Vol 3, No 4 (2012)

(1)

Available online through

ISSN 2229 – 5046

RELATED FIXED POINT THERREM

FOR TWO PAIRS OF SETVALUED MAPPINGS ON TWO UNIFORM SPACE

**Y. Rohen Singh*, L. Bishwakumar**

National Institute of Technology Manipur, Takyelpat, Pin-795004, Manipur, India

E-mail:

&

B. Fisher

Department of Mathematics, Leicester University, Leicester, LE1 7RH, England

E-mail:

(Received on: 24-02-12; Accepted on: 17-03-12)

________________________________________________________________________________________________

ABSTRACT

A

related fixed point theorem for two pairs of set-valued mappings on two complete uniform spaces is obtained. A generalization for two compact uniform spaces is also obtained.

Keywords: Fixed point, set-valued mappings, complete metric space, compact metric space, complete uniform space,

compact uniform space.

2000 Mathematics subject classification: 54H25, 47H10.

________________________________________________________________________________________________

1. INTRODUCTION

The following theorems were proved by Ranjit and Rohen [12].

Theorem 1.1: Let (X,d)and (Y,ρ)be two complete metric spaces. Let A, B be mappings of X into Y and let S, T be mappings of Y into X satisfying the inequalities

), , ( ) , ( ), , ( ) , ( max{ )

, ( ) ,

(SyTy d SAxTBx c d SyTy Ax Bx d x Sy y Ax

d ′ ′ ≤ ′ ρ ′ ′ ρ ′

d(x,x′)d(Sy,Ty′),d(Sy,SAx)d(Ty′,TBx′)}

), , ( ) , ( ), , ( ) , ( max{ )

, ( ) ,

(Ax Bx′ ρ BSy ATy′ ≤c d Sy Ty′ ρ Ax Bx′ d x′ Sy ρ y′ Ax

ρ

ρ(y,y′)ρ(Ax,Bx′),ρ(Ax,BSy)ρ(Bx′,ATy′)}

for all x,

x′

in X and y,

y

′

in Y, where 0≤c< 1. If one of the mappings A, B, S and T is continuous, then SA and TB have a unique common fixed point z in X and BS and AT have a unique common fixed point w in Y.

Further Az = Bz = w and Sw = Tw = z.

The following theorem was proved by Rohen [16].

Theorem 1.2: Let (X,d₁)and (Y,d₂)be two complete metric spaces. Let A, B be mappings of X into B(Y) and let S, T be mappings of Y into B(X) satisfying the inequalities

), , ( ) , ( ), , ( ) , ( max{ )

, ( ) ,

( ₁ ₁ ₁ ₁ ₂

1 SyTy′δ SAxTBx′ ≤c δ SyTy′δ Ax Bx′ δ x′ Syδ y′ Ax

δ

d₁(x,x′)δ₁(Sy,Ty′),δ₁(Sy,SAx)δ₁(Ty′,TBx′)}

), , ( ) , ( ), , ( ) , ( max{ )

, ( ) ,

( ₂ ₁ ₂ ₁ ₂

2 Ax By′δ BSy ATx′ ≤c δ SyTy′δ Ax Bx′ δ x′Sy δ y′ Ax

δ

d2(y,y′)δ2(Ax,Bx′),δ2(Ax,BSy)δ2(Bx′,ATy′)}

________________________________________________________________________________________________

(2)

for all x ,

x′

in X and y ,

y

′

in Y , where 0≤c< 1. If one of the mappings A, B, S and T is continuous , then SA and TB have a unique common fixed point z in X and BS and AT have a unique common fixed point w in Y . Further Az = Bz = w and Sw = Tw = z.

Before coming to our main results we recall the following definitions from Fisher and Turkoglu [15].

Let (X,U₁) and (Y,U₂) be uniform spaces. Families {d1:i I,

i _∈

being indexing set}, {d2i :i∈I} of pseudometrics on X and Y respectively, are called associated families for uniformities U₁,U₂ respectively, if families

1

β = { V₁(

i

,

r

) : i∈I,r>0},

2

β = { V₂(i,r):i∈I,r>0 }, where

} ) , ( , , : ) , {( ) ,

( 1

1 i r x x x x X d x x r

V = ′ ′∈ i ′ < , } ) , ( , , : ) , {( ) ,

( 2

2 i r y y y y Y d y y r

V = ′ ′∈ i ′ <

are sub bases for the uniformities U₁, U₂respectively. We may assume that β₁,β₂ themselves are bases by adjoining finite intersection of members ofβ₁,β₂ if necessary. The corresponding families of pseudo metrics are called an augmented associated families for U₁,U2. An associated family for U1,U2 will be denoted by D1,D2 respectively.

Let A, B be a non empty subset of a uniform space X, Y respectively. Define

= ) (

* 1 A

P sup {d1(x,x)

i _′

} , ,

:x x′∈Ai∈I

) (

* 2 B

P = sup {d2i(y,y′):y,y′∈B,i∈I}

where

* 1 1( , ): , , }

{di x x′ x x′∈Ai∈I =P ,

* 2 2( , ): , , }

{di y y′ y y′∈Bi∈I =P .

Then ( ), 2*( )

*

1 A P B

P are called an augmented diameter of A, B. Further A, B are said to be ( )<∞, 2*( )<∞

*

1 A P B

P .

Let 2X= {A: A is a non empty P1*- bounded subset of X}

2Y = {B: B is a non empty P2*- bounded subset of Y}.

For each i∈Iand A1,A2∈2X,B1,B2∈2Y, define

} ,

: sup{ ) , (

} , : sup{ ) , (

2 1 2 2

1 2

2 1 1 2

1 1

B y B y d B

B

A x A x d A

A

i i

∈ ′ ∈ =

δ δ

Let (X,U₁) and (X,U2)be uniform spaces and let U1∈ U1 and U2∈U2be arbitrary entourages.

For each A

∈

2 ,

X

B

∈

2 ,

Y define

U₁[A] = {x′∈X:(x,x′)∈U₁for some x∈A} U₂[B] = {y′∈Y:(y,y′)∈U₂for some y∈B}

The uniformities ₂U1_on₂X_and₂U2_on₂Y_{are defined by bases}

1

2β = {Ũ₁: U₁∈U₁}, ₂β2_{= {}Ũ

2: U2∈U2}

where

Ũ1= {(A1,A2)∈2 ×2 :A1×A2∈

Y X

U1}∆,

Ũ2= {(B1,B2)∈2 ×2 :B1×B2∈

Y Y

(3)

where

∆

denotes the diagonal on X×X and Y×Y. The augmented families P1,P2 also induce uniformities U1 on *

2

,

2X U on 2Ydefined by bases

}. 0 , : ) , ( {

}, 0 , : ) , ( {

* 2 * 2

* 1 * 1

> ∈ =

r I i r i V

β β

where

∆ < ∈

=

∆ < ∈

=

  } ) , ( : 2 , : ) , {( ) , (

} ) , ( : 2 , : ) , {( ) , (

2 1 2 2

1 2 1 *

2

2 1 1 2

1 2 1 *

1

r B B B

B B B r i V

r A A A

A A A r i V

i Y

i X

δ δ

Uniformities ₂U1_and * 1

U on 2Xare uniformly isomorphic and uniformities

₂

U2

and U2* on

Y

2 are uniformly

isomorphic. The spaces (2 ,U1*)

X

is thus a uniform space called the hyperspace of (X,U1). The (2 , )

* 2

U

Y

is also a uniform space called the hyperspace of (Y,U₂).

Now let {A_n:n=1,2,...} be a sequence of non empty subsets of uniform space (X,U). We say that sequence }

{A_n converge to subset A of X if

(i) each point a in A is the limit of a convergent sequence {a_n}, where a_nis in A_nfor n =1,2,……. (ii) for arbitrary ε >0, there exist an integer N such that A_n ⊆A_εfor n > N,

where A_ε =_x_∈_AU(x) = {y∈X:d_i(x,y)<ε for some x in A, i∈I}

A is then said to be a limit of the sequence{A_n}. It follows easily from the definition that if A is the limit of a sequence }

{A_n , then A is closed.

The following lemma was proved by Fisher and Turkoglu [15].

Lemma 1.3: If {A_n}and {B_n}are sequences of bounded non empty subsets of a complete uniform space (X,U)which converge to the bounded subsets A and B respectively , then sequence {δi(An,Bn)}converges to δi(A,B).

2. MAIN RESULTS

We prove the following theorems.

Theorem 2.1: Let (X,U₁) and (X,U₂)be two complete Hausdorff uniform spaces defined by

* 2 2

* 1

1, } ,{ , }

{di i∈I =P di i∈I =P , and (2 , ),(2 , 2*)

*

1 U

U Y

X

hyperspaces, let Q, R : X →2Yand S, T : Y →2X satisfying the inequalities

), , ( ) , ( ), , ( ) , ( max{ )

, ( ) ,

( 1 1 2 1 2

1 Sy Ty SQxTRx c SyTy Qx Rx x Sy y Qx

i i

i _′_δ _′ _≤ _δ _′_δ _′ _δ _′ _δ _′

δ

d1(x,x) 1(Sy,Ty), 1(Sy,SQx) 1(Ty,TRx)}

i i

i

i _′_δ _′ _δ _δ _′ _′

(1)

), , ( ) , ( ), , ( ) , ( max{ )

, ( ) ,

( 2 1 2 1 2

2 Qx Rx RSy QTy c Sy Ty Qx Rx x Sy y Qx

i i

i _′_δ _′ _≤ _δ _′_δ _′ _δ _′ _δ _′

δ

d2(y,y) 2(Qx,Rx), 2(Qx,RSy) 2(Rx,QTy)}

i i

i

i _′_δ _′ _δ _δ _′ _′

(2)

for all i∈I andx,x′ in X and y,y′in Y , where 0≤c_i <1. If one of the mappings Q, R, S, T is continuous, then SQ and TR have a unique common fixed point z in X and RS and QT have a unique fixed point w in Y. Further, Qz = Rz = w and Sw = Tw = z.

y

₂_n₊₁ inQx₂_n, a point

x

₂_n₋₁inSy₂_n₋₁, a point y₂_nin Rx₂_n₋₁and a point

x

₂_nin Ty₂_nfor n = 1, 2 …

2 1

)

i i i i

n n n n n n n n

d x

₋

x

d x

₊

x

=

δ

Sy

₋

Ty

δ

SQx

TRx

₋

max{ 1( 2 1 2 ) 2( 2 , 2 1), 1( 2 1, 2 1) 2( 2n, 2n),

i n n i n n i n n i

i Sy Ty Qx Rx x Sy y Qx

c δ ₋ δ ₋ δ ₋ ₋ δ

≤

1( 2 , 2 −1) 1( 2 −1, 2 ), 1( 2 −1, 2 ) 1( 2n, 2n+1)}

i n n i n n i n n i TRx Ty SQx Sy Ty Sy x x

d δ δ δ

max{ 1( 2 1, 2 ) 2( 2 1, 2 ),[ 1( 2n 1, 2n)]2}

i n n i n n i

i d x x d y y d x x

c − + −

=

from which it follows that

)} , ( ), , ( max{ ) ,

( 2 1 2 2 2 1 2 1 2 1 2

1 n n

i n n i i n n i x x d y y d c x x

d + ≤ + − (3)

Let U₂∈U₂be an arbitrary entourage. Since β₂is a base forU₂, there exists V₂(i,r)∈β₂such that V₂(i,r)⊆U₂.

Similarly, applying inequality (2), we get

2

2 2 2 1 2 2 1 2 2 2 1 2

[

d

i

(

y

_n

,

y

_n₊

)]

=

δ

i

(

Qx

_n₋

,

x

_n

)

δ

i

(

RSy

_n₋

,

QTy

_n

)

≤

c

_i

₂i

(

₂_n₊₁

,

y

₂_n₊₁

)}

It follows that

)} , ( ), , ( max{ ) ,

( 2 2 1 1 2 1 2 2 2 1 2

2 n n

i n n i i n n i y y d x x d c y y

d + ≤ − − (4)

We can write as

)} , ( ), , ( max{ ) ,

( 1 2 1 1 1

1 n n

i n n i i n n i x x d y y d c x x

d + ≤ + −

and )} , ( ), , ( max{ ) ,

( 1 1 1 2 1

2 n n

i n n i i n n i y y d x x d c y y

d ₊ ≤ ₋ ₋

It now follows easily by induction that

)} , ( ), , ( max{ ) , ( )} , ( ), , ( max{ ) , ( 2 1 2 1 1 1 1 2 2 1 2 1 1 1 1 y y d x x d c y y d y y d x x d c x x d i i n i n n i i i n i n n i − + + ≤ ≤

for n = 1,2,3,……... Since c_i< 1, it follows that there exists p such that d xn xm r

i _<

) , (

1 and hence (xn,xm)

∈U

1 for all

n, m

≥

p. Therefore, sequence {x_n} is a Cauchy sequence in d1i- uniformity on X. Similarly, it follows that the

sequence {y_n} is a Cauchy sequence in d2i- uniformity on Y.

Let Fp ={xn:n≥ p}for all positive integers p and let B1 be the filter basis {Fp: p = 1, 2…}. Then, since {xn} is a i

d1- Cauchy sequence for each i

∈

I, it is easy to see that the filter basis B1 is a Cauchy filter in the uniform space

(X,U₁). Since (X,U₁) is a complete Hausdorff space, the Cauchy filter B₁= {Fp} converges to a unique point z in X.

2

)

i i i i

n n n n

Sw x

SQz x

Sw Ty

SQz TRx

δ

₊

=

δ

max{ 1( , 2 ) 2( , 2 1), 1( 2 , ) 2( 2 , ), 1( , 2 ) 1( , 2n),

i n i n i n i n i n i

i Sw x Qz y x Sw y Qz d z x Swx

c δ δ ₊ δ δ δ

(5)

δ1(Sw,SQz)d1(x2n,x2n+1)}

Letting n tends to infinity, we have

1

(

, )

1

(

, )

max { (

1

, )

2

(

, ),

1

( ,

)

2

( ,

),

1

( , )

1

(

, ),

2

(

,

)

1

( , )}

i i i i i i i i i i

i

Sw z

SQz z

c

Sw z

Qz w

z Sw

w Qz d z z

Sw z

Sw SQz d z z

δ

≤

δ

c 1(Sw,z) 2(Qz,w)

i i

iδ δ

≤

and so either

Sw = z (5)

or

) , ( ) ,

( 2

1 SQz z c Qz w

i i

i _δ

δ ≤ (6)

Again, applying inequality (1), we have

), , ( ) , ( ), , ( ) , ( max{ )

, ( ) ,

( 2 1 2 1 1 2 2 2 1 1 2 1 2 2

1 n

i n i n

i n i i n

i n i

Qx w x

z Rz y Tw x c

TRz x

Tw

x δ δ δ δ δ

δ ₊ ≤ ₊ ₊

)} , ( ) , ( ), , ( ) ,

( 2 1 2 1 2 2 1 1

1 x z x Tw d x x TwTRz

d n n i

i n i n

i _δ _δ

+

), , ( ) , ( ), , ( ) , ( ), , ( ) , ( max{ )

, ( ) ,

( 1 1 2 1 2 1 1

1 zTw zTRz c zTw w Rz d z z wQz d z z zTw

i i i

i i

i i i

i _δ _δ _δ _δ _δ

δ ≤ d1(z,z) 1(Tw,TRz)}

i

i _δ

and so either

Tz = w (7)

or

) , ( ) ,

( 2

1 zTRz c wRz

i i

2

)

i i i i

n n n n

Qz y

RSw y

Qz Bx

RSw Ax

δ

₊

δ

₊

=

δ

≤ max{ 1( , 2 ) 2( , 2 +1), 1( 2 , ) 2( , ), 2( , 2 ) 2( , 2n+1),

i n i i

n i n i n i

i Sw x Qz y x Sw wQz d w y Qz y

c δ δ δ δ δ

2( , ) 2( 2n+1, 2n+1)}

i i

y y d RSw Qz

δ

)} , ( ) , ( { ) , ( ) ,

( 2 1 2

2 Qz w RSw w c Swz Qz w

i i

i _δ _δ _δ

δ ≤

and so either

Qz = w (9)

or

) , ( ) ,

( 1

2 RSww c Sw z

i i

i _δ

δ ≤ (10)

Again applying inequality (2) and letting n tend to infinity, we have

)} , ( ) , ( { ) , ( ) ,

( 2 1 2

2 w Rz wQTw c zTw w Rz

i i

i _δ _δ _δ

δ ≤

and so either

Rz = w (11)

Or 2(w,QTw) c 1(z,Tw)

i i

i _δ

(6)

If Q is continuous, then

Qz Qx y

w n

n n

n = =

=

∞ → + ∞

→ 2 1 lim 2

lim

SQz Sw=

∴

If inequality (6) holds, then

z = Sw = SQz

and so equation (5) will necessarily hold.

If R is continuous, then

Rz Rx y

w _n

n n n

= =

=

∞ → + ∞

→ 2 1 lim 2

lim

TRz Tw=

∴

z = Tw = TRz

If S is continuous, then

Sw Sy x

z _n

n n

n = =

=

∞ → ∞

→ 2 lim 2

lim

RSw Rz=

∴

w = Rz = RSw

and so equation (9) will necessarily hold .

If T is continuous, then

Tw Ty x

z _n

n n n

= =

=

∞ → ∞

→ 2 lim 2

lim

QTw Qz=

∴

w = Qz = QTw

To prove uniqueness, suppose that SQ and TR have a second common fixed point

z′

in X and RS and QT have a second fixed point

w′

in Y.

2 1

2

1( , )] [ ( , )]

[di z z′ = δi SQz TRz′

(7)

=

c

i

max{ ( , )

d z z

₁

′

δ

₂

(

Qz Rz

,

′

),[

d z z

₁

( , )] }

′

) , ( ) , ( 2

1 z z c Qz Rz

d i i

i _′ _≤ _′

∴ δ (13)

Again applying inequality (2), we have

) , ( ) , ( )] , (

[ 2 2

2

2 Qz Rz Qz Rz RSQz QTRz

i i

i _′ ₌_δ _′_δ _′

δ

c max{d1(z,z) 2(Qz,Rz), 2(Qz,Rz) 2(Qz,Rz), 2(Qz,Rz) 2(Rz,Qz)}

i i i i i i

i ′ ′ ′ ′ ′ ′

≤ δ δ δ δ δ

) , ( ) , ( 1

2 Qz Rz cd z z

i i

i ′ _≤ ′

∴δ (14)

From (13) and (14), it follows that

) , ( ) , ( ) ,

( 2 1

1 z z c Qz Rz cd z z

d i i

i i

i _′ _≤ _δ _′ _≤ _′

and so z = z′

since c_i< 1, proving the uniqueness of the fixed point z of SQ and TR. It follows similarly that w is the unique common fixed point of RS and QT.

If we let Q and R be single valued mappings of X into Y and let S and T be single valued mappings of Y into X, we obtain the following result.

Corollary 2.1: Let (X,U1)and (X,U2)be two complete Hausdorff uniform spaces. If Q, R be mappings of X into Y

and S, T be mappings of Y into X satisfying the inequalities

), , ( ) , ( ), , ( ) , ( max{ ) , ( ) ,

( 1 1 2 1 2

1 Sy Ty d SQxTRx c d SyTy d Qx Rx d x Sy d y Qx

d i i i i i

i

i _′ _′ _≤ _′ _′ _′ _′

d1(x,x)d1(Sy,Ty),d1(Sy,SQx)d1(Ty,TRx)}

i i

i

i _′ _′ _′ _′

), , ( ) , ( ), , ( ) , ( max{ ) , ( ) ,

( 2 1 2 1 2

2 Qx Rx d RSy QTy c d SyTy d Qx Rx d x Sy d y Qx

d i i i i i

i

i _′ _′ _≤ _′ _′ _′ _′

d2(y,y)d2(Qx,Rx),d2(Qx,RSy)d2(Rx,QTy)}

i i

i

i _′ _′ _′ _′

for all i∈I and

x

,

x

′

in X and

y

,

y

′

in Y , where 0≤c_i<1. If one of the mappings Q, R, S, T is continuous, then SQ and TR have a unique common fixed point z in X and RS and QT have a unique fixed point w in Y. Further, Qz = Rz = w and Sw = Tw = z.

Theorem2 .2:Let (X,U₁)and (X,U₂)be two compact Hausdorff uniform spaces defined by

* 2 2

* 1

1, } ,{ , }

{di i∈I =P di i∈I =P , and (2 , ),(2 , 2*)

*

1 U

U Y

X

hyperspaces, let Q, R : X →2Ybe continuous mappings and S, T : Y →2Xbe continuous mappings satisfying the inequalities

), , ( ) , ( ), , ( ) , ( max{ ) , ( ) ,

( 1 1 2 1 2

1 Sy Ty SQxTRx SyTy Qx Rx x Sy y Qx

i i

i

i _′_δ _′ _< _δ _′_δ _′ _δ _′ _δ _′

δ

d1(x,x) 1(Sy,Ty), 1(Sy,SQx) 1(Ty,TRx)}

i i

i

i _′_δ _′ _δ _δ _′ _′

(15) ), , ( ) , ( ), , ( ) , ( max{ ) , ( ) ,

( 2 1 2 1 2

2 Qx Rx RSy QTy Sy Ty Qx Rx x Sy y Qx

i i

i

i _′_δ _′ _< _δ _′_δ _′ _δ _′ _δ _′

δ

d2(y,y) 2(Qx,Rx), 2(Qx,RSy) 2(Rx,QTy)}

i i

i

i _′_δ _′ _δ _δ _′ _′

(16)

for all

i

∈

I

and

x

,

x

′

in X and

y

,

y

′

in Y. Then SQ and TR have a unique common fixed point z in X and RS and QT have a unique fixed point w in Y. Further, Qz = Rz = w and Sw = Tw = z.

Proof: Suppose first of all that the right hand side of inequalities (15) and (16) are never zero. Then the functions

)} , ( ) , ( ), , ( ) , ( ), , ( ) , ( ), , ( ) , ( max{ ) , ( ) , ( ) , ( 1 1 1 1 2 1 2 1 1 1 x TR y T SQx Sy y T Sy x x d Qx y Sy x x R Qx y T Sy x TR SQx y T Sy x x

f _i _i _i _i _i _i _i _i

i i ′ ′ ′ ′ ′ ′ ′ ′ ′ ′ = ′ δ δ δ δ δ δ δ δ δ and )} , ( ) , ( ), , ( ) , ( ), , ( ) , ( ), , ( ) , ( max{ ) , ( ) , ( ) , ( 2 2 2 2 2 1 2 1 2 2 y QT x R RSy Qx x R Qx y y d Qx y Sy x x R Qx y T Sy y QT RSy x R Qx x x

g _i _i _i _i _i _i _i _i

(8)

are continuous and so attain their maximum values a, b respectively . It follows from inequalities (15) and (16) that

a, b < 1. Then, with c = max {a, b} we see that the condition of theorem 2.1 are satisfied and so that theorem is proved in this case.

Now suppose that the right hand side of inequality (15) takes the value zero for points x = z and then

SQz = TRz = z

Putting Qz = Rz = w, we have

Sw = Tw = z

To prove the uniqueness, suppose that

z′

is a second distinct common fixed point of SQ and TR and RS and QT have a second distinct common fixed point in Y.

Then using inequalities (15) and (16), we have

2 1

2

1( , )] [ ( , )]

[di z z′ = δi SQz TRz′

) , ( ) ,

( 2

1 z z c Qz Rz

di ′ ≤ i ′

∴ δ (17)

and [ ( , )] 2( , ) 2( , )

2

1 Qz Rz Qz Rz RSAz QTRz

i i

i ′ ₌_δ ′_δ ′

δ

) , ( ) ,

( 1

2 Qz Rz cd z z

i

i _′ _≤ _′

∴δ (18)

From (17) and (18), we have

) , ( ) , ( ) ,

( 1

2 1

1 z z c Qz Rz c d z z

di ′ ≤ δi ′ ≤ i ′ and so z =

z′

since c < 1, proving the uniqueness of the fixed point of SQ and TR. The uniqueness of w can be proved

Similarly, This completes the proof of the theorem.

If we let Q and R be single valued mappings of X into Y and let S and T be single valued mappings of Y into X, we obtain the following result.

Corollary 2.2:Let (X,U1)and (X,U2)be two compact Hausdorff uniform spaces. Let Q, R be continuous mappings

of X into Y and S, T be continuous mappings of Y into X satisfying the inequalities

), , ( ) , ( ), , ( ) , ( max{ ) , ( ) ,

( 1 1 2 1 2

1 Sy Ty d SQxTRx d Sy Ty d Qx Rx d x Sy d y Qx

di ′ i ′ < i ′ i ′ i ′ i ′ d1(x,x)d1(Sy,Ty),d1(Sy,SQx)d1(Ty,TRx)}

i i

i

i _′ _′ _′ _′

), , ( ) , ( ), , ( ) , ( max{ ) , ( ) ,

( 2 1 2 1 2

2 Qx Rx d RSy QTy d SyTy d Qx Rx d x Sy d y Qx

di ′ i ′ < i ′ i ′ i ′ i ′ d2(y,y)d2(Qx,Rx),d2(Qx,RSy)d2(Rx,QTy)}

i i

i

i _′ _′ _′ _′

for all

i

∈

I

and

x

,

x

′

in X and

y

,

y

′

in Y for which the right-hand sides of the inequalities are positive, then SQ and TR have a unique common fixed point z in X and RS and QT have a unique fixed point w in Y. Further, Qz = Rz = w and Sw = Tw = z.

Remark: Results of [12] and [16] can be obtained by replacing metric spaces in place of uniform spaces of our results.

REFERENCES:

[1] W.J. Thorn, Topological structures, Holt, Rinehart and Winston, New York, 1966.

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[3] E. Tarafdar, An approach to fixed – point theorems on uniform spaces, Trans. Amer. Math. Soc. 191(1974), 209 – 225.

[4]D.V. Pai and P. Veeramani, Fixed point theorems for multimappings, Yokohama Math. J. 28(1980), no.1 – 2, 7 – 14.

[5] B.Fisher, Common fixed points of mappings and set-valued mappings, Rostock. Math. Kolloq. (1981), no. 18, 69 – 77.

[6] Fisher, B., Set-valued mappings on metric spaces, Fund. Math., 112 (1981), 141 – 145.

[7] S.N.Mishra and S.L. Singh, Fixed point of multivalued mapping in uniform spaces, Bull. Calcutta Math. Soc. 77 (1985), no. 6, 323 – 329.

[8] A. Ganguly, Fixed point theorems for three maps in uniform spaces, Indian J. Pure Appl. Math. 17(1986), no. 4, 476 – 480.

[9] S.N. Mishra, Fixed points of contractive type multivalued mappings, Indian J. Pure Appl. Math. 18 (1987), no. 4, 283– 289.

[10] K. Qureshi and S. Upadhya, Fixed point theorems in uniform spaces, Bull. Calcutta Math. Soc. 84 (1992), no. 1, 5 – 10.

[11] R.K. Namdeo, N.K. Tiwari, B. Fisher and Kenan Tas, Related fixed point theorems in two complete and compact metric spaces, Internat. J. Math. and Math. Sci., vol. 21, No. 3 (1998), 559 – 564.

[12] Ranjit, M. and Rohen, Y., Related fixed point theorems for two pairs of mappings on two complete and compact metric spaces , Proceedings of national seminar onanalysis, Ujjain (1999), 64 – 72.

[13] D. Turkoglu, O. Ozer, and B. Fisher, Some fixed point theorems for set-valued mappings in uniform spaces, Demonstration Math. 32 (1999), no. 2, 395 – 400.

[14] Fisher, B. and Turkoglu, D., Related fixed points for set – valued mappings on two metric spaces, Internat. J. Math. and Math. Sci., vol. 23, No. 3 (2000), 205 – 210.

[15] B. Fisher and D Turkoglu, Related fixed points for set – valued mappings on two uniform spaces, Internat. J. Math. and Math. Sci., vol. 23, No. 69 (2004), 3783 – 3791.

[16] Rohen, Y., Related fixed point theorems for two pairs of set-valued mappings on two complete and compact metric spaces, International J. of Math. Sci. and Engg. Appls. Vol. 3, no. II (2009), 137 – 145.