Some Common Fixed Points for Type(β ) Compatible Maps in an Intuitionistic Fuzzy Metric Space
- Author: Park Jong Seo
- Organization: Park Jong Seo
- Publish: International Journal of Fuzzy Logic and Intelligent Systems Volume 13, Issue2, p147~153, 25 June 2013
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ABSTRACT
Previously, Park et al. (2005) defined an intuitionistic fuzzy metric space and studied several fixed-point theories in this space. This paper provides definitions and describe the properties of type(
β ) compatible mappings, and prove some common fixed points for four self-mappings that are compatible with type(β ) in an intuitionistic fuzzy metric space. This paper also presents an example of a common fixed point that satisfies the conditions of Theorem 4.1 in an intuitionistic fuzzy metric space.
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KEYWORD
Compatible map , Type(β) compatible map , Fixed point
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Grabiec [1] demonstrated the Banach contraction theorem in the fuzzy metric spaces introduced by Kramosil and Michalek [2]. Park [3-5], Park and Kim [6] also proved a fixed-point theorem in a fuzzy metric space.
Recently, Park et al. [7] defined an intuitionistic fuzzy metric space while Park et al. [8] proved a fixed-point Banach theorem for the contractive mapping of a complete intuitionistic fuzzy metric space. Park et al. [9] defined a type(
α ) compatible map and obtained results for five mappings using a type(α ) compatibility map in intuitionistic fuzzy metric spaces. Furthermore, Park [10] introduced a type(β ) compatible mapping and proved some of the properties of the type(β ) compatibility mapping in an intuitionistic fuzzy metric space.This paper proves some common fixed points for four self-mappings that satisfy type(
β ) compatibility mapping in intuitionistic fuzzy metric space, while it also provides an example in the given conditions for an intuitionistic fuzzy metric space.First, some definitions and properties of the intuitionistic fuzzy metric space
X are provided, as follows.Let us recall ([11]) that a continuous
t ?norm is a binary operation * : [0, 1]×[0, 1] → [0, 1], which satisfies the following conditions: (a) * is commutative and associative; (b) * is continuous; (c)a * 1 =a for alla ∈ [0, 1]; (d)a *b ≤c *d whenevera ≤c andb ≤d (a ,b ,c ,d ∈ [0, 1]).Similarly, a continuous
t ?conorm is a binary operation ? : [0, 1] × [0, 1] → [0, 1], which satisfies the following conditions: (a) ? is commutative and associative; (b) ? is continuous; (c)a ? 0 =a for alla ∈ [0, 1]; (d)a ?b ≤c ?d whenevera ≤c andb ≤d (a ,b ,c ,d ∈ [0, 1]).[12] The 5?tuple (
X ,M ,N ,*,?) is said to be an intuitionistic fuzzy metric space ifX is an arbitrary set, * is a continuoust ?norm, ? is a continuoust ?conorm, andM ,N are fuzzy sets inX 2 × (0,∞), which satisfy the following conditions: for allx ,y ,z ∈X , such that(a) M(x, y, t) > 0,
(b) M(x, y, t) = 1 ⇔ x = y,
(c) M(x, y, t) = M(y, x, t),
(d) M(x, y, t) * M(y, z, s) ≤ M(x, z, t + s),
(e) M(x, y, ?) : (0,∞) → (0, 1] is continuous,
(f) N(x, y, t) > 0,
(g) N(x, y, t) = 0 ⇔ x = y,
(h) N(x, y, t) = N(y, x, t),
(i) N(x, y, t) ? N(y, z, s) ≥ N(x, z, t + s),
(j) N(x, y, ?) : (0,∞) → (0, 1] is continuous.
Note that (
M ,N ) is referred to as an intuitionistic fuzzy metric onX . The functionsM (x ,y ,t ) andN (x ,y ,t ) denote the degree of proximity and the degree of non-proximity betweenx andy with respect tot , respectively.[13] Let (
X ,d ) be a metric space. Denotea *b =ab anda ?b = min{1,a +b } for alla, b ∈ [0, 1] and letMd ,Nd be the fuzzy sets onX 2 × (0,∞), which are defined as follows :for
k ,m ,n ∈R +(m ≥ 1). Thus, (X ,Md ,Nd ,*,?) is an intuitionistic fuzzy metric space, i.e., the intuitionistic fuzzy metric space induced by the metricd .[13] Let
X be an intuitionistic fuzzy metric space.(a) {
xn } is said to be convergent to a pointx ∈X by limn →∞xn =x iffor all
t > 0.(b) {
xn } is a Cauchy sequence iffor all
t > 0 andp > 0.(c)
X is complete if every Cauchy sequence converges onX .In this paper,
X is considered to be the intuitionistic fuzzy metric space with the following condition:for all
x ,y ∈X andt > 0.[6] Let {
xn } be a sequence in an intuitionistic fuzzy metric spaceX with the condition (1). If there exists a numberk ∈ (0,1) such that for allx ,y ∈X andt > 0,for all
t > 0 andn = 1, 2, 3 …, then {xn } is a Cauchy sequence inX .[14] Let
X be an intuitionistic fuzzy metric space. If there exists a numberk ∈ (0,1) such that for allx ,y ∈X andt > 0,M(x, y, kt) ≥ M(x, y, t),
N(x, y, kt) ≤ N(x, y, t),
then
x =y .3. Properties of type(β) compatible mappings and an example
This section introduces type(
α ) and type(β ) compatible maps in an intuitionistic fuzzy metric space, and it also presents an example of the relations of type(β ) compatible maps.[14] Let
A ,B be mappings from the intuitionistic fuzzy metric spaceX into itself. These mappings are said to be compatible iffor all
t > 0, whenever {xn } ⊂X such that limn →∞Axn = limn →∞Bxn =x for somex ∈X .([10]) Let
A ,B be mappings from the intuitionistic fuzzy metric spaceX into itself. The mappings are said to be type(β ) compatible iffor all
t > 0, whenever {xn } ⊂X such that limn →∞Axn = limn →∞Bxn =x for somex ∈X .[10] Let
X be an intuitionistic fuzzy metric space andA ,B be the continuous mappings fromX into itself. Thus,A andB are compatible if they are type(β ) compatible.[10] Let
X be an intuitionistic fuzzy metric space andA ,B be mappings fromX into itself. IfA ,B are type(β ) compatible andAx =Bx for somex ∈X , thenABx =BBx =BAx =AAx .[10] Let
X be an intuitionistic fuzzy metric space andA ,B be type(β ) compatible mappings fromX into itself. Let {xn } ⊂X so limn →∞Axn = limn →∞Bxn =x for somex ∈X , then(a)limn→∞ BBxn = Ax if A is continuous at x ∈ X,
(b)limn→∞ AAxn = Bx if B is continuous at x ∈ X,
(c)ABx = BAx and Ax = Bx if A and B are continuous at x ∈ X.
Let
X = [0,∞) with the metricd defined byd (x ,y ) = |x ?y | and for eacht > 0, letMd ,Nd be fuzzy sets onX 2 × [0,∞), which are defined as followsfor all
x ,y ∈X . Clearly, (X ,Md ,Nd , *, ?) is an intuitionistic fuzzy metric space where *, ? are defined bya *b = min{a, b } anda ?b = max{a, b } for alla, b ∈ [0, 1]. Let us defineA ,B :X →X asThus,
A ,B are discontinuous atx = 1. Let {xn } ⊂X be defined byNext, we have lim
n →∞Axn = limn →∞Bxn = 1Furthermore,
and
Therefore,
A ,B are type(β ) compatible but they are not compatible.This section proves the main theorem and presents an example using the given conditions in an intuitionistic fuzzy metric space.
Let
X be a complete intuitionistic fuzzy metric space wheret *t ≥t ,t ?t ≤t for allt ∈ [0, 1]. LetA ,B ,S andT be mappings fromX into itself so:(a) AT(X) ∪ BS(X) ⊂ ST(X);
(b) there exists k ∈ (0, 1) so for all x, y ∈ X and t > 0,
M2(Ax,By,kt) * [M(Sx, Ax, kt)M(Ty,By, kt)]
*M2(Ty,By, kt) + aM(Ty,By, kt)M(Sx,By, 2kt)
≥ [pM(Sx, Ax, t) + qM(Sx, Ty, t)]M(Sx,By, 2kt),
N2(Ax,By,kt) ? [N(Sx, Ax, kt)N(Ty,By, kt)] ?N2(Ty,By, kt) + aM(Ty,By, kt)N(Sx,By, 2kt) ≤ [pN(Sx, Ax, t) + qN(Sx, Ty, t)]N(Sx,By, 2kt),
where 0 <
p ,q < 1, 0 ≤a < 1 such thatp +q ?a = 1;(c) S and T are continuous and ST = TS;
(d) the pairs (A, S) and (B, T) are type(β) compatible.
Thus,
A ,B ,S andT have a unique common fixed point inX .Proof. Letx 0 be an arbitrary point ofX . Using (a), we can construct an {xn } ⊂X as follows:ATx 2n =STx 2n +1,BSx 2n +1 =STx 2n +2,n = 0, 1, 2,…. Next, letzn =STxn . Using (b), we obtainM2(ATx2n,BSx2n+1, kt) * [M(STx2n, ATx2n, kt) ×M(TSx2n+1,BSx2n+1, kt)] *M2(TSx2n+1, BSx2n+1, kt) + aM(TSx2n+1,BSx2n+1, kt) ×M(STx2n,BSx2n+1, 2kt) ≥ [pM(STx2n+1, ATx2n, t) +qM(STx2n, TSx2n+1, t)] ×M(STx2n,BSx2n+1, 2kt),
N2(ATx2n,BSx2n+1, kt) ? [N(STx2n, ATx2n, kt) ×N(TSx2n+1,BSx2n+1, kt)] ? N2(TSx2n+1, BSx2n+1, kt) + aN(TSx2n+1,BSx2n+1, kt) ×N(STx2n,BSx2n+1, 2kt) ≤ [pN(STx2n+1, ATx2n, t) +qN(STx2n, TSx2n+1, t)] ×N(STx2n,BSx2n+1, 2kt)
and
M2(STx2n+1, STx2n+2, kt) * [M(z2n, STx2n+1, kt) ×M(z2n+1, STx2n+2, kt)] *M2(z2n+1, STx2n+2, kt) +aM(z2n+1, STx2n+2, kt)M(z2n, STx2n+2, 2kt) ≥ [pM(z2n, STx2n+1, t) + qM(z2n, z2n+1, t)] ×M(z2n, STx2n+2, 2kt),
N2(STx2n+1, STx2n+2, kt) ? [N(z2n, STx2n+1, kt) ×N(z2n+1, STx2n+2, kt)] ? N2(z2n+1, STx2n+2, kt) +aN(z2n+1, STx2n+2, kt)N(z2n, STx2n+2, 2kt) ≤ [pN(z2n, STx2n+1, t) + qN(z2n, z2n+1, t)] ×N(z2n, STx2n+2, 2kt).
Then,
M2(z2n+1, z2n+2, kt) *[M(z2n, z2n+1, kt)M(z2n+1, z2n+2, kt)] +aM(z2n+1, z2n+2, kt)M(z2n, z2n+2, 2kt) ≥ [p + q]M(z2n, z2n+1, t)M(z2n, z2n+2, 2kt),
N2(z2n+1, z2n+2, kt) ?[N(z2n, z2n+1, kt)N(z2n+1, z2n+2, kt)] +aN(z2n+1, z2n+2, kt)N(z2n, z2n+2, 2kt) ≤ [p + q]N(z2n, z2n+1, t)N(z2n, z2n+2, 2kt),
and
M2(z2n+1, z2n+2, kt)M(z2n+1, z2n+2, kt)] +aM(z2n+1, z2n+2, kt)M(z2n, z2n+2, 2kt) ≥ [p + q]M(z2n, z2n+1, t)M(z2n, z2n+2, 2kt),
N2(z2n+1, z2n+2, kt)N(z2n+1, z2n+2, kt)] +aN(z2n+1, z2n+2, kt)N(z2n, z2n+2, 2kt) ≤ [p + q]N(z2n, z2n+1, t)N(z2n, z2n+2, 2kt).
Therefore, it follows that
z2n+1M(z2n+1, z2n+2, kt) ≥ M(z2n, z2n+1, t),
N(z2n+1, z2n+2, kt) ≤ N(z2n, z2n+1, t)
for all
t > 0 andk ∈ (0, 1). In general, form = 1, 2, …, we haveM(zm+1, zm+2, kt) ≥ M(zm, zm+1, t),
N(zm+1, zm+2, kt) ≤ N(zm, zm+1, t)
Thus, {
zn } is a Cauchy sequence inX and, becauseX is complete, {zn } converges to a pointz ∈X . Since {ATx 2n }, {BSx 2n +1} are subsequences of {zn }, limn →∞ATx 2n =z = limn →∞BSx 2n +1.Let
yn =TXn ,un =Sxn forn = 1, 2,…. Thus, we haveAy 2n →z ,Sy 2n →z ,Tu 2n +1 →z andBu 2n +1 →z . Furthermore,M(AAy2n, SSy2n, t) → 1,
M(BBu2n+1, TT2n+1, t) → 1,
N(AAy2n, SSy2n, t) → 0,
N(BBu2n+1, TT2n+1, t) → 0
as
n →∞. Based on the continuity ofT and Proposition 3.4, we obtainTBu 2n +1 →Tz ,BBu 2n +1 →Tz .Next, by taking
x =y 2n ,y =Bu 2n +1 in (b), forn →∞ we obtain,M2(z, Tz, kt) * [M(z, z, kt)M(Tz, Tz, kt)] *M2(Tz, Tz, kt) + aM(Tz, Tz, kt)M(z, Tz, 2kt) ≥ [pM(z, z, t) + qM(z, Tz, t)]M(z, Tz, 2kt),
N2(z, Tz, kt) ? [N(z, z, kt)N(Tz, Tz, kt)] ?N2(Tz, Tz, kt) + aN(Tz, Tz, kt)N(z, Tz, 2kt) ≤ [pN(z, z, t) + qN(z, Tz, t)]N(z, Tz, 2kt),
then
M2(z, Tz, kt) + aM(z, Tz, 2kt) ≥ [p + qM(z, Tz, t)]M(z, Tz, 2kt), N2(z, Tz, kt) ≤ qN(z, Tz, t)N(z, Tz, 2kt).
Since
M (x ,y , ?) is nondecreasing andN (x ,y , ?) is nonincreasing for allx ,y ∈X , we obtainM(z, Tz, kt) + a ≥ p + qM(z, Tz, t),
N(z, Tz, kt) ≤ qN(z, Tz, t)
and
Thus,
z =Tz . Similarly, we havez =Sz .Next, by taking
x =y 2n andy =z in condition (b), forn →∞ we obtainM(z,Bz, kt) *M(z,Bz, kt) +aM(z,Bz, kt)M(z,Bz, 2kt) ≥ (p + q)M(z,Bz, 2kt),
N(z,Bz, kt) ? N(z,Bz, kt) +aN(z,Bz, kt)N(z,Bz, 2kt) ≤ 0.
Thus,
M(z,Bz, kt) + aM(z,Bz, kt) ≥ p + q,
N(z,Bz, kt) + aN(z,Bz, kt) ≤ 0.
Therefore,
M(z,Bz, kt) ≥ 1,
N(z,Bz, kt) ≤ 0
for all
t > 0 andk ∈ (0, 1). Thus,z =Bz . Similarly, we obtainz =Az . Therefore,z is a common fixed point ofA ,B ,S andT .Let
w be another common fixed point ofA ,B ,S andT .Using condition (b), we have
M2(z,w, kt) * [M(z, z, kt)M(w,w, kt)] *M2(w,w, kt) + aM(w,w, kt)M(z,w, 2kt) ≥ [pM(z, z, t) + qM(z,w, t)]M(z,w, 2kt),
N2(z,w, kt) ? [N(z, z, kt)N(w,w, kt)] ?N2(w,w, kt) + aN(w,w, kt)N(z,w, 2kt) ≤ [pN(z, z, t) + qN(z,w, t)]M(z,w, 2kt).
Thus,
M2(z,w, kt) +M(z,w, 2kt) ≥ (p + qM(z,w, t))M(z,w, 2kt), N2(z,w, kt) ≤ qM(z,w, t)M(z,w, 2kt),
Therefore,
M(z,w, kt) ≤ M(z,w, 2kt),
N(z,w, kt) ≥ N(z,w, 2kt),
so
Thus,
z =w . This means thatA ,B ,S andT have a unique common fixed point.Let
X be a complete intuitionistic fuzzy metric space wheret *t ≥t ,t ?t ≤t for allt ∈ [0, 1] and letA ,B be mappings fromX into itself such that:(e) A(X) ⊂ S(X),
(f) there exists k ∈ (0,1) so for all x, y ∈ X and t > 0,
M2(Ax, Ay, kt) * [M(Sx, Ax, kt)M(Sy, Ay, kt)] M2(Sy, Ay, kt) + aM(Sy, Ay, kt)M(Sx, Ay, 2kt) ≥ [pM(Sx, Ax, t) + qM(Sx, Sy, t)]M(Sx, Ay, 2kt),
N2(Ax, Ay, kt) ? [N(Sx, Ax, kt)N(Sy, Ay, kt)kt)] ?N2(Sy, Ay, kt) + aM(Sy, Ay, kt)N(Sx, Ay, 2kt) ≤ [pN(Sx, Ax, t) + qN(Sx, Sy, t)]N(Sx, Ay, 2kt),
where 0 <
p ,q < 1, 0 ≤a < 1 such thatp +q ?a = 1,(g) S is continuous,
(h) A and S are type(β) compatible.
Thus,
A andS have a unique common fixed point inX .Proof. Therefore, if we enterA =B andS =T into Theorem 4.1, all of the conditions of Theorem 4.1 are satisfied. Thus, the proof of this corollary follows from Theorem 4.1.Let
with the metric
d defined byd (x ,y ) = |x ?y | and for eacht > 0, letMd ,Nd be fuzzy sets onX 2 × [0,∞), which are defined as followsfor all
x ,y ∈X . Clearly, (X ,Md ,Nd , *, ?) is a complete intuitionistic fuzzy metric space where *, ? are defined bya *b = min{a, b } anda ?b = max{a, b } for alla, b ∈ [0, 1]. LetA ,B ,S andT be maps fromX into itself, which are defined byfor all
x ∈X . Then,Furthermore,
ST =TS andS ,T are continuous. If we takeand
t = 1, the condition (b) of Theorem 4.1 is satisfied. Moreover,A ,S are type(β ) compatible if limn →∞xn = 0 where {xn } ⊂X such that limn →∞Axn = limn →∞Sxn = 0 for some 0 ∈X .Similarly,
B ,T are type(β ) compatible. Thus,M(0,B0, kt) + aM(0,B0, kt) ≥ p + q,
N(0,B0, kt) + aN(0,B0, kt) ≤ 0.
Therefore,
M (0,B 0,kt ) ≥ 1 andN (0,B 0,kt ) ≤ 0 for all t > 0 andk ∈ (0, 1). Thus, 0 =B 0. Similarly, we obtain 0 =A 0. Therefore, 0 is a common fixed point ofA ,B ,S andT .Let w be another common fixed point of
A ,B ,S andT . Then,M2(0,w,kt) +M(0,w, 2kt) ≥ (p + qM(0,w,t))M(0,w,2kt),
M2(0,w, kt) ≤ qM(0,w, t)M(0,w, 2kt).
Therefore, because
M(0,w, kt) ≤ M(0,w, 2kt),
N(0,w, kt) ≥ N(0,w, 2kt),
Thus,
Therefore, 0 =
w . Thus,A ,B ,S andT have a unique common fixed point 0.Park et al. [7] defined an intuitionistic fuzzy metric space and Park et al. [8] proved a fixed-point Banach theorem for the contractive mapping of a complete intuitionistic fuzzy metric space. Park et al. [9] defined a type(
α ) compatible mapping and obtained results for five mappings using type(α ) compatibility in intuitionistic fuzzy metric spaces. Furthermore, Park [10] introduced type(β ) compatible mapping and proved some properties of type(β ) compatibility in an intuitionistic fuzzy metric space. In this paper, we proved some common fixed points for four self-mappings that satisfy type(β ) compatibility and we provided an example in the given conditions for an intuitionistic fuzzy metric space.This paper attempted to develop a method to provide a proof based on the fundamental concepts and properties defined in the new space. I think that the results of this paper will be extended to the intuitionistic M-fuzzy metric space and other spaces. Further research should be conducted to determine how to combine the collaborative learning algorithm with our proof method in the future.