The rationals. More generally, if \(A\) is an uncountable compact subset of a metric space then, for any \(n \in \mathbb{N}\) there is a finite subset \(S_n\) of \(A\) such that every element of \(A\) is within \(n^{-1}\) of some element of \(S_n\). Then
\[S=\bigcup_{n\in\mathbb{N}}S_n\]
Is a countable subset of \(A\) whose closure is \(A\).
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Mark Hennings
·
2 years, 10 months ago

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One example is the set \(S:= \{ x \colon x\in \mathbb{R}^2 \land \| x\| <1 \}\cap \mathbb{Q}^2 \).
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Austin Stromme
·
2 years, 10 months ago

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TopNewestThe rationals. More generally, if \(A\) is an uncountable compact subset of a metric space then, for any \(n \in \mathbb{N}\) there is a finite subset \(S_n\) of \(A\) such that every element of \(A\) is within \(n^{-1}\) of some element of \(S_n\). Then \[S=\bigcup_{n\in\mathbb{N}}S_n\] Is a countable subset of \(A\) whose closure is \(A\). – Mark Hennings · 2 years, 10 months ago

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One example is the set \(S:= \{ x \colon x\in \mathbb{R}^2 \land \| x\| <1 \}\cap \mathbb{Q}^2 \). – Austin Stromme · 2 years, 10 months ago

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