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

Asked Sep 09 '14 at 19:07

Active Sep 09 '14 at 19:38

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

edited Sep 09 '14 at 19:38

asked Sep 09 '14 at 19:07

Fin8ish

What you mean is not «that accumulation points of $A$ is $A$» but «that the set of accumulation points of $A$ is $A$». – Mariano Suárez-Álvarez Sep 09 '14 at 19:32
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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 $[0,1]$. – Ben Grossmann Sep 09 '14 at 19:35

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