$ " Let \ K\subseteq \mathbb R^n \ be \ a \ compact \ set,\ then\ the\ convex \ hull \ of\ K\ is\ also\ compact \ set\ " $
In order to prove this we use that the standard n-simpex as defined by :
$ S\ =\ \{ \ (t_1,t_2,.....,t_{n+1}) \ :\ t_i \ge 0 \ for\ every\ i\ \le (n+1) \ and\ \sum_{i=1}^{n+1} t_i \ =1 \} $
is compact subset of $ \mathbb R^{n+1} $. So my question, is why is this true?
I know that this may be obvious or silly (because I didn't find any proof of that wherever I searched) but I recently started to study convex analysis by myself and I am going through the basics now. I would appreciate it if anyone could make this a little more clear for me.
Furthermore what books would you suggest ? Thanks in advance!!
I answered this question here. For the main ingredient, basically the $n$-simplex is closed being the intersection of closed sets, and it's bounded given that it's contained in the hypercube $[0, 1]^n$.