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International Journal of Mathematics and Mathematical Sciences
Volume 2011 (2011), Article ID 795934, 13 pages
http://dx.doi.org/10.1155/2011/795934
Research Article

Fuzzy Filter Spectrum of a BCK Algebra

Department of Mathematics, Northwest University, Xi'an 710127, China

Received 14 October 2010; Accepted 22 December 2010

Academic Editor: Young Bae Jun

Copyright © 2011 Xiao Long Xin et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract

The notion of fuzzy s-prime filters of a bounded BCK-algebra is introduced. We discuss the relation between fuzzy s-prime filters and fuzzy prime filters. By the fuzzy s-prime filters of a bounded commutative BCK-algebra , we establish a fuzzy topological structure on . We prove that the set of all fuzzy s-prime filters of a bounded commutative BCK-algebra forms a topological space. Moreover, we show that the set of all fuzzy s-prime filters of a bounded implicative BCK-algebra is a Hausdorff space.

1. Introduction

BCK-algebras are an important class of logical algebras introduced by Iséki in 1966 (see [13]). Since then, a great deal of the literature has been produced on the theory of BCK-algebras. In particular, emphasis seems to have been put on the ideal and filter theory of BCK-algebras (see [4]). The concept of fuzzy sets was introduced by Zadeh [5]. At present, these ideas have been applied to other algebraic structures such as semigroups, groups, rings, ideals, modules, vector spaces, and so on (see [6, 7]). In 1991, Ougen [8] applied the concept of fuzzy sets to BCK-algebras. For the general development of BCK-algebras the fuzzy ideal theory and fuzzy filter theory play important roles (see [912]). Meng [13] introduced the notion of BCK-filters and investigated some results. Jun et al. [9, 10] studied the fuzzification of BCK-filters. Meng [13] showed how to generate the BCK-filter by a subset of Alò, and Deeba [14] attempted to study the topological aspects of the BCK-structures. They initiated the study of various topologies on BCK-algebras analogous to which has already been studied on lattices. In [15], Jun et al. introduced the notion of topological BCI-algebras and found some elementary properties.

In this paper, the topological structure and fuzzy structure on BCK-algebras are investigated together. We introduce the concept of fuzzy s-prime filters and discuss some related properties. By the fuzzy s-prime filters, we establish a fuzzy topological structure on bounded commutative BCK-algebras and bounded implicative BCK-algebras, respectively.

2. Preliminaries

A nonempty set with a constant 0 and a binary operation denoted by juxtaposition is called a BCK-algebra if for all the following conditions hold:(1), (2), (3), (4), (5) and imply .

A BCK-algebra can be (partially) ordered by if and only if . This ordering is called BCK-ordering. The following statements are true in any BCK-algebra: for all ,(6).(7).(8).(9).(10) implies and .

A BCK-algebra satisfying the identity is said to be commutative. If there is a special element 1 of a BCK-algebra satisfying for all , then 1 is called unit of . A BCK-algebra with unit is said to be bounded. In a bounded BCK-algebra , we denote by for every .

In a bounded BCK-algebra, we have(11) and .(12) implies .(13).

Now, we review some fuzzy logic concepts. A fuzzy set in is a function . We use the notation for and , called a level subset of μ, for where .

In this paper, unless otherwise specified, denotes a bounded BCK-algebra. A nonempty subset of is called a BCK-filter of if(F1),(F2) and imply for all .

Note that the intersection of a family of BCK-filters is a BCK-filter. For convenience, we call a BCK-filter of as a filter of , and write .

Let μ be a fuzzy set in . Then, μ is called a fuzzy filter of if(FF1),(FF2), for all . In this case, we write .

Note that in a bounded commutative BCK-algebra, the identity holds, then (F2) and(F3) and imply for all in coincide, and (FF2) and(FF3) coincide.

A proper filter of is said to be prime, denoted by , if, for any , implies or .

A nonconstant fuzzy filter μ of is said to be prime, denoted by , if for all .

For any fuzzy sets μ and in , we denote

Lemma 2.1. Let be a family of fuzzy filters of . Then, is a fuzzy filter of .

Proof. Let . For any , since . Then, and so, . (FF1) holds.
Moreover, for any , there exists such that
Since is arbitrary, we get . So, (FF2) holds.
Therefore, is a fuzzy filter of .

Lemma 2.2 (see, [16]). Let μ be a fuzzy filter of . For any , if , then .

Definition 2.3. Let μ be a fuzzy subset of . Then the fuzzy filter generated by μ, which is denoted by , is defined as Obviously, we get , and if , then .

Lemma 2.4. If , then .

Proof. Let , and be fuzzy filters. Then, by Lemma 2.2, and . Hence, .
Therefore, , or equivalently .
Conversely, . So .
Thus, .

Corollary 2.5. If , .

Lemma 2.6. If, .

Proof. Let . If , then from Lemma 2.2 we know . Thus, . So, .

3. Fuzzy Filter Spectrum

Definition 3.1. A nonconstant fuzzy filter μ of is said to be s-prime if for all , implies or . In this case, we write .

In this paper, we give some notations in the following.(i).(ii), where is a fuzzy subset of .(iii), where is called the complement of in .

Lemma 3.2. If is a fuzzy subset of , then . So .

Proof. Let , then and so . Hence, . Conversely, let , then . Note that , we get . Therefore, .

Theorem 3.3. Let . Then the pair is a topological space.

Proof. Consider and . Then , , and . Thus, .
Then, we prove that is closed under finite intersection.
Let and be two fuzzy filters of . We claim that . Let . Then, . Since , we have or . It follows that .
Conversely, let , then or . By Lemma 2.6, and . Thus, and so . It follows that .
Combining the above arguments we get , or equivalently, . By Corollary 2.5, and so .
Finally, let be a family of fuzzy prime filters of . We will prove that .
Let , then for any , and so . Hence, and thus .
Conversely, let , then . Thus, for any , . Hence, for all and so .
This shows that .
By Lemma 3.2, we get and so .
Furthermore, we get .
It follows that is a topological space.

Theorem 3.4. The collection of is a base of where is defined by

Proof. By Lemma 3.2, for any , , and so .
Now, we prove that ß is a base of . It is sufficient to show that for all , and , there exists such that and .
Let and . Then, and so there exists such that . Let and then . Moreover, for any , and so . Thus . This means . It follows that .
Therefore, ß is a base of .

The topological space is called fuzzy filter spectrum of , denoted by -, or for convenience.

Theorem 3.5. - is a space.

Proof. Let and . Then, or .
If , then, but . Moreover, but .
If , similarly we can get but . It follows that - is a space.

Lemma 3.6 (see [9]). Let μ be a fuzzy subset of . Then, μ is a fuzzy filter of if and only if is a filter of for each wherever .

Lemma 3.7. A non-constant fuzzy subset μ of is a fuzzy prime filter if and only if is a prime filter of for each whenever .

Proof. Let μ be a fuzzy prime filter and such that . Then by Lemma 3.6, is a filter of .
Suppose . It follows that . Since μ is prime, we have and thus or . It follows that or . Therefore is a prime filter.
Conversely, suppose that for each , is a prime filter whenever . If μ is not a fuzzy prime filter, then there exist such that . Take satisfying . Then . Since is a prime filter of , then implies or . But on the other hand, and imply and , a contradiction. It follows that μ is indeed a fuzzy prime filter.

Lemma 3.8 (see [13]). Let be a bounded commutative BCK-algebra and be a BCK-filter of . Then, is prime if and only if, for any filters , implies or .

Theorem 3.9. Let be a bounded commutative BCK-algebra and μ be a fuzzy s-prime filter. Then for each , is a prime filter of whenever and .

Proof. Let μ be a fuzzy s-prime filter and , . Then by Lemma 3.6, is a filter.
Let , be two filters such that . Define the fuzzy subset and . It is easy to see that and are fuzzy filters of . Note that
Since , then for any , and so for all . Thus . It follows from μ being a fuzzy s-prime filter that or . Without loss of generality let . Then, for any , and so . This means that . But implies and thus . Therefore, is a prime filter by Lemma 3.8.

Theorem 3.10. Let be a bounded commutative BCK-algebra. If μ is a fuzzy s-prime filter, then it is a fuzzy prime filter.

Proof. The proof follows from Lemma 3.7 and Theorem 3.9.

In general, the converse of Theorem 3.10 is not true. Let us see the following example.

Example 3.11. Let . Define the operation on as follows: , , and . It is easy to see that is a bounded commutative BCK-algebra. Define a fuzzy subset μ of by , . Clearly μ is a fuzzy prime filter of .
Moreover, we define the fuzzy filters and by for all and , . Then, we get but and . Therefore, μ is not a s-prime fuzzy filter.

Lemma 3.12. is a prime filter of if and only if is a fuzzy s-prime filter, where , and is defined by

Proof. Let be a prime filter. Then, by Lemma 3.6, we can easily see that is a fuzzy filter.
Let be two fuzzy prime filters such that , we will prove or . If it is not true, then there exist such that and . Since is prime, then . Note that , then
Thus, , a contradiction. It follows that or , and so is a fuzzy s-prime filter.
Conversely, let be a fuzzy s-prime filter. By Theorem 3.10, is also a fuzzy prime filter. Then, by Lemma 3.7, is a prime filter, where .

Corollary 3.13. is a prime filter of if and only if is a fuzzy s-prime filter.

Lemma 3.14 (see [13]). Let be a bounded implicative BCK-algebra, then and .

Lemma 3.15. Let μ be a fuzzy filter of a bounded commutative BCK-algebra . Then, for all .

Proof. Since μ is a fuzzy filter, we have for all . On the other hand, , since any fuzzy filter is order preserving. Thus, .

Lemma 3.16. If μ is a fuzzy filter of a bounded BCK-algebra , then is a filter of and is a fuzzy filter of .

Proof. Let μ be a fuzzy filter and take . Then, and so is a filter of by Lemma 3.6. Clearly is a fuzzy filter.

Lemma 3.17. Let be a bounded commutative BCK-algebra and μ be a fuzzy s-prime filter of . Then .

Proof. Suppose that . Since μ is non-constant, there exists such that . Define fuzzy subset and of by and for all . By Lemma 3.16, is a fuzzy filter and clearly is a fuzzy filter. Note that and , we get . But note that for any
Thus, for any , a contradiction. Therefore, .

Lemma 3.18. Let be a bounded implicative BCK-algebra and μ be a fuzzy s-prime filter of . Then for any , or .

Proof. By Lemma 3.14, , for all . Since μ is a fuzzy s-prime filter, we get that is a prime filter of by Theorem 3.9. Hence, implies or . Therefore, or by Lemma 3.17.

Theorem 3.19. Let be a bounded implicative BCK-algebra and μ be a fuzzy s-prime filter of . Then, for , or .

Proof. By Lemma 3.14, and then or since μ is a fuzzy s-prime filter. By Lemma 3.18, we get or . If , then by Lemma 3.15. If , then .

Lemma 3.20. Let be a bounded BCK-algebra. Then, a filter of is proper if and only if .

Proof. If , then clearly is proper.
Conversely, let be proper. If , then for any , and so . It follows that , a contradiction. Therefore, .

Lemma 3.21 (see [13]). Let be a bounded commutative BCK-algebra and be a filter of . If , then there is a prime filter of such that and .

Lemma 3.22 (see [13]). Let be a bounded implicative BCK-algebra. Then, for any , the filter , generated by , is a set of elements in satisfying .

Lemma 3.23. Let be a bounded implicative BCK-algebra and . Then, .

Proof. By Lemma 3.22, and thus .

Lemma 3.24 (see [16]). For a bounded commutative BCK-algebra , one gets(1) for all .(2), for all .(3) for all .

Theorem 3.25. Let be a bounded implicative BCK-algebra. Then,(i)if , and , then .(ii)if and , then .
Moreover, is both open or closed, where .(iii)if , where and , then .

Proof. (i) If , then and . By Theorem 3.10, μ is a fuzzy prime filter, and then . Since and , then and . It follows from Theorem 3.19 that and . Thus, . Therefore, .
Conversely, if , then and so , . Note that and , we get , and , since μ is order preserving. Thus, and , or equivalently, .
Therefore, (i) holds.
(ii) Let . Then, or . By Lemma 3.17, we have , . Therefore, or . On the other hand, by Lemma 3.14, , . Note that is a prime filter of by Theorem 3.9. If , then implies . If , then implies . Therefore, we get that or . Note that , we get and . Thus, and , and so . But or implies that or . This means, .
If , then . It follows that for all , a contradiction. Thus, . By Lemma 3.18, . Hence, and so . It follows that .
Conversely, let . Then, . Thus, . Since , then . By Lemma 3.24, and so or . If (or ), then (or ). Thus, or . It follows that .
Combining the above arguments, we get .
In order to prove is closed, we will show .
Let . Then, by Lemma 3.2. Thus, and so . Hence, . Note that we get , which implies that and so . It follows that and thus .
Conversely, let . Then, and so . By Theorem 3.19, . Note that , we get that , or equivalently, . Thus, and so .
Combining the above two sides, we get .
(iii) Let . We claim that . If this is not true, by Lemma 3.23,. By Lemma 3.21, there exists a prime filter of such that . On the other hand, by Lemma 3.12, .
Therefore, , a contradiction. It follows that .

In general, the converse of Theorem 3.25 (iii) does not hold. Let us see the following counter example.

Example 3.26. Let and -table and -table be given as follows. Then is a bounded implicative BCK-algebra and 3 is a unit. It is easy to see that is a filter of . From -table, we can see easily that is prime. So, is a fuzzy s-prime filter by Lemma 3.12. Let . Then, and so . Hence, . Therefore .

Theorem 3.27. Let be a bounded implicative BCK-algebra and for . Then, is a Hausdorff space.

Proof. Let and . We claim that . Otherwise, if , then for , and for , by Theorem 3.19, a contradiction. Thus, or . Let . Then, implies . Moreover, since . Thus and so . Therefore, . Hence,
Let . Then, , and so . Note that , we get . Moreover, we get
It follows that is a Hausdorff space.

Corollary 3.28. Let be a bounded implicative BCK-algebra. Then is a Hausdorff space.

Let be a bounded commutative BCK-algebra, be the set of all filters of , and - stand for all prime filters of .

For any subset of , we define -.

If , we denote by .

Lemma 3.29. and if , then .

Proof. Since , then implies . Then, implies .
Conversely, if , then . Hence, since implies . Therefore, .
Thus, . Similarly, we can prove that implies .

Proposition 3.30. The family forms a topology on -.

Proof. First, we get
Then, for any family ,
Finally,
Therefore, is a topology on -.

Theorem 3.31. Let be a bounded implicative BCK-algebra and the map - is defined by where is defined in Lemma 3.12. Then, is a homeomorphism.

Proof. (a) is well defined.
By Lemma 3.12, is a fuzzy s-prime filter for any -. Note that , then and so . Thus, is well defined.(b)Clearly is injective.(c) is surjective.
For any , by Theorem 3.19, or . Hence, . By Theorem 3.9, we get is a prime filter of . Thus, - and so . It follows that is surjective.(d) is continuous.
Let be an open set of . We will prove that is an open set of -. It is sufficient to prove that , since is a filter of by Lemma 3.6.
First, let , then there exists some such that and . Thus, and there exists . Hence . Therefore, and so . This shows that . It follows that .
Conversely, let , then . Hence, , and thus there exists such that . Therefore, and so . We can take such that . Then, . It follows that and so .
Combining the above two hands, we get . So is continuous.(e) is continuous. It is sufficient to prove that is an open set of for any .
We will prove that .
Thus, .
Conversely, we get
So that .
Therefore, . By Lemma 3.6, we can easily see that is a fuzzy filter of and so is an open set of . It follows that is continuous.

By Theorem 3.27 and Theorem 3.31, we get the following corollary.

Corollary 3.32. Let be a bounded implicative BCK-algebra. Then, - is a Hausdorff space.

Acknowledgment

The paper is supported by National Nature Science Foundation of Shaanxi Province (2007A19).

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