Orbit maps are proper if and only if action is proper?

Question

Let $G$ act continuously on $X$, with both $G$ and $X$ locally compact Hausdorff. One usually defines a proper action to be one for which $(g,x) \mapsto (gx,x) : G \times X \to X \times X$ is a proper map. This definition is a bit uninformative, but one can show pretty easily it is equivalent to saying that, for every compact set $K \subseteq X$, the set $\{g \in G : gK \cap K \neq \varnothing \}$ is compact in $G$.

The second characterization above is rather intuitive. It says that, for any compact set in $X$, you only have to wait a compactly long time before it is translated away from itself. Looking at things this way, we are led to consider another condition on the action which is, a priori, weaker. Namely, we can consider an action with the property that a given point only hangs around in a given compact set for a compactly long time. That is, we can consider an action such that, for each $x \in X$, the orbit mapping $g \mapsto gx : G \to X$ is a proper mapping. It is conceivable that this is actually equivalent to properness of the action, but I don't see how to prove it off the top of my head, so I pose it here as a question.

Question: Let $X$ and $G$ be locally compact Hausdorff. Suppose that, for every $x \in X$, the orbit map $g \mapsto gx : G \to X$ is proper. Does it follow that the action is proper?

I fiddled with this a bit, but didn't decide the answer. If true, I feel it should rely on some principle like the "Tube Lemma". For a little while I thought I had succeeded by applying the following claim

Claim: (False) Let $X$ be a locally compact Hausdorff space and $K$ a compact Hausdorff space. Suppose that $A \subseteq K \times X$ is a closed set with the property that, for each $k \in K$, the slice $A_k = A \cap (\{k\} \times X)$ is compact. Then $A$ is compact.

which looked plausible (roughly, a "compact union" of compact sets ought to be compact), but actually that claim is false, e.g. $A = \{ (x,1/x): 0 <x < 1\} \cup \{(0,0)\} \subseteq [0,1] \times \mathbb{R}$. I still hold out some hope that the answer to the quesion is yes, though. I feel like there is some equicontinuity or something which I am failing to exploit...? Anyway, that's quite enough half-baked speculation from me!

Answer 1

OK, following the comments of Moishe Cohen, we consider the action of $G = \mathbb{Z}$ on $X = \mathbb{R}^2 \setminus \{(0,0)\}$ given by iterating the linear transformation $\left( \begin{smallmatrix} 2 & 0 \\ 0 & \frac{1}{2}\end{smallmatrix} \right)$. The orbits of this action travel along hyperbolas, as pictured in red below.

It is pretty clear from the picture that, for fixed $v \in X$, the orbit map $n \mapsto n \cdot v$ is proper i.e. eventually escapes from any compact set (note that a compact set in $X$ needs to be bounded away from the origin).

On the other hand, we also find that this is not a proper action. Consider a compact line like $L = [0,1] \times \{1\}$. The forward iterates $n \cdot L$, represented as the blue horizontal lines in the picture below, elongate and approach the positive $x$-axis as $n \to \infty$.

Thus, we see the $n \cdot L$ never escape from certain compact sets, for instance the compact disk shaded black in the above image. So, we find that the condition of being a proper action is strictly stronger than having proper orbit maps.