# The standard n-simplex is compact set

$" Let \ K\subseteq \mathbb R^n \ be \ a \ compact \ set,\ then\ the\ convex \ hull \ of\ K\ is\ also\ compact \ set\ "$

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!!

• That's because it is closed and bounded. Bounded: note that it is contained insisde the cube $[0,1]^n$. Closed: it is the inverse image of $\{ 1 \}$ through the continuous map $(x_i) \mapsto \sum_i x_i$. May 31 '16 at 9:53
• well... you helped me a lot!! I had been stuck to this (although it was simple) !! Thank you very much! May 31 '16 at 10:14

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 $[0, 1]^n$.