On the definition of properly discontinuous action of a group

Question

A question regarding the following definition arose.

If $G$ acts properly discontinuously at $\infty$, then by the definition of a neighborhood of $\infty$ and by the above, there exists a set $\{z\in \mathbb{C}: |z|>R\}\cup\{\infty\}$, which is an open subset of the extended complex plane, that remains fixed by any element of the stabilizer $G_\infty$ and is displaced completely away from itself by any element outside of the stabilizer $G_\infty$. But the displaced neighborhood is of the form $\{z\in \mathbb{C}: |z-z_0|>R\}\cup\{\infty\}$, isn't it? How can this displaced copy of the initial neighborhood have empty intersection with the initial set $\{z\in \mathbb{C}: |z|>R\}\cup\{\infty\}$? I thought the exteriors of any two open discs intersect.

Answer 1

The displaced neighborhood won't be of that form because it won't be a neighborhood of $\infty$. If $g\in G\setminus G_\infty$, then $g(\infty)\neq\infty$, and the image of your neighborhood under $g$ will be a neighborhood of the point $g(\infty)$, not of $\infty$.

Remember, a neighborhood of a point $z_0$ is just a set that contains a disk around $z_0$ unless $z_0=\infty$, in which case it is a set that contains the complement of a disk (together with $\infty$). So when you change from being a neighborhood of $\infty$ to a neighborhood of some other point, that doesn't mean you change the center of the disk you are taking the complement of; it means you change altogether from having the complement of a disk to a disk itself.