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Can anyone help with this problem?

Q: Let $G$ be a discrete group acting properly on a manifold $M$. Prove that the action is wandering.

(The following is the definition of wandering:

Definition: Let G be a topological group acting on a topological space X. The action is called wandering if for every $x \in X$ there exists a neighborhood $U_x$ of $x$ such that the set $G_{U_x} = \{g\in G : (g\cdot U_x) \cap U_x \not = \emptyset\} $ is finite.

I have looked up related topics here, but only find very few posts. It is my first time to be introduced with the concept of wandering action. Can anyone help me with a pretty detailed explanation? Great thanks!!

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I assume that the manifold is finite dimensional.

By definition, the action of $G$ on $M$ is proper if and only if for any compact subset $C\subset M$ of $M$, $G_C=\{g\in G: g(C)\cap C\neq\phi\}$ is finite.

Let $x\in M$, since $M$ is finite dimensional, $x$ has a neighborhood $U_x$ such that the adherence $U'_x$ of $U_x$ is compact. Since the action is proper $G_{U'_x}$ is finite. You also have $G_{U_x}\subset G_{U'_x}$ is finite.

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  • $\begingroup$ Great thanks! Sir $\endgroup$
    – Xuan
    Mar 29 '16 at 23:46
  • $\begingroup$ Btw, can you briefly explain why G_c is finite? $\endgroup$
    – Xuan
    Mar 30 '16 at 15:14

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