Say $G\subseteq \mathbb{C}$ is open. Can we always choose a countable and dense subset of $G$?.
1 Answer
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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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1$\begingroup$ @seht111 Because $\mathbb{Q}$ is dense in $\mathbb{R}$. $\endgroup$– GödelDec 17, 2017 at 16:19