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Let $(X,d)$ be a metric space and $C\subset X$ be a compact subset. Let furthermore $G$ be a group that acts on $X$ proper and by isometries. Does there exist an $\epsilon >0 $ such that: Let $U=$ {$x\in X| d(x,C)<\epsilon$} and there are only finitely many $g\in G$ such that $g(U) \cap U \not= \emptyset$ ?

Cheers Helge

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  • $\begingroup$ does one need the assumption that $(X,d)$, is a metric space, where $d$ is a length metric ? $\endgroup$
    – Helge
    Commented Jun 13, 2013 at 18:12
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    $\begingroup$ Being a length metric is irrelevant. On the other hand, you ought to assume the metric space is proper (closed balls are compact), particularly since you are assuming that the action is proper (the induced map $G \times X \to X \times X$ is a proper function). In that case what you ask for is true, and is a trivial consequence of the definitions. $\endgroup$
    – Lee Mosher
    Commented Jun 13, 2013 at 21:14
  • $\begingroup$ Helge: What Lee says is correct, such actions are usually called "metrically properly discontinuous". In your setting, you probably meant "properly discontinuous" (rather than just proper). Then the assertion is correct and is a nice exercise in point-set topology although not as easy as the one for "metrically proper" actions. $\endgroup$
    – Misha
    Commented Jun 14, 2013 at 3:07
  • $\begingroup$ @Misha: Can you give some hints for the proof in the case the action is properly discontinuous? I find the result interesting because it implies that an action on some length space is metrically proper and cocompact iff it is properly discontinuous and cocompact. (I do not know if this still holds in other contexts without a properness assumption.) $\endgroup$
    – Seirios
    Commented Feb 18, 2017 at 13:44
  • $\begingroup$ @Seirios: See the answer. $\endgroup$
    – Misha
    Commented Mar 7, 2017 at 18:07

1 Answer 1

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Since this old question does seem to be of some interest, here is a proof. First of all, here is a useful (but not well-known) definition.

Definition. Let $X$ be a Hausdorff 1st countable topological space, $G\times X\to X$ a topological action of a discrete topological group $G$. Two points $x, y\in X$ are said to be dynamically related with respect to this action if there is an infinite sequence (of distinct elements) $g_n\in G$ and a sequence $x_n\to x$ in $X$ such that $g_n(x_n)\to y$.

It is a pleasant exercise to show that an action which admits a pair of dynamically related points cannot be properly discontinuous. (Consider the compact set formed by $\{x, y\}$ and the image of the sequence $(x_n)$.) The converse is also true but is not relevant for the question.

Now, to the question itself. Suppose that $G\times X\to X$ be a properly discontinuous action. (A proper action is clearly not enough, as you can consider say, $U(1)$ acting on itself.) Suppose that $\epsilon>0$ as in the question does not exist. Then for every $n$ there is $g_n\in G$ (with $g_n$'s distinct for distinct $n$'s) and points $x_n, y_n\in C$ such that $d(y_n, g_n(x_n))<1/n$. After extraction, we can assume that $x_n\to x\in C, y_n\to y\in C$. Now, it is clear that $x, y$ are dynamically related. qed

Note that we did not need the action $G\times X\to X$ to be isometric.

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