# 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?

Thanks in advance.

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You might think of the ellipsoid as a convex region then you have infinitely many vertices. And at each point you can introduce a hyperplane. IMHO, this is a good place to start with. – user13838 Oct 3 '11 at 9:08
Convexity can be thought of in different ways - what you have been asked to prove is that two possible ways of thinking about convexity are in fact equivalent. It would be useful if you could tell us the basic definition of convexity you are using, and the definition of the convex hull which you have been given - since the existence of equivalent characterisations means that people begin with different basic definitions, yet end up in the same place. – Mark Bennet Oct 3 '11 at 9:35
Sorry, @percusse, but I think I do not understand your comment. I think I am quite limited in math. I am trying to improve by doing a lot of exercises, but I am still far for understanding that relationship. :-( – Fernandez Oct 3 '11 at 10:17
@Mark, I'll try to edit my question to reflect what you are asking for. – Fernandez Oct 3 '11 at 10:18

## 1 Answer

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 $K_i$.

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