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PRIME FILTERS OF COMMUTATIVE BE-ALGEBRAS
PRIME FILTERS OF COMMUTATIVE BE-ALGEBRAS
Journal of Applied Mathematics & Informatics. 2015. Sep, 33(5_6): 579-591
Copyright © 2015, Korean Society of Computational and Applied Mathematics
  • Received : November 06, 2014
  • Accepted : January 07, 2015
  • Published : September 30, 2015
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M. SAMBASIVA RAO

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. AMS Mathematics Subject Classification : 06F35, 03G25, 08A30.
Keywords
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:
  • (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 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 ( 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 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
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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 ,
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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 m 1 , m 2 F such that am x = m 1 and bn x = m 2 . Hence
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Hence m 1 x ∈ ⟨ a ⟩. Similarly, we get m 2 x ∈ ⟨ b ⟩. Since
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we get that m 1 ∗ ( m 2 x ) ∈ ⟨ a ⟩ and m 1 ∗ ( m 2 x ) ∈ ⟨ b ⟩. Hence
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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 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
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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
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Hence
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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
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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
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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
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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
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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
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(3). By replacing y by x and z by y in (2), we get
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(4). Let x , y , z X . Then
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(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, 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 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 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 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 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 BE-algebra with condition L. For any x, y X, define a relation θ on X by
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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 . 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,
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Hence
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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:
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It can also be observed that, for any x , y X , Cx Cy if and only if Cx Cy = C 1 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 = C xy = C 1 . 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 C xy = Cx Cy = C 1 or Cy∗x = Cy Cx = 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 ofBE-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 = { G F ( X ) | a G and F G }. Clearly F . Then by the Zorn’s Lemma, 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
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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 , 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 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 ,
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= 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 ) ⊆
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. Therefore SpecF ( X ) ⊆
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. Clearly
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SpecF ( X ). Therefore
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= 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 ) ⊆
PPT Slide
Lager Image
. 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 ) ⊆
PPT Slide
Lager Image
. 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 ,…, an A such that
PPT Slide
Lager Image
Let P K ( x ). Then x P . Suppose ai P for all i = 1, 2,…, n . Since an ∗ (…( a 1 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
PPT Slide
Lager Image
. Hence K ( x ) ⊆
PPT Slide
Lager Image
, 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 =
PPT Slide
Lager Image
for some A X . Since C is compact, there exists a 1 , a 2 ,…, an A such that
PPT Slide
Lager Image
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 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 SpecF ( 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 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
PPT Slide
Lager Image
.
Proof . Let Y SpecF ( X ). Let Q Y . Then
PPT Slide
Lager Image
Q . Thus Q H (
PPT Slide
Lager Image
). Therefore H (
PPT Slide
Lager Image
) 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
PPT Slide
Lager Image
. Therefore H (
PPT Slide
Lager Image
) ⊆ H ( A ) = C . Hence H (
PPT Slide
Lager Image
) is the smallest closed set containing Y . Therefore Y =
PPT Slide
Lager Image
.
Theorem 5.13. For any commutative BE-algebra X with condition L, SpecF ( X ) is a T 1 - space if and only if every prime filter is maximal .
Proof . Assume that SpecF ( 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 SpecF ( 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 SpecF ( 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 SpecF ( 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, India-535005.
e-mail: mssraomaths35@rediffmail.com
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