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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 $x_0$.

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  • $\begingroup$ 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. $\endgroup$ – Theo Bendit Oct 30 '17 at 8:17
  • $\begingroup$ Thanks for the answer. I was talking about finite dimensions but would love to know a counterexample in infinite dimensional space. $\endgroup$ – Curious Oct 30 '17 at 8:27
  • $\begingroup$ Here's one: mathoverflow.net/questions/121526/… $\endgroup$ – Theo Bendit Oct 30 '17 at 8:35
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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 $B = S$.

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