Are $\mathbb Q$ and $\mathbb Q^2$ uniformly homeomorphic?

Question

So... This is awkward. I just found out $\mathbb Q$ and $\mathbb Q^2$ are homeomorphic! Was not expecting that. In fact, the following is a theorem

(Sierpinski). Any countable metrizable space without isolated points is homeomorphic to $\mathbb Q$.

I did not see this one coming. This means homeomorphisms are not sensitive to the linear dimension of spaces. Not always, anyway. At least $\mathbb R$ and $\mathbb R^2$ are homeomorphically distinct. Wait, does that mean the completion of two homeomorphic spaces might be distinct? :o

Anyway, if we step back from topology and into the realm of metric spaces the problem is fixed: $\mathbb Q$ and $\mathbb Q^2$ are not isometric. But isometry is far too rigid of a transformation (get it? don't laugh all at once). Can we distinguish $\mathbb Q$ and $\mathbb Q^2$ with a less restrictive isomorfism?

I mentioned this problem to a friend o'mine and he suggested an uniform homeomorphism. I put some thought into it and it sounded like a great idea! This forum dislikes multiple-question posts so you can ignore the previous ones (feel free not to do so, though), here is what I want to know: are $\mathbb Q$ and $\mathbb Q^2$ uniformly homeomorphic? I don't think so. No luck proving it, though.

Answer 1

They are not uniformly homeomorphic. This is kind of a rational version of the "removing a ball disconnects the space" argument:

Quick Note: If $A$ and $B$ are uniformly homeomorphic and $A$ is unbounded, then so is $B$. This follows directly from the existence of a uniformly continuous surjection from $B$ to $A$.

Proof $\mathbb{Q}$ and $\mathbb{Q}^2$ are not uniformly homeomorphic: Suppose $f: \mathbb{Q} \rightarrow \mathbb{Q}^2$ is a uniform homeomorphism. Then $f$ restricted to $\mathbb{Q} \setminus (-1, 1)$ must also be a uniform homeomorphism with its image, $\mathbb{Q}^2 \setminus f((-1,1))$. By our Quick Lemma, $f\left((-1, 1)\right)$ is bounded, thus must be contained in a ball of some finite radius $R$; call this ball $B_R(f(0))$.

Note that both $\mathbb{Q} \cap (-\infty, -1)$ and $\mathbb{Q} \cap (1,\infty)$ are unbounded, while $B_R(f(0))$ is bounded. Therefore by our Quick Note neither $f\left(\mathbb{Q} \cap (-\infty, -1)\right)$ nor $f\left(\mathbb{Q} \cap (1,\infty)\right)$ can be contained in $B_R(f(0))$, so we can find $x \in \mathbb{Q}$ with $x < -1$ and $f(x) \notin B_R(f(0))$, and similarly we can find $y \in \mathbb{Q}$ with $y > 1$ and $f(y) \notin B_R(f(0))$.

But $\mathbb{Q}^2 \setminus B_R(f(0))$ has the property that for any $\delta> 0$, any two points can be traversed by a series of "hops" of distance less than $\delta$; that is, we can find a finite sequence $f(x) = p_0, p_1, ..., p_{n-1}, p_n = f(y)$ such that $p_i \in \mathbb{Q}^2 \setminus B_R(f(0))$ and $d\left(p_{i-1}, p_i\right) < \delta$. By the uniform continuity of $f^{-1}$, we can pick $\delta$ small enough so that $d\left(f^{-1}(p_{i-1}), f^{-1}(p_i)\right) < 1$. And since $f((-1, 1)) \subset B_R(f(0))$, we have that $f^{-1}(p_i) \notin (-1, 1)$.

This means the sequence $x = f^{-1}(p_0), f^{-1}(p_1), ..., f^{-1}(p_n) = y$ has the properties that it starts at $x < -1$, ends at $y > 1$, and $d(p_{i-1}, p_i) < 1$, which is impossible since none of the $p_i$ lie in $(-1, 1)$. This gives us our contradiction. $\square$

Answer 2

If two metric spaces are uniformly homeomorphic then so are their completions (this generalizes to uniform spaces and their Hausdorff completions).

The completions $\Bbb R$ and $\Bbb R^2$ of $\Bbb Q$ and $\Bbb Q^2$ (with their usual metrics) are not even homeomorphic.

Therefore, $\Bbb Q$ and $\Bbb Q^2$ are not uniformly homeomorphic (as metric spaces, or equivalently as uniform spaces).