Converse to Properness of Group Action

Question

I'm working on Lee's Manifolds problem 21-1, which states the following:

Suppose a Lie group $G$ acts continuously on a manifold $M$ . Show that if the map $G \times M \rightarrow M$ defining the action is proper, then the action is a proper action. Give a counterexample to show that the converse need not be true.

I think I've got the first part sorted out by playing around with definitions and projection mappings, but I still don't have much intuition for what this really means. That lack of intuition has given me some issues trying to come up with a counterexample for the converse. A comment in this thread [https://math.stackexchange.com/questions/2434982/why-is-the-definition-of-a-proper-group-action-the-way-it-is] states that:

Also note that map $f:G \times M \rightarrow M$ is proper iff $f^{-1}[K]$ is compact for every compact 𝐾, while the mentioned map $G \times M \rightarrow M \times M$ is proper iff $f^{-1}[K] \cap (G \times K')$ is compact for every compact $K$, $K'$

which feels like it should be enough to find a counterexample, but my lack of intuition for what these actions really look like makes it impossible for me to try to apply this comment to come up with an example. Any help regarding this topic (and the converse example specifically) would be extremely welcome!

Answer 1

If $f: X \to Y$ is a map between metric spaces there is an easier way to define the notion of properness which ought to give some intuition.

A map $f: X \to Y$ is proper if, for any sequence $x_n \in X$ diverging to $\infty$ in $X$, the sequence $f(y_n)$ also diverges to $\infty$ in $Y$.

Here we say a sequence diverges to $\infty$ if it has no convergent subsequence. While I tend to prove things about properness using the general notion this gives me an intuitive handle on the concept. Exercise: prove that this is equivalent to $f$ being proper.

Hint for your question: take $G = M$. To be even more explicit, take $G = M = \Bbb R$.