I'm trying to prove the following:
$V$ topological vector space. $A,B\subseteq V$ nonempty, open subsets. Suppose that a hyperplane $H\subseteq V$ separates $A$ and $B$. Show $H$ strictly separates $A$ and $B$ and $H$ is closed
I feel like I want to apply the theorem of isolation / separation theorem, which states
If $A\subseteq V$ algebraically open, convex and $u\in V \backslash A$ $\Rightarrow \exists H $ affine hyperplane containing $u$ and strictly isolating $A$.
But honestly, I am very lost and have no clue where this is going, as my $A$ and $B$ are not convex and I cannot assume that $B\subset V\backslash A$,
Anybody that knows how I can approach this?
Cheers!