# 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:

$\overline{\left\{{x_{i}}\right\}}=\left\{{x_{i}}\right\}$ (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 $\left\{{x_{i}}\right\}^{o}=\emptyset$ and we have that $\left({\overline{\left\{{x_{i}}\right\}}}\right)^{o}=\emptyset$ 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, $D=\bigcap\limits_{i=1}^{\infty}{U_{i}}$ such that every $U$ is open. As $D$ is dense, then each $U$ is dense, because $D\subset U_{i}\Rightarrow\overline{D}=X\subset\overline{U_{i}}\quad\forall i$. But then $X-D=X-\bigcap\limits_{i=1}^{\infty}{U_{i}}=\bigcup\limits_{i=1}^{\infty}{(X-U_% {i})}$ and $\left({\overline{X-U_{i}}}\right)^{o}=\left({X-U_{i}^{o}}\right)^{o}=\left({X-% U_{i}}\right)^{o}=X-\overline{U_{i}}=\emptyset$

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

Title no countable dense subset of a complete metric space is a $G_{\delta}$ NoCountableDenseSubsetOfACompleteMetricSpaceIsAGdelta 2013-03-22 14:59:06 2013-03-22 14:59:06 gumau (3545) gumau (3545) 8 gumau (3545) Result msc 54E52