Theorem: Let $Y$ be a dense subset of a Hausdorff topological space $X$. If $Y$ is completely metrizable, then $Y$ is a $G_\delta$ set in $X$.
This is explained in detail here. The idea is to take the set of points $a\in X$ for which one can find neighborhoods whose trace on $Y$ has arbitrarily small diameter. The set of such points is a $G_\delta$ containing $Y$ and by completeness such $a$ must belong to $Y$, so $Y$ itself is a $G_\delta$ in $X$.
Note that $X$ is not even required to be metrizable here. An intructive example is the Niemytzki plane $X$ (not metrizable because it is separable but not second countable). Take $Y = \{(x,y): y>0\}$.
As a subspace of $X$, $Y$ has the usual Euclidean topology. The Euclidean distance is not complete on $Y$, but $Y$ is completely metrizable by the following equivalent metric: for $z_1=(x_1,y_1)$, $z_2=(x_2,y_2)$ as complex numbers take
$$d(z_1,z_2)=|z_1-z_2|+|\frac{1}{y_1}-\frac{1}{y_2}|$$
(Cauchy sequences in $(Y,d)$ converge in $Y$ because the term $\frac{1}{y}$ keeps the points away from the boundary.) And indeed, $Y$ is a $G_\delta$ in $X$ because it's open in $X$.
(1) Are there any good examples where $Y$ (dense) is not open in $X$?
(2) If in the theorem above we relax the assumption that $Y$ be dense in $X$, all we can conclude is that $Y$ is a $G_\delta$ in its closure $\overline{Y}$. Now if $X$ is a metric space, any closed subset is a $G_\delta$, so $Y$ would be a $G_\delta$ of a $G_\delta$ and therefore $Y$ would be a $G_\delta$ in $X$. Are there any good examples of a (non-metrizable) Hausdorff space $X$ with a completely metrizable subspace $Y$ that is not a $G_\delta$ in $X$?
Best Answer
A very natural example where $Y$ is not open in $X$ is $X=\mathbb{R}$, $Y=\mathbb{R}\setminus\mathbb{Q}$. Here is it not obvious that $Y$ actually is completely metrizable; one way to prove it is to use continued fractions to show that $Y$ is actually homeomorphic to $\mathbb{N}^{\mathbb{N}}$.
For a very simple example where $Y$ is not dense and also not $G_\delta$ in $X$, let $X=[0,1]^I$ for an uncountable set $I$ and let $Y$ be a singleton. Then $Y$ is trivially completely metrizable, but it is not $G_\delta$ since any intersection of fewer than $|I|$ basic open neighborhoods of a point of $X$ is still unconstrained on some coordinates.