Finding a nonempty perfect set with no rationals

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 $$P \subset P'$$?

• 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. Oct 17, 2019 at 3:34
• 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? Oct 17, 2019 at 11:59