# Convex hull as an infinite intersection

I have been asked to prove that the convex hull of a set $S$ equals the intersection of all possible convex sets containing $S$. As always, I have found difficulties while doing so. I'll try to describe my approximation. As we are talking about set equalities, I am trying to prove that, given an element of a set, it belongs to the other.

Edited: In this context, I'll say a set $C$ is convex if, given $p,q \in C$, then $\alpha p + (1-\alpha)q \in C, \alpha\in [0,1]$. And my definition of convex hull is: $conv(S) = \{ \theta_1 x_1+\theta_2 x_2 +\dots+\theta_k x_k : x_1,x_2,\dots,x_k \in S, \theta_1+ \theta_2+\dots+\theta_k = 1 \}$

First, if I have an element $z \in conv(S)$ then, by definition, $z$ is a convex combination $\theta_1 x_1+\dots+\theta_k x_k$ of elements of $S$. Then, as $S \subset K_i$, where $K_i$ is a convex set, we have that $z$ is also a member of $K_i$. As this holds for each $K_i$, it is then true for the intersection $\bigcap K_i$.

Problems arise when I try to prove the opposite implication, that is, if I have an element $z \in \bigcap K_i$ and I want to prove that it belongs to the convex hull. I have been trying for a while, and I am not able to do it.

Is my proof right? Any hint about the way I can try to prove the opposite implication?

The convex hull $C$ itself is a convex set containing $S$. Hence if $z$ is in the intersection of every convex set containing $S$, then $z$ is in $C$ by definition. In other words, $C$ is one of your sets $K_i$.