Does $\mathbb R^n$ have a dense countable subset not uniformly homeomorphic to $\mathbb Q^n$?

Question

Every countable dense subspace of $\mathbb R^n$ is homeomorphic to $\mathbb Q$, as is every countable metrizable space with no isolate points. However, $\mathbb Q^n$ is not uniformly homeomorphic to $\mathbb Q^m$ for $m\ne n$, as they have non-homeomorphic completions (namely $\mathbb R^n$ and $\mathbb R^m$) as pointed out by Anne Bauval in her answer to this post.

So here is what I wish to know:

For a fixed dimension $n$, is every countable dense subspace of $\mathbb R^n$ uniformly homeomorphic to $\mathbb Q^n$?

I'm quite sure this is true for $n = 1$. This is because of Cantor's Isomorphism Theorem, which states every countable dense unbounded linear order is order-isomorphic to $\mathbb Q$. I don't mean to apply the theorem, as order-isomorphisms need not to be uniform, but to adapt its constructive proof, which can be easily done to constrain distances. I haven't write a proof yet but I'm positive it can be done.

Answer 1

Theorem 1. For every two dense countable subsets $A, B\subset \mathbb R^n$ there exists a globally bilipschitz diffeomorphism $f: \mathbb R^n\to \mathbb R^n$ such that $f(A)=B$.

Here, globally bilipschitz means that there exists a constant $L\ge 1$ such that for all $x, y\in \mathbb R^n$ we have $$ L^{-1}||x-y||\le ||f(x)-f(y)||\le L||x-y||. $$

In particular, $A$ and $B$ are uniformly homeomorphic. To key to the proof of this theorem is another general result:

Theorem 2. Suppose that $M$ is a smooth manifold and $X, Y\subset M$ are dense countable subsets. Then there is a diffeomorphism $F: M\to M$ such that $F(X)=Y$.

See the references (especially the one by Dobrowolski) I gave in my answer here.

To apply this theorem, note that, without loss of generality, we may assume that $\mathbf 0\in A\cap B$.

Consider $M$ which is the unit sphere in $\mathbb R^{n+1}$. Set $e_{n+1}= (0,...,0,1)$. Let $\sigma: M\to \mathbb R^n\cup \{\infty\}$ denote the stereographic projection, where $\sigma(e_{n+1})=\infty$. Take $X=\sigma^{-1}(A\cup \{\infty\})$ and $Y:= \sigma^{-1}(B\cup \{\infty\})$.

Let $F: M\to M$ be a diffeomorphism which takes $X$ to $Y$ and such that $F(e_{n+1})=e_{n+1}, F(-e_{n+1})=-e_{n+1}$. (It is and easy consequence of Theorem 2 that such $F$ exists.) Now, set $$ f:= \sigma\circ F \circ \sigma^{-1}. $$ Then $f: \mathbb R^n\to \mathbb R^n$ is a diffeomorphism such that $f(A)=B$ and $f(\mathbf 0)=\mathbf 0$ (since $\sigma(-e_{n+1})=\mathbf 0$).

Since $f$ is a diffeomorphism, it is locally Lipschitz. To check that $f$ is globally Lipschitz, it suffices to get uniform upper bounds on the norms of the derivatives $Df, Df^{-1}$. To get such bounds one uses the fact that $F$ is a diffeomorphism at $e_{n+1}$ and computes derivatives via the change of coordinates $$ h= J \circ f \circ J, $$ where $J$ is the inversion $$ J(x)= \frac{1}{||x||^2} x, x\in \mathbb R^{n}\setminus \{\mathbf 0\}. $$ The fact that $h$ extends to a diffeomorphism at the origin implies the required uniform bounds on the norms of $Df, Df^{-1}$.