I want to find a nonempty perfect set with no rationals.

Let $\{ q_n \}$ be an enumeration of the rationals such that $\mathbb{Q} = \{ q_1, q_2 , ..., q_n ,... \}$. Define the open intervals $I_n = (q_n - \delta_n , q_n + \delta_n)$. I want to find a sequence $\{ \delta_n \}$ so that the set $\displaystyle P = \mathbb{R} \setminus \bigcup_{n=1}^\infty I_n$ is perfect and contains no rationals. $P$ is clearly closed by the DeMorgan Laws, as $\displaystyle \mathbb{R} \setminus \bigcup_{n=1}^\infty I_n = \bigcap_{n=1}^\infty (\mathbb{R} \setminus I_n)$, and the intersection of a collection of closed sets is closed. We have that if $\displaystyle \sum_{n=1}^\infty \delta_n < \infty$, then $\displaystyle\mathbb{R} \setminus \bigcup_{n=1}^\infty I_n\neq \emptyset$.

How would we pick $\delta_n$ so that $P \subset P'$?

  • $\begingroup$ If you can simply avoid having two intervals $ I_{n_1} $ and $ I_{n_2} $ that "touch" at an endpoint, then I think the set you end up with is perfect. You can recursively define the $ \delta_n $ to make this so. I commented rather than answering because I'm not 100% sure about this. $\endgroup$
    – Jake Mirra
    Oct 17, 2019 at 3:34
  • $\begingroup$ what $P'$? As stated your question is unclear (yes that construction could be done, and it would be easier done if you don't use each $q_n$, but having used some, then use the "first" $q_n$ not contained in (the union of) "previous" $I_k$ (and also pick $\delta_n$ small enough so intervals don't overlap, or don't touch at endpoints, as pointed out by @JakeMirra)), but what are you asking? $\endgroup$
    – Mirko
    Oct 17, 2019 at 11:59


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