how to prove a set is a closed half space: here is the condition

Question

Let A be a closed convex set and the complement of A is convex. prove that A is a closed half space. I guess it's like proving half space belongs to A and A belongs to halfspace, the latter is obvious, so how to prove the former. when assume not, how to use the property of convex~~ Thanks~~

Answer 1

First note, that this is only true if both $A$ and $X \setminus A$ are non empty (otherwise $A = \emptyset$, $X$ are also possible).

Let's start with finding the hyperplane bounding the space in question. Let's denote the topological vector space we are working in with $X$. As $A$ and $X \setminus A$ are convex, and $X \setminus A$ is open, there is - by the Hahn-Banach seperation thereom - an $x^* \in X^*$ and an $\alpha \in \mathbf R$, such that $$ \tag 1 \Re x^*(a) \le \alpha < \Re x^*(b), \qquad \text{all } a \in A, b \in X\setminus A $$ Hence, $A \subseteq \{\Re x^* \le \alpha\}$, and $X\setminus A \subseteq \{\Re x^* > \alpha\}$. The latter implies $$A \supseteq X \setminus \{\Re x^* > \alpha\} = \{\Re x^* \le \alpha \}$$ Hence, $A = \{\Re x^* \le \alpha\}$.