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Here is a question about the convex hull.

Let $S$ be a set and $\bar{S}$ be the closed convex hull of $S$, i.e., $\bar{S}$ is the smallest convex set that contains $S$. Then is the following claim true?

"$\forall~s\in\bar{S}$, $s$ can be written as a convex combination of a finite set of vectors in $S$"

I feel that it should be correct intuitively. But I am not sure. Is there any reference or counter example? I remember for convex hull (not closed convex hull), there is a minimal representation theorem to support this claim. But I don't know if it is also true for "closed" convex hull. I cannot perceive this difference

KevinKim
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    You say "closed convex hull", and then you say i.e. and give the definition of the convex hull. Which do you mean to ask about? The answer is yes for the convex hull but no for the closed convex hull... – David C. Ullrich Apr 01 '16 at 14:48
  • Is $S$ a set of points in $\mathbb{R}^d$ or are you allowing more general sets? If $S$ is open, then the statement you write is false, consider the open unit disk as $S$, then $\overline{S}$ is the closed disk and the points on the boundary are not convex combinations. If $S$ is a point set, then the vertices of $\overline{S}$ are in $S$. We can then triangulate (or tetrahedralize or ...) $\overline{S}$ and write any element as a convex combination of the vertices of the simplex that contains it. – Michael Burr Apr 01 '16 at 14:49
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    See Carathéodory's theorem. – user251257 Apr 01 '16 at 15:05
  • And if $S$ is compact, so is the convex hull of $S$. – user251257 Apr 01 '16 at 15:08

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