# Finding a good ordering of $\mathbb{Q}$

Oftentimes in density arguments we let $\{x_n\}$ be a dense sequence and this is sufficient to imply the desired result.

From a research question I am working on I have simplified the example/counterexample to the following problem, which I believe is perhaps a nice exercise in choice (and yet, I cannot make a good one).

Precisely, can one choose a "good" ordering of a dense set $\{x_n\} \subset (-1,1)$ and a "good" sequence $r_n>0$ such that

$\sum_n r_n <\infty$

and

$|\{x \in (-1,1):x\in B(x_n,r_n) \text{ for infinitely many } n\}|>0$

where I have used $|\cdot|$ to denote the Lebesgue measure on $\mathbb{R}$.

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Isn't this just the Borel-Cantelli lemma? – Sean Eberhard Apr 17 '13 at 11:59
Thanks for the reference Sean. What I want to prove is false! Great to know now :) – Daniel Spector Apr 17 '13 at 12:10

The Lebesgue measure of the set of $x\in(-1,1)$ such that $x\in B(x_,;r_n)$ for at least one $n>N$ is at most $2\sum_{n>N} r_n$. Your set is the intersection of these sets over all $N\in\mathbb{N}$, so that it must have measure $0$ as soon as $\sum r_n<\infty$.