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.

-
I think the answer to the question is yes, but I don't have the time to write up the argument I have in mind. It's essentially the same as the usual embedding argument of the Cantor set, except at each level of the embedding you make sure to land in one of the countably many dense open sets which intersect to your target $G_\delta$. See this link for a prototype argument: math.stackexchange.com/questions/68396 –  user83827 Nov 12 '11 at 20:45

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 $G_\delta$.

-
Thanks for comments and answer. –  Richard Nov 13 '11 at 21:45
You're welcome! –  user83827 Nov 13 '11 at 22:29

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.

-
This is certainly true if $X$ is separable, but this doesn't seem to be assumed in the question. –  user83827 Nov 12 '11 at 19:42
Oh, I see, you meant to write "completely metrizable" instead of "Polish." Now I understand. –  user83827 Nov 12 '11 at 19:54
Is it true that complete metrizable uncouuntable space contains a set homeomorphic with Cantor set? –  Richard Nov 12 '11 at 20:17
Actually it's not true (consider an uncountable discrete space), so this particular argument seems to need fixing. –  user83827 Nov 12 '11 at 20:26
I was implicitly assuming separability (hence the Polish). –  Henno Brandsma Nov 13 '11 at 6:16