Distance to the boundary

Question

Let $F$ be a closed set in $\mathbb{R}^n$ and $\Omega:=\mathbb{R}^n\setminus F$. For $x\in\Omega$, do we have ${\rm dist}(x,F)={\rm dist}(x,\partial \Omega)$?

Answer 1

$\newcommand{\dist}{\operatorname{dist}}$ Yes.

Let $y \in F$ be the point that attains the minimal distance, i.e. $\dist(x, F) = \dist(x, y)$. One should check if such point $y$ exists; Consider the ball $B_R(x)$ with property $B_R(x) \cap F \ne \varnothing$. (If it there is no such radius $R$, $F = \varnothing$ which is not interested case.) Then the distance between $x$ and $F$ is the distance between $x$ and $F \cap \overline{B_R(x)}$, which is closed and bounded so is compact, so the infimum value is attained in some points, say $y$.

Now assume that $y \not\in \partial \Omega$. Then $y \in F\setminus \partial F$, i.e. $y \in F^\circ$. So there is an open ball $U_r(y) \subset F^\circ$. Consider the element of intersection of the line $xy$ and $U_r(y)$, say $z$. (Let $z \ne y$.) Compare $\dist(x,y) $ and $\dist(x, z)$.