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It is known that a domain $\Omega$ in the complex plane is simply connected if and only if its complement in the Riemann sphere ${\mathbb C}\cup\{\infty\}$ is connected. A sufficient condition for this is that the complement of $\Omega$ in ${\mathbb C}$ has no bounded connected component. Is the reverse implication true, i.e. : if $\Omega$ is simply connected, are the connected components of ${\mathbb C}\setminus\Omega$ necessarily unbounded ?

Equivalent formulation : let $F$ be a closed subset of the complex plane with a bounded connected component. Does $F\cup \{\infty\}$ still have a bounded connected component? (This is not true is $F$ is not closed.) If this holds, is there a simple topological proof?

user1124160
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Yes, this is true. An easy argument is to use the Alexander duality, but am skeptical you are familiar with this tool. However, since you already know that $\Omega$ is simply connected if and only if it's complement is connected, note that a bounded complementary component has to be different from the one containing infinity. Hence, the complement has at least 2 complementary components, hence, is disconnected.

Moishe Kohan
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  • Thank you for your answer. I would be happy to learn how to apply Alexander duality (and in which form) to this situation. For the second part of your answer, it is not clear to me whether you are speaking of complementary components in $\mathbb C$ or $S^2$? The point of my question is that a bounded component of ${\mathbb C} \setminus \Omega$ could be a (strict) subset of the component of $S^2\setminus\Omega$ containing $\infty$. – user1124160 Nov 28 '22 at 15:41
  • In fact I just realized that you already answered this question, see Lemma 1 in your answer to this question! This is not so simple, but still elementary topology. – user1124160 Nov 29 '22 at 21:14