PRIME FILTERS OF COMMUTATIVE BE-ALGEBRAS

Abstract

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.

1. Introduction

The notion of BE-algebras 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 self-distributive 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 BE-algebra. Then we have the following:

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 self-distributive 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 non-empty 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 (X1, ∗, 1) and (X2, ◦, 1′) be two BE-algebras. Then a mapping f : X1 → X2 is called a homomorphism if f(x ∗ y) = f(x) ◦ f(y) for all x, y ∈ X1.

It it clear that if f : X1 → X2 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 non-empty subset A of a BE-algebra X, ⟨A⟩ is the smallest filter containing A.

Theorem 2.6 ([1]). If A is a non-empty subset of a self-distributive BE-algebra X, then

Let F be a filter of a BE-algebra X. Then ⟨F ∪ {a}⟩ = {x ∈ X | an ∗ 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 self-distributive, then ⟨a⟩ = {x ∈ X | a ∗ x = 1}.

 

3. Prime filters of BE-algebras

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 BE-algebra 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 BE-algebra 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 am ∗ x ∈ F and bn ∗ x ∈ F. Hence, there exists m1, m2 ∈ F such that am ∗ x = m1 and bn ∗ x = m2. Hence

Hence m1 ∗ x ∈ ⟨a⟩. Similarly, we get m2 ∗ x ∈ ⟨b⟩. Since

we get that m1 ∗ (m2 ∗ x) ∈ ⟨a⟩ and m1 ∗ (m2 ∗ x) ∈ ⟨b⟩. Hence

Since m1, m2 ∈ 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 BE-algebra 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 BE-algebras and f : X → Y a homo-morphism 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(xn ∗ a) = f(ym ∗ 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 BE-algebra. 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 BE-algebras

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 BE-algebra 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 BE-algebra. 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 BE-algebra 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 self-distributive and commutative BE-algebra. 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, letx ∈ ⟨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 self-distributive and commutative BE-algebra 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 BE-algebra 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 eitherx ∗ 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 BE-algebra 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 BE-algebra 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 BE-ordering.

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 eitherx ∗ 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 BE-algebra 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) ∈ θF 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. Sincex ∗ (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. Sincex ∗ 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 Cx be the congruence class generated by x ∈ X, i.e. Cx = {y ∈ X |x is congruent to y}. Define X/F = {Cx |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, Cx ≤ Cy if and only if Cx∗Cy = C1 is a BE-ordering on X/F .

Theorem 4.12. Let X be a commutative BE-algebra 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 Cx ∗ Cy = Cx∗y = C1. Hence Cx ≤ Cy. If y ∗ x ∈ F, then similar argument yields Cy ≤ Cx. Therefore X/F is a totally ordered set. Conversely, assume that X/F is a totally ordered set. Let x, y ∈ X. then clearly Cx ≤ Cy or Cy ≤ Cx. Hence Cx∗y = Cx ∗ Cy = C1 or Cy∗x = Cy ∗ Cx = C1. Thus, it yields x ∗ y ∈ F or y ∗ x ∈ F. Therefore F is a prime filter in X. □

 

5. The space of prime filters of BE-algebras

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 BE-algebra 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 BE-algebra and 1 ≠ a ∈ X. Then there exists a prime filter P such that a ∉ P.

Let X be a commutative BE-algebra and SpecF(X) denote the set of all prime filters of X. For any A ⊆ X, let K(A) = {P ∈ SpecF(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 BE-algebra 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 ∈ SpecF(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 ∈ SpecF(X). Suppose P ∈ K(x ∨ y). Hence x ∨ y ∉ P. If x ∈ P, thenx ∨ 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 ∈ SpecF(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 ∈ SpecF(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 ∈ SpecF(X), it is obvious.

Proposition 5.4. For any commutative BE-algebra X, = SpecF(X).

Proof. Let P ∈ SpecF(X). Since P is a proper filter, there exists a ∈ X such that a ∉ P. Hence P ∈ K(a) ⊆ . Therefore SpecF(X) ⊆ . Clearly ⊆ SpecF(X). Therefore = SpecF(X). □

Form the above proposition, it can be seen that {K(x) |x ∈ X} forms a covering of SpecF(X). Hence {K(x) |x ∈ X} is an open base for a topology on SpecF(X) which is called a hull-kernel technology . In the following, we will discuss the properties of this topology.

Lemma 5.5. Let X be a commutative BE-algebra. 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 ∈ SpecF(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 ∈ SpecF(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 BE-algebra 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 ∈ SpecF(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 ∈ SpecF(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 BE-algebra. Then for any x ∈ L, K(x) is compact in SpecF(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 a1, a2,…, an ∈ A such that

Let P ∈ K(x). Then x ∉ P. Suppose ai ∈ P for all i = 1, 2,…, n. Since an ∗ (…(a1 ∗ x)…) = 1 ∈ P and P is a filter, we get that x ∈ P, which is a contradiction. Hence ai ∉ P for some i = 1, 2,…, n. Hence P ∈ K(ai) for some ai. Therefore P ∈ . Hence K(x) ⊆ , which is a finite subcover of K(x). Hence K(x) is compact in SpecF(X). Therefore for each x ∈ X, K(x) is a compact open subset of SpecF(X).

Theorem 5.8. Let X be a commutative BE-algebra with condition L and C a compact open subset of SpecF(X). Then C = K(x) for some x ∈ X.

Proof. Let C be a compact open subset of SpecF(X). Since C is open, we get C = for some A ⊆ X. Since C is compact, there exists a1, a2,…, an ∈ A such that

Therefore C = K(x) for some x ∈ L. □

Corollary 5.9. For any commutative BE-algebra X with condition L, the set {K(x) |x ∈ X} is an open base for the prime space SpecF(X).

Theorem 5.10. Let X be a commutative BE-algebra with condition L. Then SpecF(X) is a T0-space.

Proof. Let P and Q be two distinct prime filters of X. Without loss of generality assume that P ⊈ Q. Choosex ∈ L such that x ∈ P and x ∉ Q. Hence P ∉ K(x) and Q ∈ K(x). Therefore SpecF(X) is a T0-space. □

The following corollary is a direct consequence of the above results.

Corollary 5.11. The map x ↦ K0(x) is an anti-homomorphism from X onto the lattice of all compact open subsets of SpecF(X).

For any A ⊆ X, denote H(A) = {P ∈ SpecF(X) | A ⊆ P}. Then clearly H(A) = SpecF(X) − K(A). Therefore H(A) is a closed set in SpecF(L). Also every closed set in SpecF(L) is of the form H(A) for some A ⊆ X. Then we have the following:

Theorem 5.12. The closure of any Y ⊆ SpecF(X) is given by .

Proof. Let Y ⊆ SpecF(X). Let Q ∈ Y . Then ⊆ Q. Thus Q ∈ H( ). Therefore H() is a closed set containing Y . Let C be any closed set in SpecF(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 BE-algebra X with condition L, SpecF(X) is a T1-space if and only if every prime filter is maximal.

Proof. Assume that SpecF(X) is a T1-space. Let P be a prime filter of X. Suppose there exists a proper filter Q of X such that P ⊆ Q. Since SpecF(X) is a T1-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 P1 and P2 be two distinct elements of SpecF(X). Hence by the assumption, both P1 and P2 are maximal filters in X. Hence P1 ⊈ P2 and P2 ⊈ P1. Then there exists a, b ∈ X be such that a ∈ P1 − P2 and b ∈ P2 − P1. Hence P1 ∈ K(b) − K(a) and P2 ∈ K(a) − K(b). Therefore SpecF(X) is a T1-space. □