Take the tour ×  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?   share|improve this question add comment
1 Answer  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 c