An exercise that was left in class asked to prove that a free and wandering group action $G$ on $T_2$ and locally compact space $X$ is properly discontinuous. My solution to this problem seems to not use the fact that $X$ is locally compact. My proof is that given any $x$ in $X$ since the action is wandering there exist some open neighborhood $U$ of $x$ such that ($U$|$U$) is finite. Then for each $g$ in ($U\mid U$) that is not the identity and using the fact that $X$ is $T_2$ and since $gx$ is not equal to $x$ since the group action is free I can create two disjoint open sets $U_g$ and $V_g$ that separate $x$ and $gx$. Since G is a group action we have that $g$ is a continuous function so $g^{-1}(V_g)$ is also open for each $g$. Taking the finite intersection of $U$ and all $U_g$ and all $g^{-1}(V_g)$ I can create an open neighborhood $A$ of $x$ such that for any non identity element $g$ in $G$ $A$ intersected with $g(A)$ is empty so it's properly discontinuous. What's the error in this proof or is the locally compact condition not necessary? Also is it implicit that the group action on a topological space is continuous?
Proof: a group action on a $T_2$ space that is free and wandering is properly discontinuous
algebraic-topologygeneral-topology
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These properties are not equivalent. Here's a counterexample: Let $X=\mathbb R^2\smallsetminus\{(0,0)\}$, and define an action of $\mathbb Z$ on $X$ by $n\cdot (x,y) = (2^n x, 2^{-n} y)$. This is properly discontinuous by your definition, but it's not a proper action. The subset $K \times K \subseteq X\times X$ is compact, where $K = \{(x,y): \max(|x|,|y|)=1\}$, but $\rho^{-1}(K\times K)$ contains the sequence $(n, (2^{-n},1))$, which has no convergent subsequence.
I think one reason for your confusion is that different authors give different definitions of "properly discontinuous." Topologists concerned primarily with actions that determine covering maps often give the definition you gave:
(i) Every $x \in X$ has a neighborhood $U$ such that $gU \cap U \neq \emptyset$ implies $g = e$.
This is necessary and sufficient for the quotient map $X\to X/G$ to be a covering map. However, in order for the action to be proper (and thus for the quotient space to be Hausdorff), an additional condition is needed:
(ii) If $x,x'\in X$ are not in the same $G$-orbit, then there exist neighborhoods $U$ of $x$ and $U'$ of $x'$ such that $gU\cap U' = \emptyset$ for all $g\in G$.
When $X$ is a locally compact Hausdorff space and $G$ is a discrete group acting freely on $X$, the action is proper if and only if both conditions (i) and (ii) are satisfied. Differential geometers, who are typically concerned with forming quotient spaces that are manifolds, are more apt to define "properly discontinuous" to mean both (i) and (ii) are satisfied.
Because of this ambiguity (and because the term "properly discontinuous" leads to oxymoronic phrases such as "a continuous properly discontinuous action"), Allen Hatcher in his Algebraic Topology coined the term covering space action for an action satisfying condition (i). I've adopted that terminology, and I use free and proper action for an action satisfying (i) and (ii) (at least for locally compact Hausdorff spaces). I sincerely hope the term properly discontinuous will eventually die out.
You can find more about these issues in the second editions of my books Introduction to Topological Manifolds (Chapter 12) and Introduction to Smooth Manifolds (Chapter 21).
The behavior of the action can change in different connected components, so in general you have no way to guarantee that the fibers of points in distinct components are homeomorphic.
As a simple example consider $Y=\{a,b,c\}$ with the discrete topology and a two element group $G$, where the non-trivial element swaps $a$ and $b$. Then $Y/G$ has two points, and the fibers are the non-homeomorphic sets $\{a,b\}$ and $\{c\}$.
Edit: If in the definition of covering you don't require the fibers to be homeomorphic, your argument does indeed work, i.e. the lemma is valid also for a disconnected $Y$.
Note that some authors may require connectedness of the base space in the definition of covering, since some properties will be only locally constant over the base space (e.g. the degree, as shown in the above example).
Best Answer
I think your proof is correct. The point is the definition of a properly discontinuous action. Your understanding seems to be the standard, but in https://en.wikipedia.org/wiki/Group_action_(mathematics) you find another definition: