Krein–Milman existence of extreme points

Let $\mathcal{P}$ be the set of nonempty faces of $K$ that are closed in $K$, ordered by $F\preccurlyeq G$ when $F\supseteq G$. It is a nonempty poset because $K\in \mathcal{P}$.

Let $\mathcal{C}$ be a chain in $\mathcal{P}$. If $\mathcal{C}=\mathrm{\varnothing }$, then $K$ is an upper bound. Otherwise every finite subfamily of $\mathcal{C}$ has intersection equal to its inclusion-smallest member and hence nonempty. Its members are closed in compact $K$, so [F3] gives a nonempty intersection $H:={\bigcap }_{F\in \mathcal{C}}F$, and $H$ is closed and convex.

If $x,y\in K$, $0<t<1$, and $(1-t)x+ty\in H$, then this combination lies in every $F\in \mathcal{C}$; since each $F$ is a face, $x,y$ lie in every $F$ and therefore in $H$. Thus $H$ is a face, hence belongs to $\mathcal{P}$, and $F\supseteq H$ for every $F\in \mathcal{C}$, so $H$ is an upper bound in the reverse-inclusion order.

By [F4], using AC as declared in [F5], $\mathcal{P}$ has a maximal element $M$; equivalently, $M$ is an inclusion-minimal nonempty closed face of $K$.

Suppose $p,q\in M$ are distinct. AC supplies HB by [F6], so [F2] gives $f\in X^{\mathrm{\prime }}$ with $u(p):=\mathrm{Re}f(p)\mathrm{\ne }\mathrm{Re}f(q)=:u(q)$. The restriction $u{\mathrm{\mid }}_M$ is continuous, real-valued, and affine.

By [F1], the minimizer set $N$ of $u{\mathrm{\mid }}_M$ is a nonempty compact face of $M$, hence a face of $K$. It is closed in $M$ and $M$ is closed in $K$, so it is closed in $K$ and lies in $\mathcal{P}$. Since $u(p)\mathrm{\ne }u(q)$, at least one of $p,q$ is not a minimizer, so $N⊊M$, contradicting the inclusion-minimality of $M$.

Hence $M$ is a singleton, say $M=\{x\}$. Since $M$ is a face of $K$, the singleton characterization in the face definition makes $x$ an extreme point of $K$.

The empty-chain case in step 2.1 and the nonempty-chain construction in steps 2.1–3.1 verify every chain hypothesis of Zorn; steps 4.1–7.1 then produce the required extreme point.