Definition of a metric space with bounded growth

Question

Does anyone know the definition of a metric space with bounded growth?

I was reading a paper by Roe titled Hyperbolic groups have finite asymptotic dimension, where he writes a definition, but I think there is something missing.

Let $X$ be a geodesic metric space. Say that $X$ has bounded growth if for each $s>0$ there is a number $N_s$ such that each ball of radius $S+s$ in $X$ can be covered by at most $N_s$ balls of radius S.

But he doesn't mention anything about $S$. Is it fixed? is also for any $S>0$.

Answer 1

The number $N_s$ must be independent of $S$; however we should restrict $S$ to $S\ge s$. Allowing arbitrary positive $S$ would make all nontrivial geodesic spaces fail the condition. E.g., in $\mathbb R$ we need at least $(S+s)/S$ balls of radius $S$ to cover a ball of radius $S+s$; and $(S+s)/S\to\infty$ as $S\to 0$.

The answer appears to be too short, so let's use this definition to see how it works. Roe says

Since $X$ is geodesic, one may take for $N_s$ the supremum (if finite) of the cardinalities of $s$-separated subsets in balls of radius $2s$.

Proof. Fix a point $a$ and consider closed balls $B(a,S+s) = \{x:d(x,a)\le r\}$. Let $Z$ be a maximal $s$-separated subset of $B(a,2s)$. Then the union of balls $B(z,s)$, $z\in Z$, covers $B(a,2s)$. Since $X$ is geodesic, every point of $B(a,S+s)$ is at distance at most $S-s$ from a point of $B(a,2s)$. Therefore, $B(a,S+s)\subset \bigcup_{z\in Z} B(z,S )$.