Is the the derived set of irrational numbers $\subset \mathbb R$ an empty set?

Let cl denotes the closure, $A'$ denotes the derived set, int denotes interior, and bd the boundary, it is easy to see the following properties:

$A'$= int$A \ \cup$ bd(int($A$)) = cl$($int$(A))$ ?

Since the interior of irrational numbers are empty, so the derived set is also empty?

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    $\begingroup$ What makes you think $A'=\operatorname{int}(A)\cup\operatorname{bd}(\operatorname{int}(A))$? $\endgroup$ – bof Oct 28 '17 at 4:19

I assume you use the topology derive by the canonical metric in $R$.then the derive set of the irrational number $\overline{R-Q}$ is the whole space $R$.This is just because the irrational number is dense in $R$,or you can view this as $int(Q)=\emptyset$.


Since $\mathbb Q$ is dense in $\mathbb R$, we have $\overline{\mathbb Q} = \mathbb R \Rightarrow \mathbb{I}^{\circ} = \mathbb R \setminus\overline{\mathbb Q} = \emptyset$. So $\overline{\mathbb{I}^{\circ}}=\overline{\emptyset} = \emptyset$.

$\mathbb I$ is also dense in $\mathbb R$, so $\mathbb I' = \mathbb R$.

Then the statement $A' = \overline{A^{\circ}}$ is false.


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