Union of Balls around Rationals

Let $q_3\in \Bbb Q \cap (5/6,1).$

Recursively, for $j\in \Bbb Z^+$ let $q_{3(j+1)} \in \Bbb Q \cap (q_{3j}+\frac {1}{3j}-\frac {1}{6(j+1)}, q_{3j}+\frac {1}{3j}).$

Then $0<q_{3i}< q_{3j}$ when $i<j,$ and we have $\cup_{j=1}^n(-\frac {1}{3j}+q_{3j},\frac {1}{3j}+q_{3j}\supset [1,1+ \sum_{j=1}^n\frac {1}{6j}).$

So $\cup_{j\in \Bbb Z^+}(-\frac {1}{3j}+q_{3j},\frac {1}{3j}+q_{3j})\supset [1,\infty).$

Similarly we can find a discrete $\{q_{3j-1}:j\in \Bbb Z^+\}\subset \Bbb Q$ such that $q_{3j+2}<q_{3j-1}<0$ and $\cup_{j\in \Bbb Z^+}(-\frac {1}{3j-1}+q_{3j-1},\frac {1}{3j-1}+q_{3j-1})\supset (-\infty,-1].$

Let $q_1=0.$

Since the set $S=\{q_1\}\cup \{q_{3j}:j\in \Bbb Z^+\}\cup \{q_{3j-1}:j\in \Bbb Z^+\}$ is discrete, the set $\Bbb Q$ \ $S$ is infinite so we can enumerate $\Bbb Q$ \ $S=\{q_{3j+1}:j\in \Bbb Z^+\}.$

And we have $\cup_{j\in \Bbb Z^+}(-1/j+q_j,1/j+q_j)=\Bbb R.$

We can also enumerate $\Bbb Q$ in a different way, to make $\cup_{n\in \Bbb N}(-1/n+q_n,1/n+q_n)$ a set of finite measure.