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Let $(X,\| \cdot \|)$ be a finite dimensional normed space. Show that if $S\subseteq X$ is compact, then the $\text{Conv(S)}$ is also compact.

I used the Caratheodory's theorem to show that $\text{Conv(S)}=\bigcup_{n=1}^{\dim(X)+1} T_n(S)$, now I need to show that $T_n(S)$ is closed and bounded but I'm stuck. Is this the right way to prove it? If it is how do can I proceed?

$T_n(S):=\{x\in X : \sum_{i=1}^n a_i \cdot x_i$ for some $a \in \Delta^{n-1}, \{v_1,...,v_n\} \subseteq S\}$ where $\Delta^{n-1}$ is the $n-1$ dimensional simplex.

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  • $\begingroup$ What is $T_n(S)$in your question? $\endgroup$ – polmath Oct 17 '14 at 11:30
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Hint: consider the continuous map $f : \Delta_{n-1} \times S^{n} \rightarrow \mathrm{co}(S)$ defined by $f(\lambda_1, \ldots, \lambda_n, x_1, \ldots, x_n) = \sum_{i=1}^n \lambda_i \cdot x_i$, where $n = \dim X + 1$, and $\Delta_{n-1}$ is the simplex $\{ (\lambda_1, \ldots, \lambda_n) \in \mathbb{R}_+ : \sum_i \lambda_i = 1 \}$.

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  • $\begingroup$ Does this works if S is not finite? $\endgroup$ – Rby Oct 18 '14 at 12:57
  • $\begingroup$ Sure, S compact is enough (but you need X with finite dimension). $\endgroup$ – polmath Oct 18 '14 at 13:53
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let $V=\{(t_1 ,t_2 , ... ,t_n )\in\mathbb{R}^{n+1} : t_1 , t_2 , ....,t_{n+1} \geq 0 \wedge t_1 + t_2 +...+t_{n+1} =1 \}.$ By Caratheodory theorem the function $T: V\times S^{n+1}\to \mbox{conv} (S) ,$ $$T(t_1 ,t_2 , ... t_{n+1} , x_1 , x_2 , ..., x_{n+1} ) =\sum_{j=1}^{n+1} t_j x_j $$ is suriection. And since $V\times S^{n+1}$ is compact then so is $T(V\times S^{n+1} ) =\mbox{conv} (S) .$

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