E Hausdorff topological space, G acts properly discontinous

Question

Let $E$ be a Hausdorff topological space, $G$ a homeomorphism group that acts on $E$ properly discontinous, i.e. $\forall e\in E$ exists a neighborhood $U$ of $e$ such that $gU\cap U = \emptyset $ for all $g \neq 1_{G}$.

Then prove that the orbits are closed and discrete.

I'm interested in proving orbits are closed. If G is finite the problem is easy.

What about the case G infinite ? Any hint ?

Answer 1

Let the orbit of $x$ not be closed, and hence have a limit point $y$ outside. Now, pick the neighborhood $V$ of $y$ given by the definition of proper discontinuity. For some $g$, $gx \in V$. Since the space is Hausdorff and $y \neq gx$, $V\setminus \{gx\}$ is a neighborhood of $y$. Thus, there is some $hx \in V\setminus \{gx\}$, in particular $h \neq g$. Then $V \cap gh^{-1}V \neq \varnothing$. Hence contradiction.

Answer 2

Let $h : G \times E \to E \times E$ be defined as $h(g,x) = (gx,x)$. Proper discontinuity says that $h$ is proper.

I have a partial solution - it works with sequential characterizations. Assume $E$ is first-countable.

Suppose there is an $x \in E$ whose orbit $O_x$ is not closed. Then there exists a $y$ outside $O_x$ and a sequence $g_1x, g_2x, \ldots$ in the orbit converging to $y$. Let $S = \{y\} \cup \{g_1x, g_2x, \ldots\}$. Then $S$ is compact and $S \times \{x\}$ is compact. Its inverse image under $h$ is at least $\{g_1, g_2, \ldots\} \times \{x\}$. The inverse image is infinite (this follows from $E$ being Hausdorff). Therefore it is not compact, contradicting proper discontinuity.

Suppose there is an $x \in E$ whose orbit $O_x$ is not discrete. I assume this means that the subspace topology on $O_x$ is not the discrete topology. Then there is a subset of $O_x$ which is not closed in $O_x$, and we can repeat the previous argument.

I'm not sure how to generalize this, working with neighbourhoods directly or with nets.