# Convex hull is the minimal convex set containing $X$

How one can prove that convex hull is the minimal convex set containing $X$?

We need to show that for each convex set $M$ if $X\subseteq M$ then $conv(X)\subseteq M$.

I am thinking of proof by contradiction. Let $x\in conv(X)$ but $x \notin M$, then we can separate $x$ from $M$. How to get the contradiction?

Lets define convex hull in this way:

$conv(X) = \{x\ |\ x=\alpha_1x_1+\dots\alpha_kx_k,\ \alpha_i\ge0(1\le i\le k),\ x_i\in X,\ \alpha_1+\dots\alpha_k=1,\ k\in\mathbb N \}$

-
This is what I think of as the definition of convex hull. What definition are you using? –  Chris Eagle Mar 2 '13 at 20:33
You need to add that $M$ is convex. –  1015 Mar 2 '13 at 20:37
First pick your definition among the four equivalent properties here: en.wikipedia.org/wiki/Convex_hull –  1015 Mar 2 '13 at 20:38
–  Zev Chonoles Mar 2 '13 at 20:38