A note on closedness of algebraic sum of sets

Abstract

In this note we generalize the fact that in topological vector spaces the algebraic sum of closed set A and compact set B is closed. We also prove some conditions that are equivalent to reflexivity of Banach spaces.2010 Mathematics Subject Classification. 46A22. 46N10 Keywords. Closed convex sets. InroductionLet X be a Hausdorff topological vector space. By C(X) we denote the family of all closed subsets of X and by K(X) the family of all compact subset of X. For a nonempty subsets A, B ⊂ X we define the algebraic sum (Minkowski sum ) as followIt is wll known that if A, B are closed sets then A + B need not to be closed, but also it is known that if A ∈ C(X) and B ∈ K(X) then A + B ∈ C(X). In this note we prove the last result with some more abstract point of view. Reflexive spacesIn this section we give some results which show the connections of closedness of algebraic sum of sets with reflexivity of Banach spaces.Theorem 2.1. Let X be a Banach space and let B be the closed unit ball in X. Then X is a reflexive Banach space if and only if for every closed convex and bounded subset A of X the algebraic sum A + B is closed.Proof. (Necessity.) Assume that X is a reflexive Banach space, then every closed bounded and convex set is a weakly compact. Since algebraic sum of two weakly compact sets is again weakly compact we conclude that A + B is a weakly compcact set, and therefore closed. (Sufficiency.) Assume that X is not reflexive Banach space. Then by Theorem of James there exists a continuous linear functional f : X → R, such that ||f || = 1 and f (x) < 1 for every x ∈ B. Let A = {x ∈ X : f (x) 1, ||x|| 2}Then A is a closed bounded and convex set but A + B is not closed.q.e.d.

Order Cancellation Law in a Semigroup of Closed Convex Sets

“…If the space X is not reflexive, then we can always find two bounded closed convex sets, with not closed Minkowski sum. In fact, possibility of finding such two bounded closed convex sets is equivalent to non-reflexivity of the space X, see [24]. Theorem 3.15.…”

Order cancellation law in the family of bounded convex sets

“…Then there exists subsets A, B ∈ C(X) with trivial reccesion cones such that the sum A +B is not closed.If the space X is not reflexive then we can always find two bounded closed convex sets, with not closed Minkowski sum. In fact, possibility of finding such two bounded closed convex sets is equivalent to non-reflexivity of the space X[23].Proof. Let X be an infinitely dimensional Banach space.…”