# Hahn-Banach separation theorem where one of the sets is neither open nor closed.

Let $$X$$ be a convex set in the real Hilbert space with countable basis, $$0\notin X$$. Is it true that there is a hyperplane containing $$0$$ and disjoint with $$X$$?

For an open $$X$$ this would be a direct consequence of the Hahn-Banach separation theorem.

In the plane, let $$X = \{(x,y): x>0\} \cup \{(0,y):y>0\} .$$ Note $$X$$ is convex, and $$(0,0) \notin X$$. Now try to find a line through $$(0,0)$$ disjoint from $$X$$.
What you can find is the line $$x=0$$, such that $$X$$ is in the half-plane $$x \ge 0$$ and $$(0,0)$$ is in the opposite half-plane $$x \le 0$$.
• Thanks, that answers the question. The follow up will be can you always find a hyperplane such that $X$ is in one closed half-space, and $0$ is in the other? As you've pointed out you can do it in your example. – freddy Sep 25 '18 at 13:28
• @freddy: en.wikipedia.org/wiki/Hahn–Banach_theorem#Geometric_Hahn–Banach_theorem would require that $X$ has nonempty interior. – GEdgar Sep 25 '18 at 13:37