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.


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.

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    $\begingroup$ When you show the first direction, i.e. $C$ is contained in $H$, you use the premise that $H$ is the collection of closed half-spaces that contain $C$. that premise is not part of the premise OPs question. He just says "can be written as an intersection of half spaces " but does not say that $C$ is contained in $H$. so how can you use that? $\endgroup$ – makansij Oct 1 '18 at 22:13
  • $\begingroup$ The proof assumes there is at least one half space that contains $C$. A counterexample is $C=\mathbb{R}^n$. $\endgroup$ – Michael Jul 6 '20 at 10:05
  • $\begingroup$ Of course some people define $\cap_{H \in \phi}H=\mathbb{R}^n$. This assumes we somehow know we are working in $\mathbb{R}^n$ (the color green is a point that is “in each one” of the sets in the collection of no sets). $\endgroup$ – Michael Jul 6 '20 at 10:09

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