@MISC{Bertsekas10section2.1:, author = {Dimitri P. Bertsekas}, title = {SECTION 2.1: Extreme Points}, year = {2010} }

Share

OpenURL

Abstract

Show by example that the set of extreme points of a nonempty compact set need not be closed. Hint: Consider a line segment C1 = { (x1, x2, x3) | x1 = 0, x2 = 0, −1 ≤ x3 ≤ 1} and a circular disk C2 = { (x1, x2, x3) | (x1 −1) 2 +x 2 2 ≤ 1, x3 = 0} , and verify that the set conv(C1 ∪ C2) is compact, while its set of extreme points is not closed. Solution: For the sets C1 and C2 as given in the hint, the set C1 ∪C2 is compact, and its convex hull is also compact by Prop. 1.2.2. The set of extreme points of conv(C1 ∪ C2) is not closed, since it consists of the two end points of the line segment C1, namely (0, 0, −1) and (0,0, 1), and all the points x = (x1, x2, x3) such that x ̸ = 0, (x1 − 1) 2 + x 2 2 = 1, x3 = 0. 2.2 (Krein-Milman Theorem) Show that a convex and compact subset of ℜ n is equal to the convex hull of its extreme points. Solution: By convexity, C contains the convex hull of its extreme points. To show the reverse inclusion, we use induction on the dimension of the space. On the real line, a compact convex set C is a line segment whose endpoints are the extreme points of C, so every point in C is a convex combination of the two endpoints. Suppose now that every vector in a compact and convex subset of ℜ n−1 can be represented as a convex combination of extreme points of the set. We will show that the same is true for compact and convex subsets of ℜ n. † This set of exercises will be periodically updated as new exercises are added. Many of the exercises and solutions given here were developed as part of my earlier convex optimization book [BNO03] (coauthored with Angelia Nedić and Asuman Ozdaglar), and are posted on the internet of that book’s web site. The contribution of my coauthors in the development of these exercises and their solutions is gratefully acknowledged. Since some of the exercises and/or their solutions have been modified and also new exercises have been added, all errors are my sole responsibility. 2 x1 α x