A discrete group $\Gamma$ is said to act properly discontinuously on a smooth manifold $M$ if the action is smooth and satisfies the following two conditions:
(i) Each $x \in M$ has a neighborhood $U$ such that $\{h \in \Gamma ; hU \cap U \neq \emptyset\}$ is finite;
(ii) If $x,y \in M$ are not in the same orbit, then there are neighborhoods $U, V$ of $x,y$ such that $U \cap \Gamma V = \emptyset$.
Consider
(i') The isotropy group $\Gamma_x$ of each $x \in M$ is finite, and each $x$ has a neighborhood $U$ such that $hU \cap U = \emptyset$ if $h \not\in \Gamma_x$ and $hU=U$ if $h \in \Gamma_x$.
Question: Prove (i) and (i') are equivalent. I know one side is trivial, what about the other side?