Every proper and cobounded action is acylindrical.

Question

I am trying to show every proper and cobounded action is acylindrical. Given any $\epsilon> 0$, we need to show there exist $R, N > 0$, s.t. for every two points $x, y$ with $ d(x, y)\ge R$, there are at most $N$ elements $g\in G$ satisfying $d(x, g x)\le\epsilon$ and $d(y, g y) \le\epsilon$. Since the action is proper, each orbit $G\cdot x$ is discrete, then in the compact set $\bar{B_\epsilon}(x)$, the intersection $G\cdot x\cap\bar{B_\epsilon}(x)$ is finite. Also, by the proper action, the stabilizers are also finite. So for given 2 points $x,y$, the number of elements $g\in G$ satisfying $d(x, g x)\le\epsilon$ and $d(y, g y) \le\epsilon$, say $N_{x,y}$, is finite. But how do we get a universal $N$ to bound all the $N_{x,y}$ when $x,y$ are apart enough? In particular, I am not sure how to apply the cobounded condition.

An action of a group $G$ on a metric space $S$ is called acylindrical if for every $\epsilon> 0$ there exist $R, N > 0$ such that for every two points $x, y$ with $ d(x, y)\ge R$, there are at most $N$ elements $g\in G$ satisfying $d(x, g x)\le\epsilon$ and $d(y, g y) \le\epsilon$.

Answer 1

Update: The proof below assumes that the space $X$ is proper, i.e., that closed balls are compact. I think the statement is false without this statement. See the comments below.

Since the action is assumed to be proper, this is in a sense a stronger condition than an acylindrical action. So, it makes sense to try to focus on a single point $x$ and then to relate a second point $y$ by choosing a point in its orbit that is sufficiently close to $x$. Since the action is cobounded, there will be a point that is, say, $R$-close to $x$, where $R$ does not depend on $x$ or $y$. The $N$ will come from the finite number of group elements that fail to move an $R$-ball centered at $x$ off of itself. Really, as we'll see, we'll use a $2R + 2\epsilon$ ball centered at $x$.

Let $G_{x,\epsilon} = \{g \in G \mid \overline{g.B(x,\epsilon)} \cap \overline{B(x,\epsilon)} \neq \emptyset\}$. Let $N_{x, \epsilon}$ be the cardinality of $G_{x,\epsilon}$, which is finite for each $\epsilon \geq 0$ because the action is proper.

Here's an outline of what to show:

1. There is an $R > 0$ such that for any $x,y \in X$, the distance between their orbits is at most $R$. This is the definition of cobounded.

2. Suppose $y \in X$ and $\epsilon \geq 0$. Prove that $hG_{y,\epsilon}h^{-1} = G_{h.y,\epsilon}$. It's the same proof as for showing the analogous statement for point stabilizers.

3. Fix a point $x \in X$ and let $y$ be any point in $X$. Let $\epsilon \geq 0$. Prove that $N_{y,\epsilon} \leq N_{x,2R + 2\epsilon}$. This latter number is the $N$ you seek. The main idea is that if $g \in G_{y,\epsilon}$ and $h.y$ is $R$-close to $x$, then $hgh^{-1}.(h.y)$ is $\epsilon$-close to $h.y$ and so $hgh^{-1}.x$ is $2R + 2\epsilon$-close to $x$; therefore $hgh^{-1} \in G_{x,2R + 2\epsilon}$. Conjugation by $h$ defines a bijection $G \to G$ and so there are at most $N_{x,2R + 2\epsilon}$ such elements $g \in G$.