This stems from exercise 6, section 81 in Munkres. Let $X$ be a locally compact Hausdorff space; let $G$ be a group of homeomorphisms of $X$ such that the action of $G$ is fixed-point free. Suppose that for each compact subspace $C$ of $X$, there are only finitely many elements $g$ of $G$ such that the intersection $C\cap g(C)$ is nonempty. Then the action of $G$ is properly discontinuous, and $X/G$ is locally compact Hausedorff.
The first part of the exercise wants me to show that for any compact $C$ in $X$, the union of g(C) for $g\in G$ is closed in X. The hint states that if U is a neighborhood of x with $\bar{U}$ compact, then $\bar{U}\cup C$ intersects $g(\bar{U}\cup C)$ only for finitely many g.
I'm not too sure where he wants me to go with that hint. What I do know is that since $X$ is Hausdorff, any compact subspace of $X$ is closed. Furthermore, since each $g$ is a homeomorphism, they are all continuous. Therefore $g(C)$ is also a compact subset of $X$ (which is also closed) in $X$ for all $g$. Now the issue is that we are not guaranteed that the union of of closed sets is closed. I am not sure how the $U$ in the hint will help me resolve this issue.
Thanks,
-akt
There is a very useful
Proposition: Let $\mathscr{A}$ be a locally finite(1) family in a topological space $E$. Then we have
$$\overline{\bigcup_{A\in\mathscr{A}} A} = \bigcup_{A\in \mathscr{A}} \overline{A}.$$
(Proof left as an exercise.)
In particular, the union of a locally finite family of closed sets is closed.
The hint in particular tells you that the family $\mathscr{G} = \{ g(C) : g\in G\}$ is locally finite.
(1) A family $\mathscr{A}$ is locally finite if every $x\in E$ has a neighbourhood $U$ such that $U\cap A \neq \varnothing$ only for finitely many $A\in\mathscr{A}$.