Giới thiệu Sau khi giấy Khamsi [8] theo nguyên tắc KKM-bản đồ trong không gian hyperconvex số liệu, một số tác giả đã thiết lập các ứng dụng chính của nguyên tắc này, chẳng hạn như Kỳ Fan minimax bất bình đẳng, định lý điểm cố định, định lý minimax, định lý điểm cân bằng, . | Vietnam Journal of Mathematics 33. V Í e It ini ai m J o mt r im ai I of MATHEMATICS VAST 2005 Further Applications of the KKM-Maps Principle in Hyperconvex Metric Spaces Le Anh Dung Department of Mathematics Hanoi Univesity of Education 136 Xuan Thuy Road Hanoi Vietnam Abstract. 1. Introduction . Le Anh Dung 2. Preliminaries Definition 1. A metric space is said to be hyperconvex if for any collection of points .of and any collection of nonnegative reals . such that . . . . for all. then . . . G the closed ball centered at with radius . . R x . . .R2 x. . Example. . Definition 2. A set in a metric space is said to be admissible if it is an intersection of some closed balls. The collection of all admissible sets in a metric space is denoted by A . Definition 3. The admissible hull of a set A in a metric space denoted by ad is the smallest admissible set containing A . . Theorem be a hyperconvex metric space C be an arbitrary subset of and. be a KKM-map such that is closed for every C . Then the the finite intersection property. Definition 4. A set A in a metric space is said to be sub-admissible if for each finite subset D of A we have ad A . Further Applications of the KKM-Maps Principle in Hyperconvex Metric Spaces Definition 5. Let be an admissible set in a metric space. A function R is said to be quasi-convex or quasi-concave if for each R the set . respectively . is sub-admissible. Definition 6. A multivalued mapping from a topological space into a topological space is said to be upper semicontinuous usc at a point 0 of if for any open set containing 0 there exists a neighborhood of 0 such that. 3. Ky Fan Inequality and a Minimax Theorem Theorem . Let be a hyperconvex metric space a nonempty compact admissible set of . for all .is closed for all C . for all . Then there exists 0 such that 0. Proof. . 1 1. i 1 . 1 n .0. y
