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This question already has an answer here:

Is there a non-empty subset of $\mathbb{R} $ like $A$ such that the set of accumulation points of $A$ is $A$ and $A\cap\mathbb{Q}=\emptyset\,$?

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marked as duplicate by Omnomnomnom, Asaf Karagila general-topology Sep 9 '14 at 19:17

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  • $\begingroup$ What you mean is not «that accumulation points of $A$ is $A$» but «that the set of accumulation points of $A$ is $A$». $\endgroup$ – Mariano Suárez-Álvarez Sep 9 '14 at 19:32
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    $\begingroup$ Another example (in addition to those found in the link) is one similar to the generalized Cantor set in which we remove progressively smaller neighborhoods of rational numbers in $[0,1]$. $\endgroup$ – Omnomnomnom Sep 9 '14 at 19:35

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