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 $bd(S)$
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.