# Must every convex compact set have extreme points?

I am reading a paper and there is such description as title. Why?

I have an example: $(0,1)$. This is a convex set but not closed, so I cannot find an extreme point. However if convex and compact,

I read some related problems:

1. Exposed point of a compact convex set
There must be at least one exposed point. But an extreme point is not necessary equal to an exposed point.
2. Convex hull of extreme points
A convex hull $P$ of finite points. Then $P$ is the convex hull of its extreme points.

It seems there is a requirement "finite points" to guarantee the topic?

• – A.Γ.
Jul 2 '16 at 22:10
• Yes, that is what I want. Jul 2 '16 at 22:13
• What is the setting here, are we talking about Euclidean space $\mathbb{R}^n$ or a normed space, or a locally convex topological space?
– user147263
Jul 3 '16 at 2:53
• Actually the paper is about positive semidefinite matrices with unity trace and rank one, which form extreme points Jul 3 '16 at 21:33

In a finite dimensional space (which is the case here, according to a comment), the existence of an extreme point of a compact convex set $K$ is easy to prove. Take any point $x\in K$ at which the norm $\|x\|$ is maximized. If there is $y\ne 0$ such that $x\pm y \in K$, then $$2\|x\|^2 \ge \|x+y\|^2 + \|x-y\|^2 = 2\|x\|^2+2\|y\|^2> 2\|x\|^2$$ a contradiction.

• What is this contradicting? That $x$ is not and extreme point? How do I know there is a $y$ such that $x \pm y \in K$? Mar 2 '20 at 19:40