# Countability of the wobbling group of a bounded geometry metric space

Let $$(X, d)$$ be a uniformly discrete metric space of bounded geometry, that is, $$\sup_{x \in X} |B_r(x)| < \infty$$ for every $$r \geq 0$$ and there is a uniform $$\delta > 0$$ such that $$d(x, y) \geq \delta$$ for all $$x \neq y \in X$$.

Let $$G$$ be the so-called wobbling group of $$X$$, that is, the set of all bijections $$f \colon X \rightarrow X$$ such that

$$\sup_{x \in X} d(x, f(x)) < \infty.$$

Question: is $$G$$ countable?

• For $X=\mathbf{Z}$ it has continuum cardinal.
– YCor
Dec 8, 2021 at 16:04
• @Ycor Does it? I've been told otherwise. Do you have a proof of it? Dec 8, 2021 at 16:06
• Yes, I'm going to post an answer.
– YCor
Dec 8, 2021 at 16:06
• In $\mathbf{Z}$ just exchange any prescribed set of even numbers to their successor. Morality: never believe what you're told :)
– YCor
Dec 8, 2021 at 16:19

Say that a metric space $$X$$ is hyperdiscrete if the distance map $$X\times X\to\mathbf{R}_{\ge 0}$$ is a proper map, i.e. for every $$R$$, the set of $$(x,y)$$ such that $$d(x,y)\le R$$ is finite. For instance, with the Euclidean distance, $$\mathbf{N}$$ is not hyperdiscrete, but the set $$\{0,1,4,\dots\}$$ of squares in $$\mathbf{N}$$ is hyperdiscrete. The terminology is from my book with P. de la Harpe, Example 3.A.14(7) (ArXiv link, EMS link).

Let $$X$$ be a proper discrete metric space (i.e., every bounded subset is finite— note this forces $$X$$ to be countable) and $$W(X)$$ its wobbling group (= permutations of $$X$$ at bounded distance to identity).

Proposition. Equivalent statements:

1. $$W(X)$$ is countable
2. $$W(X)$$ has cardinal $$<$$ continuum
3. $$W(X)$$ is reduced to finitely supported permutations
4. $$X$$ is hyperdiscrete.

Suppose (4) fails. Then there exists an injective sequence $$(x_n)_{n\ge 0}$$ with $$d(x_n,x_0)$$ tending to infinity, such that $$d(x_{2n},x_{2n+1})$$ bounded. Then there is an injective homomorphism from the group $$(\mathbf{Z}/2\mathbf{Z})^\mathbf{N}$$ into $$W(X)$$, mapping $$(a_n)$$ the permutation exchanging $$x_{2n}\leftrightarrow x_{2n+1}$$ whenever $$a_n=1$$, and identity elsewhere. So (2) fails. Hence (2) implies (4).

The implications (3)$$\Rightarrow$$(1)$$\Rightarrow$$(2) are trivial.

Finally, if $$X$$ is hyperdiscrete and $$f$$ is a permutation at distance $$\le R$$ to identity, then since $$\{(x,y):d(x,y)\le R\}$$ is finite, its projection to $$X$$ is finite, and $$f$$ has to be identity elsewhere, so $$f$$ is finitely supported. Hence (4) implies (3).

• Awesome, thanks! I was wondering indeed that it should not be countable, glad to know for sure. Dec 8, 2021 at 16:24