 Solution

The finite-dimensional Krein-Milman theorem says that a nonempty compact convex set $K$ is the convex hull of its extreme points. A point is extreme if a proper convex combination of two points of $K$ equals it only when both points equal it.

Prove the theorem by induction on the dimension of the affine hull. Dimension zero is immediate. In positive dimension work inside that affine hull, where $K$ has interior. Every boundary point $x$ lies on a supporting hyperplane. To see this, choose exterior points $y_j\to x$ and nearest points $p_j\in K$. Convexity and the minimum-distance condition give $(y_j-p_j)\cdot (z-p_j)\le 0$ for $z\in K$. A subsequence of the normalized nonzero vectors $y_j-p_j$ converges to a unit vector $v$, while $p_j\to x$, giving $v\cdot (z-x)\le 0$. The resulting face $F=\{z\in K:v\cdot (z-x)=0\}$ has smaller affine dimension. Its extreme points are extreme in $K$, since equality in the supporting inequality forces both endpoints of any decomposition to lie in $F$. Induction expresses $x$ as a convex combination of extreme points of $K$.

Every interior point is a convex combination of the two endpoints of a line segment through it maximal in $K$; compactness gives the endpoints, and both are boundary points. Combining their convex decompositions completes the proof. In the generalization requested, a nonempty weak-star compact convex subset $K$ of the dual unit ball of a separable Banach space satisfiesThe closure is now essential; separability makes this compact weak-star setting metrizable but the general Krein-Milman theorem holds more broadly in locally convex spaces.