no countable dense subset of a complete metric space is a $G_{\delta }$

Let $(X,d)$ be a complete metric space with no isolated points, and let $D\subset X$ be a countable dense set. Then $D$ is not a$G_{\delta }$ set (http://planetmath.org/G_deltaSet).

Proof

First, we will prove that $D$ is first category. Then, supposing that $D$ is a $G_{\delta }$, we will conclude that $X-D$ must be first category. But then so must be $X$, which is absurd because $X$ is complete.

1) $D$ is a first category set:

By hypothesis $D={\left\{x_i\right\}}_{i\in N}$. Let’s see that each singleton is nowhere dense if $X$ has no isolated points:

$\overline{\left\{x_i\right\}}=\left\{x_i\right\}$ (trivially). Suppose that ${\left\{x_i\right\}}^o=\left\{x_i\right\}$. Then there is a ball aisolating the point. Absurd ($X$ has no isolated points). Then ${\left\{x_i\right\}}^o=\mathrm{\varnothing }$ and we have that ${\left(\overline{\left\{x_i\right\}}\right)}^o=\mathrm{\varnothing }$ so every singleton is nowhere dense and $D$ is of first category because it is a countable union of nowhere dense sets.

2) Suppose $D$ is a $G_{\delta }$, that is, $D=\bigcap _{i=1}^{\mathrm{\infty }}{U}_i$ such that every $U$ is open. As $D$ is dense, then each $U$ is dense, because $D\subset U_i\Rightarrow \overline{D}=X\subset \overline{U_i}\mathit{\hspace{1em}}\forall i$. But then$X-D=X-\bigcap _{i=1}^{\mathrm{\infty }}{U}_i=\bigcup _{i=1}^{\mathrm{\infty }}(X-U_i)$ and ${\left(\overline{X-U_i}\right)}^o={\left(X-U_i^o\right)}^o={\left(X-U_i\right)}^o=X-\overline{U_i}=\mathrm{\varnothing }$

which implies that $X-D$ is of first category. Then $D\bigcup (X-D)=X$ is of first category. Absurd, because $X$ is complete.