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Let $S$ be a finite subset of $\mathbb R^n$ , we know that $x \in S$ is a vertex of $Conv (S)$ , the convex hull or convex polytope of $S$ , iff $x \notin Conv\Big(S$ \ $\{x\}\Big)$ ; then is the no. of vertices of $Conv(S)$ finite ? Is it true that $Conv(S)=Conv (V)$ , where $V$ is the set of vertices of $Conv (S) $ ?

( I can obviously prove the one side $Conv(V) \subseteq Conv(S) $ as $V \subseteq S$ )

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Hint: Suppose $y\in\text{conv}(S)$, then we want to show $y\in\text{conv}(V)$. By definition there are appropriate $a_s$ for all $s\in S$ such that $y=\sum a_ss$. Moreover, for each $s\in S$, there are appropriate $\lambda_{s,v}$ for all $v\in V$ such that $a_s=\sum\lambda_{s,v}v$.

Why is this true? How does it help?

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