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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.

Markus Scheuer
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MathRookie2204
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  • What do you mean by the $D_\infty$ in $\hat {\mathbb C}$ given a domain $D$ in the $\mathbb C$? An equivalent criteria for simply connectedness is that $\hat{\mathbb C} \setminus D$ is connected, so, I am not so sure what you mean by $D_\infty$. – ash Sep 26 '23 at 06:48
  • $D_{\infty}$ is the image of $D$ in the Riemann sphere $S$ under the stereographic projection. – MathRookie2204 Sep 26 '23 at 08:13
  • $D\subseteq \mathbb C\subset \mathbb C_\infty$ so $ D$ does not contain $\infty$ and the standard homeomorphism from $\mathbb C_\infty$ to the sphere sends $\infty$ to the north pole, so the statement about the north pole is superfluous in standard treatments. (Someone could choose a different homeomorphism though; I suggest working directly with $\mathbb C_\infty$.) For (iii) your line $\Im(z)=\frac{1}{2}$ is not compact in $\mathbb C$ so it cannot be the image of a closed curve hence it tells you nothing about whether or not $D$ is simply connected. – user8675309 Sep 26 '23 at 15:51
  • I understand your argument @user8675309. – MathRookie2204 Sep 26 '23 at 17:33
  • But how to see that $D$ is simply connected, as in Wiki ? – MathRookie2204 Sep 26 '23 at 17:34
  • One way thinking about simply connectedness of $D$ is that $D$ has no holes. Intuitively speaking, no holes translates to $\mathbb C \setminus D$ does not have any bounded connected components. It is easy to see from this criterion that any horizontal strip is simply connected. – ash Sep 26 '23 at 20:00
  • If you are looking for proofs of equivalent criterions of simply connectedness, I suggest you take a look at Rudin's Real and Complex Analysis Theorem 13.11 in the Third edition. – ash Sep 26 '23 at 20:03
  • The North pole is irrelevant here: it was asked and answered several times on MSE that an open connected subset of the 2-sphere is simply connected if and only if it's complement is connected. For instance https://math.stackexchange.com/questions/330479/complement-is-connected-iff-connected-components-are-simply-connected?rq=1 – Moishe Kohan Sep 27 '23 at 01:37
  • Also https://math.stackexchange.com/questions/1219757/connected-closed-subset-in-a-plane – Moishe Kohan Sep 27 '23 at 01:43

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