Isolated points in a complete metric space

Question

I need to prove that if $X$ is a countable (infinite) complete metric space, then $X$ has infinitely many isolated points.

I have read that the Baire Category theorem implies that $X$ should have at least one isolated point, but I have no idea how to show the set of isolated points is infinite.

Thanks for any help!

Answer 1

Suppose that it has a finite number of isolated points $x_1,..,x_n$, $Y=X-\{x_1,..,x_n\}$ is still complete. Every element $y\in Y$ as an empty interior. You can write $Y=\cup_{n\in\mathbb{N}}\{y_n\}$, Baire implies that $\cup_{n\in\mathbb{N}}\{y_n\}=Y$ has an empty interior. Contradiction.

Answer 2

Well, as Moishe points out in the comments above, if $X$ has only finitely-many isolated points, then we can remove them all, and obtain a complete, countably-infinite metric space (say $Y$) with no isolated points.

But this is impossible! Let $f:\Bbb N\to Y$ be a bijection, and consider the sets $O_n:=Y\setminus\bigl\{f(n)\bigr\}.$ Each $O_n$ is open (obviously) and dense (since $f(n)$ is not isolated in $Y$), so since $Y$ is a complete metric space, then BCT tells us that $\bigcap_{n\in\Bbb N}O_n$ is dense in $Y,$ but this is absurd, since $\bigcap_{n\in\Bbb N}O_n=\emptyset.$