YOUNG BAE JUN, KYUNG HO KIM, AND EUN HWAN ROH
Received 17 May 2005; Revised 30 March 2006; Accepted 4 April 2006
A condition for a strong hyperK-ideal to be a strong implicative hyperK-ideal is given. Homomorphic images and inverse images of strong implicative hyperK-ideals are con-sidered.
Copyright © 2006 Hindawi Publishing Corporation. All rights reserved.
1. Introduction
The study of BCK-algebras was initiated by K. Is´eki in 1966 as a generalization of the concept of set-theoretic difference and propositional calculus. The hyperstructure the-ory (called also multialgebras) is introduced in 1934 by Marty [6] at the 8th congress of Scandinavian Mathematicians. Around the 40’s, several authors worked on hypergroups, especially in France and in the United States, but also in Italy, Russia, Japan, and Iran. Hyperstructures have many applications to several sectors of both pure and applied sci-ences. Recently in [2] Borzoei et al. and Borzoei and Zahedi [3] constructed the hyper K-algebras, and studied (positive implicative) hyperK-ideals in hyperK-algebras. In [4] Jun and Roh studied strong hyperK-ideals in hyperK-algebras. Also Jun et al. [5] intro-duced the notion ofs-implicative hyperK-ideal of a hyperK-algebra, and investigated related properties. As a continuation of [5], in this paper, we find a condition for a strong hyperK-ideal to be a strong implicative hyperK-ideal. We also consider homomorphic images and inverse images of strong implicative hyperK-ideals.
2. Preliminaries
We include some elementary aspects of hyperK-algebras that are necessary for this paper, and for more details we refer to [2,7]. LetH be a nonempty set endowed with a hyper operation “◦,” that is,◦is a function from H×H toᏼ∗(H)=ᏼ(H)\ {∅}. For two subsetsAandBofH, denote byA◦Bthe seta∈A,b∈Ba◦b.
By a hyperI-algebra we mean a nonempty setHendowed with a hyper operation “◦” and a constant 0 satisfying the following axioms:
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International Journal of Mathematics and Mathematical Sciences Volume 2006, Article ID 86316, Pages1–6
(H1) (x◦z)◦(y◦z)< x◦y, (H2) (x◦y)◦z=(x◦z)◦y, (H3)x < x,
(H4)x < yandy < ximplyx=y,
for allx,y,z∈H, wherex < y is defined by 0∈x◦y and for everyA,B⊆H,A < B is defined by∃a∈Aand∃b∈Bsuch thata < b. If a hyperI-algebra (H,◦, 0) satisfies an additional condition:
(H5) 0< xfor allx∈H,
then (H,◦, 0) is called a hyperK-algebra (see [2]).
In a hyperI-algebraH, the following hold (see [2, Proposition 3.4]): (a1) (A◦B)◦C=(A◦C)◦B,
(a2)x◦(x◦y)< y, (a3)x◦y < z⇔x◦z < y, (a4)A◦B < C⇔A◦C < B, (a5) (x◦z)◦(x◦y)< y◦z, (a6) (A◦C)◦(B◦C)< A◦B, (a7)A◦(A◦B)< B,
(a8)A < A,
(a9)A⊆BimpliesA < B,
for allx,y,z∈Hand for all nonempty subsetsA,B, andCofH.
A nonempty subsetIof a hyperK-algebraHis called a weak hyperK-ideal ofH(see [2]) if it satisfies
(I1) 0∈I,
(I2) (for allx,y∈H) (x◦y⊆I, y∈I⇒x∈I).
A nonempty subsetIof a hyperK-algebraHis called a hyperK-ideal ofH(see [2]) if it satisfies (I1) and
(∀x,y∈H) (x◦y < I, y∈I=⇒x∈I). (2.1)
Note that every hyperK-ideal is a weak hyperK-ideal (see [1]). A nonempty subsetI of a hyperK-algebraHis called a strong hyperK-ideal ofH(see [4]) if it satisfies (I1) and
(∀x,y∈H) (x◦y)∩I= ∅, y∈I=⇒x∈I. (2.2)
Note that every strong hyperK-ideal is a hyperK-ideal (see [4]).
A nonempty subsetIof a hyperK-algebraHis called a weak implicative hyperK-ideal ofH(see [1]) if it satisfies (I1) and
(∀x,y,z∈H) (x◦z)◦(y◦x)⊆I,z∈I=⇒x∈I. (2.3)
A nonempty subsetI of a hyperK-algebraHis called an implicative hyperK-ideal of H(see [1]) if it satisfies (I1) and
(∀x,y,z∈H) (x◦z)◦(y◦x)< I,z∈I=⇒x∈I. (2.4)
Table 3.1
◦ 0 a b c
0 {0} {0} {0} {0}
a {a} {0} {0} {0}
b {b} {b} {0} {b}
c {c} {a,b} {0,a} {0,b}
3. Strong implicative hyperK-ideals
In what follows letH denote a hyper K-algebra unless otherwise specified. In [5], Jun et al. introduced the notion ofs-implicative hyperK-ideal ofHas follows.
Defintion 3.1 [5]. A nonempty subsetIofHis called ans-implicative hyperK-ideal ofH if it satisfies (I1) and
(∀x,y,z∈H) (x◦z)◦(y◦x)∩I= ∅,z∈I=⇒x∈I. (3.1)
In this paper it is called a strong implicative hyperK-ideal ofH.
Example 3.2. LetH= {0,a,b,c}be a hyperK-algebra with the Cayley table (Table 3.1). ThenI:= {0,a,c}is a strong implicative hyperK-ideal ofH. ButI:= {0,a}is not a strong implicative hyperK-ideal ofHsince ((c◦a)◦(0◦c))∩I= {a}anda∈Ibutc∈I. Theorem 3.3. If{Iλ|λ∈Λ}is a family of strong implicative hyperK-ideals of H, then
λ∈ΛIλis a strong implicative hyperK-ideal ofH.
Proof. Clearly 0∈λ∈ΛIλ. Letx,y,z∈Hbe such that ((x◦z)◦(y◦x))∩(λ∈ΛIλ)= ∅ andz∈λ∈ΛIλ. Then ((x◦z)◦(y◦x))∩Iλ= ∅andz∈Iλfor allλ∈Λ. By using (3.1), we havex∈Iλfor allλ∈Λ, and hencex∈λ∈ΛIλ. Henceλ∈ΛIλis a strong implicative
hyperK-ideal ofH.
Lemma 3.4 [5]. Every strong implicative hyperK-ideal ofHis a strong hyperK-ideal. Proposition 3.5. LetIbe a strong hyperK-ideal ofH, andA,B⊆H. If (A◦B)∩I= ∅ andB⊆I, thenA < I.
Proof. SinceA◦B=a∈A,b∈Ba◦band (A◦B)∩I= ∅, there existst∈a◦b for some a∈A,b∈Bsuch thatt∈I. Hence (a◦b)∩I= ∅. SinceIis a strong hyperK-ideal and b∈B⊆I, we havea∈I. Therefore we getA < I.
Now we give a condition for a strong hyperK-ideal to be a strong implicative hyper K-ideal.
Theorem 3.6. LetIbe a strong hyperK-ideal ofHsuch that
(∀x,y∈H) x◦(y◦x)< I=⇒ x∈I. (3.2)
Table 3.2
◦ 0 a b
0 {0} {0} {0}
a {a} {0,a} {a}
b {a,b} {0,a} {0,a}
Proof. Letx,y,z∈H be such that ((x◦z)◦(y◦x))∩I= ∅and z∈I. Then we have ((x◦(y◦x))◦z)∩I= ∅. It follows fromProposition 3.5 that x◦(y◦x)< I so from (3.2) thatx∈I. ThereforeIis a strong implicative hyperK-ideal ofH. Theorem 3.7. Every strong implicative hyperK-idealIofHsatisfies (3.2).
Proof. LetI be a strong implicative hyper K-ideal of H and letx,y∈H be such that x◦(y◦x)< I. The inclusionx◦(y◦x)⊆(x◦0)◦(y◦x) implies that (x◦0)◦(y◦x)< I.
Hencex∈Isince 0∈I.
Corollary 3.8. LetI be a nonempty subset of H. Then I is a strong implicative hyper K-ideal ofHif and only ifIis a strong hyperK-ideal ofH that satisfies the condition (3.2). Theorem 3.9. Every strong implicative hyperK-ideal is a weak implicative hyperK-ideal.
Proof. Straightforward.
In general, a weak implicative hyperK-ideal may not be a strong implicative hyper K-ideal as seen in the following example.
Example 3.10. LetH= {0,a,b}be a hyperK-algebra with the Cayley table (Table 3.2). ThenI:= {0,b}is a weak implicative hyperK-ideal ofH, butIis not a strong implicative hyperK-ideal since ((a◦0)◦(a◦a))∩I= ∅and 0∈Ibuta∈I.
Our future work will focus on finding conditions for a weak implicative hyperK-ideal to be a strong implicative hyperK-ideal.
4. Homomorphisms of hyperK-algebras
Defintion 4.1 [7]. LetH1andH2be two hyperK-algebras. A mappingf :H1→H2is said
to be a homomorphism if it satisfies (i) f(0)=0,
(ii) (for allx,y∈H1) (f(x◦y)=f(x)◦f(y)).
Moreover if f is 1−1 (or onto), we say that f is a monomorphism (or epimorphism). And if f is both 1−1 and onto, we say that f is an isomorphism.
Proposition 4.2. Let f :H1→H2be a homomorphism of hyperK-algebras. Then
∀A,B⊆H1 A < B=⇒f(A)< f(B)
. (4.1)
Proof. LetA,B⊆H1be such that A < B. Then∃a∈A,∃b∈Bsuch thata < b, that is,
Theorem 4.3. Let f :H1→H2 be a homomorphism of hyperK-algebras. IfI is a strong
hyperK-ideal ofH2, thenf−1(I) is a strong hyperK-ideal ofH1.
Proof. LetI be a strong hyperK-ideal ofH2. Clearly 0∈ f−1(I). Letx,y∈H1 be such
that (x◦y)∩f−1(I)= ∅andy∈ f−1(I). Then we have
∅ = f(x◦y)∩f−1(I)⊆f(x◦y)∩ff−1(I)⊆f(x)◦f(y)∩I,
(4.2)
and so (f(x)◦f(y))∩I= ∅and f(y)∈ f(f−1(I))⊆I. SinceIis a strong hyperK-ideal
ofH2, we have f(x)∈Iand sox∈f−1(I). Therefore f−1(I) is a a strong hyperK-ideal
ofH1.
Theorem 4.4. Let f :H1→H2 be a homomorphism of hyperK-algebras. IfI is a strong
implicative hyperK-ideal ofH2, thenf−1(I) is a strong implicative hyperK-ideal ofH1.
Proof. The proof is similar to the proof ofTheorem 4.3by some modification. Theorem 4.5. Let f :H1→H2 be a homomorphism of hyperK-algebras. Then kerf :=
{x∈H1|f(x)=0}is a strong hyperK-ideal ofH1.
Proof. First we show that{0}is a strong hyperK-ideal ofH2. To do this, letx,y∈H1be
such that (x◦y)∩ {0} = ∅andy∈ {0}. Theny=0 and so 0∈x◦0 since (x◦0)∩ {0} = ∅. Thus we havex <0. By (H4) and (H5), we getx=0∈ {0}. This shows that{0}is a strong hyperK-ideal ofH2. It follows fromTheorem 4.3that kerf = f−1({0}) is a strong
hyperK-ideal ofH1.
In general, a strong hyperK-ideal{0}may not be a strong implicative hyperK-ideal. For example, consider the hyperK-algebraH of Example 3.2. Clearly {0} is a strong hyperK-ideal ofH, while it is not a strong implicative hyperK-ideal ofHsince ((c◦0)◦ (b◦c))∩ {0} = ∅and 0∈ {0}butc∈ {0}.
Theorem 4.6. Let f :H1→H2be a homomorphism of hyperK-algebras. Iff is onto andI
is a strong hyperK-ideal ofH1which contains kerf, then f(I) is a strong hyperK-ideal of
H2.
Proof. LetI be a strong hyper K-ideal ofH1. Clearly 0∈ f(I). Let x,y∈H2 be such
that (x◦y)∩f(I)= ∅andy∈ f(I). Sincey∈ f(I) and f is onto, there are y1∈Iand
x1∈H1such thaty=f(y1) andx=f(x1). Thus
∅ =(x◦y)∩f(I)=f(x1◦y1)∩f(I), (4.3)
and so there existsa∈H2such thata∈f(x1◦y1) anda∈f(I). It follows that there are
a1∈x1◦y1andb1∈Isuch thata= f(a1) anda= f(b1) so that
0∈a◦a=fa1
◦fb1
=fa1◦b1
, (4.4)
which implies that f(c)=0 for somec∈a1◦b1. Hencec∈kerf ⊆I and so (a1◦b1)∩
I= ∅. Now sinceI is a strong hyperK-ideal ofH1andb1∈I, we geta1∈I. Thus (x1◦
y1)∩I= ∅, which implies thatx1∈I. Therebyx=f(x1)∈f(I), and so f(I) is a strong
Theorem 4.7. Let f :H1→H2be a homomorphism of hyperK-algebras. If f is onto and
I is a strong implicative hyperK-ideal of H1 which contains kerf, then f(I) is a strong
implicative hyperK-ideal ofH2.
Proof. The proof is similar to the proof ofTheorem 4.6. The following theorem is straightforward, and so we omit the proof.
Theorem 4.8. Let f :H1→H2be an epimorphism of hyper K-algebras. Then there is a
one to one correspondence between the set of all strong (implicative) hyperK-ideals ofH1
containing kerf and the set of all strong (implicative) hyperK-ideals ofH2.
References
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Young Bae Jun: Department of Mathematics Education, Gyeongsang National University, Jinju 660-701, South Korea
E-mail address:[email protected]
Kyung Ho Kim: Department of Mathematics, Chungju National University, Chungju 380-702, South Korea
E-mail address:[email protected]
Eun Hwan Roh: Department of Mathematics Education, Chinju National University of Education, Jinju 660-756, South Korea
