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.

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.
