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).