TITLE: Equivalence to properly discontinuous action
QUESTION [8 upvotes]: 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).

REPLY [5 votes]: 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.”