Let $A_0$ be a compact set (closed and totally bounded in some metric space) and consider a sequence of sets $A_n=\{x:d(x,A_0)<1/n\}$. For each $n$, $A_0\subset B_n\subset A_n$ is compact.
$$B=\bigcup_{n\ge1}B_n$$
Is there any condition under which $B$ is also compact?
Thank's!
Best Answer
$B$ is always compact. Let $\mathscr{U}$ be an open cover of $B$. $A_0\subseteq B$, and $A_0$ is compact, so some finite $\mathscr{U}_0\subseteq\mathscr{U}$ covers $A_0$. Let $V=\bigcup\mathscr{U}_0$; $V$ is an open nbhd of the compact set $A_0$, so there is an $n\in\Bbb Z^+$ such that $A_n\subseteq V$. Let $K=\bigcup_{k=1}^nB_k$; then $K$ is a compact subset of $B$, so some finite $\mathscr{U}_1\subseteq\mathscr{U}$ covers $K$, and $\mathscr{U}_0\cup\mathscr{U}_1$ is a finite subset of $\mathscr{U}$ covering $B$.