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Let $X$ be a topological $2$-sphere. Let $D_1, D_2, \dots, D_n \subset X$ be a finite family of closed topological disks (i.e. sets homeomorphic to the closed unit disk). Let $\mathcal{U} = \bigcup_{1 \le i \le n} \text{int}(D_i)$ be the union of the interiors of these disks.

That the following two statements are equivalent should be clear, but I am having a hard time proving it rigorously:

a) Two points $x,y$ are in the same connected component $K$ of the complement of $\mathcal{U}$.

b) $\mathcal{U}$ contains no Jordan curve separating $x$ and $y$.

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  • $\begingroup$ Are these disks assumed to be pairwise disjoint? $\endgroup$
    – Wlod AA
    Commented Sep 8, 2022 at 23:18
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    $\begingroup$ It's not even necessary to assume that your open set is a union of the interiors of finitely many topological closed disks. This has to do with Alexander duality. $\endgroup$
    – Tom Goodwillie
    Commented Sep 9, 2022 at 0:00
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    $\begingroup$ @WlodAA they are not, if they were, the complement would always be path connected. I am aware that some kind of Alexander duality should be lurking in the background but I am not able to rephrase the equivalence in homological terms. $\endgroup$
    – Leon Staresinic
    Commented Sep 9, 2022 at 8:27

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