 Solution

A sufficient geometric form of the Hahn-Banach separation theorem is this: if $C$ is a nonempty closed convex set in a real normed vector space and $x\notin C$, there exists a continuous linear functional $\ell$ such thatWe use this form without proving it, as permitted. We will prove the Krein-Milman theorem for a nonempty compact convex set $K$: $K$ is the closed convex hull of its extreme points.

First prove the existence of extreme points by the minimal compact face argument. A face of a convex set $K$ is a convex subset $F$ such that, whenever $ta+(1-t)b\in F$ with $a,b\in K$ and $0<t<1$, both $a$ and $b$ belong to $F$. Consider all nonempty compact faces of $K$, ordered by reverse inclusion. The set $K$ itself is such a face. A chain has a nonempty compact intersection, by the finite intersection property and compactness of $K$. This intersection is convex and a face, since membership in every face forces the same membership of endpoints. It is an upper bound in the reverse order. The Zorn lemma therefore gives a minimal nonempty compact face $F$.

If $F$ contained distinct $a,b$, geometric separation of one point from the other would give a continuous real linear $\ell$ with $\ell (a)\ne \ell (b)$. Compactness supplies a maximum $m={\max }_F\ell$. The set $F^{\prime }=\{v\in F:\ell (v)=m\}$ is a nonempty proper compact convex subset. It is also a face of $K$: if a strict convex combination belongs to $F^{\prime }$, first the face property of $F$ puts both endpoints in $F$, and then linearity and maximality of $m$ force both their values to equal $m$. Thus $F^{\prime }$ contradicts minimality of $F$. Hence $F=\{e\}$, whose face property says precisely that $e$ is an extreme point of $K$. The same argument starts with any nonempty compact face, so every such face contains an extreme point of $K$.

Let $E$ be the set of all extreme points of $K$ and $C=\overline{\mathrm{conv}}E$. It is nonempty, and $C\subseteq K$ because $K$ is closed and convex. Suppose $x\in K\setminus C$. Geometric separation gives a continuous real linear $\ell$ with $\ell (x)>{\sup }_C\ell$. Let $m={\max }_K\ell$ and consider the maximizer face $F=\{v\in K:\ell (v)=m\}$. It is nonempty, compact and a face of $K$, so the previous argument gives an extreme point $e\in F$. But $e\in E\subseteq C$, whilea contradiction. Therefore every point of $K$ belongs to the closed convex hull of its extreme points, provingNeither completeness of the ambient normed space nor finite dimensionality was used; compactness is the key hypothesis.