1
$\begingroup$

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

$\endgroup$
3
$\begingroup$

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 $1/b$.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.