Does a dense $G_\delta$ subset of a complete metric space without isolated points contain a perfect set? Let $(X,d)$ be a complete metric space without isolated points. Is it true that each dense $G_\delta$ subset of $X$ contains a nonempty perfect set (i.e. closed without isolated points)?
Thanks.
 A: Without the assumption of separability, it remains true that every $G_\delta$ subset is completely metrizable, but merely being uncountable completely metrizable does not ensure the existence of a perfect subset.  But such a perfect set may be found within any dense $G_\delta$ by a slight modification of the usual embedding argument that uncountable Polish spaces contain copies of the Cantor set.
Fix open dense sets $U_n$ which intersect to your dense $G_\delta$.  We build inductively for each finite binary string $s$ a nonempty open set $X_s$ satisfying the following conditions:
$X_{s0} \cap X_{s1} = \emptyset$
$\mathrm{cl}(X_{s0}), \mathrm{cl}(X_{s1}) \subseteq X_s$
$X_s \subseteq U_{|s|}$
$\mathrm{diam}(X_s) \leq 2^{-|s|}$,
where  $|s|$ is the length of $s$.  The first two conditions can be met since the space has no isolated points, the third is by density of each $U_n$, and the fourth is no problem.  By completeness of the metric, for each infinite binary string $\sigma \in 2^\mathbb{N}$ the intersection $\bigcap_{s \subseteq \sigma} X_s$ (where $s$ ranges over initial segments of $\sigma$) is a singleton.  This allows us to build a continuous embedding of the Cantor set by setting $f(\sigma)$ to be the unique element of this singleton.  Finally, the third condition in our construction ensures that each point in the image of this embedding is an element of each $U_n$, and thus lands in our target $G_\delta$.
A: Yes, as this set is an uncountable Polish space in it's own right (using a different metric), and all such spaces contain a Cantor set.
