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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.

A point $x$ of a convex set $K$ is an extreme point if $x=(1-t)a+tb$ with $a,b\in K$ and $0<t<1$ forces $a=b=x$. It is enough to test midpoints: a nontrivial interior point of a segment is the midpoint of a smaller nontrivial segment.

Here is a proof of the Krein-Milman theorem. The empty-set case is immediate, so take $K\ne \varnothing$. A face of a convex set is a convex subset $F\subseteq K$ such that an interior convex combination lying in $F$ has both endpoints in $F$. Consider nonempty compact faces. They include $K$. Every inclusion chain has nonempty intersection, because its closed subsets of the compact set $K$ have the finite intersection property. The intersection remains a compact face. The Zorn lemma, ordered by reverse inclusion, therefore supplies a minimal nonempty compact face $F$.

If $F$ contains two distinct points, the Hausdorff locally convex hypotheses and the allowed separation theorem supply a continuous real linear functional $\ell$ taking different values there. Its maximizer set in $F$ is nonempty and compact. It is convex, and if a convex combination attains the maximum, both endpoint values attain it too; hence it is a face of $F$. A face of a face is a face of $K$, by applying the defining endpoint property twice. This maximizer face is proper because $\ell$ is nonconstant, contradicting minimality. Thus $F$ is a singleton, whose point is extreme in $K$. The same minimal compact face argument inside any nonempty compact face shows that each such face contains an extreme point of $K$.

Let $H=\overline{\mathrm{conv}}(\mathrm{ext}K)$. Since $K$ is compact, hence closed in the Hausdorff space, and convex, $H\subseteq K$. It is nonempty, closed and convex. If $p\in K\setminus H$, the allowed strict separation theorem supplies a continuous linear functional withThe maximizer face of $\ell$ on $K$ contains an extreme point $e$ by the preceding argument. Then $e\in H$ but $\ell (e)={\max }_K\ell \ge \ell (p)>{\sup }_H\ell$, a contradiction. ThereforeAll closures refer to the given locally convex topology, and compactness is what justified both maxima and chain intersections.

For the real l-infinity sequence space, if $|x_j|<1$ for any coordinate, choose $0<ϵ\le 1-|x_j|$. The two distinct sequences $x\pm ϵe_j$ belong to the closed unit ball and have midpoint $x$, so $x$ is not extreme. Conversely, if every $x_j$ is one or minus one, writing $x=(1-t)a+tb$ for unit-ball sequences forces $a_j=b_j=x_j$ at every coordinate, because an endpoint of $[-1,1]$ cannot be a nontrivial convex average of its points. ThusFor the space of sequences converging to zero, every unit-ball sequence has some coordinate with $|x_j|<1$, since its coordinates tend to zero. The same opposite single-coordinate perturbations still belong to $c_0$, givingThese are extreme points of real sequence-space unit balls. There is no conflict with Krein-Milman: the norm-closed unit ball of $c_0$ is not compact in the norm topology, as the coordinate vectors have mutual distance one.