TITLE: Convex hull of a compact in $R^n$ is compact
QUESTION [9 upvotes]: Show that if $A \subset \mathbb{R}^n$ is compact then the convex hull of $A$ is compact.
Solution: $A \subset \mathbb{R}^n$ is compact then is limited and closed, therefore $A \subset B_{[0,r]}$ (Ball closed center $0$ and radius $r$) for some $r>0$. Since $B_{[0,r]}$ is convex and $Conv(A)$ is the least convex that contains $A$: $Conv(A) \subset B_{[0,r]}$. It remains to show that $Conv (A)$is closed, how do I do this?

REPLY [12 votes]: A theorem of Caratheodory states that each element of the convex hull of
$A$ is a convex combination of $n+1$ elements of $A$ (since $A\subseteq\Bbb R^n$).
So the convex hull is the image of a compact set $A^{n+1}\times\Delta_n$
under a continuous map $F$, and so is compact. Here $\Delta_n=\{(t_0,\ldots,t_n):t_j\ge0,t_0+\cdots+t_n=1\}$ and $F$ is defined by
$$F(a_0,\ldots,a_n;t_0,\ldots,t_n)=\sum_i t_i a_i.$$