Denition.9. Let a A; t 0; 1]. Then by a fuzzy point a t we mean the fuzzy subset of A given below: a t (x) = t if x = a 0 otherwise Denition.101]. A f

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1 Some Properties of F -Spectrum of a Bounded Implicative BCK-Algebra A.Hasankhani Department of Mathematics, Faculty of Mathematical Sciences, Sistan and Baluchestan University, Zahedan, Iran abhasan@hamoon.usb.ac.ir, Fax: Abstract: Compactness, connectedness and irreducible properties of the space X of the fuzzy prime spectrum of a given bounded implicative BCK-algebra are discussed. Key Words: Fuzzy(prime) ideal, (Integral domain)implicative BCK-algebra, Fuzzy prime spectrum. 1. Introduction In 1966, Imai and Iseki 6] introduced a new notion called a BCK-algebra. In 3] a topology on the set of all fuzzy prime ideal of a commutative BCK-algebra Y was de- ned. Now in this paper some properties of this topological space over a bounded implicative BCK-algebra are discussed. 1. preliminaries Denition.1. 6] An algebra < Y;;0 > of type < ; 0 > is said to be a BCK-algebra provided the following axioms are satised: (1) (x y) (x z) z y () x (x y) y (3) x x (4) 0 x (5) x y; y x ) x = y for all x; y; z Y, where x y is dened by x y =0. Denition.8] A BCK-algebra Y is said to be commutative if x(xy)=y(yx);8x; y Y: For all x; y, y (y x) is denoted by x ^ y. Denition.38]. A BCK-algebra Y is said to be implicative if it satises the identity x (y x) =x; 8x; y Y: Denition.48]. A BCK-algebra Y is said to be bounded if there is an element e Y satisfying x e, for all x Y. In this case, for all x Y, e x is denoted by Nx. Denition.59]. A subset J of a BCKalgebra Y is said to be an ideal if (i) 0J (ii) x y J;y J ) xj;8x; y J. Denition.6see 7]. A proper ideal P of a commutative BCK-algebra Y is said to be s- prime if x ^ y P ) x P or y P; 8x; y Y: Denition.7(see7]). A proper ideal P of a commutative BCK-algebra Y is said to be prime of Y if for all ideals A; B of Y such that AB P, then either A P or B P, where AB = fa ^ b : a A; b Bg. Denition.813]. Let A be a non-empty set. A fuzzy subset of A is a function from A into 0; 1]. If and be two fuzzy subsets of A, then by we mean (x) (x), for all x A. 1

2 Denition.9. Let a A; t 0; 1]. Then by a fuzzy point a t we mean the fuzzy subset of A given below: a t (x) = t if x = a 0 otherwise Denition.101]. A fuzzy subset of a BCK-algebra Y is said to be ideal if (i) (0) (x); 8x Y (ii) (x) minf(x y);(y)g;8x; y Y. Denition.11 3]. Let and be two fuzzy subsets of a BCK-algebra Y. Then the fuzzy subset is dened by: (x) = sup fmin((y);(z))g: x=y^z Denition.13]. A non-constant fuzzy ideal of a commutative BCK-algebra Y is called prime if, for all fuzzy ideals and such that, then either or. Denition.13(see4]). A non-constant fuzzy ideal of a commutative BCK-algebra Y is called s-prime if for all x; y Y (x ^ y) =(x)or(y): Denition.14. If is a fuzzy subset of a non-empty set A. For t 0; 1], the set t = fx A : (x) tg is called a level subset of. Lemma.153]. Let be a fuzzy prime ideal of a commutative BCK-algebra Y. Then 0 = fx Y : (x) =(0)g is a prime ideal of Y. Denition.16. Let be a fuzzy subset of a non-empty set A, where is \ in the index set. We dene the fuzzy subset and as follows: \ (i) ( )(x) =inf (x); 8x A. (ii) ( )(x) =sup (x); 8x A. Denition.17. If is a fuzzy subset of a BCK-algebra Y. Then the ideal generated by which is denoted by <>is dened as follows: <>= \ f:; is a fuzzy ideal of Y g: Notation. Let Y be a commutative BCKalgebra. (i) X = f : is a fuzzy prime ideal of Y g. (ii) V () = f X : g; where is any fuzzy ideal of Y. (iii) X() = XnV (); the complement ofv() in X. Clearly V (< >)= V();for all fuzzy subset of Y. Theorem.183]. Let Y be a commutative BCK-algebra and = fx() : is a fuzzy ideal of Y g. Then the pair (X; ) is a topological space, it is called F- spectrum of Y and denoted by F, spec(y ), for convenience by X. Lemma.19.8] Any implicative BCKalgebra is commutative. 1. Main results From now on Y is a bounded implicative BCK-algebra. Denition 3.1. Let 0; 1). Then by X we mean the set f X : (e) =g: Lemma 3..(i) Each proper ideal I of Y is contained in a maximal ideal (1]). (ii)an ideal I of Y is proper i e 6 I (1]). (iii) Let I be a proper ideal of Y. Then I is a prime ideal i I is a maximal ideal (5]). Lemma 3.3. Let 0; 1), (; 1] and fx i : i g Y, where \ is an index set. If V (< (x i ) >) X = ; (1) then for any prime ideal P of Y fx i : i g 6P: Proof. Let P be a prime ideal of Y. Dene the fuzzy subset by : (x)= 1 xp otherwise. One observes that is a fuzzy prime ideal of Y and (e) =. Now the conclusion follows from (1). Lemma 3.4.(i)Let be a fuzzy prime ideal. Then for each x Y either (x) =

3 (0) or (x) =(e) (3]). (ii)with the assumptions in Lemma 3.3, if V (< n i=1 (x i ) >) \ X ; where n N, then fx i : i =1;; :::; ng 0 : Proof.(ii) Since V (< n i=1 (x i ) >), we have (x i ), for all i = 1; :::; n. Hence (x i ) > = (e) for all i = 1; :::; n. Now the proof follows from (i). Lemma 3.5.Let K (0; 1]; = supft : t Kg and fx i : i g Y. Then fx((xi ) t ):;tkg= fx((x i ) ):g: Proof. Similar to the proof of Lemma 3.7 of ]. Recall that a subset A of a topological space is compact i every covering of A by open sets is reducible to a nite subcover of A. Theorem 3.6.Let 0; 1). Then X is compact. Proof. First we show that B = fx(x ) \ X : x Y; (; 1]g constitutes a base for X. To do this let W be an open set in X. Hence there is X() such that W = X() T X. Now let W. Then 6. In other words there is x Y such that (x) > (x). Choose = (x). Hence we have = (x) > (x) (e) = and X(x ) T X. Clearly X(x ) \ X X() \ X : Thus by Lemma 3 of 10], B is a base for X. Now let C = fx((x i ) t ) \ X : i ;tk (; 1]g be any covering of X. Let = supft : t Kg. Then by Lemma 3.4 we have \ X =X(< (x i ) >) X = X \ X(< X n(v < (x i ) >) X c ]= (x i ) > \ X ): \ That is V (< (x i ) >) X =.By Lemma 3.3, we get that fx i : i g 6P,for all prime ideal P of Y.Hence from Lemmas 3.(i), 3.(iii), < fx i : i g >= Y: Thus there exists n N such that (:::((e x i1 ) x ) :::) x in =0; where i j, for j =1; :::; n. Now if V(< n j=1 (x ij ) >) \ X. Then by Lemma 3.4(ii), fx ij : j = 1; ; :::; ng 0. Therefore e 0, which is a contradiction, thus V (< n j=1 (x ij ) ) \ X = : Consequently T X is covered by fx((x ij ) ) X : j =1;; :::; ng. Lemma 3.7.(i) Let J be a subset of Y. Then the set I = fx Y : 9a 1 ; :::; a n J satisfying (:::((x a 1 ) a ) :::) a n =0gis the minimal ideal containing J. I is called the ideal generated by J and will be denoted by I =< J> (9]). (ii)if f i : i g is any family of fuzzy ideals of Y. Then X( i )=X( i ) (3]): (iii)if ; (0; 1];, then X(x ) X(x ),and V (x ) V (x ),for all x Y. (iv) Let be a fuzzy ideal of Y. Then we have xy)(y)(x);8x; y Y (4]): Theorem 3.8.The topological space X is compact. Proof. Let fx((x i ) t ):;t K (0; 1]g be any covering of X and = supft : t Kg. Dene the family f n g nn of fuzzy prime ideals of Y as follows n (x) = 1 xp 1, 1 n otherwise, where P is an arbitrary prime ideal of Y. Thus for any n N there exist t n K and i n such that n X((x in ) tn ), which implies that 3

4 t n > 1, 1,and hence =1. On the other hand n from Lemmas 3.5 and 3.7(iii) we have: V (< (x i ) >)= Now similar to the proof of Lemma 3.3 we get that fx i : i g 6 P, for all prime ideal of Y. Hence by Lemma 3., < fx i : i g >= Y. Thus by Lemma 3.7(i), there exists n N such that (:::((e x i1 ) x ) :::) x in =0; where i j, for j =1; :::; n. Now we show that V (< (x ij ) >)= () On the contrary, let V (< (x ij ) >). Then by Denition.10(ii) and Lemma 3.7(iv), (x) = 1, for all x Y, which is a contradiction. Therefore () holds. Consequently X = X(< (x ij ) >)= X((x ij ) ); by Lemma 3.7(ii). Lemma 3.98]. (x y) y = x y, for all x; y Y: In particular x ^ Nx = (e x) ((e x) x) = (e x) (e x) = 0, for all x in Y. Lemma 3.10.(i) P is an s-prime ideal of Y i P is a prime ideal of Y (11]). (ii) P is a prime ideal of Y i P is a fuzzy prime ideal,(where P is the characteristic function of P ) (3]). Lemma 3.113]. Let be a fuzzy prime ideal of Y, then (0) = 1. Lemma 3.13]. Let x; y Y and (0; 1]. Then X(x ) \ X(y )=X((x ^ y) ): Recall that a topological space is disconnected i it can be expressed as the union of two nonempty disjoint closed subsets. Theorem Let Y 6= f0;eg: Then the space X is disconnected proof. Let 0 6= x 6= e. First we show that V (x 1 ) and V ((Nx) 1 ) are non-empty(x 1 and (Nx) are two fuzzy points). By Lemma 3., there is a prime ideal P of Y such that <x>p. Now from Lemma 3.10 wehave P V(x 1 ). That is V (x 1 ) 6=.Since 0 6= x; ex 6= e and similarly we can prove that V ((Nx) 1 ) 6=. Now by Lemmas 3.1,3.9,3.11 we have X(x 1 ) \ X((Nx) 1 )=X((x ^ Nx) 1 ); = X(0 1 ); = ; : Hence V (x 1 ) S V ((Nx) 1 ) = X. We shall show that V (x 1 ) \ V ((Nx) 1 )=: (3) On the contrary, we get that (x) =(Nx)=1, for some X. Now from Denition.10(ii) we obtain that (e) minf(nx);(x)g = 1. In other words (x) =1,for all x X, which isa contradiction. Consequently (3) holds. Denition 3.14.A bounded commutative BCK-algebra (e 6= 0) is called an integral domain if x ^ y = 0 implies that x =0ory= 0, for all x; y in this BCK-algebra. Lemma 3.15.Let Y be an integral domain. Then X(x t ) = if and only if x = 0, for all x Y and t (0; 1]: Proof. Since Y is integral domain, f0g is an s-prime ideal. Thus by Lemma 3.10(i), f0g is a prime ideal. Hence from Lemma 3.10(ii), f0g is a fuzzy prime ideal. Now since f0g 6 X(x t ), we have x t f0g and f0g (x) =t>0. That is x =0. The proof of converse follows from Lemma Recall that a topological space is irreducible i the intersection of any two non-empty basic open sets is non-empty. Theorem 3.16.Let Y be an integral domain.then the topological space X T is irreducible. Proof.Let x; y Y and X(x t ) X(y t ) =, where t (0; 1]. Then X((x^y) t )=,by Lemma 3.1. Hence from Lemma 3.15 ;X(x t ) = or X(y t )=. Refrences 1] :J. Ahsan and A.B. Thaheem, on ideals in BCK-algebra, Math. seminar Notes, 5(1977) ] : H.Hadji-Abadi and M.M. Zahedi, Some results on fuzzy prime spectrum of a ring, Fuzzy sets and Systems 77(1996)

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