I want to prove the following result:
Let $\Gamma$ be a discrete group acting continuously on a topological manifold $M$. Then, the action of $\Gamma$ on $M$ is proper if and only if the following conditions hold:
- For each $p \in M$, there is an open set $p \in U \subseteq M$ such that $g \cdot U \cap U = \emptyset$ for all but finitely many $g \in \Gamma$.
- If $p, p' \in M$ are in distinct orbits (under the action of $\Gamma$), then there are open sets $U, U'$ containing $p, p'$ such that for all $g \in \Gamma$ we have $g \cdot U \cap U' = \emptyset$.
Prior to this, I have already used the following result:
Theorem: Let $\Gamma$ be a discrete group and $M$ be a topological manifold. Let $\Gamma$ act continuously on $M$. Then, the action of $\Gamma$ on $M$ is proper if and only if for any $p, p' \in M$ we have open sets $U, U' \subseteq M$ containing $p, p'$ such that $\overline{U}$ and $\overline{U'}$ are compact and the set $\left\lbrace g \in \Gamma | g \cdot U \cap U' \neq \emptyset \right\rbrace$ is finite.
In proving the result I want, first, we assume that the action is proper. Then, point (1) is satisfied due to the theorem. In proving (2), let $g \in \Gamma$ be such that $g \cdot U \cap U' \neq \emptyset$. Then, $g \cdot p \in U'$ (which is a neighbourhood of $p'$). I want to use this to conclude that $p$ and $p'$ are in the same orbit. However, I am unable to get a hint at this!
Similarly, when proving the converse, if $p$ and $p'$ are in distinct orbits, then the set $\left\lbrace g \in \Gamma | g \cdot U \cap U' \neq \emptyset \right\rbrace = \emptyset$ and hence finite. However, if they are in the same orbit, then I know that it is required to use point (1), but again, I fail to see how!
Any suggestions and hints in this direction will be helpful and much appreciated!