Supporting hyperplane theorem and halfspaces

Question

Does the supporting hyperplane theorem imply that closure of any convex set can be expressed as intersection of halfspaces(possibly infinitely many halfspaces). The statement of supporting hyperplane theorem is: For any nonempty convex set $C$, and any $x_0 \in boundary(C)$, there exists a supporting hyperplane to $C$ at $x_0$.

Answer 1

By definition, the convex set $S$ is contained in one of the halfspaces bounded by its supporting hyperplane. So $S$ lies in any intersection of these halfspaces.

By the separating hyperplane theorem, any point $a$ outside of $S$ can be separated, such that the halfspace that contains $S$ does not contain $a$. So the intersection of all such subspaces leaves you with $S$.

That is, for the set of points $S^c = \{x:x \notin S\}$, there exists an intersection of halfspaces $B$ such that $S^c\cap B = \emptyset$ and $S \subseteq B$. Thus $B = S$.