# Some examples of compact convex sets

Due to reasons (connected to incomplete von-neumann morgenstern orders if you must know) I’ve been thinking about the Krein-Milman theorem recently, and I realised that I had some misconceptions in my head that were obviously false when I thought about them for five minutes. Here are some examples to illustrate its boundaries.

The Krein-Milman theorem says that every compact convex subset of a locally convex vector space is the closed convex hull of its extreme points.

Consider $$\mathbb{R}^n$$ with the normal euclidean norm. Then the unit ball $$\overline{B}(0, 1)$$ is a compact convex set with infinitely many extreme points.

The same is true of any $$l^p$$ norm with $$1 < p < \infty$$, because such spaces are strictly convex.

It is not true for $$p = 1$$ (whose extreme points are the points $$\pm e_n$$), or for $$l^\infty$$ (whose extreme points are the set of points with $$x_i = \pm 1$$), both of whose unit balls have finitely many extreme points.

Moreover you cannot drop the “closed” part. Not every point is a convex combination of finitely many extreme points:

Consider $$A \subseteq l^\infty$$ defined by $$A = \{x : |x_n| \leq \frac{1}{n}\}$$. This is convex (because it’s a product of convex sets) and compact (it’s closed, so complete, and you can construct $$\epsilon$$-nets explicitly). Then the extreme points are the set of all points $$\{x : |x_n| = \frac{1}{n}\}$$.

Let $$y = \sum\limits_n \lambda_i x_i$$ be a finite combination of extreme points. Then $$n y_i$$ can only take the $$2^n$$ (and in particular finitely many) values $$\sum\limits_n \pm \lambda i$$.

Consider $$y_n = 2^{-n}$$. Then $$y \in A$$ and $$n y_n$$ takes infinitely many values and thus $$y$$ cannot be a combination of finitely many extreme points of $$A$$.

Note that it is true that every point in a compact convex set is a sum $$\sum\limits_{n=0}^\infty p_n x_n$$ where $$\sum p_n = 1$$ and the $$x_n$$ are extreme points, you just need the sum to be infinite.

Edit: Actually I think it’s not, though I don’t yet have a counterexample. In general you may need a full integral over a probability measure on the extreme points. I’m going to have a thing about how to construct such.

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