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In $\mathbb{R}^n$, I know that if $A$ is a convex set and $b$ in the boundary of $A$. Then we can separate $A$ and $b$, which means there exists $f \in \mathbb{R}^n$ such that $f\cdot x \ge f \cdot b$ for all $x \in A$.

I also know that if $A$ and $B$ are two disjoint convex sets. Then we can separate them (just apply the above result for the convex set $A - B$ and the point $0$).

My question is: If $A$ and $B$ are two convex sets with disjoint interiors (by interior, I mean topology interior), can we separate $A$ and $B$?

I think that this is true. My approach is just take a hyperplane separate $B$ and the interior of $A$ (they are disjoint), then prove that this hyperplane also separates $A$ and $B$. Which is: Let $f \in \mathbb{R}^n$ and $\alpha \in \mathbb{R}$ such that $f \cdot a \ge \alpha \ge f \cdot b$ for all $a \in \operatorname{Int}A$, $b \in B$. But I stuck with proving that $f\cdot a \ge \alpha$ for all $a\in A$.

This is true if $\overline{\operatorname{Int}A} = \overline{A}$. But is this true for any convex set $A \in \mathbb{R}^n$?

Thanks very much for any help!

Du Phan
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    The interiors of $A$ and $B$ (supposed nonempty) are convex and disjoint. A hyperplane that separates these interiors also separates $A$ and $B$. – Christian Blatter May 25 '14 at 17:42

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If $A$ and $B$ are two convex sets with disjoint interiors (by interior, I mean topology interior), can we separate $A$ and $B$?

Counterexample in Did's style: $$\mathsf{X}$$

(Or, take any convex set and cross it by a line). So you need the interiors to be nonempty. with this assumption your argument works, because

$\overline{\operatorname{Int}A} = \overline{A}$ for every convex set $A\subset \mathbb R^n$ with nonempty interior.

Proof: fix an open ball $B(a,r)$ contained in $A$. For every $x\in A$ and every $\epsilon>0$, the ball $B(x+\epsilon(a-x),\epsilon r)$ is contained in $A$, because it's contained in the convex hull of $x$ and $B(a,r)$. Since $\epsilon$ can be arbitrarily small, $x\in \overline{\operatorname{Int}A}$. This proves $A\subseteq \overline{\operatorname{Int}A}$, from which $\overline{A}\subseteq \overline{\operatorname{Int}A}$ follows.

Somewhat related: Why does a convex set have the same interior points as its closure?

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