Existence of special enumeration of the rationals

Question

I am looking for an enumeration of $\mathbb Q$ which has a specific property. It is inspired by Problem 4.13(ii)(b) from the book Measures, Integrals and Martingales by René L. Schilling (second edition).

Let $q_1,q_2,\dots$ be an enumeration of $\mathbb Q$. For all $\epsilon>0$, let $$C(\epsilon):=\cup_{n\in\mathbb N}[q_n-\epsilon2^{-n},q_n+\epsilon2^{-n}).$$ Let $C:=\cap_{\epsilon>0}C(\epsilon)$. Now clearly $\mathbb Q\subseteq C$. Is it possible to find an enumeration such that $\mathbb Q=C$?

Answer 1

Νο.

Note that $\bigcap_{\epsilon>0}^{\infty}C(\epsilon)=\bigcap_{m=1}^{\infty}C(\frac{1}{m})$

So you would have that $$\Bbb{Q}\subseteq\bigcap_{m=1}^{\infty}\bigcup_{n=1}^{\infty}(q_n-\frac{1}{2^nm},q_n+\frac{1}{2^nm}) \subseteq C=\Bbb{Q}$$

So $\Bbb{Q}$ is a $G_{\delta}$ set which contradicts Baire's Category Theorem.

Answer 2

Let $(q_n)$ be any enumeration of the rationals. Now define closed intervals $[a_n, b_n]$ with the following properties: $[a_n, b_n] \subseteq C(1/n)$, $q_n \notin [a_n, b_n]$, and $[a_{n+1}, b_{n+1}] \subseteq [a_n, b_n]$. Then $\bigcap_n [a_n, b_n]$ is nonempty, it is contained in $C$, and it is disjoint from $\mathbb{Q}$. So no such enumeration can exist.