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ON FUZZY DOT SUBALGEBRAS OF BCH-ALGEBRAS
SUNG MIN HONG, YOUNG BAE JUN, SEON JEONG KIM, and GWANG IL KIM
(Received 21 December 2000)
Abstract.We introduce the notion of fuzzy dot subalgebras in BCH-algebras, and study its various properties.
2000 Mathematics Subject Classification. 06F35, 03G25, 03E72.
1. Introduction. In [4], Hu and Li introduced the notion of BCH-algebras which are a generalization of BCK/BCI-algebras. In 1965, Zadeh [6] introduced the concept of fuzzy subsets. Since then several researchers have applied this notion to various mathematical disciplines. Jun [5] applied it to BCH-algebras, and he considered the fuzzification of ideals and filters in BCH-algebras. In this paper, we introduce the notion of a fuzzy dot subalgebra of a BCH-algebra as a generalization of a fuzzy subalgebra of a BCH-algebra, and then we investigate several basic properties related to fuzzy dot subalgebras.
2. Preliminaries. A BCH-algebra is an algebra(X,∗,0)of type(2,0)satisfying the following conditions:
(i) x∗x=0,
(ii) x∗y=0=y∗ximpliesx=y,
(iii) (x∗y)∗z=(x∗z)∗yfor allx, y, z∈X.
In any BCH-algebraX, the following hold (see [2]):
(P1) x∗0=x,
(P2) x∗0=0 impliesx=0, (P3) 0∗(x∗y)=(0∗x)∗(0∗y).
A BCH-algebraXis said to bemedialifx∗(x∗y)=yfor allx, y∈X. A nonempty subsetSof a BCH-algebraXis called asubalgebraofXifx∗y∈Swheneverx, y∈ S. A map f from a BCH-algebraXto a BCH-algebraY is called ahomomorphismif f (x∗y)=f (x)∗f (y)for allx, y∈X.
We now review some fuzzy logic concepts. A fuzzy subset of a setXis a function µ:X→[0,1]. For any fuzzy subsetsµandνof a setX, we define
µ⊆ν⇐⇒µ(x)≤ν(x) ∀x∈X, (µ∩ν)(x)=min
µ(x), ν(x)
∀x∈X. (2.1)
Letf:X→Y be a function from a setXto a setY and letµ be a fuzzy subset ofX.
The fuzzy subsetνofY defined by
ν(y):=
supx∈f−1(y)µ(x) iff−1(y)= ∅,∀y∈Y ,
0 otherwise, (2.2)
is called theimageofµ underf, denoted byf [µ]. Ifν is a fuzzy subset ofY, the fuzzy subsetµ ofXgiven byµ(x)=ν(f (x))for allx∈Xis called thepreimageof νunderf and is denoted byf−1[ν].
A fuzzy relationµ on a set X is a fuzzy subset ofX×X, that is, a mapµ:X× X→[0,1]. A fuzzy subsetµ of a BCH-algebraXis called afuzzy subalgebraofXif µ(x∗y)≥min{µ(x), µ(y)}for allx, y∈X.
3. Fuzzy product subalgebras. In what follows letXdenote a BCH-algebra unless otherwise specified.
Definition3.1. A fuzzy subsetµ of X is called afuzzy dot subalgebraofX if µ(x∗y)≥µ(x)·µ(y)for allx, y∈X.
Example3.2. Consider a BCH-algebraX= {0, a, b, c}having the following Cayley table (see [1]):
∗ 0 a b c
0 0 0 0 0
a a 0 0 a
b b c 0 c
c c 0 0 0
Define a fuzzy setµinXbyµ(0)=0.5,µ(a)=0.6,µ(b)=0.4,µ(c)=0.3. It is easy to verify thatµis a fuzzy dot subalgebra ofX.
Note that every fuzzy subalgebra is a fuzzy dot subalgebra, but the converse is not true. In fact, the fuzzy dot subalgebraµinExample 3.2is not a fuzzy subalgebra since
µ(a∗a)=µ(0)=0.5<0.6=µ(a)=min
µ(a), µ(a)
. (3.1)
Proposition3.3. Ifµis a fuzzy dot subalgebra ofX, then µ(0)≥
µ(x)2
, µ 0n∗x
≥
µ(x)2n+1, (3.2)
for allx∈Xandn∈Nwhere0n∗x=0∗(0∗(···(0∗x)···))in which0occursn times.
Proof. Sincex∗x=0 for allx∈X, it follows that µ(0)=µ(x∗x)≥µ(x)·µ(x)=
µ(x)2
(3.3) for allx∈X.The proof of the second part is by induction onn. Forn=1, we have µ(0∗x)≥µ(0)·µ(x)≥(µ(x))3for allx∈X. Assume thatµ(0k∗x)≥(µ(x))2k+1for
allx∈X. Then µ
0k+1∗x
=µ 0∗
0k∗x
≥µ(0)·µ 0k∗x
≥ µ(x)2
·
µ(x)2k+1
=
µ(x)2(k+1)+1
. (3.4)
Henceµ(0n∗x)≥(µ(x))2n+1for allx∈Xandn∈N.
Proposition3.4. Letµbe a fuzzy dot subalgebra ofX. If there exists a sequence {xn}inXsuch thatlimn→∞(µ(xn))2=1, thenµ(0)=1.
Proof. According toProposition 3.3,µ(0)≥(µ(xn))2for eachn∈N. Since 1≥ µ(0)≥ lim
n→∞(µ(xn))2=1,it follows thatµ(0)=1.
Theorem3.5. Ifµandνare fuzzy dot subalgebras ofX, then so isµ∩ν.
Proof. Letx, y∈X, then (µ∩ν)(x∗y)=min
µ(x∗y), ν(x∗y)
≥min
µ(x)·µ(y), ν(x)·ν(y)
≥ min
µ(x), ν(x)
· min
µ(y), ν(y)
=
(µ∩ν)(x)
·
(µ∩ν)(y) .
(3.5)
Henceµ∩νis a fuzzy dot subalgebra ofX.
Note that a fuzzy subsetµofXis a fuzzy subalgebra ofXif and only if a nonempty level subset
U (µ;t):=
x∈X|µ(x)≥t
(3.6) is a subalgebra ofXfor everyt∈[0,1]. But, we know that ifµis a fuzzy dot subalgebra ofX, then there existst∈[0,1]such that
U (µ;t):=
x∈X|µ(x)≥t
(3.7) is not a subalgebra ofX. In fact, ifµis the fuzzy dot subalgebra ofXinExample 3.2, then
U (µ; 0.4)=
x∈X|µ(x)≥0.4
= {0, a, b} (3.8)
is not a subalgebra ofXsinceb∗a=c∉U (µ; 0.4).
Theorem3.6. Ifµis a fuzzy dot subalgebra ofX, then U (µ; 1):=
x∈X|µ(x)=1
(3.9) is either empty or is a subalgebra ofX.
Proof. If x and y belong to U (µ; 1), then µ(x∗y)≥ µ(x)·µ(y)= 1. Hence µ(x∗y)=1 which impliesx∗y∈U (µ; 1). Consequently,U (µ; 1)is a subalgebra ofX.
Theorem 3.7. LetX be a medial BCH-algebra and let µ be a fuzzy subset ofX such that
µ(0∗x)≥µ(x), µ
x∗(0∗y)
≥µ(x)·µ(y), (3.10) for allx, y∈X. Thenµis a fuzzy dot subalgebra ofX.
Proof. SinceXis medial, we have 0∗(0∗y)=yfor ally∈X. Hence µ(x∗y)=µ
x∗
0∗(0∗y)
≥µ(x)·µ(0∗y)≥µ(x)·µ(y) (3.11) for allx, y∈X. Thereforeµis a fuzzy dot subalgebra ofX.
Theorem3.8. Letg:X→Y be a homomorphism of BCH-algebras. Ifν is a fuzzy dot subalgebra ofY, then the preimageg−1[ν]ofνundergis a fuzzy dot subalgebra ofX.
Proof. For anyx1, x2∈X, we have g−1[ν]
x1∗x2
=ν g
x1∗x2
=ν g
x1
∗g x2
≥ν g
x1
·ν g
x2
=g−1[ν]
x1
·g−1[ν]
x2
. (3.12)
Thusg−1[ν]is a fuzzy dot subalgebra ofX.
Theorem3.9. Letf:X→Y be an onto homomorphism of BCH-algebras. Ifµ is a fuzzy dot subalgebra ofX, then the imagef [µ]ofµunderfis a fuzzy dot subalgebra ofY.
Proof. For anyy1, y2∈Y, letA1=f−1(y1), A2=f−1(y2), andA12=f−1(y1∗ y2).Consider the set
A1∗A2:=
x∈X|x=a1∗a2for somea1∈A1, a2∈A2
. (3.13)
Ifx∈A1∗A2, thenx=x1∗x2for somex1∈A1andx2∈A2so that f (x)=f
x1∗x2
=f x1
∗f x2
=y1∗y2, (3.14)
that is,x∈f−1(y1∗y2)=A12.HenceA1∗A2⊆A12. It follows that f [µ]
y1∗y2
= sup
x∈f−1(y1∗y2)
µ(x)= sup
x∈A12
µ(x)
≥ sup
x∈A1∗A2
µ(x)≥ sup
x1∈A1, x2∈A2
µ
x1∗x2
≥ sup
x1∈A1, x2∈A2
µ x1
·µ x2
.
(3.15)
Since·:[0,1]×[0,1]→[0,1] is continuous, for every ε >0 there exists δ >0 such that if ˜x1 ≥supx1∈A1µ(x1)−δ and ˜x2 ≥supx2∈A2µ(x2)−δ, then ˜x1·x˜2≥ supx1∈A1µ(x1)·supx2∈A2µ(x2)−ε.Choosea1∈A1and a2∈A2 such thatµ(a1)≥
supx1∈A1µ(x1)−δandµ(a2)≥supx2∈A2µ(x2)−δ.Then µ
a1
·µ a2
≥ sup
x1∈A1
µ x1
· sup
x2∈A2
µ x2
−ε. (3.16)
Consequently,
f [µ]
y1∗y2
≥ sup
x1∈A1, x2∈A2
µ x1
·µ x2
≥ sup
x1∈A1
µ x1
· sup
x2∈A2
µ x2
=f [µ]
y1
·f [µ]
y2
,
(3.17)
and hencef [µ]is a fuzzy dot subalgebra ofY.
Definition3.10. Letσ be a fuzzy subset ofX. Thestrongest fuzzyσ-relationon Xis the fuzzy subsetµσ ofX×Xgiven byµσ(x, y)=σ (x)·σ (y)for allx, y∈X.
Theorem3.11. Letµσ be the strongest fuzzyσ-relation onX, whereσ is a fuzzy subset ofX. Ifσ is a fuzzy dot subalgebra ofX, thenµσ is a fuzzy dot subalgebra of X×X.
Proof. Assume thatσ is a fuzzy dot subalgebra ofX. For anyx1, x2, y1, y2∈X, we have
µσ
x1, y1
∗ x2, y2
=µσ
x1∗x2, y1∗y2
=σ x1∗x2
·σ y1∗y2
≥ σ
x1
·σ x2
· σ
y1
·σ y2
= σ
x1
·σ y1
· σ
x2
·σ y2
=µσ
x1, y1
·µσ
x2, y2
,
(3.18)
and soµσ is a fuzzy dot subalgebra ofX×X.
Definition3.12. Letσbe a fuzzy subset ofX. A fuzzy relationµonXis called a fuzzyσ-product relationifµ(x, y)≥σ (x)·σ (y)for allx, y∈X.
Definition3.13. Letσbe a fuzzy subset ofX. A fuzzy relationµonXis called a left fuzzy relationonσ ifµ(x, y)=σ (x)for allx, y∈X.
Similarly, we can define a right fuzzy relation onσ. Note that a left (resp., right) fuzzy relation onσ is a fuzzyσ-product relation.
Theorem3.14. Letµbe a left fuzzy relation on a fuzzy subsetσofX. Ifµis a fuzzy dot subalgebra ofX×X, thenσ is a fuzzy dot subalgebra ofX.
Proof. Assume that a left fuzzy relationµonσis a fuzzy dot subalgebra ofX×X.
Then
σ x1∗x2
=µ
x1∗x2, y1∗y2
=µ x1, y1
∗ x2, y2
≥µ x1, y1
·µ x2, y2
=σ x1
·σ x2
(3.19)
for allx1, x2, y1, y2∈X. Henceσ is a fuzzy dot subalgebra ofX.
Theorem3.15. Letµ be a fuzzy relation onX satisfying the inequalityµ(x, y)≤ µ(x,0)for allx, y∈X. Givenz∈X, letσzbe a fuzzy subset ofXdefined byσz(x)= µ(x, z)for allx∈X. Ifµ is a fuzzy dot subalgebra ofX×X, thenσz is a fuzzy dot subalgebra ofXfor allz∈X.
Proof. Letz, x, y∈X, then
σz(x∗y)=µ(x∗y, z)=µ(x∗y, z∗0)
=µ
(x, z)∗(y,0)
≥µ(x, z)·µ(y,0)
≥µ(x, z)·µ(y, z)=σz(x)·σz(y),
(3.20)
completing the proof.
Theorem3.16. Letµ be a fuzzy relation onX and letσµ be a fuzzy subset ofX given byσµ(x)=infy∈Xµ(x, y)·µ(y, x)for allx∈X. Ifµis a fuzzy dot subalgebra ofX×Xsatisfying the equalityµ(x,0)=1=µ(0, x)for allx∈X, thenσµis a fuzzy dot subalgebra ofX.
Proof. For anyx, y, z∈X, we have
µ(x∗y, z)=µ(x∗y, z∗0)=µ
(x, z)∗(y,0)
≥µ(x, z)·µ(y,0)=µ(x, z), µ(z, x∗y)=µ(z∗0, x∗y)=µ
(z, x)∗(0, y)
≥µ(z, x)·µ(0, y)=µ(z, x).
(3.21)
It follows that
µ(x∗y, z)·µ(z, x∗y)≥µ(x, z)·µ(z, x)
≥
µ(x, z)·µ(z, x)
·
µ(y, z)·µ(z, y) (3.22) so that
σµ(x∗y)=inf
z∈Xµ(x∗y, z)·µ(z, x∗y)
≥
z∈Xinfµ(x, z)·µ(z, x) ·
z∈Xinfµ(y, z)·µ(z, y)
=σµ(x)·σµ(y).
(3.23)
This completes the proof.
Definition3.17(see Choudhury et al. [3]). Afuzzy mapffrom a setXto a setY is an ordinary map fromXto the set of all fuzzy subsets ofYsatisfying the following conditions:
(C1) for allx∈X, there existsyx∈Xsuch that(f (x))(yx)=1, (C2) for allx∈X,f (x)(y1)=f (x)(y2)impliesy1=y2.
One observes that a fuzzy mapffromXtoY gives rise to a unique ordinary map µf :X×X→I, given byµf(x, y)=f (x)(y).One also notes that a fuzzy map fromX toY gives a unique ordinary mapf1:X→Y defined asf1(x)=yx.
Definition3.18. A fuzzy mapffrom a BCH-algebraXto a BCH-algebraYis called afuzzy homomorphismif
µf
x1∗x2, y
= sup
y=y1∗y2
µf
x1, y1
·µf
x2, y2
(3.24)
for allx1, x2∈Xandy∈Y.
One notes that iff is an ordinary map, then the above definition reduces to an ordinary homomorphism. One also observes that if a fuzzy mapf is a fuzzy homo- morphism, then the induced ordinary mapf1is an ordinary homomorphism.
Proposition3.19. Letf:X→Y be a fuzzy homomorphism of BCH-algebras. Then (i) µf(x1∗x2, y1∗y2)≥µf(x1, y1)·µf(x2, y2)for allx1, x2∈Xandy1, y2∈Y. (ii) µf(0,0)=1.
(iii) µf(0∗x,0∗y)≥µf(x, y)for allx∈Xandy∈Y.
(iv) ifY is medial andµf(x, y)=t=0, thenµf(0, yx∗y)=tfor allx∈X and y∈Y, whereyx∈Y withµf(x, yx)=1.
Proof. (i) For everyx1, x2∈Xandy1, y2∈Y, we have µf
x1∗x2, y1∗y2
= sup
y1∗y2=y˜1∗y˜2
µf
x1,y˜1
·µf
x2,y˜2
≥µf
x1, y1
·µf
x2, y2
.
(3.25)
(ii) Letx∈Xandyx∈Y be such thatµf(x, yx)=1.Using (I) and (i), we get µf(0,0)=µf
x∗x, yx∗yx
≥µf x, yx
·µf x, yx
=1 (3.26)
and soµf(0,0)=1.
(iii) The proof follows from (i) and (ii).
(iv) Assume thatY is medial andµf(x, y)=t=0 for allx∈Xandy∈Y, and let yx∈Y be such thatµf(x, yx)=1. Then
µf
0, yx∗y
=µf
x∗x, yx∗y
≥µf
x, yx
·µf(x, y)
=t=µf(x, y)=µf
x∗0, yx∗
yx∗y
≥µf
x, yx
·µf
0, yx∗y
=µf
0, yx∗y ,
(3.27)
and henceµf(0, yx∗y)=t. This completes the proof.
Acknowledgement. This work was supported by Korea Research Foundation Grant (KRF-99-005-D00003).
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Sung Min Hong: Department of Mathematics, Gyeongsang National University, Chinju,660-701, Korea
E-mail address:[email protected]
Young Bae Jun: Department of Mathematics Education, Gyeongsang National Uni- versity, Chinju660-701, Korea
E-mail address:[email protected]
Seon Jeong Kim: Department of Mathematics, Gyeongsang National University, Chinju660-701, Korea
E-mail address:[email protected]
Gwang Il Kim: Department of Mathematics, Gyeongsang National University, Chinju 660-701, Korea
E-mail address:[email protected]
