# 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, $Ex(K)\ne \mathrm{\varnothing}$.
We consider that $\mathcal{A}=\{A\subset K:A\subset K,$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 $$. The ordering relation $$ 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

$$\mathcal{C}=\{{A}_{i}:i\in I\}$$ |

is a chain of $\mathcal{A}$.Witout loss of
generality we take $A={\bigcap}_{i\in I}{A}_{i}$ and then $A\ne \mathrm{\varnothing}$. $\mathcal{C}$ has the property of finite intersections^{}
and it is consist of closed sets. So we have that ${\bigcap}_{i\in I}{A}_{i}\ne \mathrm{\varnothing}$. 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 ${x}^{*}(x)\ne {x}^{*}(y)$, witout loss of
generality we have that $$. $A$ 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 $B=\{z\in A:{x}^{*}(z)=\alpha \}\ne \mathrm{\varnothing}$.
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 $B\u228aA$ that contradicts to the fact that $A$ is minimal^{} extreme subset of $\mathcal{A}$.

From the above two steps we have that $Ex(K)\ne \mathrm{\varnothing}$.

Step3

$K=\overline{c}o(Ex(K))$ where $\overline{c}o(Ex(K))$ denotes the closed convex hull^{} of extreme points of $K$.

Proof3

Let $L=\overline{c}o(Ex(K))$. Then $L$ is closed subset of $K$, therefore it is compact, and convex clearly by the definition.
We suppose that $L\u2acbK$. Then there exist $x\in K-L$. Let use Hahn-Banach theorem^{}(geometric form).
There exist ${x}^{*}\in {X}^{*}$ such as $$. Let $\alpha =sup\{{x}^{*}(y):y\in K\}$, $B=\{y\in K:{x}^{*}(y)=\alpha \}$. Similar^{} to
Step2 $B$ is extreme subset of $K$. $B$ is compact and from step1 and step2 we have that $Ex(B)\ne \mathrm{\varnothing}$. It is true that
$Ex(B)\subset Ex(K)\subset L$. Now let $y\in Ex(B)$ then ${x}^{*}(y)=\alpha $ and if $$. 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 |