I am trying to understand the Krein Milman theorem from Rudin's Functional Analysis, Second Edition (page no 75). It has been shown that if $K $ is a compact convex subset of a topological vector space $X$, then every compact extreme set $S$ has a non empty intersection with the set of all extreme points $E (K)$ of $K$. Herefrom how can I say that $K$ is a superset of the closure of the convex hull of $E (K ) $?

  • $\begingroup$ Can you define "extreme set"? In particular, I would think that an extreme set for $K$ is (by definition?) a subset of extreme points of $K$, but that would make your statement about "has a non empty intersection" somewhat obvious (perhaps the key aspect of that statement is just that extreme points exist?) $\endgroup$ – Michael Apr 30 '17 at 8:14
  • $\begingroup$ Of course, it seems you are asking about the "easy" direction, so I think that can be done without knowing the definition of the term "extreme set" used in your middle sentence: $$E(K) \subseteq K \implies Conv(E(K)) \subseteq Conv(K) = K \implies Cl(Conv(E(K)) \subseteq Cl(K) = K$$ where $Conv(K)=K$ holds because $K$ is convex; $Cl(K)=K$ holds if you can assume that the closure of a compact set is the same set back. $\endgroup$ – Michael Apr 30 '17 at 8:23
  • $\begingroup$ I have also got this. But just a little confusion about why $Cl(K)=K$? Is a convex compact set closed always? @ Michael $\endgroup$ – Anupam Apr 30 '17 at 8:36
  • $\begingroup$ For $\mathbb{R}^n$, a set is compact if and only if closed and bounded. For more general spaces related-but-not-exactly-the-same results hold. Can you define "extreme set" and "closed set"? $\endgroup$ – Michael Apr 30 '17 at 8:42
  • $\begingroup$ So, you cannot define "extreme set" and "closed set" ? $\endgroup$ – Michael Apr 30 '17 at 8:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.