# The Krein Milman theorem

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 $E (K )$?

• 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?) – Michael Apr 30 '17 at 8:14
• 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. – Michael Apr 30 '17 at 8:23
• I have also got this. But just a little confusion about why $Cl(K)=K$? Is a convex compact set closed always? @ Michael – Anupam Apr 30 '17 at 8:36
• 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"? – Michael Apr 30 '17 at 8:42
• So, you cannot define "extreme set" and "closed set" ? – Michael Apr 30 '17 at 8:53