# Properly discontinuous action on a non-locally compact space

Let me begin with some definitions in order to avoid confusion.

An action of a group $G$ on a space $X$ is proper if the map $G \times X \to X \times X$ given by $(g, x) \mapsto (x, gx)$ is proper, i.e. the preimage of any compact set is again a compact set. Such an action of a discrete group is called a properly discontinuous action. An action is free is for any $x \in X$ the equality $gx=x$ implies $g=1$.

On the other hand, there is the following property of an action: for any $x \in X$, there exists its neighbourhood $V$ such that $gV \cap V = \emptyset$ for any nontrivial $g \in G$. (Sometimes such an action is called a wandering one.)

Now, if $X$ is a locally compact space, then a free and properly discontinuous action has the above property. The question is what happens if $X$ isn't locally compact? Can one salvage this implication, or is there a counterexample? If not in full generality, perhaps something can be said e.g. under the additional assumption that $X$ is a CW complex and the action is cellular?

By the way: I've seen the wonderful post at mathoverflow, but it doesn't seem to have an answer for my question.

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What map $G\times X\to X\times X$? Didn't you mean $G\times X\to X$? –  tomasz Dec 11 '12 at 0:26
@tomasz : It is correct as written. The map is (g,x) $\mapsto$ (gx, x). –  Dylan Wilson Dec 11 '12 at 15:28