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Say $G\subseteq \mathbb{C}$ is open. Can we always choose a countable and dense subset of $G$?.

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    $\begingroup$ In fact every subset of $\mathbb C$ has a finite or countable dense subset. This follows from the Lindelof property of $\mathbb C.$ $\endgroup$
    – zhw.
    Dec 17, 2017 at 18:00

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Let $K:=\{a+bi:a,b\in\mathbb{Q}\}$. You know that $K$ is dense in $\mathbb{C}$. Then $K\cap G$ is dense in $G$.

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  • $\begingroup$ why is K dense in $\mathbb{C}$? $\endgroup$
    – seht111
    Dec 17, 2017 at 16:18
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    $\begingroup$ @seht111 Because $\mathbb{Q}$ is dense in $\mathbb{R}$. $\endgroup$
    – Gödel
    Dec 17, 2017 at 16:19

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