Show that $\mathbb{N},$ the set of natural numbers, is a locally
compact metric space with the metric $d(x,y) = |x-y|$ for all $x$ and
$y$ in $\mathbb{N}.$
My attempt:
For any $n \in \mathbb{N},$ $\{ n \}$ is a neighbourhood of $n.$
Clearly $\{ n \}$ is bounded.
Since $\{ n \}^c = \mathbb{N} \setminus \{n \}$ is a subset of $\mathbb{N},$ we have $\{ n \}^c$ is open.
Therefore, $\{ n \}$ is closed in $\mathbb{N}.$
By Heine-Borel theorem, $\{ n \}$ is compact.
Hence, $\mathbb{N}$ is locally compact.
Is my proof correct?
Best Answer
Yes you are right. You could say any finite set is compact. so any singleton set is compact.