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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?

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  • $\begingroup$ Duplicate? math.stackexchange.com/questions/1384579/… $\endgroup$ – A.Γ. Jul 2 '16 at 22:10
  • $\begingroup$ Yes, that is what I want. $\endgroup$ – Denny Jul 2 '16 at 22:13
  • $\begingroup$ What is the setting here, are we talking about Euclidean space $\mathbb{R}^n$ or a normed space, or a locally convex topological space? $\endgroup$ – user147263 Jul 3 '16 at 2:53
  • $\begingroup$ Actually the paper is about positive semidefinite matrices with unity trace and rank one, which form extreme points $\endgroup$ – Denny Jul 3 '16 at 21:33
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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.

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