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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.

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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 $x \le 0$.

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  • $\begingroup$ 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. $\endgroup$ – freddy Sep 25 '18 at 13:28
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    $\begingroup$ @freddy: en.wikipedia.org/wiki/Hahn–Banach_theorem#Geometric_Hahn–Banach_theorem would require that $X$ has nonempty interior. $\endgroup$ – GEdgar Sep 25 '18 at 13:37

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