# Fuzzy Filter Spectrum of a BCK Algebra

Authors: Xiao Long Xin; Wei Ji; Xiu Juan Hua
Journal: International Journal of Mathematics and Mathematical Sciences
 (2011)

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## 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-algebraX, we establish a fuzzy topological structure on X. 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.


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## Body

## 1. Introduction

BCK-algebras are an important class of logical algebras introduced by Iséki in 1966 (see [1–3]). 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 [9–12]). 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 setX with a constant 0 and a binary operation denoted by juxtaposition is called a BCK-algebra if for all x,y,z∈X the following conditions hold:(1)
((xy)(xz))(zy)=0,(2)
(x(xy))y=0,(3)
xx=0,(4)
0x=0,(5)
xy=0 and yx=0 imply x=y.A BCK-algebra can be (partially) ordered byx≤y if and only if xy=0. This ordering is called BCK-ordering. The following statements are true in any BCK-algebra: for all x,y,z,(6)
x0=x.(7)
(xy)z=(xz)y.(8)
xy≤x.(9)
(xy)z≤(xz)(yz).(10)
x≤y implies xz≤yz and zy≤zx.A BCK-algebraX satisfying the identity x(xy)=y(yx) is said to be commutative. If there is a special element 1 of a BCK-algebra X satisfying x≤1 for all x∈X, then 1 is called unit of X. A BCK-algebra with unit is said to be bounded. In a bounded BCK-algebra X, we denote 1x by x* for every x∈X.In a bounded BCK-algebra, we have(11)
1*=0 and 0*=1.(12)
y≤x implies x*≤y*.(13)
x*y*≤yx.Now, we review some fuzzy logic concepts. A fuzzy set inX is a function μ:X→[0,1]. We use the notation Xμ for {x∈X∣μ(x)=μ(1)} and μt, called a level subset of μ, for {x∈X∣μ(x)≥t} where t∈[0,1].In this paper, unless otherwise specified,X denotes a bounded BCK-algebra. A nonempty subset F of X is called a BCK-filter of X if(F1)
1∈F,(F2)
(x*y*)*∈F and y∈F imply x∈F for all x,y∈X.Note that the intersection of a family of BCK-filters is a BCK-filter. For convenience, we call a BCK-filter ofX as a filter of X, and write  F<FX.Letμ be a fuzzy set in X. Then, μ is called a fuzzy filter of X if(FF1)
μ(1)≥μ(x),(FF2)
μ(x)≥min{μ(x*y*)*,μ(y)}, for all x,y∈X. In this case, we write μ<FFX.Note that in a bounded commutative BCK-algebra, the identityx*y*=yx holds, then (F2) and(F3)
(yx)*∈F and y∈F imply x∈F for all x,y in X coincide, and (FF2) and(FF3)
μ(x)≥min{μ(yx)*,μ(y)} coincide.A proper filterF of X is said to be prime, denoted by F<PFX, if, for any x,y∈X, x∨y∈F implies x∈F or y∈F.A nonconstant fuzzy filterμ of X is said to be prime, denoted by μ<FPFX, if μ(x∨y)≤max{μ(x),μ(y)} for all x,y∈X.For any fuzzy setsμ and ν in X, we denote
(2.1)μ⊆ν⟺μ(x)≤ν(x),∀x,yinX,μ∩ν(x)=min⁡{μ(x),ν(x)},∀x∈X,μ∪ν(x)=max⁡{μ(x),ν(x)},∀x∈X,⋂α∈Ωμα(x)=inf⁡α∈Ωμα(x),∀x∈X,⋃α∈Ωμα(x)=sup⁡α∈Ωμα(x),∀x∈X,μη(x)=sup⁡x=y∨z{min⁡{μ(y),η(z)}}.Lemma 2.1.
Let{ηα∣α∈Ω} be a family of fuzzy filters of X. Then, ⋂α∈Ωηα is a fuzzy filter of X.Proof.
Letx∈X. For any α∈Ω, ηα(1)≥ηα(x) since ηα<FFX. Then, infα∈Ωηα(1)≥infα∈Ωηα(x) and so, ⋂α∈Ωηα(1)≥⋂α∈Ωηα(x). (FF1) holds.
Moreover, for anyɛ>0, there exists α(ɛ)∈Ω such that
(2.2)⋂α∈Ωηα(x)+ε=inf⁡α∈Ωηα(x)+ɛ≥ηα(ε)(x)≥min⁡{ηα(ε)((x*y*)*),ηα(ε)(y)}≥min⁡{inf⁡α∈Ωηα(x*y*)*,inf⁡α∈Ωηα(y)}=min⁡{⋂α∈Ωηα(x*y*)*,⋂α∈Ωηα(y)}.
Sinceɛ is arbitrary, we get ⋂α∈Ωηα(x)≥min{⋂α∈Ωηα(x*y*)*,⋂α∈Ωηα(y)}. So, (FF2) holds.
Therefore,⋂α∈Ωηα is a fuzzy filter of X.Lemma 2.2 (see, [16]).
Letμ be a fuzzy filter of X. For any x,y∈X, if x≤y, then μ(x)≤μ(y).Definition 2.3.
Letμ be a fuzzy subset of X. Then the fuzzy filter generated by μ, which is denoted by 〈μ〉, is defined as
(2.3)〈μ〉=⋂{η:μ⊆η,η<FFX}.
Obviously, we get μ⊆〈μ〉, and if μ<FFX, then μ=〈μ〉.Lemma 2.4.
Ifμ,η<FFX, then μη=μ∩η.Proof.
Letx∈X, x=a∨b and μ,η be fuzzy filters. Then, by Lemma 2.2, μ(a)≤μ(a∨b)=μ(x) and η(b)≤η(a∨b)=η(x). Hence, min{μ(a),η(b)}≤μ∩η(x).
Therefore,μη≤μ∩η(x), or equivalently μη⊆μ∩η.
Conversely,μη(x)=supx=y∨z{min{μ(y),η(z)}}≥min{μ(x),η(x)}=μ∩η(x). So μη⊇μ∩η.
Thus,μη=μ∩η.Corollary 2.5.
Ifμ,η<FFX, μη<FFX.Lemma 2.6.
Ifη<FFX, μη⊆η.Proof.
Letη<FFX. If x=y∨z, then from Lemma 2.2 we know η(z)≤η(x). Thus, μη(x)=supx=y∨z{min{μ(y),μ(z)}≤supx=y∨z{η(z)}≤η(x). So, μη⊆η.


## 3. Fuzzy Filter Spectrum

Definition 3.1.
A nonconstant fuzzy filterμ of X is said to be s-prime if for all θ,σ<FFX, θσ⊆μ implies θ⊆μ or σ⊆μ. In this case, we write μ<FSPX.In this paper, we give some notations in the following.(i)
F(X)={μ∣μ<FSPX}.(ii)
V(θ)={μ∈F(X)∣θ⊆μ}, where θ is a fuzzy subset of X.(iii)
F(θ)=F(X)∖V(θ)={μ∈F(X)∣θ⊈μ}, where F(X)∖V(θ) is called the complement of V(θ) in F(X).Lemma 3.2.
Ifσ is a fuzzy subset of X, then V(〈σ〉)=V(σ). So F(σ)=F(〈σ〉).Proof.
Letμ∈V(σ), then σ⊆μ and so 〈σ〉⊆μ. Hence, μ∈V(〈σ〉). Conversely, let μ∈V(〈σ〉), then 〈σ〉⊆μ. Note that σ⊆〈σ〉⊆μ, we get μ∈V(σ). Therefore, V(σ)=V(〈σ〉).Theorem 3.3.
Letζ={F(θ)∣θ<FFX}. Then the pair (F(X),ζ) is a topological space.Proof.
Considerθ0=0 and θ1=1. Then θ0, θ1<FFX,F(θ0)=∅ and F(θ1)=F(X). Thus, F(X),∅∈ζ.
Then, we prove thatζ is closed under finite intersection.
Letη and θ be two fuzzy filters of X. We claim that V(θ)∪V(η)=V(θη). Let τ∈V(θη). Then, θη⊆τ. Since τ∈F(X), we have θ⊆τ or σ⊆τ. It follows that τ∈V(θ)∪V(η).
Conversely, letτ∈V(θ)∪V(η), then θ⊆τ or η⊆τ. By Lemma 2.6, θη⊆θ and θη⊆η. Thus, θη⊆τ and so τ∈V(θη). It follows that V(θ)∪V(η)⊆V(θη).
Combining the above arguments we getV(θ)∪V(η)=V(θη), or equivalently, F(θ)∩F(η)=(F(X)∖V(θ))∩(F(X)∖V(η))=(F(X)∖(V(θ)∪V(η))=(F(X)∖V(θη))=F(θη). By Corollary 2.5, θη<FFX and so F(θ)∩F(η)=F(θη)∈ζ.
Finally, let{θα∣α∈Ω} be a family of fuzzy prime filters of X. We will prove that ⋂α∈ΩV(θα)=V(⋃α∈Ωθα).
Letμ∈⋂α∈ΩV(θα), then for any α∈Ω, μ∈V(θα) and so θα⊆μ. Hence, ⋃α∈Ωθα⊆μ and thus μ∈V(⋃α∈Ωθα).
Conversely, letμ∈V(⋃α∈Ωθα), then ⋃α∈Ωθα⊆μ. Thus, for any α∈Ω, θα⊆⋃α∈Ωθα⊆μ. Hence, μ∈V(θα) for all α∈Ω and so μ∈⋂α∈ΩV(θα).
This shows that⋂α∈ΩV(θα)=V(⋃α∈Ωθα).
By Lemma3.2, we get V(⋃α∈Ωθα)=V(〈⋃α∈Ωθα〉) and so ⋂α∈ΩV(θα)=V(〈⋃α∈Ωθα〉).
Furthermore, we get⋃α∈ΩF(θα)=⋃α∈Ω(F(X)∖V(θα))=F(X)∖⋂α∈ΩV(θα)=F(X)∖V(〈⋃α∈Ωθα〉)=F(〈⋃α∈Ωθα〉)∈ζ.
It follows that(F(X),ζ) is a topological space.Theorem 3.4.
The collection(3.1)ß={F(xβ)∣x∈X,β∈(0,1]}
of ζ is a base of ζ where xβ<FFX is defined by
(3.2)xβ(t)={β,t=x,0,t≠x.Proof.
By Lemma3.2, for any x∈X, β∈(0,1], F(xβ)=F(〈xβ〉) and so F(xβ)∈ζ.
Now, we prove that ß is a base ofζ. It is sufficient to show that for all F(θ)∈ζ, and μ∈F(θ), there exists F(xβ)∈β such that μ∈F(xβ) and F(xβ)⊆F(θ).
LetF(θ)∈ζ and μ∈F(θ). Then, θ⊈μ and so there exists x∈X such that μ(x)<θ(x). Let θ(x)=β and then μ∈F(xβ). Moreover, for any σ∈V(θ), σ(x)≥θ(x)=β=xβ(x) and so xβ⊆σ. Thus σ∈V(xβ). This means V(θ)⊆V(xβ). It follows that F(xβ)⊆F(θ).
Therefore, ß is a base ofζ.The topological space(F(X),ζ) is called fuzzy filter spectrum of X, denoted by FF-spec(X), or F(X) for convenience.Theorem 3.5.
FF-spec(X) is a T0 space.Proof.
Letμ,η∈F(X) and μ≠η. Then, μ⊈η or η⊈μ.
Ifμ⊈η, then, η∉V(μ) but μ∈V(μ). Moreover, η∈F(μ) but μ∉F(μ).
Ifη⊈μ, similarly we can get μ∈F(η) but η∉F(η). It follows that FF-spec(X) is a T0 space.Lemma 3.6 (see [9]).
Letμ be a fuzzy subset of X. Then, μ is a fuzzy filter of X if and only if μt is a filter of X for each t∈[0,1] wherever μt≠∅.Lemma 3.7.
A non-constant fuzzy subsetμ of X is a fuzzy prime filter if and only if μt is a prime filter of X for each t∈[0,1] whenever μt≠∅.Proof.
Letμ be a fuzzy prime filter and t∈[0,1] such that μt≠∅. Then by Lemma 3.6, μt is a filter of X.
Supposex∨y∈μt. It follows that μ(x∨y)≥t. Since μ is prime, we have max{μ(x),μ(y)}≥μ(x∨y)≥t and thus μ(x)≥t or μ(y)≥t. It follows that x∈μt or y∈μt. Therefore μt is a prime filter.
Conversely, suppose that for eacht∈[0,1], μt is a prime filter whenever μt≠∅. If μ is not a fuzzy prime filter, then there exist x,y∈X such that μ(x∨y)>max{μ(x),μ(y)}. Take t satisfying μ(x∨y)>t>max{μ(x),μ(y)}. Then x∨y∈μt. Since μt is a prime filter of X, then x∨y∈μt implies x∈μt or y∈μt. But on the other hand, μ(x)≤max{μ(x),μ(y)}<t and μ(y)≤max{μ(x),μ(y)}<t imply x∉μt and y∉μt, a contradiction. It follows that μ is indeed a fuzzy prime filter.Lemma 3.8 (see [13]).
LetX be a bounded commutative BCK-algebra and F be a BCK-filter of X. Then, F is prime if and only if, for any filters A,B, F=A∩B implies F=A or F=B.Theorem 3.9.
LetX be a bounded commutative BCK-algebra and μ be a fuzzy s-prime filter. Then for each t∈[0,1], μt is a prime filter of X whenever μt≠∅ and μt≠X.Proof.
Letμ be a fuzzy s-prime filter and t∈[0,1], μt≠∅. Then by Lemma 3.6, μt is a filter.
LetA, B be two filters such that μt=A∩B. Define the fuzzy subset θ=tχA and σ=tχB. It is easy to see that θ and σ are fuzzy filters of X. Note that
(3.3)θ∩σ(x)=θ⋅σ(x)={t,x∈A∩B,0,x∉A∩B.
Sinceμt=A∩B, then for any x∈A∩B=μt, μ(x)≥t=θ∩σ(x) and so μ(x)≥θ∩σ(x) for all x∈X. Thus μ⊇θ∩σ. It follows from μ being a fuzzy s-prime filter that θ⊆μ or σ⊆μ. Without loss of generality let θ⊆μ. Then, for any x∈A, θ(x)=tχA(x)=t≤μ(x) and so x∈μ(t). This means that A⊆μt. But μt=A∩B implies μt⊆A and thus μt=A. Therefore, μt is a prime filter by Lemma 3.8.Theorem 3.10.
LetX 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 Lemma3.7 and Theorem 3.9.In general, the converse of Theorem3.10 is not true. Let us see the following example.Example 3.11.
LetX={0,1}. Define the operation * on X as follows: 0*0=0, 0*1=0, 1*0=1 and 1*1=0. It is easy to see that 〈X;*,0〉 is a bounded commutative BCK-algebra. Define a fuzzy subset μ of X by μ(0)=0, μ(1)=1/2. Clearly μ is a fuzzy prime filter of X.
Moreover, we define the fuzzy filtersσ and θ by σ(x)=1/2 for all x∈X and θ(1)=1, θ(0)=0. Then, we get θσ⊆μ but σ⊈μ and θ⊈μ. Therefore, μ is not a s-prime fuzzy filter.Lemma 3.12.
F is a prime filter of X if and only if χFα is a fuzzy s-prime filter, where α∈[0,1), and χFα is defined by
(3.4)χFα(x)={1,x∈Fα,x∉F.Proof.
LetF be a prime filter. Then, by Lemma 3.6, we can easily see that χFα is a fuzzy filter.
Letθ,σ be two fuzzy prime filters such that θσ⊆χFα, we will prove θ⊆χFα or σ⊆χFα. If it is not true, then there exist x,y∈X∖F such that θ(x)>α and σ(y)>α. Since F is prime, then x∨y∉F. Note that θσ⊆χFα, then
(3.5)α<min⁡{θ(x),σ(y)}≤min⁡{θ(x∨y),σ(x∨y)}=θ∩σ(x∨y)=θσ(x∨y)≤χFα(x∨y).
Thus,x∨y∈F, a contradiction. It follows that θ⊆χFα or σ⊆χFα, and so χFα is a fuzzy s-prime filter.
Conversely, letχFα be a fuzzy s-prime filter. By Theorem 3.10, χFα is also a fuzzy prime filter. Then, by Lemma 3.7, (χF)t=F is a prime filter, where α≤t≤1.Corollary 3.13.
F is a prime filter of X if and only if χF is a fuzzy s-prime filter.Lemma 3.14 (see [13]).
LetX be a bounded implicative BCK-algebra, then x∧x*=0 and x∨x*=1.Lemma 3.15.
Letμ be a fuzzy filter of a bounded commutative BCK-algebra X. Then, μ(0)=min{μ(x),μ(x*)} for all x∈X.Proof.
Sinceμ is a fuzzy filter, we have μ(0)≥min{μ(x*0)*,μ(x)}=min{μ(x*),μ(x)} for all x∈X. On the other hand, μ(0)≤min{μ(x*),μ(x)}, since any fuzzy filter is order preserving. Thus, μ(0)=min{μ(x),μ(x*)}.Lemma 3.16.
Ifμ is a fuzzy filter of a bounded BCK-algebra X, then μ1={x∈X∣μ(x)=μ(1)} is a filter of X and χμ1 is a fuzzy filter of X.Proof.
Letμ be a fuzzy filter and take t=μ(1). Then, μt=μ1 and so μt=μ1 is a filter of X by Lemma 3.6. Clearly χμ1 is a fuzzy filter.Lemma 3.17.
LetX be a bounded commutative BCK-algebra and μ be a fuzzy s-prime filter of X. Then μ(1)=1.Proof.
Suppose thatμ(1)<1. Since μ is non-constant, there exists a∈X such that μ(a)<μ(1). Define fuzzy subset θ and σ of X by
(3.6)θ(x)={1,μ(x)=μ(1),0,otherwise.
and σ(x)=μ(1) for all x∈X. By Lemma 3.16, θ(x)=χμ1is a fuzzy filter and clearly σ is a fuzzy filter. Note that θ(1)=1>μ(1) and σ(a)=μ(1)>μ(a), we get θσ⊈μ. But note that for any x,y∈X(3.7)min⁡{θ(x),σ(x)}≤{σ(y),x∈μ10,x∉μ1≤{μ(1),x∈μ1μ(x),x∉μ1=μ(x)≤μ(x∨y).
Thus,θσ(x)=supx=y∨z{min{θ(y),σ(z)}}≤supx=y∨z{μ(y∨z)}=sup{μ(x)}=μ(x) for any x∈X, a contradiction. Therefore, μ(1)=1.Lemma 3.18.
LetX be a bounded implicative BCK-algebra and μ be a fuzzy s-prime filter of X. Then for any x∈X, μ(x)=1 or μ(x*)=1.Proof.
By Lemma3.14, x∨x*=1, for all x∈X. Since μ is a fuzzy s-prime filter, we get that μ1 is a prime filter of X by Theorem 3.9. Hence, x∨x*=1∈μ1 implies x∈μ1 or x*∈μ1. Therefore, μ(x)=μ(1)=1 or μ(x*)=μ(1)=1 by Lemma 3.17.Theorem 3.19.
LetX be a bounded implicative BCK-algebra and μ be a fuzzy s-prime filter of X. Then, for x∈X, μ(x)=μ(1)=1 or μ(x)=μ(0).Proof.
By Lemma3.14, x∨x*=1 and then μ(1)=μ(x∨x*)=μ(x) or μ(x*) since μ is a fuzzy s-prime filter. By Lemma 3.18, we get μ(x)=1 or μ(x*)=1. If μ(x*)=1, then μ(0)=min{μ(x),μ(x*)}=μ(x) by Lemma 3.15. If μ(x*)≠1, then μ(x)=1=μ(1).Lemma 3.20.
LetX be a bounded BCK-algebra. Then, a filter F of X is proper if and only if 0∉F.Proof.
If0∉F, then clearly F is proper.
Conversely, letF be proper. If 0∈F, then for any x∈X, (x*0*)*=(x*1)*=0*=1∈F and so x∈F. It follows that F=X, a contradiction. Therefore, 0∉F.Lemma 3.21 (see [13]).
LetX be a bounded commutative BCK-algebra and F be a filter of X. If x∈X∖F, then there is a prime filter A of X such that F⊆A and x∉A.Lemma 3.22 (see [13]).
LetX be a bounded implicative BCK-algebra. Then, for any a∈X, the filter 〈a〉, generated by a, is a set of elements x in X satisfying a≤x.Lemma 3.23.
LetX be a bounded implicative BCK-algebra and a≠0. Then, 〈a〉≠X.Proof.
By Lemma3.22, 0∉〈a〉 and thus 〈a〉≠X.Lemma 3.24 (see [16]).
For a bounded commutative BCK-algebraX, one gets(1)
x**=x for all x∈X.(2)
x*∧y*=(x∨y)*, x*∨y*=(x∧y)* for all x,y∈X.(3)
x*y*=yx for all x,y∈X.Theorem 3.25.
LetX be a bounded implicative BCK-algebra. Then,(i)
ifβ1,β2∈(0,1], β=min{β1,β2} and x,y∈X, then F(xβ1)∩F(yβ2)=F((x∨y)β).(ii)
ifβ∈(μ(0),1] and x,y∈X, then F(xβ)∪F(yβ)=F((x∧y)β).
Moreover,F(xβ) is both open or closed, where μ∈F(X).(iii)
ifF(xβ)=F(X), where x∈X and β∈(0,1], then x=0.Proof.
(i) Ifμ∈F(xβ1)∩F(yβ2), then μ(x)<β1 and μ(y)<β2. By Theorem 3.10, μ is a fuzzy prime filter, and then μ(x∨y)≤max{μ(x),μ(y)}. Since μ(x)<β1 and μ(y)<β2, then μ(x)≠μ(1) and μ(y)≠μ(1). It follows from Theorem 3.19 that μ(x)=μ(0) and μ(y)=μ(0). Thus, μ(x∨y)≤max{μ(x),μ(y)}=μ(0)=min{μ(x),μ(y)}<min{β1,β2}=β. Therefore, μ∈F((x∨y)β).
Conversely, ifμ∈F(x∨y)β, then μ(x∨y)<β and so μ(x∨y)<β1, μ(x∨y<β2). Note that x≤x∨y and y≤x∨y, we get μ(x)≤μ(x∨y)<β1, and μ(y)≤μ(x∨y)<β2, since μ is order preserving. Thus, μ∈F(xβ1) and μ∈F(yβ2), or equivalently, μ∈F(xβ1)∩F(yβ2).
Therefore, (i) holds.
(ii) Letμ∈F(xβ)∪F(yβ). Then, μ(x)<β or μ(y)<β. By Lemma 3.17, we have μ(1)=1≥β>μ(x), μ(1)=1≥β>μ(y). Therefore, x∉μ1 or y∉μ1. On the other hand, by Lemma 3.14, x∨x*=1∈μ1, y∨y*=1∈μ1. Note that μ1=μ1 is a prime filter of X by Theorem 3.9. If x∉μ1, then x∨x*∈μ1 implies x*∈μ1. If y∉μ1, then y∨y*∈μ1 implies y*∈μ1. Therefore, we get that x*∈μ1 or y*∈μ1. Note that x∧y≤x,y, we get x*≤(x∧y)* and y*≤(x∧y)*. Thus, μ(x*)≤μ((x∧y)*) and μ(y*)≤μ((x∧y)*), and so max{μ(x*),μ(y*)}≤μ((x∧y)*). But μ(x*)=μ(1) or μ(y*)=μ(1) implies that 1=max{μ(x*),μ(y*)}≤μ((x∧y)*) or μ((x∧y)*)=μ(1). This means, (x∧y)*∈μ1.
Ifx∧y∈μ1, then μ(0)≥min{μ(((x∧y)*0)*),μ(x∧y)}=min{μ((x∧y)*),μ(x∧y)}=μ(1). It follows that μ(x)=μ(1) for all x∈X, a contradiction. Thus, x∧y∉μ1. By Lemma 3.18, μ(x∧y)=μ(0). Hence, μ(x∧y)<β and so μ∈F((x∧y)β). It follows that F(xβ)∪F(yβ)⊆F((x∧y)β).
Conversely, letμ∈F((x∧y)β). Then, μ(x∧y)<β≤1=μ(1). Thus, x∧y∉μ1. Since (x∧y)∨(x∧y)*=1∈μ1, then (x∧y)*∈μ1. By Lemma 3.24, (x∧y)*=x*∨y* and so x*∈μ1 or y*∈μ1. If x*∈μ1 (or y*∈μ1), then μ(0)≥min{μ((x*0)*),μ(x)}=min{μ(x*),μ(x)}=μ(x) (or μ(0)≥μ(y)). Thus, x∉μ1 or y∉μ1. It follows that F((x∧y)β)⊆F(xβ)∪F(yβ).
Combining the above arguments, we getF(xβ)∪F(yβ)=F((x∧y)β).
In order to proveF(xβ) is closed, we will show F(xβ)=V((x*)β).
Letμ∈F(xβ). Then, μ∈F(〈xβ〉) by Lemma 3.2. Thus, 〈xβ〉⊈μ and so μ(x)<β≤1=μ(1). Hence, x∉μ1. Note that x∨x*=1∈μ1 we get x*∈μ1, which implies that μ(x*)=μ(1)=1≥β and so (x*)β⊆μ. It follows that μ∈V((x*)β) and thus F(xβ)⊆V((x*)β).
Conversely, letμ∈V((x*)β). Then, (x*)β⊆μ and so μ(x*)≥β>μ(0). By Theorem 3.19, μ(x*)=μ(1). Note that μ(0)≥min{((x*0)*),μ(x)}=min{μ(x*),μ(x)}=μ(x), we get that μ(x)=μ(0)<β, or equivalently, 〈xβ〉⊈μ. Thus, μ∈F(xβ) and so V((x*)β)⊆F(xβ).
Combining the above two sides, we getF(xβ)=V((x*)β).
(iii) LetF(xβ)=F(X). We claim that x=0. If this is not true, by Lemma 3.23,〈x〉≠X. By Lemma 3.21, there exists a prime filter P of X such that 〈x〉⊆P. On the other hand, by Lemma 3.12, χP∈F(X)=F(xβ).
Therefore,〈x〉⊈P, a contradiction. It follows that x=0.In general, the converse of Theorem3.25 (iii) does not hold. Let us see the following counter example.Example 3.26.
LetX={0,1,2,3} and *-table and ∨-table be given as follows. 							(3.8)*012300000110102220033210∨012300123111332232333333
Then (x;*,0) is a bounded implicative BCK-algebra and 3 is a unit. It is easy to see that P={1,3} is a filter of X. From ∨-table, we can see easily that P is prime. So, χP is a fuzzy s-prime filter by Lemma 3.12. Let β=1/2. Then, 3β⊆μ and so 〈3β〉⊆μ. Hence, μ∉F(3β). Therefore F(3β)≠F(X).Theorem 3.27.
LetX be a bounded implicative BCK-algebra and Xα={μ∈F(X)∣μ(0)=α} for α∈[0,1). Then, Xα is a Hausdorff space.Proof.
Letμ,σ∈Xα and μ≠σ. We claim that μ1≠σ1. Otherwise, if μ1=σ1, then for x∈μ1=σ1, μ(x)=μ(1)=1=σ(1)=σ(x) and for x∉μ1=σ1, μ(x)=μ(0)=α=σ(0)=σ(x) by Theorem 3.19, a contradiction. Thus, μ1⊈σ1 or σ1⊈μ1. Let μ1⊈σ1. Then, x∈μ1∖σ1 implies x*∈σ1. Moreover, μ(0)≥min{μ(x*0)*,μ(x)}=min{μ(x*),μ(x)}=μ(x*) since μ(x)=μ(1)=1. Thus μ(x*)=μ(0)≠μ(1) and so x*∉μ1. Therefore, x*∈σ1∖μ1. Hence,
(3.9)σ(x)=α=μ(x*),μ(x)=1=σ(x*).
Lett∈(α,1]. Then, (xt)(x)=t>σ(x), and so σ∈F(xt). Note that ((x*)t)(x*)=t>α=μ(x*), we get μ∈F((x*)t). Moreover, we get
(3.10)F(xt)∩F((x*)t)=F((x∨x*)t)(byTheorem3.25(i))=F(1t)(byLemma3.14)=∅(byLemma3.17).
It follows thatXα is a Hausdorff space.Corollary 3.28.
LetX be a bounded implicative BCK-algebra. Then X0={μ∈F(X)∣μ(0)=0} is a Hausdorff space.LetX be a bounded commutative BCK-algebra, L(X) be the set of all filters of X, and F-spec(X) stand for all prime filters of X.For any subsetA of X, we define S(A)={P∈F-spec(X)∣A⊈P}.IfA={a}, we denote S({a}) by S(a).Lemma 3.29.
S(A)=S(〈A〉) and if A⊆B, then S(A)⊆S(B).Proof.
SinceA⊆〈A〉, then A⊈P implies 〈A〉⊈P. Then, P∈S(A) implies P∈S(〈A〉).
Conversely, ifP∈S(〈A〉), then 〈A〉⊈P. Hence, A⊈P since A⊆P implies 〈A〉⊆P. Therefore, P∈S(A).
Thus,S(A)=S(〈A〉). Similarly, we can prove that A⊆B implies S(A)⊆S(B).Proposition 3.30.
The familyT(X)={S(F)∣F∈L(X)} forms a topology on F-spec(X).Proof.
First, we get(3.11)S(1)=S(〈1〉)={P∈F-spec(X)∣〈1〉⊈P}=∅∈T(X),S(X)={P∈F-spec(X)∣X⊈P}=F-spec(X)∈T(X).
Then, for any family{S(Fα)}α∈Ω,
(3.12)⋃α∈ΩS(Fα)={P∈F-spec(X)∣Fα⊈PFor someFα}={P∈F-spec(X)∣⋃α∈ΩFα⊈P}={P∈F-spec(X)∣〈⋃α∈ΩFα〉⊈P}=S(〈⋃α∈ΩFα〉)∈T(X).
Finally,(3.13)S(F1)∩S(F2)={P∈F-spec(X)∣F1∩F2⊈P}=S(F1∩F2)∈T(X).
Therefore,T(X) is a topology on F-spec(X).Theorem 3.31.
LetX be a bounded implicative BCK-algebra and the map f:F-spec(X)→Xα is defined by f(P)=χPα where χPα is defined in Lemma 3.12. Then, f is a homeomorphism.Proof.
(a)f is well defined.
By Lemma3.12, χPα is a fuzzy s-prime filter for any P∈F-spec(X). Note that 0∉P, then χPα(0)=α and so χPα∈Xα. Thus, f is well defined.(b)
Clearlyf is injective.(c)
f is surjective.
For anyμ∈Xα, by Theorem 3.19, μ(x)=μ(1)=1 or μ(x)=μ(0)=α. Hence, μ=χμ1α. By Theorem 3.9, we get μ1 is a prime filter of X. Thus, μ1∈F-spec(X) and so f(μ1)=χμ1α=μ. It follows that f is surjective.(d)
f is continuous.
LetFα(θ)=F(θ)∩Xα be an open set of Xα. We will prove that f-1(Fα(θ)) is an open set of F-spec(X). It is sufficient to prove that f-1(Fα(θ))=⋃α≤t≤1S(θt), since θt is a filter of X by Lemma 3.6.
First, letP∈⋃α<t<1S(θt), then there exists some t such that α<t<1 and P∈S(θt). Thus, θt⊈P and there exists x∈θt∖P. Hence θ(x)≥t>α=χPα(x). Therefore, θ⊈χPα and so χPα∈Fα(θ). This shows that f(P)∈Fα(θ). It follows that P∈f-1(Fα(θ)).
Conversely, letP∈f-1(Fα(θ)), then f(P)=χPα∈Fα(θ). Hence, θ⊈χPα, and thus there exists x∈X such that χPα<θ(x). Therefore, x∉P and so χPα=α<θ(x). We can take t such that α<t1<θ(x). Then, x∈θt1∖P. It follows that θt1⊈P and so P∈S(θt1)⊆⋃α<t<1S(θt).
Combining the above two hands, we getf-1(Fα(θ))=⋃α<t<1S(θt). So f is continuous.(e)
f-1 is continuous. It is sufficient to prove that f(S(F)) is an open set of Xα for any F∈L(X).
We will prove thatf(S(F))=Fα(χFα). 							(3.14)μ∈f(S(F))⟹∃P∈S(F)such thatf(P)=μ=χPα⟹F⊈P,f(P)=μ=χPα⟹∃x∈F∖P,μ=χPα⟹χPα<1=χFα(x)⟹χFα⊈χPα=μ⟹μ∈Fα(χFα).
Thus,f(S(F))⊆Fα(χFα).
Conversely, we get(3.15)μ∈Fα(χFα)⟹χFα⊈μ,∃P∈F-spec(X)such thatf(P)=μ=χPα.⟹∃x∈Xsuch thatχFα(x)>μ(x)=χPα⟹F⊈P⟹P∈S(F)⟹μ=f(P)∈f(S(F)).
So thatFα(χFα)⊆f(S(F)).
Therefore,f(S(F))=Fα(χFα). By Lemma 3.6, we can easily see that χFα is a fuzzy filter of X and so Fα(χFα)=F(χFα)∩Xα is an open set of Xα. It follows that f-1 is continuous.By Theorem3.27 and Theorem 3.31, we get the following corollary.Corollary 3.32.
LetX be a bounded implicative BCK-algebra. Then, F-spec(X) is a Hausdorff space.



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*Source: 795934-2011-01-24.xml*