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Is it true that every open subset of $\mathbb{C}$ is a countable union of disjoint connected open sets?

That is if $A\subset\mathbb{C}$ then $A=\bigcup _{i=1}^{\infty}U_{i}$ where $U_{i}$ are pairwise disjoint, connected, open sets?

Mazzer
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  • If you allow $U_i = \varnothing$ then yes, otherwise you'd need to allow for open sets with finitely many components. The point is that $\mathbb{C}$ is second countable and locally connected. – Daniel Fischer Jul 12 '15 at 19:52
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    The notion of connected components means that all you have to prove is the "countable" union. In each connected component, choose a complex number whose real and imaginary parts are rational numbers. This is an injection from the set of connected components into $\Bbb Q^2$; hence the set of connected components is countable. – Greg Martin Jul 12 '15 at 19:56
  • So given the rest, countable can be show as you say @GregMartin, however my problem is with the pairwise disjoint, connected, open sets part. Please have patients with me, i haven't read any topology. – Mazzer Jul 12 '15 at 20:08
  • Sorry, I did not read these posts before writing my answers; I did not mean to ripoff from the comments. – Gary. Jul 12 '15 at 20:14
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    Define an equivalence relation $\sim$ on $A$ by: $x\sim y$ if there exists a connected subset of $A$ containing both $x$ and $y$. The equivalence classes under this relation are the connected components of $A$. (It's a good exercise to check everything in those two sentences.) See https://en.wikipedia.org/wiki/Connected_space#Connected_components – Greg Martin Jul 12 '15 at 22:08

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Yes, $\mathbb C$ is homeomorphic to $\mathbb R^2$ , and so it is 2nd countable. This implies that every connected component is the countable union of open sets. And every open set can have at most countably-many components.

Gary.
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