We know that if $M$ is a topological space and $G$ a group action on $M$ then we say that the action is properly discontinuous if for all $x\in M$ there is $U \ni x$ such that for every $g\neq e \in G$, $gU \cap U = \emptyset.$
So, I was asked to show that a certain action is properly discontinuous. My friend said that is enough to show that for every $x \in M$ then $gx \neq x.$ Is this true? Why?
What does mean in practice mean that $gU \cap U = \emptyset$? It does mean in fact that $gx \neq x$ for all $x$, but, to show that $gx\neq x \forall x \in U$ does not necessarily implies that $gU \cap U = \emptyset$, right?
For example, I could have $x,y \in U$ such that $gx = y.$ Or this does not occur?
Thanks