[Math] Does a dense $G_\delta$ subset of a complete metric space without isolated points contain a perfect set

Question

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.

Answer 1

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 $G_\delta$.