# 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:

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

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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

In my answer here: Proof that the Convex Hull of a finite set S is equal to all convex combinations of S I have given a proof that goes from the definition of the convex hull as the intersection of all convex sets containing the set of points to some of the other equivalent definitions, so you will work your way backwards if you are starting with a different equivalent definition. See the comment by julien.

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