# Equivalence to properly discontinuous action

Let $X$ be a metric space and let $G$ be a group of homeomorphisms $X \to X$ acting on $X$. We say $G$'s action is properly discontinuous in case for every $x \in X$ and compact $K \subseteq X$, there are at most finitely many $g \in G$ such that $g(x) \in K$. Equivalently (and this is not hard to show), $G \cdot x$ is discrete and $G_x$ finite for any $x$.

Why is it the case that $G$ acts properly discontinuously if and only if for any compact $K$, $g(K) \cap K \ne \emptyset$ for only finitely many $g$? One direction is relatively easy, but I just cannot seem to prove the "only if". The best I've been able to do is for finite $K$ (which is kind of the next best thing when you're stumped on proving something for compact sets, I guess).

• Do you assume that $X$ is locally compact? Otherwise, I do not understand how you prove equivalenc of proper discontinuity and discreetness of orbits and finiteness of stabilizers. Sep 26, 2013 at 20:57
• @Joshua Ciappara : are you sure about the definition? what I know is that if $G$ is a discret group acting continuously on a Housdorff space $X$. The action is properly discontinuous if the mapping $(x,g) \mapsto (x,xg)$ is proper. If $X$ is locally compact, properness is equivalent to that every inverse image of a compact of $X \times X$ is compact in $X \times G$. Sep 28, 2013 at 19:34
• I think this is the most general definition, which is actually aquivalent to your second condition. Please can you give us the reference that you taken the definition from it. Sep 28, 2013 at 19:41
• The question appears as an exercise in Svetlana Katok's "Fuchsian groups". Maybe it's an error. If that's the case, I'll accept a counterexample as an answer, of course. Sep 28, 2013 at 23:44
• How did you prove that discreteness of the orbits and finiteness of the stabilizers imply proper discontinuity? I was only able to do that by assuming the orbits are also closed. Sep 13, 2021 at 14:35

I taken a look in Svetlana Katok's "Fuchsian groups", and I'm really confused.

For instance, it is proved in that book that $G$ acts properly discontinuously on $X$ iff every point $x \in X$ has a neighborhood $U_x$ such that $U_x \cap gU_x$ is not empty for only finitely many $g \in G$. (I don't understand the "only if" part, and I don't think that this holds in general).

Although, let be $G$ the group of all the (continuous) transformations $f_n$, $n \in \mathbb{Z}$, of the plane without the origine $(0,0)$, with $f_n:(x,y) \mapsto (2^nx,2^{-n}y)$.

One can easily see that for each point $(x,y)$, a disc with this point as a center and a sufficiently small radius intersects the orbit of $(x,y)$ only in one point, thus the orbit of each point is a discret subset, and clearly G acts freely on our set, thus the stabilizer of each point is finite.

On the other hand, consider the segment $K = \{(t, 1-t)| 0 \leq t \leq 1 \}$, clearly $K$ is compact, $K$ contains for each $n>0$ the element $P_n =(\frac{2^n-1}{2^{2n}-1}, \frac{2^n (2^n-1)}{2^{2n}-1})$, and $f_n(P_n) \in K\cap f_nK$. Therefore $K \cap f_nK$ is not empty for infinitely many elements $f_n$ of $G$.

see the errata posted here: http://www.personal.psu.edu/sxk37/errata.pdf

where katok writes:

p.27 l.9: Replace “metric” by “locally compact metric”

p.27 l.10: Replace “homeomorphisms” by “isometries”.

p.27 l.-5-l.-3 Replace “It is clear from the definition that a group $G$ acts properly discontinuously on $X$ if and only if each orbit is discrete and the stabilizer of each point is finite.” by “Since $X$ is locally compact, a group $G$ acts properly discontinuously on $X$ if and only if each orbit has no accumulation point in $X$, and the order of the stabilizer of each point is finite. The first condition, however, is equivalent to the fact that each orbit of $G$ is discrete. For, if $g_n(x) \to s \in X$, then for any $\varepsilon > 0$, $\rho(g_n(x),\,g_{n+1}(x))< \varepsilon$ for sufficiently large $n$, but since $g_n$ is an isometry, we have $\rho(g^{-1}_ng_{n+1}(x),\,x)< \varepsilon$, which mplies that $x$ is an accumulation point for its orbit $Gx$, i.e. $Gx$ is not discrete.”

Suppose there is a compact $K$ such that $g(K) \cap K \neq 0$ for infinitely many $g$. Denote by $L_k := g_k(K) \cap K$ where $g_k, k\in \mathbb N$ are among those elements of $G$ for which the inequality holds. Now pick any sequence $y_k \in L_k$. Because $K$ is compact, we have a convergent subsequence $y_{k_n} \to y$. But because every $y_{k_n} \in g_{k_n}(K)$ there must also be a sequence $x_{k_n}$ in $K$ such that $g_{k_n} \cdot x_{k_n} = y_{k_n}$ and so again by compactness there is a convergent subsequence $x_{k_{n_m}} \to x$ of $x_{k_n}$. But then $g_{k_{n_m}}\cdot x \to y$ is not discrete.

• Can you elaborate on why $g_{k_{n_m}} \cdot x \to y$? Sep 27, 2013 at 1:02
• You get this from combining the two convergences $x_{k_{n_m}} \to x$ and $g_{k_{n_m}} \cdot x_{k_{n_m}} = y_{k_{n_m}} \to y$. Is that fine with you are do you want me to fill in the epsilon-deltas? Sep 27, 2013 at 8:04
• Correct me if I'm wrong, but you'd need equicontinuity of the $g_{n_k}$ in $x$ for that, wouldn't you? Sep 27, 2013 at 10:35
• @Daniel: indeed, seems you are right, the answer is not complete as it stands. Thank you. Sep 27, 2013 at 15:46