Union of Balls around Rationals Let $(r_n)_{n \ge 1}$ be an enumeration of the rationals. Consider the union $A := \cup_n (r_n-\frac{1}{n^2},r_n+\frac{1}{n^2})$. It is unclear a-priori whether $A$ covers the real line, since although the rationals are dense in the reals, the $\frac{1}{n^2}$'s might shrink too fast. However, using measure theory, it is very easy to see $A$ does not cover much: indeed, $m(A) \le \sum_n m((r_n-\frac{1}{n^2},r_n+\frac{1}{n^2})) = \sum_n \frac{2}{n^2} = \frac{\pi^2}{3}$. 
Since this argument relies much on the convergence of $\sum_n \frac{1}{n^2}$, I am wondering whether $B := \cup_n (r_n-\frac{1}{n},r_n+\frac{1}{n})$ covers the whole real line, or what portion of it? Does the amount covered depend on the enumeration we choose?

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. 
