# Supporting hyperplane theorem and halfspaces

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 $x_0$.

• I'm assuming you're talking about finite-dimensions? In infinite dimensions, this theorem doesn't hold (although it does hold for $x_0$ in a norm-dense subset of the boundary of $C$). It also holds under the assumption that $C$ has non-empty interior. Oct 30 '17 at 8:17
• Thanks for the answer. I was talking about finite dimensions but would love to know a counterexample in infinite dimensional space. Oct 30 '17 at 8:27
• Here's one: mathoverflow.net/questions/121526/… Oct 30 '17 at 8:35

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 $B = S$.