Is $\cup_{k=1}^\infty (r_k-\frac{1}{k}, r_k+\frac{1}{k}) = \mathbb{R}$? Let $r_k$ be the rational numbers in $\mathbb{R}$. 
(1).Is $\cup_{k=1}^\infty (r_k-\frac{1}{k^2}, r_k+\frac{1}{k^2}) = \mathbb{R}$?
(2).Is $\cup_{k=1}^\infty (r_k-\frac{1}{k}, r_k+\frac{1}{k}) = \mathbb{R}$?
(1).Because $m(\mathbb{R})=+\infty, \sum_{k=1}^\infty \frac{1}{k^2}<+\infty$, so $\mathbb{R} \setminus\cup_{k=1}^\infty (r_k-\frac{1}{k^2}, r_k+\frac{1}{k^2})\neq \Phi  $
(2) What about (2)?

For $k=2n+1(n \in N)$ we put $q_{2k}=\sum_{i=1}^{n-1}1/(2(2i+1))+1/(4(2n+1))$;
For $k=2n(n \in N)$ we put $q_{2k}=-\sum_{i=1}^{n-1}1/(4i)-1/(8n)$;
Let $f:N\setminus \{ 2k:k \in N\} \to Q \setminus \{q_{2k}:k\in N\}$ be a bijection. We put $q_n=f(n)$ for $n \in N\setminus \{ 2k:k \in N\}$. Then for the sequence $(q_n)_{n \in N}$ we have $R=\cup_{n \in N}(q_n-1/n,q_n+1/n)$. 
