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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
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1 Answer 1

up vote 6 down vote accepted

The answer is no.

First, the ordering and density hypothesis are irrelevant (you do not use the ordering, and the density can be managed independently of the measure assumption we are trying to satisfy).

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 $\sum r_n<\infty$.

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