6

Is it true or false?

If $A\subset \Bbb R^n$ is convex and $\Bbb R^n\setminus A$ is convex then $A=\emptyset$, $A=\Bbb R^n$ or $\partial A$ is a hyperplane.

I remember vaguely this fact from university, but I'm not sure if it is true or not. (My guts say to me that it is true, but...)

Hints or references are welcome.

To try it myself, I have considered two points $P\in A$ and $Q\in\Bbb R^n\setminus A$. Then there is a "first" point $X$ ($\sup$ can be used to formalize this idea) in the segment $PQ$ such that $PX$ is not in $A$. (Or a "last" point $X$ such that $PX$ is in $A$). But I still have to prove that all these points $X$ are in a hyperplane.

Is it true for infinite dimensional vector spaces?

ajotatxe
  • 65,084
  • 1
    The closure of a convex set is convex. The boundary of a set $A$ equals to $\overline{A} \cap \overline{A^c}$. So such a set has convex boundary. – Crostul Jan 05 '17 at 17:47

1 Answers1

6

Your claim follows from the following well-known property of convex sets:

Let $A\subset\mathbb{R}^n$ be a convex set and $x\in \partial A$. Then there exists a hyperplane $H$ such that $x\in H$ and $A$ lies in one of the half spaces for $H$.

Google: Hyperplane separation theorem.

Ievgen Bondarenko
  • 361
  • 1
  • 7