Does every metrizable space admit a completion in which the original space is Borel?

Question

Let $X$ be a metrizable space. Does $X$ admit a compatible metric $d$ such that if $(\hat X,\hat d)$ is the completion of $X$ with respect to $d$, then $X$ is a Borel subset of $\hat X$? I'm particularly interested in separable $X$, but if something can be said in general that would also be nice.

Answer 1

A Bernstein set will give a counterexample. Recall that a Bernstein set is a subset $X \subseteq \mathbb{R}$ such that for every uncountable perfect $P \subseteq \mathbb{R}$ both $X \cap P$ and $X \cap (\mathbb{R} \setminus P)$ are non-empty.

If $X$ is a Bernstein set (with the subspace topology inherited from $\mathbb{R}$) then $X$ is uncountable and has no subspace homeomorphic to the Cantor set. On the other hand, every uncountable Borel subset of a Polish space does have a subspace homeomorphic to the Cantor set. Thus $X$ cannot be a Borel subset of any Polish space.

Now if $d$ is any compatible metric on $X$ then $(X, d)$ is a separable metric space, so its completion is a Polish space in which $X$ is necessarily non-Borel.