Convex sets as intersection of half spaces I want to prove that any closed convex sets can be written as an intersection of half spaces using only the separation theorem as a pre-requisite. I'm getting a feel that I need to show two sets are subsets of each other, but not being able to understand how exactly to go about it.
 A: I think that your approach should work.
Let $C\subseteq \mathbb{R}^n$ be a closed, convex set. Let $\mathcal{H}$ be the collection of closed, half-spaces that contain $C$. You would like to show that 
$$C = \bigcap_{H\in \mathcal{H}}H.$$
First we can show $C \subseteq \bigcap_{H\in \mathcal{H}}H.$ Let $x\in C$. By the definition of $\mathcal{H}$, any $H\in \mathcal{H}$ satisfies $C\subseteq H$. Hence $x\in H$ for any $H\in \mathcal{H}$ and therefore $x\in \bigcap_{H\in \mathcal{H}}H.$ This gives us the desired inclusion. 
It is left to show that $C \supseteq \bigcap_{H\in \mathcal{H}}H.$ We prove this using the contrapositive, that is we will show that if $x\not\in C$ then $x\not\in \bigcap_{H\in \mathcal{H}}H.$ So choose $x$ such that $x\not\in C$. Since $C$ is closed and convex, there is a hyperplane that strictly separates $x$ from $C$. This hyperplane defines a half space $H$ containing $C$. Hence $x\not\in H$ implying that $x\not\in \bigcap_{H\in \mathcal{H}}H$. This proves the desired inclusion. 
