$d(x,E^c) > \frac{1}{j}$ is compact if $E$ is open.

Question

It is claimed in an analysis text that

Let $E \subseteq \Bbb R^n$ be an open set. Then $$K_j := \{ x \, :\, d(x, E^c) \ge 1/j \}$$ is a compact set.

How does one see this? I guess it is closed by $d:\Bbb R^n \rightarrow \Bbb R$, $x \mapsto d(x,E^c)$. But how does one show this is bounded?

Answer 1

It's true if $E$ is a bounded open set. Then $K_j$ is a subset of $U$ (hence bounded) and then closedness plus Heine-Borel do the rest.

Answer 2

This is false. Take $n=j=1$ and $E=(0,\infty)$. Then $K_j$ is unbounded and therefore not compact.