International Journal of Engineering Trends and Technology- Volume4Issue1- 2013
Semicompatibility and Fixed point theorem in Fuzzy Metric
Space by using Control Function
Amardeep Singh*, Alok Asati **, Bharat Singh***
*Govt. Motilal Vigyan Mahavidyalaya, Bhopal (India).
**Jagadguru Dattatray College of Technology, Indore (India)
*** School Of Computer & Electronics, IPS Academy, Indore (India)
Abstract—
This paper presents some common fixed
point theorem for semicompatiable mapping in
fuzzy metric space. Our result generalizes the result
of Cho. [1].
. Mathematical Subject Classification: Primary
47H10, Secondary 54H25.
III.
M (x, y, t) = 1 for all t > 0 if and
only if x=y,
IV.
M (x, y,.) : [0,  )  [0, 1] is
left continuous,
V.
M (x, y, t)  M(y, z, s)  M (x, z,
t + s),
VI.
lim t  M (x,y,t ) = 1,
Keywords: Common Fixed Points, FuzzyMetric
Then M is called a fuzzy metric on X. Then M(x, y,
Space, Semi-compatible map, Control function.
t) denotes the degree of nearness between x and y
with respect to t.
Defination 2.3 [18] Let (X, d) be a metric space.
I. INTRODUCTION
Denote a  b = ab for all a, b  [0, 1] and let M be
fuzzy sets then the fuzzy matric space x2  [0,  ]
AFTER THE INTRODUCTION OF A FUZZY SET BY
t
ZADEH [16] SEVERAL RESEARCHER CONDUCTED
on defined as follows M :(x, y, t) =
.
t  d ( x, y )
ON GENERALIZATION OF THE CONCEPT OF A FUZZY
Definition 2.4 [18] Let (X, M,  ) be a fuzzy metric
SET.THE CONCEPT OF FUZZY METRIC SPACE HAS
space. Then
BEEN INTRODUCED AND GENERALIZATION BY
MANY WAYS [3, 4].GEORGE AND VEERAMANI [18] (i) a sequence {xn} in X is said to converges to x in
X if for each  >0 and each t > 0, there exists n0 
MODIFIED THE CONCEPT OF FUZZY METRIC SPACE
INTRODUCED BY KRAMOSIL AND MICHALEK [19], N such that M(xn, x, t) > 1-  for all n  n0 .
(ii) A sequence {xn} in X is said to be Cauchy if for
CHO [1, 2] ET, AL INTRODUCED THE CONCEPT OF
SEMICOMPATIBILITY OF MAPS.
each  > 0 and each t > 0, there exists n0  N such
that M (xn, xm, t) > 1 −  for all n,m  n0 .
2. Preliminaries
Definition 2.1 [15] A binary operation  : [0, 1] × (iii) A fuzzy metric space in which every Cauchy
sequence is convergent is said to be complete.
[0, 1] → [0, 1] is a continuous t norm if  is an
Defination 2.5 [1] A pair (F, G) of self maps of an
abelian topological monoid with unit 1 such that a
fuzzy metric space
 b ≤ c  d whenever a ≤ c and b ≤ d, a, b, c, d 
(X, M,  ) are said to be semicompatible of
[0, 1].
Definition 2.2 [19] A Triplet (X, M,  ) is called a
lim n   FGxn=Gz where {xn} is a sequence in X
fuzzy metric space if x is an arbitrary (non-empty)
such that limn   Fxn= limn   Gxn= z it follows
set,  is a continuous t-norm and M is a fuzzy set
that (F, G) is semi compatible and Fx=Gx then
on x2  [0,  ] satisfying the following conditions
FGx=GFx.
for each x, y, z  x and t, s > 0
I.
M (x, y, 0) = 0,
II.
M (x, y, t) = M(y, x, t),
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International Journal of Engineering Trends and Technology- Volume4Issue1- 2013
Lemma 2.1 [6] Let (X, M,  ) be a fuzzy metric
space. Then for all x, y  X, M(x, y, .) is a nondecreasing function.
Lemma 2.2 [2] Let (X, M,  ) be a fuzzy metric
space. If there exists
k   (0, 1) such that for all x, y  X, M(x, y, kt) ≥
M(x, y, t)  t > 0, then x = y.
Lemma 2.3 [8] Let {xn} be a sequence in a fuzzy
metric space (X, M,  ). If there exists a number k 
(0, 1) such that M (xn+2, xn+1, kt) ≥ M (xn+1, xn, t) 
t > 0 and n  N then {xn} is a Cauchy sequence in
X.
Lemma 2.4 [12] The only t-norm  satisfying r  r
≥ r for all r  [0, 1] is the minimum t-norm, that is a
 b = min {a, b} for all a, b  [0, 1].
M (Px2n, Qx2n+1, qt) ≥  [min {M (ABx2n, STx2n+1,
t), M (Px2n, ABx2n,
M (Qx2n 1, STx2 n 1, t )  M ( Px2 n , STx 2 n 1, t )
t),
,M
2
(Qx2n+1, STx2n+1, t)}].
M(y2n+1, y2n+2,qt) ≥  [min { M(y2n, y2n+1, t),M(y2n+1,
y2n,
M ( y2 n  2 , y2n 1 , t )  M ( y2 n 1 , y2 n 1 , t )
t) ,
,M(y2n+2,y2n+
2
1,t)}]
M(y2n+1, y2n+2, qt) ≥  [min { M(y2n, y2n+1,
t),M(y2n+1, y2n,
M ( y2 n  2 , y2 n 1 , t )  1)
t) ,
,M(y2n+2,y2n+1,t)}]
2
From lemma 2.1 we have
3. Main Result
M (y2n+1, y2n+2,qt) ≥  [ M(y2n, y2n+1, t)].
Similarly, we have
Theorem 3.1 Let (X, M,  ) be a complete fuzzy
M (y2n+2, y2n+3, qt) ≥  [M(y2n+1, y2n+2, t)]
metric space and let A, B, S, T,P and Q be
Thus,
we have
mappings from X into itself such that the following
M (yn+1, yn+2, qt) ≥ 
conditions are satisfied :
[
M(y
,
y
,
t)]
n
n+1
(a) P(X)  ST(X), Q(X)  AB(X);
for n = 1, 2, 3,...
(b) AB is continious;
t
(c) PB = BP, ST = TS, AB = BA, QT = TQ;
M (yn, yn+1, t) ≥  [M (yn, yn-1, )]
(d) (P, AB) and (Q, ST) are pairs of
q
semicompatible maps;
t
M (yn, yn+1, t) ≥  [M (yn-1, yn-2, 2 )]
(e) There exists q  (0, 1) such that for every x, y
q
 X and t > 0
 M (Px, Qy, qt) ≥  [min {M (ABx, STy, t), M (Px,
------------------------ABx, t),
≥ [M (yn-1, ynM (Qy, STy, t )  M ( Px.STy, t )
, M (Qy, STy,t) }] ;
t
2, n )]  1 as n → ∞
2
q
Where  is continuous function from [0, 1] to [0, 1]
and hence M (yn, yn+1, t)  1 as n → ∞ for any
such that  (a)> a for each o<a<1 and
t >0 for each ε > 0 and t > 0, we can choose n0  N
 (a) =1 and  (0) = 0.
such that M(yn, ym, t) > 1 - ε for all n > n0.
Then A, B, S, T, P and Q have a unique common
For m, n  N, we suppose m ≥ n. Then we have
fixed point in X.
t
M (yn, ym, t) ≥ M (yn, yn+1,
)
Proof: Let x0  X. From (a) there exist x1, x2  X
mn
such that Px0 = STx1 and Qx1 = ABx2 .Using by
t
inductively we get a sequences { xn} and {yn} in X
≥ M (yn+1, yn+2,
)
mn
such that Px2n-2= STx2n-1 = y2n-1 and Qx2n-1 = ABx2n
t
= y2n for n = 1, 2, 3, ... .
≥ M (ym-1, ym,
)
mn
Step 1 Put x = x2n and y = x2n+1 in (e), we get
(m  n )times


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International Journal of Engineering Trends and Technology- Volume4Issue1- 2013
≥ (1 - ε)  (1 - ε)  ... 
(1 - ε)
≥ (1 - ε)
and hence {yn} is a Cauchy sequence in X.
Since (X, M,  ) is complete, {yn} converges to
some point z  X. Also its subsequences converges
to the same point.i.e. z  X
i.e., {Qx2n+1} → z and {ST2n+1} → z ;{Px2n} → z
and {ABx2n} → z --(1)
as (P, AB) is semicompatible therefore we have
lim n   P (AB)xn=ABz
Step 2
Put x=ABx2n and y=x2n+1 in equation (e)
M(PABx2n,Qx2n+1,qt)   [min
(ABABx2n,STx2n+1,t),M(PABx2n, ABABx2n,t),
M (Qx2n 1.STx2n  , t )  M ( PABx2 n .STx2 n 1 , t )
,M(Qx2n
2
+1,STx2n+1,t)}]
Taking n   we get
M (ABz,z, qt)   [min
{M(ABz,z,t),M(ABz, ABz,t),
M ( z , z, t )  M ( ABz, z , t )
,M(z,z,t)}]
2
  [min {M
1  M ( ABz, z, t )
(ABz,z,t),1,
,1}]
2
M (ABz,z, qt)   [M(ABz,z,t)]
M (ABz,z, qt)   [M(ABz,z,t)]>
M(ABz,z,t)
There fore using lemma 2.2 we get
ABz=z
Step 3
Put x=z and y= x2n+1 in Equation (e)
M(Pz,Qx2n+1,qt)   [min
{M(ABz,STx2n+1,t),M(Pz,ABz,t),
M (Qx2 n 1.STx2n  , t )  M ( Pz , STx2 n 1 , t )
,M(Qx2n+1,S
2
Tx2n+1,t)}]
Taking n   we get and using equation (1) we get
M(Pz,z,qt)   [min
M ( z , z , t )  M ( Pz , z , t )
{M(z,z,t),M(Pz,z,t),
,M(z,z,t)
2
}]
ISSN: 2231-5381
  [min {1, M
(Pz,z,t),
1  M ( Pz , z , t ) ,1
}]
2
  [M (Pz,z,t)]
M (Pz, z, qt)   [M (Pz, z,
t)] >M (Pz, z, t)
there fore using lemma 2.2 we get
Pz=z Therefore ABz=Pz=z
Step 4
Put x=Bz and y= x2n+1 in Equation (e)
M(PBz,Qx2n+1,qt)   [min
{M(ABBz,STx2n+1,t),M(PBz,ABBz,t),
M (Qx2n 1.STx 2 n 1  , t )  M ( PBz , STx2 n 1, t )
,M(Qx2n+1
2
,STx2n+1,t)}]
As BP=PB, AB=BA so we have
P (Bz) = B (Pz) =Pz and
(AB)(Bz)= (BA) (Bz) = B (ABz) =Bz. Taking
n   and using equation (1)
M(Bz,z,qt)   [min
M ( z , z , t )  M ( Bz , z , t )
{M(Bz,z,t),M(Bz,Bz,t),
,M(z
2
,z,t)}]
  [min
1  M ( Bz, z , t )
{M(Bz,z,t),
,1}]
2
M (Bz,z,qt)   [M
(Bz,z,t)] >M (Bz,z,t)
Therefore using lemma 2.2 we get
Bz=z and therefore also we have ABz=z and this
imlies that Bz=z
Therefore Az=Bz=Pz=z --- (A-1)
Step 5
As P(x)  ST(x) there exsists u  X such that
z=Pz=Stu
Put x= x2n and y= u in Equation (e)
M(Px2n,Qu,qt)   [min
{M(ABx2n,STu,t),M(Px2n,ABx2n ,t),
M (Qu, STu , t )  M ( Px2 n , STu , t )
,M(Qu,STu,t)}]
2
Taking n  
M(z,Qu,qt)   [min {M(z,z,t),M(z,z,t),
M (Qu , z, t )  M ( z , z , t )
, M(Qu,z,t)}]
2
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International Journal of Engineering Trends and Technology- Volume4Issue1- 2013
  [min {1, 1,
M (Qu , z , t )  1
, M (Qu,z,t)}]
2
M(z,Qu,qt)   [ M(z,Qu,t)] >M(z,Qu,t)
Therefore using Lemma 2.2 we get
Qu =z hence STu=z=Qu Since (Q, ST) is semi
compatible therefore
QSTu=STQu
Thus Qz =STz.
Step 6
Put x= x2n and y= z in Equation (e)
M(Px2n,Qz,qt)   [min
{M(ABx2n,STz,t),M(Px2n,ABx2n ,t),
M (Qz , STz , t )  M ( Px 2 n , STz, t )
,M(Qz,STz,t)}]
2
Taking n  
M(z,Qz,qt)   [min
M (Qu , z , t )  M ( z , z , t )
{M(z,Qz,t),M(z,z,t),
,
2
M(Qz,z,t)}]
  [min { M(z,Qz,t),1,
M (Qz, z , t )  1
, M (Qz,z,t)}]
2
  [min {M (z,Tz,t),1,
M (Tz , z , t )  1
, 1}]
2
M (z,Tz,qt)   [M(z,Tz,t)] >
M (z,Tz,t )
Therefore using Lemma 2.2 we get Tz=z
Now STz=Tz=z implies Sz =z
Hence Sz=Tz=Qz=z ------- (A-2)
 Combining equation (A-1 & A-2)
Az=Bz=Pz=Qz=Tz=Sz=z Hence z is the common
fixed point of A,B,S,T,P,Q.
Uniqness : Let u be another common fixed point
of A,B,S,T,P, and Q then
Au=Bu=Pu=Qu=Tu=Su=u put x=z and y=u in (e)
We get
M (Pz, Qu, qt) ≥  [min {M (ABz, STu, t), M (Pz,
ABz, t),
M (Qu, STu , t )  M ( Pz.STu , t )
, M (Qu, STu,t)}]
2
taking n   and using this equation and we get
M (Pz, Qu, qt) ≥  [min {M (z, u, t), M (z,
M (u , u , t )  M ( z.u , t )
z, t),
, M (u,u,t)}]
2
M (z, Qz, qt)   [M
(z,Qz,t)]> M (z,Qz,t)]
Therefore using Lemma 2.2 we get
Qz =z.
Step7
Put x= x2n and y= Tz in Equation (e)
M(Px2n,QTz,qt)   [min
{M(ABx2n,STTz,t),M(Px2n,ABx2n ,t),
M (QTz , STTz , t )  M ( Px 2 n , STTz , t )
,M(QTz,STTz,t
2
)}]
as QT=TQ and ST=TS we have QTz =TQz=Tz and
ST (Tz) =T (STz)=TQz=Tz.
Taking n   and we get
M(z,Tz,qt)   [min
M (Tz ,Tz , t )  M ( z , Tz , t )
{M(z,Tz,t),M(z,z,t),
,
2
M(Tz,Tz,t)}]
ISSN: 2231-5381
M (Pz, Qu, qt) ≥  [min {M (z, u, t),
1  M ( z.u , t )
1,
, 1}]
2
M (Pz, Qu, qt) ≥  [M (z, u, t)] >M (z, u,
t)]
Therefore by using lemma (2.2) We get z=u There
fore z is the unique common fixed point of
semicompatable A,B,S,T,P,Q.
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