Let $M$ be a connected (smooth Riemannian) manifold which admits a universal cover $\tilde{M}$. Let $\Gamma$ be the group of deck transformations on $\tilde{M}$. I want to show that $\Gamma$ acts properly discontinuously on $\tilde{M}$. The definition for properly discontinuous that I'm using is as follows:
Definition 1: A group $G$ acts properly discontinously on a (smooth) manifold $N$ (it is already enough to require $N$ being a locally compact Hausdorff space) if $G$ acts by homeomorphisms and for any compact set $K \subset N$ the set $\{ \gamma \in \Gamma \mid \gamma(K) \cap K \neq \emptyset \}$ is finite.
In literature however, one sometimes finds the following definition, which for the sake of distinction, I will call properly discontinuous$^{TypeB}$ (compare Munkres' book on Topology):
Definition 2: A group $G$ acts properly discontinuously$^{TypeB}$ on a (smooth) manifold $N$ if it acts by homeomorphisms and for every $x \in N$ there is an open neighborhood $U$ of $x$ in $N$ such that $g(U) \cap U = \emptyset$ for all $g \in G \setminus\{id\}$.
It is easy enough to show that the group of deck transformations $\Gamma$ acts properly discontinuously$^{TypeB}$ on $\tilde{M}$ (i.e. using Definition 2).
I was unable to prove however, that properly discontinuous$^{TypeB}$ implies properly discontinuous (i.e. Definition 2 implies Definition 1). The proofs I've found are all either rather indirect (i.e. with many intermediate steps) or use methods that I don't understand. Compare for example the treatment here: https://mathoverflow.net/questions/55726/properly-discontinuous-action
So, I'm looking for either an immediate proof that properly discontinuous$^{TypeB}$ implies properly discontinuous or a direct proof that $\Gamma$ acts properly discontinuously on $\tilde{M}$ (using Definition 1).
Thanks in advance for any help!
