2
$\begingroup$

$ " 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!!

$\endgroup$
2
  • 2
    $\begingroup$ 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 $(x_i) \mapsto \sum_i x_i$. $\endgroup$
    – Crostul
    May 31 '16 at 9:53
  • $\begingroup$ well... you helped me a lot!! I had been stuck to this (although it was simple) !! Thank you very much! $\endgroup$ May 31 '16 at 10:14
3
$\begingroup$

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 $[0, 1]^n$.

$\endgroup$
1
  • $\begingroup$ If I 'd seen it eariler , I wouldn't have posted this ... :) Thanks a lot!! $\endgroup$ Jun 1 '16 at 9:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.