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?



Countability of the wobbling group of a bounded geometry metric space
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](https://arxiv.org/abs/1403.3796), [EMS link](https://www.ems-ph.org/books/book.php?proj_nr=209)).


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).


