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:

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


1 Answer 1


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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