# badly approximated numbers on the real line form a meagre set

Let $S$ be the set of real numbers $x$ such that there exist infinitely many (reduced) rational numbers $p/q$ such that $$\left\vert x-\frac{p}{q}\right\vert <\frac{1}{q^8}.$$ I would like to prove that $\mathbb{R}\setminus S$ is a meagre set (i.e. union of countably many nowhere dense sets). I have no idea about how to prove this, as I barely visualise the problem in my mind. I guess that the exponent $8$ is not just a random number, as it seems to me that with lower exponents (perhaps $2$?) the inequality holds for infinitely many rationals for every $x\in\mathbb{R}$.

Could you help me with that?

Thanks.

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The idea is to transform the quantifiers into unions/intersections. For example, let $T$ be the same as $S$ but dropping the infinitely many assumption.
Consider $A_{\frac p q}=(-\frac 1 {q^8}+\frac p q, \frac p q +\frac 1{q^8})$ then $T=\bigcup_{\frac p q\in\mathbb Q}A_{\frac p q}$. Thus, $T$ is a countable union of open sets ($\mathbb Q$ is countable). The same idea applies to $S$.
For $S$ you take the intersection of sets $T_n$ defined similarly to $T$ but with the restriction $q > n$, so $S$ is a $G_\delta$. – Robert Israel Mar 1 '12 at 18:17