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

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.

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 $$Y_c$$.