Equivalent conditions for covering space actions I'm trying to reconcile two different sets of conditions on a group action $G$ acting on $X$ to make it a covering space action. Hatcher uses this condition (changed notation for consistency):

Each $x \in X$ has a neighbourhood $U$ such that all the images $g(U)$ for varying $g \in G$ are disjoint. In other words, $g_1(U) \cap g_2(U) \neq \varnothing$ impliles $g_1 = g_2$.

The other comes from Schlag and is similar to Wikipedia's description of a properly discontinuous action:

Let $K \subset \Omega$ be compact. Then the cardinality of $\{g \in G : g(K) \cap K \neq \varnothing\}$ is finite.

Does one of these imply the other?
 A: 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.
