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I have an intuition about convex, closed, bounded sets but I can't really find a way to prove whether it's right or wrong.

Let $\Sigma$ be a convex set, that means, that given any $A,B \in \Sigma$, and $\lambda \in [0,1]$, $C = \lambda A + (1 - \lambda) B \in \Sigma$. This set $\Sigma$ is closed and bounded under some metric $\| \|$. I'd like to prove that, the set $G = \{ A \in \Sigma: \nexists B,C \in \Sigma \text{ and } \lambda \in (0,1) \text{ s.t. } A = \lambda B + (1-\lambda) C \} \neq \emptyset $.

In plain words, that there exists elements that cannot be expressed as a convex combination of other elements in $\Sigma$.

My intuition comes from polygons where it's easy to see that their vertices can't be expressed as convex combinations of other points within the polygon.

Sam
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    I think you're talking about extreme points. IIRC: In any locally convex Hausdorff topological vector space, any compact convex set has extreme points. The locally convex assumption is needed, as an example of James Roberts shows (see this paper for some results). – David Mitra Dec 17 '13 at 12:28
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    per my last comment, compactness is also crucial. For example, the closed unit ball of the space $c_0$ has no extreme points. See this. (So your conjecture is false in the infinite dimensional case. I'm not sure about finite dimensional spaces.) – David Mitra Dec 17 '13 at 12:44
  • @DonAntonio, it is not enough to be in the boundary. If we imagine a square in $R^2$. Let us concentrate on a side excluding its extremes. Every point is in the boundary (since the side of the square is part of the boundary) but it can be expressed as a convex combination of the extremes of this side. – Sam Dec 17 '13 at 12:48
  • @DavidMitra, I'm having a look at your doc. It does seem that this is what I'm after. I do agree in your claim about the compactness, because I came from the $R^2$ intuition, I've (wrongly) assumed that closed+bounded = compact. – Sam Dec 17 '13 at 12:58
  • @Sam, correct. Thanks. – DonAntonio Dec 17 '13 at 12:58
  • By the way, the theorem mentioned in my first comment is known as the Krein-Milman Theorem. – David Mitra Dec 17 '13 at 13:00
  • Thanks a lot. Let me go through this, and I'll post a summary later. It's always a good exercise to see if I fully understand the proof. – Sam Dec 17 '13 at 13:05

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