Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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|cite|improve this question

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 $S$.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.