A union of balls centered at all the rationals on $[0,1]$ with decreasing radius $r_n \to 0$ relation with $[0,1]$ Given a strictly decreasing sequence $r_n$ such that $r_n\in [0,1]$, $r_1 < \frac{1}{2}$, and $\lim\limits_{n\to \infty} r_n=0$.

Question: Does there exist a bijective sequence $x:\mathbb N\to \mathbb Q\cap[0,1]$ such that $$\bigcup_{n\in \mathbb N}J(x_n,r_n)\ne [0,1]$$ where $J(a,b)=(a-b,a+b)\cap[0,1]$ and $x_n$ denotes $x(n)$?

 A: We fix an irrational number $x$ in $(.9,1)$ and we build an enumeration $(x_n)_{n\geqslant1}$ of $\mathbb Q\cap[0,1]$ such that, for every $n\geqslant1$, the interval $(x_n-r_n,x_n+r_n)$ avoids $x$. To ease the presentation, we start from an enumeration $(a_n)_{n\geqslant1}$ of $\mathbb Q\cap[.4,1]$ and an enumeration $(b_n)_{n\geqslant1}$ of $\mathbb Q\cap[0,.4)$. 
First, we choose the radius $r_{\phi(n)}$ associated to each $a_{n}$. Set $\phi(1)=\inf\{k\geqslant1\mid r_k\lt |x-a_1|\}$ and note that $\phi(1)$ is finite since $|x-a_1|\ne0$ and that $(a_1-r_{\phi(1)},a_1+r_{\phi(1)})$ does not contain $x$ since $r_{\phi(1)}\lt |x-a_1|$.
Once every $r_{\phi(k)}$ for $1\leqslant k\leqslant n$ is chosen, set $\phi(n+1)=\inf\{k\geqslant\phi(n)+2\mid r_k\lt |x-a_{n+1}|\}$. Note that $\phi(n+1)$ is finite since $|x-a_{n+1}|\ne0$ and that $(a_{n+1}-r_{\phi(n+1)},a_{n+1}+r_{\phi(n+1)})$ does not contain $x$ since $r_{\phi(n+1)}\lt |x-a_{n+1}|$. Note also that the index set $\mathbb N\setminus\phi(\mathbb N)$ is infinite since, for every $n\geqslant1$, $\phi(n+1)\geqslant\phi(n)+2$.
Once $\phi$ is determined, enumerate the infinite set $\mathbb N\setminus\phi(\mathbb N)$ by $\psi$ and decide that, for every $n\geqslant1$, $r_{\psi(n)}$ is the radius associated to $b_n$. Note that$(b_{n}-r_{\psi(n)},b_{n}+r_{\psi(n)})$ does not contain $x$ since $b_{n}+r_{\psi(n)}\lt.4+.5=.9$ and $x\gt.9$.
To conclude, for every $n\geqslant1$, if $n=\phi(k)$, define $x_n=a_k$ and if $n=\psi(k)$, define $x_n=b_k$. Then $(x_n)_{n\geqslant1}$ enumerates $\mathbb Q\cap[0,1]$ and no interval $(x_n-r_n,x_n+r_n)$ contains $x$ hence their union is not $[0,1]$.
Using the same idea, one can build an enumeration $(x_n)_{n\geqslant1}$ of $\mathbb Q\cap[0,1]$ such that the union of the intervals $(x_n-r_n,x_n+r_n)$ avoids any finite set of irrational numbers chosen in advance. Using a slight modification of this idea, for every positive $\varepsilon$ chosen in advance, one can build an enumeration $(x_n)_{n\geqslant1}$ of $\mathbb Q\cap[0,1]$ such that the measure of the union of the intervals $(x_n-r_n,x_n+r_n)$ is less than $\frac12r_1+\varepsilon$.
A: Let's give this a try.  I see that Did already gave an answer, but this should also work.
First, let's break the $(r_i)$ sequence into two infinite subsequences that partition the original sequence.  Informally, we're making a nice sequence and a rubbish sequence.  The nice sequence will supply radii that fail to cover all irrationals in $[\frac{1}{2}, 1]$, and the rubbish sequence will just be used to cover $[0, \frac{1}{2})$ excessively.
Our nice sequence: Let $s_i$ be the first element of $(r_i)$  less than or equal to $\frac{1}{2^{3+i}}$ which hasn't already been used in our sequence (i.e. $s_i \neq s_j$ for any $j<i$).  In particular, notice that $\sum s_i \le \frac{1}{8}$.
The rubbish sequence: $(t_i)$ is just the strictly decreasing sequence that results from the removal of the sequence $(s_i)$ from our original sequence.
Now, for each $t_i$, pick a rational $y_i$ in the interval $(\frac{1}{2}-t_{i-1}, \frac{1}{2}-t_i)$. (Treat $t_0$ as $\frac{1}{2}$.)
Notice that $y_i \in [0, \frac{1}{2})$, so $(\mathbb{Q}\cap[0,1])\setminus(y_i)_1^\infty$ is countably infinite (and there's a bijection $\phi: (s_i) \rightarrow (\mathbb{Q}\cap[0,1])\setminus(y_i)_1^\infty$).  Let $\bar{x}: (r_i)_1^\infty \rightarrow \mathbb{Q}\cap[0,1]$ be the bijection that sends $t_i$ to $y_i$ and $s_i$ to $\phi(s_i)$, and let $x_n = \bar{x}(r_n)$.  Then $(x_n) = \mathbb{Q}\cap [0,1]$, and, since $J(t_i, y_i) \subset [0, \frac{1}{2})$ for all $i$, it's enough to show $[\frac{1}{2}, 1] \not\subset \bigcup J(\phi(s_i), s_i)$.
Consider the Lebesgue measure $m$ of $\bigcup J(\phi(s_i), s_i)$:
\begin{align*}
m(\bigcup J(\phi(s_i), s_i)) &\le \sum_1^\infty m(J(\phi(s_i), s_i))\\
&\le 2\sum_1^\infty s_i \le \frac{1}{4}.
\end{align*}
So $[\frac{1}{2}, 1] \not\subset \bigcup J(\phi(s_i), s_i)$, and $\bigcup J(x_i, r_i)$ fails to cover $[0, 1]$.
