∈
Spec_{F}
(
X
) be an arbitrary prime filter. Let
P
∈
K
(
F
) ∩
K
(G). Then
F
⊈
P
and
G
⊈
P
. Then there exists
x
∈
F
and
y
∈
G
such that
x
∉
P
and
y
∉
P
. Since
P
is prime, we get
x
∨
y
∉
P
. Since
F
and
G
are filters, we get that
x
∨
y
∈
F
∩
G
. Hence
F
∩
G
⊈
P
. Then
P
∈
K
(
F
∩
G
). Therefore
K
(
F
)∩
K
(G) ⊆
K
(
F
∩
G
). The opposite inclusion is obvious. Therefore
K
(
F
)∩
K
(G) =
K
(
F
∩
G
). □
Lemma 5.6.
Let F be a filter of a commutative BEalgebra X and x
∈
X. Then x
∈
F if and only if K
(
x
) ⊆
K
(
F
).
Proof
. Let
F
be a filter of a commutative
BE
algebra
X
and
x
∈
X
. Assume that
x
∈
F
. Let
P
∈
Spec_{F}
(
X
) be such that
P
∈
K
(
x
). Then we get that
x
∉
P
. Hence
F
⊈
P
. Therefore
P
∈
K
(
F
).
Conversely, assume that
K
(
x
) ⊆
K
(
F
). Suppose
x
∉
F
. Then by Theorem 5.1, there exists
P
∈
Spec_{F}
(
X
) such that
x
∉
P
and
F
⊆
P
. Hence, we get that
P
∈
K
(
x
) and
P
⊈
K
(
F
). Therefore
K
(
x
) ⊊
K
(
F
), which is a contradiction. Hence, it concludes that
x
∈
F
. □
Theorem 5.7.
Let X be a commutative BEalgebra. Then for any x
∈
L, K
(
x
)
is compact in Spec_{F}
(
X
).
Proof
. Let
x
∈
X
. Let
A
⊆
X
be such that
K
(
x
) ⊆
. Let
F
be the filter generated by
A
. Suppose
x
∉
F
. Then there exists a prime filter
P
of
X
such that
F
⊆
P
and
x
∉
F
. Hence
P
∈
K
(
x
) ⊆
. Therefore
y
∉
P
for some
y
∈
A
, which is a contradiction (because of
y
∈
A
⊆
F
⊆
P
). Hence
x
∈
F
. Then there exist
a
_{1}
,
a
_{2}
,…,
a_{n}
∈
A
such that
Let
P
∈
K
(
x
). Then
x
∉
P
. Suppose
a_{i}
∈
P
for all
i
= 1, 2,…,
n
. Since
a_{n}
∗ (…(
a
_{1}
∗
x
)…) = 1 ∈
P
and
P
is a filter, we get that
x
∈
P
, which is a contradiction. Hence
a_{i}
∉
P
for some
i
= 1, 2,…,
n
. Hence
P
∈
K
(
a_{i}
) for some
a_{i}
. Therefore
P
∈
. Hence
K
(
x
) ⊆
, which is a finite subcover of
K
(
x
). Hence
K
(
x
) is compact in
Spec_{F}
(
X
). Therefore for each
x
∈
X
,
K
(
x
) is a compact open subset of
Spec_{F}
(
X
).
Theorem 5.8.
Let X be a commutative BEalgebra with condition L and C a compact open subset of Spec_{F}
(
X
)
. Then C
=
K
(
x
)
for some x
∈
X
.
Proof
. Let
C
be a compact open subset of
Spec_{F}
(
X
). Since
C
is open, we get
C
=
for some
A
⊆
X
. Since
C
is compact, there exists
a
_{1}
,
a
_{2}
,…,
a_{n}
∈
A
such that
Therefore
C
=
K
(
x
) for some
x
∈
L
. □
Corollary 5.9.
For any commutative BEalgebra X with condition L, the set
{
K
(
x
) 
x
∈
X
}
is an open base for the prime space Spec_{F}
(
X
).
Theorem 5.10.
Let X be a commutative BEalgebra with condition L. Then Spec_{F}
(
X
)
is a T
_{0}

space
.
Proof
. Let
P
and
Q
be two distinct prime filters of
X
. Without loss of generality assume that
P
⊈
Q
. Choose
x
∈
L
such that
x
∈
P
and
x
∉
Q
. Hence
P
∉
K
(
x
) and
Q
∈
K
(
x
). Therefore
Spec_{F}
(
X
) is a
T
_{0}
space. □
The following corollary is a direct consequence of the above results.
Corollary 5.11.
The map x
↦
K
_{0}
(
x
)
is an antihomomorphism from X onto the lattice of all compact open subsets of Spec_{F}
(
X
).
For any
A
⊆
X
, denote
H
(
A
) = {
P
∈
Spec_{F}
(
X
) 
A
⊆
P
}. Then clearly
H
(
A
) =
Spec_{F}
(
X
) −
K
(
A
). Therefore
H
(
A
) is a closed set in
Spec_{F}
(
L
). Also every closed set in
Spec_{F}
(
L
) is of the form
H
(
A
) for some
A
⊆
X
. Then we have the following:
Theorem 5.12.
The closure of any Y
⊆
Spec_{F}
(
X
) is given by
.
Proof
. Let
Y
⊆
Spec_{F}
(
X
). Let
Q
∈
Y
. Then
⊆
Q
. Thus
Q
∈
H
(
). Therefore
H
(
) is a closed set containing
Y
. Let
C
be any closed set in
Spec_{F}
(
X
). Then
C
=
H
(
A
) for some
A
⊆
X
. Since
Y
⊆
C
=
H
(
A
), we get that
A
⊆
P
for all
P
∈
Y
. Hence
A
⊆
. Therefore
H
(
) ⊆
H
(
A
) =
C
. Hence
H
(
) is the smallest closed set containing
Y
. Therefore
Y
=
.
Theorem 5.13.
For any commutative BEalgebra X with condition L, Spec_{F}
(
X
)
is a T
_{1}

space if and only if every prime filter is maximal
.
Proof
. Assume that
Spec_{F}
(
X
) is a
T
_{1}
space. Let
P
be a prime filter of
X
. Suppose there exists a proper filter
Q
of
X
such that
P
⊆
Q
. Since
Spec_{F}
(
X
) is a
T
_{1}
space, there exists two basic open sets
K
(
x
) and
K
(
y
) such that
P
∈
K
(
x
) −
K
(
y
) and
Q
∈
K
(
y
) −
K
(
x
). Since
P
∉
K
(
y
), we get
y
∈
P
⊂
Q
, which is a contradiction to that
Q
∈
K
(
y
). Hence
P
is a maximal filter.
Conversely, assume that every prime filter is a maximal filter. Let
P
_{1}
and
P
_{2}
be two distinct elements of
Spec_{F}
(
X
). Hence by the assumption, both
P
_{1}
and
P
_{2}
are maximal filters in
X
. Hence
P
_{1}
⊈
P
_{2}
and
P
_{2}
⊈
P
_{1}
. Then there exists
a
,
b
∈
X
be such that
a
∈
P
_{1}
−
P
_{2}
and
b
∈
P
_{2}
−
P
_{1}
. Hence
P
_{1}
∈
K
(
b
) − K(
a
) and
P
_{2}
∈ K(
a
) −
K
(
b
). Therefore
Spec_{F}
(
X
) is a
T
_{1}
space. □
BIO
M. Sambasiva Rao received his M.Sc. and Ph.D. degrees from Andhra University, Andhra Pradesh, India. Since 2002 he has been at M.V.G.R. College of Engineering, Vizianagaram. His research interests include abstract algebra, implication algebras and Fuzzy Mathematics.
Department of Mathematics, MVGR College of Engineering, Chintalavalasa, Vizianagaram, Andhra Pradesh, India535005.
email: mssraomaths35@rediffmail.com
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