# Is there a computable homeomorphism between two different Cartesian powers of the computable real numbers?

It's well know that it is surprisingly difficult to prove that $\mathbb{R}^n$ and $\mathbb{R}^m$ are not homeomorphic for $n\neq m$. Commonly proofs go through Brouwer's fixed point theorem, which is 'computably false' for dimensions greater than one: let $K$ denote the computable real numbers. For $n>1$, there are computable functions from $\left( [0,1]\cap K \right)^n$ to itself with no computable fixed point.

That by itself may not seem very bad, since the function may still extend to a continuous function on $[0,1]^n$ and have an incomputable fixed point, but a corollary of this is that there is a computable retraction of $\left( [0,1]\cap K \right)^n$ onto its boundary. Such a function clearly can't be extended to a continuous function on $[0,1]^n$. So we can see that the topological behavior of $K^n$, even when restricted to computable functions, is very different from the topological behavior of $\mathbb{R}^n$.

On the other hand, $K$ is homeomorphic to $\mathbb{Q}$ (although not computably so) and $\mathbb{Q}^n$ is homeomorphic to $\mathbb{Q}^m$ for any $n$ and $m$, so $K^n$ is homeomorphic to $K^m$ for any $n$ and $m$.

So the question is: Is there a computable homeomorphism between $K^n$ and $K^m$ for some $n\neq m$? If there is one I would assume we need $n,m>1$.

• For context, do you know if $\mathbf{Q}$ is computably homeomorphic to $\mathbf{Q}^2$? – YCor Jun 20 '18 at 19:03
• @YCor Yes, they are; the usual back-and-forth (with requirements) construction of a homeomorphism is totally effective. – Noah Schweber Jun 20 '18 at 19:21