badly approximated numbers on the real line form a meagre set

Question

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.

Answer 1

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 $S$.

Answer 2

Here is a more detailed proof. (BTW, 8 is completely arbitrary. It could be any number greater than or equal to 2. For generality, I will replace 8 by $n \ge 2$ from now on.)

Let $$T_q = \mathbb{R} - \bigcup_{p \text{ is relatively prime to }q} \left(\frac{p}{q}-\frac{1}{q^n}, \frac{p}{q}+\frac{1}{q^n}\right).$$

Clearly, $T_q$ is closed for any $q$.

Now for any natural number $Q$, define $$R_Q = \bigcap_{q\ge Q} T_q.$$

$R_Q$ is clearly closed. It is further nowhere dense, as shown here.

Now observe that $$\mathbb{R} - S = \bigcup_{Q=1}^\infty R_Q$$ and we get the desired result.