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Let the polytope defined by $$S:=co \left\{ x_1,x_2,...,x_k \right\}$$ where $x_1,x_2,...,x_k \in \mathbb{R^n}$ and $co \left \{... \right \}$ is the convex Hull. Prove that S i closed.

I tried the following. I want to show that $S=cl(S)$

I've proved that for all set $S \subseteq cl(S)$. Now I want to prove that $cl(S)\subseteq S$.

I know that $cl(S) = int(S) \cup bd(S)$ So, taking $x \in cl(S)$.

If $x \in int(S)$ then its clear that $x \in S$.

If $x \in bd(S)$ is not clear but intuitively the border is the convex combination between $x_i$ and $x_j$ (in pairs) and $i,j=1,2,...,k$. I don't know how to rite it formally and how to prove that my intuition about the border is actually $bd(S)$

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  • $\begingroup$ I would have thought that the border would be the convex combination of up to $n$ of the $x_i$s $\endgroup$
    – Henry
    Nov 21 '15 at 0:41
  • $\begingroup$ And how to prove that actually is the border? By definition? $\endgroup$ Nov 21 '15 at 1:37
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    $\begingroup$ If you assume $y$ is a limit point then there is an infinite sequence of points $y_n$ that converge to $y$, and all $y_n$ points can be written as a convex combination of $\{x_1, ..., x_k\}$. Can you finish the argument? Do you know Bolzano-Wierstrass? $\endgroup$
    – Michael
    Nov 21 '15 at 3:37
  • $\begingroup$ I like this approach. Thank you $\endgroup$ Nov 21 '15 at 17:02
  • $\begingroup$ As se sequence is convergent then is bounded. For B-W this sequence has a convergent subsequence. How to prove that the limit is a actually a convex combination? $\endgroup$ Nov 21 '15 at 17:10
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In fact, the convex hull $\mathrm{conv}\{x_1,\ldots,x_k\} := \{\sum_{1 \le j \le k}t_kx_k | t \in \mathbb{R}^n, t \ge 0, \sum_{1 \le j \le n}t_j = 1\}$ is compact (in the usual euclidean topology)!

Step 1: The simplex $\Delta_n := \{t \in \mathbb{R}^k | t \ge 0, \sum_{1 \le j \le k}t_j = 1\}$ is compact. Indeed it closed, being the intersection of closed sets, namely the orthant $\mathbb{R}^k_+$ and the hyperplane $H:= \{t \in \mathbb{R}^k | \sum_{1 \le j \le k}t_j = 1\}$. Next, $\Delta_n$ is a subset of the hypercube $[0, 1]^k$, and is therefore bounded.

Step 2: The map $g: \Delta_k \rightarrow \mathbb{R}^k$, $t \mapsto \sum_{1 \le j \le k}t_kx_k$ is continuous. Thus $\mathrm{conv}\{x_1,\ldots,x_k\} := g(\Delta_k)$ is the continuous image of a compact set and is therefore compact.

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