The equidistant set of a closed set 4 This question is related to this (The equidistant set of a closed set 1). For a set $F$ in $\mathbb{R}^m$ ($m>1$) and $r>0$, let $F_r$ be the set of all points in $\mathbb{R}^m$ having a distance of $r$ to $F$. Suppose $F$ is a connected compact set whose complement in $\mathbb{R}^m$ is also connected. My question is: is it possible that for some $r>0$  the complement of $F_r$ have an infinite number of components?
 A: Edited answer.
Let $(x_n)_{n>0}$ be a decreasing sequence covergent to zero with all terms $0<x_n\leq 1$ (It can be $x_n=\frac{1}{n}$ for instance). Moreover let $x_0=0$.
Let $I_n$ be the segment joining $(x_n,2),(x_n,1)$ and $I_n'$ be the segment joining $(x_n,-2),(x_n,-1)$.
Consider a broken line $L$ going (in a given order) through points
$$(1,1), (1,2), (-1,2), (-1,-2), (1,-2), (1,-1) $$
and define
$$F:= L\cup \bigcup_{n\geq 0}(I_n\cup I_n')$$
For $r=1$ the complement of the set $F_r$ has infinitely many components.
Previous answer, which gives the construction of $F$ such that $F_r$ has infinitely many components.
Let $B_n$ be an open ball centered at $\left(\frac{1}{n},0\right)$ of radius $1$  and $S_n$ be its sphere for each $n\in\mathbb{N}$.
$$F:=\bigcup_{n\in\mathbb{N}}\left(S_n\setminus\bigcup_{m\in\mathbb{N}}B_m\right)\cup\left\{(0,1),(0,-1)\right\}$$
$F$ has all required properties (it is actually a curve) and for each $n\in\mathbb{N}$ the set $\left\{\left(\frac{1}{n},0\right)\right\}$ is a component of $F_r$ for $r=1$.
