Equivalent definitions of properly discontinuous action

Question

Let $G$ be a subgroup of isometries of a metric space $X$. I want to prove that the following statements are equivalent:

1. For a non-empty compact subset $K$ of $X$, $g(K) \cap K \neq \phi$ for only finitely many $g\in G$.

2. Each orbit $Gx$ of $X$ is locally finite i. e. any compact subset $K$ of $X$ intersects only finitely many elements of $Gx$.

I am following the book $\textbf{Fuchsian Groups}$ by $\textbf{Svetlana Katok}$. The second statement is definition of properly discontinuous action according to this book. While in Wikipedia a properly discontinuous action is defined as the first statement. I want to show that both the definitions are equivalent but couldn't get any lead. How to prove this?

Answer 1

First of all, the correct interpretation of the definition given in Katok's book is:

Definition. Given an isometric $G$-action on a metric space $X$, the orbit $Gx$ is said to be locally finite if for every compact $K\subset X$, the subset $$ \{g\in G: gx\in K\}\subset G $$ is finite. In other words, if we equip $G$ with the discrete topology, then the orbit map $o_x: G\to X, o_x(g)=gx$, is proper.

Example. Take any infinite group $G$ acting on itself via left multiplication. Consider the set $X:= G\sqcup \{p\}$; we extend the action of $G$ to $X$ by requiring that every $g\in G$ fixes $p$. Equip $X$ with the discrete metric ($d(x,y)=1$ iff $x\ne y$). This metric is obviously $G$-invariant. Then for every compact (equivalently, finite) subset $K\subset X$ and every $x\in X$, the intersection $$ Gx\cap K $$ is a finite subset of $X$. However, the action of $G$ on $X$ is not proper in the usual sense (since $p$ has infinite stabilizer). Thus, assuming local finiteness as stated in your question is not enough for proper discontinuity of an isometric action.

Theorem. Let $X$ be a metric space and $G$ is a group acting isometrically on $X$. Then the following are equivalent:

1. $G$ acts on $X$ properly discontinuously. (I.e. if we equip $G$ with the discrete topology, the action $G\times X\to X$ is proper.)

2. $G$ acts on $X$ such that all $G$-orbits are locally finite in the sense of Definition.

Proof. (a) Suppose 1 holds, but 2 fails, i.e. there is a compact $K$ and a point $x\in X$ such that $$ S=\{g\in G: gx\in K\} $$
is infinite. Then the subset $T\subset G$ consisting of products $gh^{-1}$, $g, h\in S$, isinfinite. Moreover, for each $t\in T$, $tK\cap K\ne \emptyset$. This contradicts the assumption 1. (Note that this part did not require an isometric action, just a continuous $G$-action.)

(b) Suppose that 2 holds but 1 fails. Consider $K\subset X$ such that $$ \{g\in G: gK\cap K\ne \emptyset\} $$ is infinite. Thus, there exists a sequence $x_n\in K$ and an infinite sequence of pairwise distinct elements $g_n\in G$ such that $g_n(x_n)=y_n\in K$. By compactness of $K$, after passing to a subsequence, we can assume that $x_n$ converges to some $x\in K$ and $y_n$ converges to some $y\in K$. Since the $G$-action is isometric, $g_n(x)$ also converges to $y$ (just use the triangle inequality and $G$-invariance of the metric). Now, form the compact subset $K\subset X$ equal to $$ \{y\}\cup \{y_n: n\in {\mathbb N}\}. $$ (I will leave it to you to verify compactness: Prove that every sequence in $K$ contains a convergent subsequence.) We then have $g_n(x)\in K$ for all $n\in {\mathbb N}$, contradiction the assumption 2. qed