Let D be the set of all numbers $x \in [0,1]$ such that there exist infinitely many $p,q \in \mathbb{N} \backslash \{0\}$ such that $$|x-\frac{p}{q}| \leq \frac{2019}{q^3} $$ For each $q \in \mathbb{N} \backslash \{0\}$ set $$A_q=\bigcup_{p=1}^{q} \left[ \frac{p}{q}-\frac{2019}{q^3},\frac{p}{q}+\frac{2019}{q^3}\right]$$
- Explain why $\mathbb{Q} \cap (0,1] \subseteq D$
- Show that $\lambda(A_q \cap [0,1]) \leq \frac {4038}{q^2}$ for all $q \geq 1$.
My problem is with the second question. Can't we just simplify $A_q$ to $$A_q=\left[ \frac{1}{q}-\frac{2019}{q^3},1+\frac{2019}{q^3}\right] = \left[ \frac{q^2-2019}{q^3},1+\frac{2019}{q^3}\right]?$$ Because as we take the union over the p's, the left bound just increases as well as the right bound, so the left bound is smallest when $p=1$ and the right bound is largest when $p=q$. In this case, for say $q\geq100$, the left bound of the interval is positive so $A_q \cap [0,1]) = \left[ \frac{q^2-2019}{q^3},1 \right]$. Thus $$\lambda(A_q \cap [0,1]) = 1-\frac{q^2-2019}{q^3} \geq \frac {4038}{q^2}$$ which is a contradiction.