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It is well known that $\mathbb{Q}$ is not a $G_\delta$. In fact no countable dense subset of $\mathbb{R}$ is a $G_\delta$. We order the rationals in a sequence: $\mathbb{Q}=\lbrace r_k\rbrace$, and for each integer $n\ge1$ define $$O_n=\bigcup_{k=1}^\infty(r_k-2^{-(k+n)},r_k+2^{-(k+n)}).$$ $O_n$ is open with measure less than $2^{-n+1}$. Then $$O=\bigcap_{n=1}^\infty O_n$$ is a $G_\delta$ of measure $0$ containig the rationals. We know that $O$ must be uncountable, so that there are uncountably many irrationals in $O\setminus\mathbb{Q}$.

Now the questions. What can we say about any $x\in O\setminus\mathbb{Q}$, apart from the fact that $x$ is irrational? Is it possible to say something about how well it is approximated by rationals? What is the dependence of $O$ on the chosen ordering of $\mathbb{Q}$?

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    $\begingroup$ For a given irrational $a\in O$, isn't it clear that you can change your sequence $r_k$ so that $a\not\in O$? Since all you have to do is remove $a$ from one of the sets $O_n$... $\endgroup$ – Thierry Zell Feb 6 '11 at 2:37
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    $\begingroup$ They will be Liouville ( en.wikipedia.org/wiki/Liouville_number ) $\endgroup$ – Not Mike Feb 6 '11 at 2:56
  • $\begingroup$ @Michael: and in particular transcendental. $\endgroup$ – Anton Petrunin Feb 6 '11 at 4:49
  • $\begingroup$ Though Thierry has obviously misread the question, his observation holds with "$a\in O$" and "include $a$ into every" as well, so if nothing is known about the enumeration, nothing can be said about any individual irrational in $O$. Of course, if the denominator of $r_k$ is bounded by a known function of $k$, we have the approximation speed restrictions but this is sort of tautological. $\endgroup$ – fedja Feb 6 '11 at 14:37

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