Cover of the positive real numbers by intervals

Question

For which real numbers $x$ and $y$ does the following hold?: $$ \bigcup_{\frac{a}{b} \in \mathbb{Q}^+} \left[\frac{a}{b},\frac{a}{b}+\frac{1}{a^x b^y}\right] \ = \ \mathbb{R}^+ $$

Answer 1

Here is the solution for $x>0$ and arbitrary $y$.

If $x+y\geq 2$, then the covering won't work. Indeed, consider $\alpha=n+\sqrt{2}$ for large natural number $n$. An elementary argument shows that $|\alpha-a/b|>1/3b^2$ for all $a,b>0$. If $\alpha$ lies in the union of your intervals, then $1/3b^2<|\alpha-a/b|\leq 1/a^xb^y$. Since in particular $|\alpha-a/b|<1$, $a/b>n,a>nb$, thus giving $$\frac{1}{3b^2}<\frac{1}{a^xb^y}<\frac{1}{n^xb^{x+y}}\leq\frac{1}{n^xb^2}$$ which fails for large $n$.

Now consider $x+y<2$. Take any $\alpha\in\mathbb R^+$ which we may assume is irrational. A variation of Dirichlet's theorem on diophantine approximations shows that there are infinitely many fractions $a/b$ such that $a/b<\alpha<a/b+1/b^2$. We have $$\frac{1}{b^2}=\frac{1}{b^{2-x-y}b^xb^y}<\frac{1}{\alpha^xb^xb^y}=\frac{1}{(\alpha b)^xb^y}<\frac{1}{a^xb^y}$$ for large enough $b$, since $2-x-y>0$ and $\alpha b>a$. Thus $\alpha\in[a/b,a/b+1/a^xb^y]$.