This is a homework question from a Fractals & Chaos University course:

Find subsets $A,B \subset \Bbb{R}$ (with the usual topology) such that:

i) $\partial (\partial A)=\partial A$

ii)$\partial (\partial B)\subsetneq\partial B$

Attempt at solutions i) take $A=\emptyset$ then $\partial\emptyset=\emptyset$ so $\partial(\partial\emptyset)=\partial\emptyset$

ii) I don't know how to start looking for a Set $B$. I have a feeling I can take $B$ as a subset of the rationals.

Thanks in advance. :)

  • $\begingroup$ Have you looked at any examples besides the empty set? $\endgroup$
    – Carsten S
    Oct 23, 2013 at 20:10
  • 1
    $\begingroup$ Yes, you can take $B=\mathbb{Q}$. $\endgroup$
    – njguliyev
    Oct 23, 2013 at 20:11
  • $\begingroup$ @CarstenSchultz No, my lecturer was only looking for one example. But, I guess, I could have $\Bbb{R}$ or maybe even any 'clopen' set. $\endgroup$ Oct 23, 2013 at 20:29
  • $\begingroup$ Actually, I guess that your lecturer posed the problem to make you think about this. What I wanted to say is that if you determine $\mathfrak d\mathfrak d A$ for several sets $A$ you may get an idea of what's going on. $\endgroup$
    – Carsten S
    Oct 23, 2013 at 20:40

1 Answer 1


To close the question:

One can take $A=\emptyset$ to satisfy i) and take $B=\mathbb Q$ (or any subset of the rationals) to satisfy ii).

With thanks to njguliyev and Carsten Schultz.


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