TITLE: A union of balls centered at all the rationals on $[0,1]$ with decreasing radius $r_n \to 0$ relation with $[0,1]$
QUESTION [8 upvotes]: 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 $x(n)$?

REPLY [3 votes]: 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 $[0, 1]$.