Open Access

On the existence and stability of solutions of a mixed general type of variational relation problems

Journal of Inequalities and Applications20142014:337

https://doi.org/10.1186/1029-242X-2014-337

Received: 11 April 2014

Accepted: 15 August 2014

Published: 2 September 2014

Abstract

In this paper, we introduce a mixed general type of variational relation problems, and establish the existence theorem of solutions of mixed general types of variational relation problems. Moreover, we study the stability of a solution set of mixed general types of variational relation problems. We prove that most of mixed general types of variational relation problems (in the sense of Baire category) are essential and, for any mixed general type of variational relation problems, there exists at least one essential connected component of its solution set.

MSC:49J53, 49J40.

Keywords

mixed general types of variational relation problems existence generic stability essential connected components

1 Introduction

It is well known that the equilibrium problems are unified models of several problems, namely optimization problems, saddle point problems, variational inequalities, fixed point problems, Nash equilibrium problems etc. Recently, Luc [1] introduced a more general model of equilibrium problems, which is called a variational relation problem (VR). The stability of the solution set of variational relation problems was studied in [2, 3]. Further studies of variational relation problems have been done (see [412]). Recently, Agarwal et al. [13] presented a unified approach in studying the existence of solutions for two types of variational relation problems, which encompass several generalized equilibrium problems, variational inequalities and variational inclusions investigated in the recent literature. Balaj and Lin [14] established existence criteria for the solutions of two very general types of variational relation problems.

Motivated and inspired by research works mentioned above, we introduce mixed general types of variational relation problems, which is a mixed structure of two general types of variational relation problems in [14]. Moreover, we study the stability of a solution set of mixed general types of variational relation problems.

2 Mixed general types of variational relation problems

In [14], let X, Y be convex sets in two Hausdorff topological vector spaces, Z be a topological space, S 1 , S 2 : X X , T : X Y , P : X Z be set-valued mappings with nonempty values, and R ( x , y , z ) be a relation linking elements x X , y Y and z Z . Balaj and Lin [14] established existence criteria for the solutions of the following variational relation problems:

(VRP1) Find ( x , y ) X × Y such that x S 1 ( x ) , y T ( x ) and, u S 2 ( x ) , z P ( x ) for which R ( u , y , z ) holds.

(VRP2) Find ( x , y ) X × Y such that x S 1 ( x ) , y T ( x ) and R ( u , y , z ) holds u S 2 ( x ) and z P ( x ) .

In this paper, we introduce mixed general types of variational relation problems. Let X, Y be convex sets in two Hausdorff topological vector spaces, Z be a topological space, S : X × Y X , T : X × Y Y , H : X × Y X , G : X × Y Y , P : X × Y Z be set-valued mappings with nonempty values, and R ( x , y , z ) , Q ( y , x , z ) be two relations linking elements x X , y Y and z Z . A mixed general type of variational relation problems (MGVR) consists in finding ( x , y ) X × Y such that x S ( x , y ) , y T ( x , y ) and
u H ( x , y ) , z P ( x , y ) s.t. R ( u , y , z )  holds , Q ( v , x , z )  holds , v G ( x , y ) , z P ( x , y ) .

Remark 2.1 Balaj and Lin [14] established existence criteria for the solutions of two very general types of variational relation problems. The mixed general type of variational relation problems is a combination of (VRP1) and (VRP2), and (VRP1) and (VRP2) are some special cases of (MGVR).

Theorem 2.1 Assume that
  1. (i)

    X, Y, Z are three nonempty, compact and convex subsets of three Hausdorff linear topological spaces;

     
  2. (ii)

    C = { ( x , y ) X × Y : x S ( x , y ) } and D = { ( x , y ) X × Y : y T ( x , y ) } are closed in X × Y ;

     
  3. (iii)

    P is continuous with nonempty compact values;

     
  4. (iv)

    for any ( x , y ) X × Y , co H ( x , y ) S ( x , y ) , co G ( x , y ) T ( x , y ) , H, G have open fibers, and R ( x , , ) , Q ( , y , ) are closed;

     
  5. (v)

    for any fixed y Y , any finite subset { u 1 , , u n } of X and any x co { u 1 , , u n } , there are i { 1 , , n } and z P ( x , y ) such that R ( u i , y , z ) holds;

     
  6. (vi)

    for any fixed x X , any finite subset { v 1 , , v n } of Y and any y co { v 1 , , v n } , there is i { 1 , , n } such that Q ( v i , x , z ) holds for any z P ( x , y ) .

     

Then (MGVR) has at least one solution.

Proof Define A : X × Y X and B : X × Y Y as follows:
A ( x , y ) = { u X : R ( u , y , z )  does not hold  z P ( x , y ) } , B ( x , y ) = { v Y : Q ( v , x , z )  does not hold  z P ( x , y ) } .

As R ( x , , ) , Q ( , y , ) are closed for any ( x , y ) X × Y , and P is continuous with nonempty compact values, by Propositions 3.1, 3.3 of [14], A, B have open fibers.

Suppose that there exists ( x , y ) X × Y such that x co A ( x , y ) , then there is a finite subset { u 1 , , u n } of A ( x , y ) such that x co { u 1 , , u n } . By (v), there are i 0 { 1 , , n } and z P ( x , y ) such that R ( u i 0 , y , z ) holds, which contradicts the fact that u i A ( x , y ) for any i { 1 , , n } , i.e., R ( u i , y , z ) does not hold for any z P ( x , y ) and any i { 1 , , n } . Hence x co A ( x , y ) for any ( x , y ) X × Y .

Suppose that there exists ( x , y ) X × Y such that y co B ( x , y ) , then there is a finite subset { v 1 , , v n } of B ( x , y ) such that y co { v 1 , , v n } . By (vi), there is i 0 { 1 , , n } such that Q ( v i 0 , x , z ) holds, which contradicts the fact that v i B ( x , y ) for any i { 1 , , n } , i.e., there is z P ( x , y ) such that Q ( v i , x , z ) does not hold for any i { 1 , , n } . Hence y co B ( x , y ) for any ( x , y ) X × Y .

Define A : X × Y X and B : X × Y Y as follows:
A ( x , y ) = { A ( x , y ) H ( x , y ) if  ( x , y ) C , H ( x , y ) if  ( x , y ) C , B ( x , y ) = { B ( x , y ) G ( x , y ) if  ( x , y ) D , G ( x , y ) if  ( x , y ) D .
For any u X , A 1 ( u ) = [ H 1 ( u ) A 1 ( u ) ] [ ( ( X × Y ) C ) H 1 ( u ) ] is open in X × Y . Similarly, B 1 ( v ) is open for any v Y . Hence, A 1 ( u ) , B 1 ( v ) are open for any ( u , v ) X × Y , and x co A ( x , y ) , y co B ( x , y ) for any ( x , y ) X × Y . By Theorem 3 of [15], there exists ( x , y ) X × Y such that A ( x , y ) = and B ( x , y ) = , which implies that x S ( x , y ) , y T ( x , y ) and
u H ( x , y ) , z P ( x , y ) s.t. R ( u , y , z )  holds , Q ( v , x , z )  holds , v G ( x , y ) , z P ( x , y ) .

 □

Theorem 2.2 Assume that
  1. (i)

    X, Y, Z are three nonempty, compact and convex subsets of three normed linear topological spaces;

     
  2. (ii)

    S, T are continuous with nonempty convex compact values;

     
  3. (iii)

    P is continuous with nonempty compact values;

     
  4. (iv)

    R ( , , ) and Q ( , , ) are closed;

     
  5. (v)

    for any fixed y Y , any finite subset { u 1 , , u n } of X and any x co { u 1 , , u n } , there are i { 1 , , n } and z P ( x , y ) such that R ( u i , y , z ) holds;

     
  6. (vi)

    for any fixed x X , any finite subset { v 1 , , v n } of Y and any y co { v 1 , , v n } , there is i { 1 , , n } such that Q ( v i , x , z ) holds for any z P ( x , y ) .

     
Then there exists ( x , y ) X × Y such that x S ( x , y ) , y T ( x , y ) and
u S ( x , y ) , z P ( x , y ) s.t. R ( u , y , z )  holds , Q ( v , x , z )  holds , v T ( x , y ) , z P ( x , y ) .
Proof For any n, define S n : X × Y X , T n : X × Y Y , H n : X × Y X and G n : X × Y Y by
S n ( x , y ) = ( S ( x , y ) + cl V n ) X , T n ( x , y ) = ( T ( x , y ) + cl V n ) Y , H n ( x , y ) = ( S ( x , y ) + V n ) X , G n ( x , y ) = ( T ( x , y ) + V n ) Y ,
where V n = { x X : x < 1 n } , V n = { y Y : y < 1 n } . Since S, T are continuous, H n , G n have open fibers, and Graph ( S n ) , Graph ( T n ) are closed in X × Y , by Theorem 2.1, there exist x n S n ( x n , y n ) , y n T n ( x n , y n ) and
u H n ( x n , y n ) , z P ( x n , y n ) s.t. R ( u , y n , z )  holds , Q ( v , x n , z )  holds , v G n ( x n , y n ) , z P ( x n , y n ) .
Since X, Y are nonempty and compact, without loss of generality, we assume that ( x n , y n ) ( x , y ) . Let m > 0 be arbitrarily fixed and n 0 > 0 such that V n 0 + V n 0 + cl V n 0 V m . Since S is continuous and ( x n , y n ) ( x , y ) , there is N 1 > 0 such that x x n V n 0 and S ( x n , y n ) S ( x , y ) + V n 0 for any n > N 1 . Therefore, for any n > max { N 1 , n 0 } ,
x = x x n + x n x x n + S ( x n , y n ) + cl V n V n 0 + S ( x , y ) + V n 0 + cl V n 0 S ( x , y ) + V m .

Hence x m > 0 ( S ( x , y ) + V m ) = cl S ( x , y ) = S ( x , y ) . Similarly, y T ( x , y ) .

Suppose that there exists u S ( x , y ) such that R ( u , y , P ( x , y ) ) does not hold. Since R ( , , ) is closed, there exists k > 0 such that R ( u + V k , y + V k , P ( x , y ) + V k ) does not hold. Since S is continuous, there exists a sequence { u n } convergent to u with u n S ( x n , y n ) S ( x n , y n ) + V n H n ( x n , y n ) . Since P is continuous, there exists N 0 > 0 such that, for any n > N 0 , u n u + V k , y n y + V k , P ( x n , y n ) P ( x , y ) + V k , which implies that u n H n ( x n , y n ) and R ( u n , y n , P ( x n , y n ) ) does not hold. It is a contradiction.

Suppose that there exist v T ( x , y ) and z P ( x , y ) such that Q ( v , x , z ) does not hold. Since T, P are continuous, there exist two sequences { v n } and { z n } convergent to v and z with v n T ( x n , y n ) T ( x n , y n ) + V n G n ( x n , y n ) and z n P ( x n , y n ) . As Q ( , , ) is closed, there exists N 1 > 0 such that, for any n > N 1 , Q ( v n , x n , z n ) does not hold, which is a contradiction. This completes the proof. □

3 Generic stability analysis

Let X, Y, Z be three nonempty, compact and convex subsets of three normed linear topological spaces. Denote by the collection of all (MGVR) such that all conditions of Theorem 2.2 hold. For each q M , denote by F ( q ) the solution set of q. Thus, a correspondence F : M X × Y is well defined. For each q , q M , define the distance on by
ρ ( q , q ) = sup ( x , y ) X × Y h X ( S ( x , y ) , S ( x , y ) ) + sup ( x , y ) X × Y h Y ( T ( x , y ) , T ( x , y ) ) + sup ( x , y ) X × Y h Z ( P ( x , y ) , P ( x , y ) ) + H ( Gr ( R ) , Gr ( R ) ) + H ( Gr ( Q ) , Gr ( Q ) ) ,

where Gr ( R ) = { ( x , y , z ) X × Y × Z : R ( x , y , z )  holds } , Gr ( Q ) = { ( x , y , z ) X × Y × Z : Q ( y , x , z )  holds } , h X ( h Y , h Z ) is the Hausdorff distance defined on X (Y, Z), and H is the Hausdorff distance defined on X × Y × Z .

Definition 3.1 Let q M . An ( x , y ) F ( q ) is said to be an essential point of F ( q ) if, for any open neighborhood N ( x , y ) of ( x , y ) in X × Y , there is a positive δ such that N ( x , y ) F ( q ) for any q M with ρ ( q , q ) < δ . q is said to be essential if each ( x , y ) F ( q ) is essential.

Definition 3.2 Let q M . A nonempty closed subset e ( q ) of F ( q ) is said to be an essential set of F ( q ) if, for any open set U, e ( q ) U , there is a positive δ such that U F ( q ) for any q M with ρ ( q , q ) < δ .

Definition 3.3 Let q M . An essential subset m ( q ) F ( q ) is said to be a minimal essential set of F ( q ) if it is a minimal element of the family of essential sets in F ( q ) ordered by set inclusion. A connected component C ( q ) of F ( q ) is said to be an essential component of F ( q ) if C ( q ) is essential.

Remark 3.1 (1) It is easy to see that the problem q M is essential if and only if the mapping F : M X × Y is lower semicontinuous at q. (2) For two closed e 1 ( q ) e 2 ( q ) F ( q ) , if e 1 ( q ) is essential, then e 2 ( q ) is also essential.

Lemma 3.1 (17.8 Lemma, 17.11 Closed Graph Theorem of [16])

(i) The image of a compact set under a compact-valued upper semicontinuous set-valued mapping is compact. (ii) A correspondence with compact Hausdorff range space is closed if and only if it is upper hemicontinuous and closed-valued.

Lemma 3.2 ([17])

If X, Y are two metric spaces, X is complete and F : X Y is upper semicontinuous with nonempty compact values, then the set of points, where F is lower semicontinuous, is a dense residual set in X.

Theorem 3.1 ( M , ρ ) is a complete metric space.

Proof Let { q n } n = 1 be any Cauchy sequence in , then, for any ε > 0 , there is N > 0 such that ρ ( q n , q m ) < ε for any n , m > N , that is, for any n , m > N ,
sup ( x , y ) X × Y h X ( S n ( x , y ) , S m ( x , y ) ) + sup ( x , y ) X × Y h Y ( T n ( x , y ) , T m ( x , y ) ) + sup ( x , y ) X × Y h Z ( P n ( x , y ) , P m ( x , y ) ) + H ( Gr ( R n ) , Gr ( R m ) ) + H ( Gr ( Q n ) , Gr ( Q m ) ) ε .
  1. (1)

    Clearly, we consult Proposition 3.1 of [18]. There are S : X × Y X , T : X × Y Y and P : X × Y Z such that S, T are continuous with nonempty convex compact values, and P is continuous with nonempty compact values.

     
  2. (2)
    There exist two closed subsets A, B of X × Y × Z such that Gr ( R n ) A and Gr ( Q n ) B . Denote q = ( S , T , P , R , Q ) , where
    R ( x , y , z )  holds iff ( x , y , z ) A , Q ( y , x , v )  holds iff ( x , y , z ) B .
     
Clearly R ( , , ) and Q ( , , ) are closed.
  1. (3)

    Suppose the existence of y Y , finite subset { u 1 , , u n } of X and x co { u 1 , , u n } such that R ( u i , y , P ( x , y ) ) does not hold for any i { 1 , , n } , which implies ( u i , y , P ( x , y ) ) Gr ( R ) = , i { 1 , , n } . Since q m q for enough large m, ( u i , y , P m ( x , y ) ) Gr ( R m ) = , i { 1 , , n } , i.e., R m ( u i , y , P m ( x , y ) ) does not hold for any i { 1 , , n } , which is a contradiction.

     

Suppose the existence of x X , finite subset { v 1 , , v n } of Y, y co { v 1 , , v n } and z P ( x , y ) such that Q ( v i , x , z ) does not hold for any i { 1 , , n } , which implies ( x , v i , z ) Gr ( Q ) , i { 1 , , n } . Since q m q , there exists a sequence { z m } convergent to z with z m P m ( x , y ) . Hence, for enough large m, ( x , v i , z m ) Gr ( Q m ) , i { 1 , , n } , i.e., Q m ( v i , x , z m ) does not hold for any i { 1 , , n } , which is a contradiction. Hence q M and ( M , ρ ) is complete. □

Theorem 3.2 The mapping F : M X × Y is upper semicontinuous with nonempty compact values.

Proof The desired conclusion follows from Lemma 3.1 as soon as we show that Graph ( F ) is closed. Denote q n = ( S n , T n , P n , R n , Q n ) and q = ( S , T , P , R , Q ) . Let { ( q n , x n , y n ) M × X × Y } n = 1 be a sequence converging to ( q , x , y ) such that ( x n , y n ) F ( q n ) for any n. Then x n S n ( x n , y n ) and y n T n ( x n , y n ) ,
u S n ( x n , y n ) , z P n ( x n , y n ) s.t. R n ( u , y n , z )  holds , Q n ( v , x n , z )  holds , v T n ( x n , y n ) , z P n ( x n , y n ) .

Clearly, x S ( x , y ) and y T ( x , y ) .

Suppose the existence of u S ( x , y ) such that R ( u , y , P ( x , y ) ) does not hold, then ( u , y , P ( x , y ) ) Gr ( R ) = . Since S, P are continuous, q n q and
h X ( S n ( x n , y n ) , S ( x , y ) ) h X ( S n ( x n , y n ) , S ( x n , y n ) ) + h X ( S ( x n , y n ) , S ( x , y ) ) 0 , h Z ( P n ( x n , y n ) , P ( x , y ) ) h Z ( P n ( x n , y n ) , P ( x n , y n ) ) + h Z ( P ( x n , y n ) , P ( x , y ) ) 0 ,

then S n ( x n , y n ) S ( x , y ) and P n ( x n , y n ) P ( x , y ) . Thus, there exists a sequence { u n } convergent to u with u n S n ( x n , y n ) such that, for enough large n, ( u n , y n , P n ( x n , y n ) ) Gr ( R n ) = , i.e., u n S n ( x n , y n ) and R n ( u n , y n , P n ( x n , y n ) ) does not hold, which is a contradiction.

Suppose the existence of v T ( x , y ) and z P ( x , y ) such that Q ( v , x , z ) does not hold, then ( x , v , z ) Gr ( Q ) . Similarly, T n ( x n , y n ) T ( x , y ) and P n ( x n , y n ) P ( x , y ) . Thus, there exist two sequences { v n } and { z n } convergent to u and z with v n T n ( x n , y n ) and z n P n ( x n , y n ) . Hence, for enough large n, ( x n , v n , z n ) Gr ( Q n ) , i.e., v n T n ( x n , y n ) , z n P n ( x n , y n ) and Q n ( v n , x n , z n ) does not hold, which is a contradiction. Hence ( x , y ) F ( q ) . □

Theorem 3.3 (i) There exists a dense residual subset G of such that q is essential for each q G . (ii) For any q M , there exists at least one minimal essential subset of F ( q ) .

Proof The proofs are similar to those of Theorems 3.3 and 3.4 of [19]. Here, we do not repeat the process. □

4 Existence of essential connected components

In this section, let P 0 : X × Y Z be fixed. Assume that (i)  P 0 is continuous with nonempty compact values; (ii) if W 1 W 2 = , P 0 ( W 1 ) P 0 ( W 2 ) = ; (iii)  P 0 1 ( z ) for any z Z . Denote by M 0 the collection of ( S , T , P 0 , R , Q ) mixed general types of variational relation problems such that all conditions of Theorem 2.2 hold. Clearly M 0 M . For convenience in the later presentation, for any subset A of X, denote A c = { x X : x A } .

Lemma 4.1 ([20])

Let C, D be two nonempty, convex and compact subsets of a linear normed space E. Then h ( C , λ C + μ D ) h ( C , D ) , where h is the Hausdorff distance defined on E, and λ , μ 0 , λ + μ = 1 .

Lemma 4.2 ([21])

Let ( Y , ρ ) be a metric space, K 1 and K 2 be two nonempty compact subsets of Y, V 1 and V 2 be two nonempty disjoint open subsets of Y. If h ( K 1 , K 2 ) < ρ ( V 1 , V 2 ) : = inf { ρ ( x , y ) | x V 1 , y V 2 } , then
h ( K 1 , ( K 1 V 2 ) ( K 2 V 1 ) ) h ( K 1 , K 2 )

where h is the Hausdorff metric defined on Y.

Theorem 4.1 For any q M 0 , every minimal essential subset of F ( q ) is connected.

Proof For each fixed q M 0 , let m ( q ) F ( q ) be a minimal essential subset of F ( q ) . If m ( q ) is not connected, then there are two nonempty compact subsets c 1 ( q ) , c 2 ( q ) and two disjoint open subsets V 1 , V 2 of X × Y such that m ( q ) = c 1 ( q ) c 2 ( q ) and V 1 c 1 ( q ) , V 2 c 2 ( q ) . Since m ( q ) is a minimal essential set of F ( q ) , neither c 1 ( q ) nor c 2 ( q ) is essential. There exist two open sets O 1 c 1 ( q ) , O 2 c 2 ( q ) such that, for any δ > 0 , there exist q 1 , q 2 M 0 with
ρ ( q , q 1 ) < δ , ρ ( q , q 2 ) < δ , F ( q 1 ) O 1 = , F ( q 2 ) O 2 = .
Denote W 1 = V 1 O 1 , W 2 = V 2 O 2 , we know that W 1 , W 2 are open, W 1 c 1 ( q ) , W 2 c 2 ( q ) and we may assume that V 1 W ¯ 1 , V 2 W ¯ 2 . Denote
G 1 = X × Y × ( P 0 ( W 2 c ) ) c , G 2 = X × Y × ( P 0 ( W 1 c ) ) c .

Since P 0 is continuous with nonempty compact values, and W 1 c , W 2 c are nonempty compact in X × Y , by Lemma 3.1, P 0 ( W 1 c ) , P 0 ( W 2 c ) are nonempty compact in Z. Thus G 1 , G 2 are open in X × Y × Z .

To prove by contraposition that ( P 0 ( W 1 c ) ) c ( P 0 ( W 2 c ) ) c = , suppose the existence of z Z such that z ( P 0 ( W 1 c ) ) c ( P 0 ( W 2 c ) ) c , which implies that z P 0 ( W 1 c ) and z P 0 ( W 2 c ) , i.e., W 1 c P 0 1 ( z ) = , W 2 c P 0 1 ( z ) = . It follows that P 0 1 ( z ) W 1 and P 0 1 ( z ) W 2 , which contradicts the fact that W 1 W 2 = .

Denote inf { d ( a , b ) | a G 1 , b G 2 } = ε > 0 . Since m ( q ) is essential, and m ( q ) ( W 1 W 2 ) , there exists 0 < δ < ε such that F ( q ) ( W 1 W 2 ) for any q M 0 with ρ ( q , q ) < δ . Since m ( q ) is a minimal essential set of F ( q ) , neither c 1 ( q ) nor c 2 ( q ) is essential. Thus, for δ 32 > 0 , there exist two q 1 , q 2 M 0 such that
F ( q 1 ) W 1 = , F ( q 2 ) W 2 = , ρ ( q 1 , q ) < δ 32 , ρ ( q 2 , q ) < δ 32 .
Thus ρ ( q 1 , q 2 ) < δ 16 . Next, define q = ( S , T , P 0 , R , Q ) as follows:
S ( x , y ) = λ ( x , y ) S 1 ( x , y ) + μ ( x , y ) S 2 ( x , y ) , T ( x , y ) = λ ( x , y ) T 1 ( x , y ) + μ ( x , y ) T 2 ( x , y ) , A = [ Gr ( R 1 ) G 2 ] [ Gr ( R 2 ) G 1 ] , B = [ Gr ( Q 1 ) G 2 ] [ Gr ( Q 2 ) G 1 ] , R ( u , y , z )  holds iff ( u , y , z ) A , Q ( v , x , z )  holds iff ( x , v , z ) B ,
where
λ ( x , y ) = d ( ( x , y ) , W ¯ 2 ) d ( ( x , y ) , W ¯ 1 ) + d ( ( x , y ) , W ¯ 2 ) , ( x , y ) X × Y , μ ( x , y ) = d ( ( x , y ) , W ¯ 1 ) d ( ( x , y ) , W ¯ 1 ) + d ( ( x , y ) , W ¯ 2 ) , ( x , y ) X × Y .
Easily, we check that (i)  S , T are continuous with nonempty compact convex values. (ii) Since Gr ( R 1 ) and Gr ( R 2 ) are closed in X × Y × Z , A is closed in X × Y × Z , which implies that R ( , , ) is closed. Similarly, Q ( , , ) is closed. (iii) Suppose the existence of y Y , finite subset { u 1 , , u n } X and x co { u 1 , , u n } such that R ( u i , y , P 0 ( x , y ) ) does not hold for any i { 1 , , n } , i.e., ( u i , y , P 0 ( x , y ) ) Gr ( R ) = , i { 1 , , n } . Since
Gr ( R ) = [ Gr ( R 1 ) G 2 ] [ Gr ( R 2 ) G 1 ]
and W 1 W 2 = , without loss of generality, we may assume that ( x , y ) W 1 c , which implies P 0 ( x , y ) P 0 ( W 1 c ) . Since ( u i , y , P 0 ( x , y ) ) [ Gr ( R 2 ) G 1 ] = for all i { 1 , , n } , that is,
( y , u i , P 0 ( x , y ) ) [ Gr ( R 2 ) ( X × Y × P 0 ( W 1 c ) ) ] = , i { 1 , , n } ,

then ( u i , y , P 0 ( x , y ) ) Gr ( R 2 ) = , i { 1 , , n } , i.e., R 2 ( u i , y , P 0 ( x , y ) ) does not hold for any i { 1 , , n } , which is a contradiction.

(iv) Suppose the existence of x X , finite subset { v 1 , , v n } Y , y co { v 1 , , v n } and z P 0 ( x , y ) such that Q ( v i , x , z ) does not hold for any i { 1 , , n } , i.e., ( x , v i , z ) Gr ( Q ) , i { 1 , , n } . Since
Gr ( Q ) = [ Gr ( Q 1 ) G 2 ] [ Gr ( Q 2 ) G 1 ]
and W 1 W 2 = , without loss of generality, we may assume that ( x , y ) W 1 c , which implies z P 0 ( x , y ) P 0 ( W 1 c ) . Since ( x , v i , z ) [ Gr ( Q 2 ) G 1 ] for all i { 1 , , n } , that is, ( x , v i , z ) [ Gr ( Q 2 ) ( X × Y × P 0 ( W 1 c ) ) ] for all i { 1 , , n } , then
( x , v i , z ) Gr ( Q 2 ) , i { 1 , , n } ,

that is, there is z P 0 ( x , y ) such that Q 2 ( v i , x , z ) does not hold for any i { 1 , , n } , which is a contradiction. Hence q M 0 .

(v) By Lemmas 4.1, 4.2,
ρ ( q , q ) = sup ( x , y ) X × Y h X ( S ( x , y ) , S ( x , y ) ) + sup ( x , y ) X × Y h Y ( T ( x , y ) , T ( x , y ) ) + H ( Gr ( R ) , Gr ( R ) ) + H ( Gr ( Q ) , Gr ( Q ) ) sup ( x , y ) X × Y h X ( S ( x , y ) , S 1 ( x , y ) ) + sup ( x , y ) X × Y h X ( S 1 ( x , y ) , S ( x , y ) ) + sup ( x , y ) X × Y h Y ( T ( x , y ) , T 1 ( x , y ) ) + sup ( x , y ) X × Y h Y ( T 1 ( x , y ) , T ( x , y ) ) + H ( Gr ( R ) , Gr ( R 1 ) ) + H ( Gr ( R 1 ) , Gr ( R ) ) + H ( Gr ( Q ) , Gr ( Q 1 ) ) + H ( Gr ( Q 1 ) , Gr ( Q ) ) < δ .

Thus q M 0 and ρ ( q , q ) < δ .

Since ( F ( q ) W 1 ) ( F ( q ) W 2 ) = F ( q ) ( W 1 W 2 ) , without loss of generality, we assume F ( q ) W 1 . Then there exists ( x ¯ , y ¯ ) F ( q ) W 1 such that ( x ¯ , y ¯ ) W 1 , x ¯ S ( x ¯ , y ¯ ) , y ¯ T ( x ¯ , y ¯ ) , and
u S ( x ¯ , y ¯ ) , z P 0 ( x ¯ , y ¯ ) s.t. R ( u , y ¯ , z )  holds , Q ( v , x ¯ , z )  holds , v T ( x ¯ , y ¯ ) , z P 0 ( x ¯ , y ¯ ) .
It follows from ( x ¯ , y ¯ ) W 1 that S ( x ¯ , y ¯ ) = S 1 ( x ¯ , y ¯ ) , T ( x ¯ , y ¯ ) = T 1 ( x ¯ , y ¯ ) , P 0 ( x ¯ , y ¯ ) P 0 ( W 1 ) and P 0 ( x ¯ , y ¯ ) P 0 ( W 1 c ) = . Therefore,
u S 1 ( x ¯ , y ¯ ) , z P 0 ( x ¯ , y ¯ ) s.t. R 1 ( u , y ¯ , z )  holds , Q 1 ( v , x ¯ , z )  holds , v T 1 ( x ¯ , y ¯ ) , z P 0 ( x ¯ , y ¯ ) .

Then ( x ¯ , y ¯ ) F ( q 1 ) W 1 , which is a contradiction. This completes the proof. □

Theorem 4.2 For any q M 0 , there exists at least one essential connected component of  F ( q ) .

Proof By Theorem 4.1, there exists at least one connected minimal essential subset m ( q ) of F ( q ) . Thus, there is a component C of F ( q ) such that m ( q ) C . It is obvious that C is essential by Remark 3.1(2). This completes the proof. □

Declarations

Acknowledgements

This research is supported the Chen Guang Project sponsored by the Shanghai Municipal Education Commission and Shanghai Education Development Foundation (no. 13CG35), and open project of Key Laboratory of Mathematical Economics (SUFE), Ministry of Education (no. 201309KF02).

Authors’ Affiliations

(1)
School of Economics, Shanghai University of Finance and Economics
(2)
Key Laboratory of Mathematical Economics (SUFE), Ministry of Education

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© Yang; licensee Springer. 2014

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