Covering $\mathbb{R}$ by intervals centered at rationals I am trying to solve the problem that asks the following.
For a given sequence $(a_j)$ of positive real numbers, show that there exists an enumeration of rationals $\{r_1,r_2,\ldots\}$ such that $\bigcup_{j=1}^\infty (r_j-a_j,r_j+a_j)=\mathbb{R}$ if and only if $\sum_{j=1}^{\infty}a_j=\infty$.
I showed with ease that if $\sum_{j=1}^{\infty}a_j<\infty$, it cannot cover $\mathbb{R}$. But I am stuck with the converse: I know that there are enumerations such that even when $\sum_{j}a_j=\infty$, the union does not cover $\mathbb{R}$. But how do I show that there always exists one?
 A: Let $0<\epsilon_j<\min\{2^{-j},a_j/2\}$ be such that $0< b_j=a_j-\epsilon_j\in\mathbb{Q}$.
We have that $\sum b_j=\infty$ iff $\sum a_j=\infty$ because $\sum\varepsilon_j\le 1$
You can cover $\mathbb{R}$ with closed intervals $I_j$ of lenght $2b_j$ such that all their endpoints are at rational numbers: $I_1=(0,2b_1)$, $I_2=(-2b_2,0)$, $I_3=(2b_1,2b_1+2b_3)$ etcetera (for some sequences $b_j$ this does not work but you can always find a way)
Choose your $r_j$ to be the middle point of the interval $I_j$ (it is a ratonal number) 
Then $I_j=[r_j-b_j,r_j+b_j]\subseteq(r_j-a_j,r_j+a_j)$ so that
$$\mathbb{R}=\bigcup\,[r_j-b_j,r_j+b_j]\subseteq\bigcup\,(r_j-a_j,r_j+a_j)$$
$\sum a_j=\infty\implies\exists\{r_1,r_2,\dots\}=\mathbb{Q}$ such that $\mathbb{R}=\bigcup\,(r_j-a_j,r_j+a_j)$
Edit for a shorter proof: Cover the real line with closed intervals $I_j$ of lenght $a_j$ (this is possible because $\sum a_j=\infty$) then choose $r_j$ to be any rational number in the interior of $I_j$, then $I_j\subseteq(r_j-a_j,r_j+a_j)$ and the result follows
