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I encountered this question in Abbott's Understanding Analysis. The problem asks to construct a nonempty perfect set with no rationals. It starts with enumerating the rationals $\mathbb{Q}=\{r_1,r_2,...\}$. At first we start with an open set $O:=\bigcup_1^\infty V_{\epsilon_n}(r_n)$ where $\epsilon_n:=1/2^n$ and $V_{\epsilon_n}(r_n)$ is open neighborhood of $r_n$ with radius $\epsilon_n$. Then $F:=O^c$ is obviously closed and nonempty (since lengths of the open intervals add up to 2) and $F$ obviously contains only irrationals also $F$ does not contain any open intervals since it does not contain any rational so $F$ is totally disconnected as well. But it is not possible to know that $F$ is perfect, since it may have isolated points. How do I modify the above construction to make $F$ as a perfect set??

Edit: I have seen similar questions here but with different constructions of different sets. Here the question asks for a very specific modification of a given construction.


marked as duplicate by Eric Towers, Asaf Karagila general-topology Jun 21 at 8:01

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  • $\begingroup$ Yes I have seen similar questions. But the context is different. There are indeed many proofs possible using many fancy constructions. This problem asks a modification of a very specific construction. $\endgroup$ – Arpan Das Jun 21 at 7:50
  • $\begingroup$ Infact one can easily construct a cantor set like thing by avoiding the $n-th$ rational at the $n-th$ stage of construction by removing large enough open sets from the $n-1$ th stage. $\endgroup$ – Arpan Das Jun 21 at 7:53
  • $\begingroup$ Every uncountable closed set contains a perfect subset. That is, in effect, Cantor's first step towards proving CH. $\endgroup$ – Asaf Karagila Jun 21 at 8:04
  • $\begingroup$ Every uncountable Borel set contains a subspace homeomorphic to the Cantor set. The irrationals are a Borel set. $\endgroup$ – Henno Brandsma Jun 21 at 9:36