Non-homeomorphic computable metric spaces whose computable points are computably homeomorphic

Question

This is a follow-up of sorts to an earlier question on mine in that it should be easier to construct a positive example of this, if it exists.

To be clear about definitions, a computable metric space is a metric space $(X,d)$ with a dense sequence $\{q_n\}_{n<\omega}$ such that $d(q_n,q_m)$ is a uniformly computable family of computable real numbers. The points $q_n$ are called rational points. Given a computable metric space $X$ and a point $x\in X$, a name of $x$ is any function $f:\omega\rightarrow \omega$ satisfying $d(q_{f(n)},q_{f(m)})\leq\frac{1}{n}+\frac{1}{m}$ for all $n,m<\omega$ and such that $q_{f(n)}\rightarrow x$ as $n\rightarrow \infty$. A point $x\in X$ is computable if it is a name that is a computable function. Given a computable metric space $X$, we'll denote the set of computable points in $X$ as $X_{c}$.

Given a pair of computable metric spaces $X$ and $Y$, a Turing functional $A \mapsto \Phi_e^A$ induces a partial function $F_e$ from $X$ to $Y$ by the rule $F_e(x)=y$ iff for any name $f$ of $x$, $\Phi_e^f$ is a name of $y$. One can check that any such function is automatically continuous on its domain. A computable homeomorphism between $X$ and $Y$ is a pair of Turing functionals $\Phi_e$ and $\Phi_i$ such that $F_e$ is a total bijection from $X$ to $Y$ and $F_i$ is its inverse from $Y$ to $X$. $X$ and $Y$ are said to be computably homeomorphic if there exists a computable homeomorphism between them.

Now we can state the question:

Does there exist a pair of complete computable metric spaces $X$ and $Y$ with no isolated points such that $X$ and $Y$ are not homeomorphic, but $X_c$ and $Y_c$ are computably homeomorphic?

The completeness stipulation is necessary because technically the definition of a computable metric space allows it to be any arbitrary subset of its completion that contains its rational points. The no isolated points stipulation is necessary because there is a relatively easy example, building off of the Kleene tree, of a compact computable metric space whose rational points are all isolated and with the property that the only computable points are rational. You can easily give a functional mapping the computable points in this space to a non-compact discrete space. On the other hand, since $X_c$ and $Y_c$ are countable metric spaces with no isolated points, they are both classically homeomorphic to $\mathbb{Q}$, so there is no obvious proof that a computable homeomorphism can't exist. The proof of this is effective if you actually have a list of the elements of the space, but you can show that in a complete computable metric space with no isolated points you can computably find a point not on any given list of points in the space.

Answer 1

I think that maybe I should have thought about this a little harder before posting, because it seems like there is a fairly easy positive answer:

Let $X=[0,1]$ with a standard enumeration of the rational numbers. Let $f:[0,1]_c \rightarrow \mathbb{R}$ be a computable function that is unbounded (an easy example is to let $f(x)=\frac{1}{g(x)}$ where $g:[0,1]\rightarrow [0,1]$ is a computable function whose zeroset contains no computable points). Then let $Y_0 = [0,1]_c$ with the metric $d(x,y)=|x-y|+|f(x)-f(y)|$ and let $Y$ be the metric completion of $Y_0$. $X$ and $Y$ are obviously not homeomorphic because $X$ is compact and $Y$ is not. On the other hand, the identity function is a computable homeomorphism between $X_c$ and $Y_c$.