We have the following result in complex analysis.
Let $D$ be a domain (open connected set) in $\mathbb{C}$ and $D_{\infty}$ be the set corresponding to $D$ on the Riemann Sphere $S.$ Then, $D$ is simply connected if $D_{\infty}^c=S{\setminus}D_{\infty}$ is connected and contains the north pole.
I have a doubt understanding this result.
My thoughts:
Consider $D=\{z \in \mathbb{C}: 0 < \Im(z)<1 \}.$
(i) We can see that $D$ is an open connected set. I think that the north pole in $S$ in contained in $D_{\infty},$ meaning that $D$ is not simply connected.
(ii) However, suppose that $\infty \in D_{\infty},$ we need a open ball around (neighbourhood of $\infty$) that is completely contained in $D_{\infty}.$ But this is not possible as $\infty$ can be approached via the Imaginary axis also. So, $\infty \notin D_{\infty}.$ Hence, $D$ is simply connected.
The problem is this: Does $\infty \in D_{\infty}$ or not ??
One more point that I want to add is this. Say we take the definition of simple connected set as:
A path-connected set $S \subset\mathbb{C}$ is called simply connected if any closed loop in $S$ can be contracted to a point.
(iii) For the same set $D,$ that we took previously, we can take the line $\Im(z)=\frac{1}{2},$ which is mapped to the circle under the stereographic projection. And that cannot be shrunk to a point while remaining in $D$ only. So, $D$ is not simply connected.
But Wikipedia says $D$ is simply connected.
What is happening here? I am confused as to which one to follow. Please help.