proof of Krein-Milman theorem

The proof is consist of three steps for good understanding. We will show initially that the set of extreme points of $K$ , $Ex(K)$ is non-empty, $Ex(K)\neq\emptyset$ . We consider that $\mathcal{A}=\{A\subset K\colon A\subset K,$ extreme $\}$ .
Step1
The family set $\mathcal{A}$ ordered by $\subset$ has a minimal element, in other words there exist $A\in\mathcal{A}$ such as $\forall B\in\mathcal{A},B\subset A$ we have that $B=A$ .
Proof1
We consider $A<B\Leftrightarrow B\subset A,\forall A,B\in\mathcal{A}$ . The ordering relation $<$ is a partially relation on $\mathcal{A}$ . We must show that $A$ is maximal element for $\mathcal{A}$ . We apply Zorn’s lemma.We suppose that

$\mathcal{C}=\{A_{i}\colon i\in I\}$

is a chain of $\mathcal{A}$ .Witout loss of generality we take $A=\bigcap_{i\in I}A_{i}$ and then $A\neq\emptyset$ . $\mathcal{C}$ has the property of finite intersections and it is consist of closed sets. So we have that $\bigcap_{i\in I}A_{i}\neq\emptyset$ . It is easy to see that $A\in\mathcal{A}$ . Also $A\subset A_{i}$ , for any $i\in I$ , so we have that $A>A_{i}$ , for any $i\in I$ .
Step2
Every minimal element of $\mathcal{A}$ is a set which has only one point.
Proof2
We suppose that there exist a minimal element $A$ of $\mathcal{A}$ which has at least two points, $x,y\in A$ . There exist $x^{*}\in X^{*}$ such as $x^{*}(x)\neq x^{*}(y)$ , witout loss of generality we have that $x^{*}(x)<x^{*}(y)$ . $A$ is compact set (closed subset of the compact $K$ ). Also there exist $\alpha\in\mathbb{R}$ such that $\alpha=\sup_{z\in A}x^{*}(z)$ and $B=\{z\in A\colon x^{*}(z)=\alpha\}\neq\emptyset$ . It is obvious that $B$ is an extreme subset of $A$ , $B$ is an extreme subset of $K$ , $B\in\mathcal{A}$ . $x\notin B$ since $B\in\mathcal{A}$ and $B\subsetneq A$ that contradicts to the fact that $A$ is minimal extreme subset of $\mathcal{A}$ .
From the above two steps we have that $Ex(K)\neq\emptyset$ .
Step3
$K=\bar{c}o(Ex(K))$ where $\bar{c}o(Ex(K))$ denotes the closed convex hull of extreme points of $K$ .
Proof3
Let $L=\bar{c}o(Ex(K))$ . Then $L$ is closed subset of $K$ , therefore it is compact, and convex clearly by the definition. We suppose that $L\subsetneqq K$ . Then there exist $x\in K-L$ . Let use Hahn-Banach theorem(geometric form). There exist $x^{*}\in X^{*}$ such as $\sup_{w\in L}x^{*}(w)<x^{*}(x)$ . Let $\alpha=\sup\{x^{*}(y)\colon y\in K\}$ , $B=\{y\in K\colon x^{*}(y)=\alpha\}$ . Similar to Step2 $B$ is extreme subset of $K$ . $B$ is compact and from step1 and step2 we have that $Ex(B)\neq\emptyset$ . It is true that $Ex(B)\subset Ex(K)\subset L$ . Now let $y\in Ex(B)$ then $x^{*}(y)=\alpha$ and if $y\in Lx^{*}(y)<x^{*}(x)\leq\alpha$ . That is a contradiction.

Title	proof of Krein-Milman theorem
Canonical name	ProofOfKreinMilmanTheorem
Date of creation	2013-03-22 15:24:40
Last modified on	2013-03-22 15:24:40
Owner	georgiosl (7242)
Last modified by	georgiosl (7242)
Numerical id	6
Author	georgiosl (7242)
Entry type	Proof
Classification	msc 46A03
Classification	msc 52A07
Classification	msc 52A99