Let $G$ be a finite group acting on some Hausdorff space $X$ such that the action is free. I have some feeling that the action of $G$ is always properly discontinuous, i.e. for all $x\in X$, there is an open neighbourhood $U$ of $x$ such that $g\cdot U\cap U=\emptyset$ for all $g\neq e_G\in G$. Is this true, or under what additional conditions might this be true?
Finite group action on Hausdorff space
general-topologygroup-theory
Related Solutions
The action of a finite group $G$ on a Hausdorff space $X$ is always discontinuous.
Let $x_0 \in X$, and consider the orbit $Gx_0 = \{ x_0, x_1,\dotsc, x_k\}$. Since $X$ is Hausdorff, we can find open neighbourhoods $V_{\kappa}$ of $x_{\kappa}$ such that $V_{\kappa} \cap V_{\lambda} = \varnothing$ for $\kappa \neq \lambda$. For each $g \in G$, we can find an open neighbourhood $W_g$ of $x_0$ such that $gW_g \subset V_{\kappa}$ if $gx_0 = x_{\kappa}$ since $x \mapsto gx$ is continuous. (For example $W_g = V_0 \cap g^{-1}V_{\kappa}$.) Then
$$U = V_0 \cap \bigcap_{g \in G} W_g$$
is an open neighbourhood of $x_0$, and we have $gU \subset V_{\kappa}$ whenever $gx_0 = x_{\kappa}$, in particular
$$gU \cap U \neq \varnothing \iff gx_0 = x_0.$$
On a non-Hausdorff space, the action of a finite group need not be discontinuous. As an extreme example, take an indiscrete space, then the only discontinuous action is the trivial action.
If $X$ is Hausdorff, an action of a finite group is also proper: Given a compact $K\subset X\times X$, let $K_i = \pi_i(K) \subset X$. Then $K_1,K_2$ are compact, $K \subset K_1 \times K_2$, and $\theta^{-1}(K)$ is a closed subset of $\theta^{-1}(K_1 \times K_2)$. But
$$\theta^{-1}(K_1\times K_2) = \bigcup_{g\in G} \{g\} \times (K_1 \cap g^{-1}K_2)$$
is the union of finitely many compact sets, hence compact.
Note that when speaking of the Schlag definition, we require the space be locally compact.
I first note that a Hatcher action is not necessarily a Schlag action. I'm stealing the proof of this another question.
Consider $X = \mathbb{R}^2 \setminus \{(0, 0)\}$ and the action of $\mathbb{Z}$ on this space given by $n \cdot (x, y) = (2^n x, 2^{-n} y)$. It's easy to show that this action is a Hatcher covering space action. For suppose we have $(x, y) \in X$. Then either $x \neq 0$ or $y \neq 0$. In the first case, WLOG suppose $x > 0$ and let $U = (x/\sqrt{2}, \sqrt{2}x) \times \mathbb{R}$; in the second case, WLOG suppose $y > 0$ and let $U = \mathbb{R} \times (y / \sqrt{2}, \sqrt{2} y)$. However, this action fails to meet Schlag's definition. For consider the compact set $K = \{(x, 1) : x \in [0, 1]\} \cup \{(1, y) : y \in [0, 1]\}$. Then for every $n \geq 0$, we have $p_n = (2^{-n}, 1) \in K$. And we have $n \cdot p_n = (1, 2^{-n}) \in K$. Then $n(K) \cap K \neq \emptyset$ for all $n \geq 0$. This clearly contradicts Schlag's condition.
On the other hand, an action meeting Schlag's condition may not meet Hatcher's. In particular, consider the action of $\mathbb{Z}_2$ on $\mathbb{R}$ defined as $x \cdot y = y$. This clearly meets the condition laid out in Schlag, as for every compact $K$, the set $\{g \in \mathbb{Z}_2 : g(K) \cap K \neq \emptyset\}$ is finite. However, it definitely fails to meet Hatcher's condition since for all $U$, $0(U) = 1(U)$.
So these are two completely different conditions.
Best Answer
Try to prove the following observation: If $g⋅x ≠ x$, then there is $U_g$ a neighborhood of $x$ such that $g ⋅ U_g ∩ U_g = ∅$. Moreover, for every smaller neighborhood $V$ of $x$, the same holds.