Let $X$ be locally compact metric space, and $X^*$ be its one-point compactification, with the extra point being $\{\infty\}$. Then, the following are equivalent –
- $X$ is separable
- $X = \bigcup\limits_{n=1}^\infty K_n$, where $K_n$ is compact and $K_n \subseteq K_{n-1}^\circ \ \forall n$
- $X^*$ is metrizable
I've been able to prove $3 \implies 1$, but have failed to prove $2\implies 3$ or $1\implies 2$.
For $1\implies 2$, I think we have to use separability to get a countably dense subset, and use local compactness to get a countable collection of compact subsets $L_n$. I tried to define $K_n = \bigcup\limits_{i=1}^n L_i$, but this choice did not work.
For $2\implies 3$, I'm pretty sure we have to use the following metrization theorem –
$X$ is metrizable iff every $x\in X$ has a countable neighbourhood system $\{U_n^x\}$ such that –
- $y\in U_n^x \implies U_n^y\subseteq U_{n-1}^x$
- $y\not\in U_{n-1}^x \implies U_n^y\cap U_n^x = \varnothing$
The choice $U_n^\infty = \{\infty\}\cup K_n^c$, where $A^c$ is the complement of $A$, seems like the most obvious choice. But, I've not been able to get a solution.
Any help is appreciated!
Best Answer
$\newcommand{\cl}{\operatorname{cl}}$$(3)\implies(1)$ is even easier than that: $X^*$ is a compact metrizable space, so it’s separable.
For $(1)\implies(2)$, note that $X$ is a separable metric space, so $X$ is second countable. Show that it has a countable base $\mathscr{B}=\{B_n:n\in\Bbb N\}$ of open sets with compact closures. Let $K_0=\cl B_0$. Cover $K_0\cup\cl B_1$ with members of $\mathscr{B}$, take a finite subcover, let $U_1$ be its union, and let $K_1=\cl U_1$. Continue in this vein.
For $(2)\implies(3)$ I would use the Urysohn metrization theorem.