Let $\mathbb{N}$ be the set of positive integers. Given a set $A\subseteq \mathbb{N}$ we let the *(upper) density* of $A$ be defined by $$\mu^+(A) = \lim\sup\_{n\to\infty}\frac{|A\cap\{1,\ldots,n\}|}{n}.$$


If $\alpha\in\mathbb{R}$, we say $q\in\mathbb{N}$ is *good for approximating $\alpha$* if there is $p\in\mathbb{Z}$ such that $$|\alpha - \frac{p}{q}|< \frac{1}{q^2},$$


and denote the set of those positive integers by $G\_\alpha$. The [approximation theorem of Dirichlet](https://en.wikipedia.org/wiki/Dirichlet%27s_approximation_theorem) states that $G\_\alpha$ is infinite for any $\alpha\in\mathbb{R}$.


**Question.** Given $\delta\in[0,1]$, is there $\alpha\in\mathbb{R}$ such that $\mu^+(G\_\alpha) = \delta$?



Density of the set of numbers that are "good approximators" to a given real in the sense of Dirichlet's approximation theorem
If $\alpha$ is irrational, then the sequence $\alpha,2\alpha,\ldots$ is equidistributed modulo 1 (Weyl's theorem). Thus the inequality $|q\alpha-p|<1/q$ holds for $q$ of density $0$. If $\alpha=a/b$ ($a,b$ are coprime) is rational, then the density of your numbers equals $1/b$.


