What convexity really asks for
You met the convex set back in the foundations, but here we lean on it as a structural property, not a picture. A set C in R^n is convex if for every pair of points x, y in C the whole segment {(1 - t)x + t y : 0 <= t <= 1} stays inside C. That single clause is astonishingly rigid: it forbids dents, holes, and pinches all at once. A disk, a slab, a half-space, a single point, all of R^n — convex; an annulus or a star — not.
Two adjectives matter for the deep theorems. A convex set is a convex body when it is compact (closed and bounded in R^n) and has nonempty interior — a solid blob, not a lower-dimensional sliver. The convex body is the natural arena for the rung's later stars: Brunn-Minkowski, mixed volumes, lattice points. Convexity is also stable under the operations you care about: an arbitrary intersection of convex sets is convex (each constraint carves away an extra slice), and the image under an affine map is convex.
The convex hull and Caratheodory's economy
Given any set S, the smallest convex set containing it is its convex hull conv(S) — the intersection of all convex sets that swallow S, equivalently the set of all finite convex combinations sum t_i x_i with t_i >= 0 and sum t_i = 1, x_i in S. Picture stretching a rubber band around a scatter of nails: it snaps to the outermost ones and the interior nails become irrelevant. The hull of three non-collinear points is a filled triangle; the hull of the grid points {0,1}^n is the unit cube.
How many points do you actually need in one of those combinations? In R^n, never more than n + 1. That is Caratheodory's theorem: every point of conv(S) is a convex combination of at most n + 1 points of S. The plane needs at most three (a point inside a polygon sits inside some triangle of its vertices); 3-space needs at most four. The bound n + 1 is exactly the number of vertices of a simplex — a dimension-counting fact, not an accident, and it makes the hull a computable object rather than an infinite tangle.
Supporting and separating hyperplanes
Now press a flat board against the body from outside. A supporting hyperplane of a convex body K at a boundary point p is a hyperplane H through p such that K lies entirely in one closed half-space bounded by H. The defining fact — proved by projecting onto the nearest point of K — is that every boundary point of a convex body has at least one supporting hyperplane. At a smooth point that hyperplane is unique (the tangent plane); at a corner of a cube a whole fan of them works, which is exactly how convexity tolerates edges and vertices.
Push the same idea between two bodies and you get the separating hyperplane theorem: two disjoint convex sets can be separated by a hyperplane, with one set in each closed half-space. If one set is compact and the other closed, you can even separate them strictly with a slab of positive width between. This is the geometric heart of duality everywhere it appears — the Hahn-Banach theorem in functional analysis, Farkas' lemma in linear programming, the existence of a price that clears a market in economics are all this one picture wearing different clothes.
supporting at p: <a, x> <= <a, p> for all x in K, a != 0 separating A | B: <a, x> <= c <= <a, y> for all x in A, y in B
Extreme points and the Krein-Milman theorem
Call a point e of a convex set K extreme if it is not the midpoint — indeed not the interior of any segment — lying in K: you cannot write e = (1 - t)x + t y with 0 < t < 1 and x, y in K distinct from e. Extreme points are the corners: every vertex of a square, every point of a circle's boundary. The flat edge of the square is not extreme, because its points split into two neighbours. A polytope has finitely many extreme points, its vertices; a disk has infinitely many.
The Krein-Milman theorem is the stunning converse: a compact convex set is the convex hull of its extreme points. The corners alone regenerate the whole body. In R^n this is concrete — a polytope is conv(its vertices), and you can throw the edges and faces away. The full theorem lives in a locally convex topological vector space, where 'extreme points exist at all' is itself the content, and the closure of the hull is what you recover. State the hypothesis, never the slogan: compact and convex are both load-bearing — drop compactness and an open disk has no extreme points whatsoever.
Why this rung leans on these three
These tools are the scaffolding for everything that follows in the rung. The next guide studies polytopes face by face — and a face is exactly the intersection of K with a supporting hyperplane, while vertices are the extreme points Krein-Milman hands you. Brunn-Minkowski and mixed volumes (guide 3) measure convex bodies, and their proofs slice bodies by hyperplanes. Minkowski's lattice theorem (guide 4) hunts lattice points inside symmetric convex bodies. And Helly, Radon and Caratheodory (guide 5) are the combinatorial trio whose first member you just met.
Let us walk the cleanest argument once, to show the pieces interlock rather than just coexist. Here is why a linear functional attains its max on a compact convex set at an extreme point — the bridge between Krein-Milman and optimization.
- K is compact and <a, .> is continuous, so the maximum value c is attained; let F = {x in K : <a, x> = c} be the face where it is achieved.
- F is itself compact and convex (it is K intersected with a supporting hyperplane), so by Krein-Milman F has an extreme point e.
- Check that e is also extreme in the whole of K: if e were the interior of a segment in K, applying <a, .> would force both endpoints to also hit the max c, putting that segment inside F — contradicting that e is extreme in F.
- Therefore the maximum of <a, .> over K is attained at a genuine extreme point of K, and you never need to inspect the interior or the flat faces.