We are asked if there exists an enumeration $\{r_n\}$ of the rationals such that the set $\displaystyle \bigcup_{n=1}^{\infty}(r_n-\frac{1}{n}, r_n+\frac{1}{n})$ has a non empty complement.
I believe the answer is yes. Consider an enumeration $\{r_1,r_2,\dots\}$ where if $i$ is not a power of $2$, then $r_i \in (0, 1)$. In that case, $\displaystyle \bigcup_{n\text{ is not power of 2}}(r_n-\frac{1}{n}, r_n+\frac{1}{n})\subseteq(-\frac{1}{3},\frac{4}{3})$ and so the measure of this union (which is measurable since it is a countable union of measurable sets) is at most $\frac{5}{3}$.
Now look at $\displaystyle \bigcup_{n = 1}^\infty(r_{2^n}-\frac{1}{2^n}, r_{2^n}+\frac{1}{2^n})$. Specifically, look at its measure. Again this is a measurable set, and its measure is less or equal to the sum of the measures of the subintervals in the union, meaning $\displaystyle m(\bigcup_{n = 1}^\infty(r_{2^n}-\frac{1}{2^n}, r_{2^n}+\frac{1}{2^n})) \leq \sum_{n=1}^{\infty}\frac{1}{2^{n-1}}= 2$
To sum up, in this enumeration of the rationals, the union we are originally asked about is a measurable set of measure at most $2+\frac{5}{3}$, hence its complement is also measurable and has infinite measure, so it's for sure not empty.
Is this solution correct? And furthermore, can we actually find such an enumeration of the rationals?
Best Answer
Yes.
There are $\aleph_0$
If you desire to be more specific: Let us assume you have an enumeration of $\Bbb Q=\{q_1,q_2,\ldots\}$. Then you can let $$a_n=n\text{th rational in }(0,1)= q_{\min\{\,k\in\Bbb N\colon\left|(0,1)\cap\{q_1,\ldots, q_k\}\right|=n\,\}}$$ and $$b_n=n\text{th rational not in }(0,1)= q_{\min\{\,k\in\Bbb N\colon\left|\{q_1,\ldots, q_k\}\setminus(0,1)\right|=n\,\}}$$ and notice that $\Bbb N\to \Bbb Q\cap (0,1)$, $n\mapsto a_n$ and $\Bbb N\to \Bbb Q\setminus (0,1)$, $n\mapsto b_n$ are bijections. Finally, let $$r_n=\begin{cases}a_k&\text{if }n=2^k\\ b_{n-\lfloor\log_2 n\rfloor}&\text{otherwise}\end{cases} $$