Krein–Milman theorem

Remarks

Call a nonempty closed subset $F \subseteq K$ a face if every open segment of $K$ that meets $F$ has both endpoints in $F$. By Zorn's lemma and compactness, every face contains a minimal face. A minimal face is a single point, since otherwise a continuous functional that separates two of its points would cut out a smaller face; that point is extreme. If $\overline{\operatorname{conv}}(\operatorname{ext} K)$ were smaller than $K$, the Hahn–Banach separation theorem would give a continuous functional whose real part has its maximum on $K$ only on a face disjoint from $\operatorname{ext} K$. But that face contains a minimal face, which is an extreme point of $K$, a contradiction.

For the square $[0, 1]^2$ in $\R^2$ the theorem says that the square is the convex hull of its four corners. A standard application: the closed unit ball of a dual Banach space is weak*-compact by the Banach–Alaoglu theorem, so it has extreme points. The unit ball of $c_0$ has no extreme points, so $c_0$ is not isometrically isomorphic to the dual of any Banach space. Compactness is needed: the closed half-plane $\set{(x, y) : y \ge 0}$ in $\R^2$ is closed and convex but has no extreme points.

Going deeper

For a student

The theorem is used to prove that certain Banach spaces are not dual spaces, by showing that their unit balls lack extreme points. In measure theory and probability, the extreme points of compact convex sets of measures or states are often the "pure" or "ergodic" objects. The theorem shows that they generate everything. Choquet theory refines the closed convex hull into an integral representation over the extreme points. The proof combines Zorn's lemma with the Hahn–Banach separation theorem.