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?
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:
- $W(X)$ is countable
- $W(X)$ has cardinal $<$ continuum
- $W(X)$ is reduced to finitely supported permutations
- $X$ 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).
