Wikipedia claims the following:
In mathematics, an action of a group G on a topological space X is cocompact if the quotient space X/G is a compact space or, equivalently, if there is a compact subset K of X such that
the image of K under the action of G covers X.
My question is: Isn't this wrong? It is evident that the existence of such a subset K ensures cocompactness, but I am doubting the other direction. How could one possibly choose K? Taking an arbitrary transversal (or its closure) does not work, and I do not see what else could be a candidate.
By the way: Wikipedia points to a specific page in the Handbook of Geometric Topology. This page, however, contains only the definition of a cocompact space, not the claimed equivalence.
Best Answer
I agree with @Bugs that some extra assumptions are needed, although I do not have a counter-examples either.
Here is an argument assuming that $X$ is locally compact. For each $y\in Y=X/G$ choose (arbitrarily) a point $y'\in X$ which projects to $y$ (you'd need axiom of choice here). Now, for each $y'$ pick an open neighborhood $U_{y'}$ whose closure in $X$ is compact. Projecting the open sets $U_{y'}$ to $Y$ yields an open covering by the sets $U_y$ (images of $U_{y'}$'s). By compactness, there is a finite set $\{y_1,...,y_n\}\subset Y$ so that the sets $U_{y_i}$ cover $Y$. Then the union $$ \bigcup_{i=1}^n cl(U_{y_i})\subset X $$ is compact and projects onto $Y$.