If $A$ and $B$ are disjoint convex sets, is it possible to write $A=\bigcup_{n\in\mathbb{N}}A_n$ where: 1) each $A_n$ is a convex set and 2) The distance between $A_n$ and $B$, $d(A_n, B)=\inf\{\|x-y\|:x\in A_n, y\in B\}$ is greater than zero for each $n\in\mathbb{N}$?

Motivation: The separating hyperplane theorem states that if $A$ and $B$ are disjoint convex sets, there is some hyperplane separating them; one way of phrasing this is to say that there exists a vector $c$ such that $$\sup_{x\in A}c\cdot x\leq \inf_{x\in B} c\cdot x.$$

Every proof of this theorem I've seen proceeds as follows. First prove a lemma: $$\textrm{Lemma. If $B$ is convex and $y\not\in B$, there is a vector $c$ such that $c\cdot y \leq \inf _{x\in B} c\cdot x$}.$$ Then, consider $A-B=\{a-b:a\in A, b\in B\}$. This is a convex set which doesn't contain 0, so an appeal to the lemma yields the separating hyperplane.

I'm interested in a more "natural" proof (using term "natural" loosely), mostly because the trick of considering $A-B$ doesn't feel like something I would ever come up with just trying to attack the problem geometrically. In particular, if the distance between the two sets $d(A,B)$ is greater than 0 a more geometric proof comes out of considering a hyperplane bisecting the line between two closest points on the (closures) of $A$ and $B$. If in general we could approximate $A=\bigcup A_n$ where each $A_n$ is convex and $d(A_n, B)>0$ then we could take the limit of the corresponding hyperplanes.

  • $\begingroup$ Thinking about it some more, I think something like $A_\epsilon=\{x\in A: \textrm{$\epsilon$ ball around $x$ is a subset of $A$}\}$ might work... $\endgroup$ – Justin Jan 17 '16 at 18:33

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