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 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 $\mu^+(G_\alpha) = \delta$?


1 Answer 1


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 $1/b$.


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