0
$\begingroup$

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?

$\endgroup$
  • 1
    $\begingroup$ What makes you think $A'=\operatorname{int}(A)\cup\operatorname{bd}(\operatorname{int}(A))$? $\endgroup$ – bof Oct 28 '17 at 4:19
2
$\begingroup$

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 $int(Q)=\emptyset$.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.