TITLE: Does there exist a real number a given distance from each rational number?
QUESTION [6 upvotes]: Let $r_n$ be an enumeration of the rational numbers and let $a_n$ be a sequence of positive real numbers that converges to zero. Does there exist $x\in \mathbb{R}$ such that $|x-r_n|>a_n$ for all $n$? 

This problem was inspired by an easier version of the problem where we assume the stronger condition that $\displaystyle\sum_{n=1}^\infty a_n$ converges. I have a simple solution in this particular case but I will spoiler it in case anyone would like to try this too. 

 Let $\Omega$ denote the set of $x\in \mathbb{R}$ that do not satisfy the   given property; we claim that $\Omega \neq \mathbb{R}$ and so an   $x\in \mathbb{R}$ with given property exists. Indeed, $\Omega=\{x\in\mathbb{R} \ | \ \exists \ n\in \mathbb{N} \ \mathrm{such \ that} \ |x-r_n| \leq a_n\}=\displaystyle\bigcup_{n=1}^\infty \ [r_n-a_n,r_n+a_n]$
 whose Lebesgue measure $\lambda(\Omega)\leq \displaystyle\sum_{n=1}^\infty \lambda([r_n-a_n,r_n+a_n])=\displaystyle\sum_{n=1}^\infty 2a_n < \infty$, so $\Omega \neq \mathbb{R}$.

Unfortunately it is very specific to this particular case so I doubt it helps with the general case, which I have no idea how to solve. I assume that, unless I'm missing something obvious, it uses some deeper theory (irrationality measure?) that I have not learned. Ideas for the general case or alternative (more elementary) solutions to the easier case would be appreciated.

REPLY [3 votes]: The answer is no. Consider the following open cover of $\Bbb R$:
$$\mathcal U=\{ (H_{2n} , H_{2n+4})  \ | n \in \Bbb N_{\ge 1}\} \cup \{ (-H_{2n+4} , -H_{2n})  \ | n \in \Bbb N_{\ge 1}\} \cup \{ (-3, 3) \}$$
where $$H_n = \sum_{j=1}^n \frac 1j$$
is the $n$-th harmonic number. Note that for all $\varepsilon >0$ only finitely many elements $U \in \mathcal U$ satisfy $\lambda (U) >  \varepsilon$.
Let $\{ U_{n} \}_{n \ge 1}$ be an enumeration of it indexed by positive natural numbers. This means that $a_n= \lambda (U_n)$ converges to $0$.
Let $\{ r_n \}_{n \ge 1}$ be an enumeration of rational numbers such that for all $n$
$$r_{n} \in U_n$$
holds. Recall that $a_n$ is rational, and it converges to $0$.
Now, for all $x \in \Bbb R$ there exists $n \ge 1$ such that $x \in U_n$: thus $|x-r_n| < a_n$.