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Assume $O_1$ and $O_2$ are two open connected sets of $\mathbb{C}$ such that $O_1\cup O_2=\mathbb{C}$.

Is it true that $O_1\cap O_2$ is connected ?

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Yes. For open subsets of $\mathbb{C}$ connectedness agrees with pathwise connectedness (this is true for any locally pathwise connected space).

The Mayer-Vietoris-sequence for reduced singular homology gives us an exact sequence

$$H_1(\mathbb{C}) \to \tilde{H}_0(O_1 \cap O_2) \to \tilde{H}_0(O_1) \oplus \tilde{H}_0(O_2) .$$

But $H_1(\mathbb{C}) = 0$ and $\tilde{H}_0(O_k) = 0$, hence $\tilde{H}_0(O_1 \cap O_2) = 0$. This implies that $O_1 \cap O_2$ has only one path component.

Note the $H_0(X)$ is always a free abelian group whose basis can be identified with the set of path components of $X$.

This generalizes as follows: Let $X$ be connected with $H_1(X) = 0$ and $O_1,O_2$ two open path connected subsets such that $O_1 \cup O_2 = X$. Then $O_1 \cap O_2$ is path connected.

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  • $\begingroup$ Very nice, Thanks ! $\endgroup$ – user111 Sep 29 '18 at 14:19

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