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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).

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    $\begingroup$ 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. $\endgroup$ Sep 26, 2013 at 20:57
  • $\begingroup$ @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 $X \times G$. $\endgroup$ Sep 28, 2013 at 19:34
  • $\begingroup$ 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. $\endgroup$ Sep 28, 2013 at 19:41
  • $\begingroup$ 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. $\endgroup$
    – Mr. Chip
    Sep 28, 2013 at 23:44
  • $\begingroup$ 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. $\endgroup$ Sep 13, 2021 at 14:35

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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 $G$.

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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.”

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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.

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  • $\begingroup$ Can you elaborate on why $g_{k_{n_m}} \cdot x \to y$? $\endgroup$
    – Mr. Chip
    Sep 27, 2013 at 1:02
  • $\begingroup$ 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? $\endgroup$
    – Marek
    Sep 27, 2013 at 8:04
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    $\begingroup$ Correct me if I'm wrong, but you'd need equicontinuity of the $g_{n_k}$ in $x$ for that, wouldn't you? $\endgroup$ Sep 27, 2013 at 10:35
  • $\begingroup$ @Daniel: indeed, seems you are right, the answer is not complete as it stands. Thank you. $\endgroup$
    – Marek
    Sep 27, 2013 at 15:46

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