Convergent subsequences of $\{ n \alpha \bmod 1 \}$

Question

It's well known that if $\alpha \not\in \mathbb{Q}$, then the sequence $\{n \alpha \bmod 1\}_{n \geq 0}$ is dense in the torus $\mathbb{T} = \mathbb{R}/\mathbb{Z}$. Does every convergent subsequence necessarily have zero density? To be a bit more precise, if the sequence $\{n_j \alpha \bmod 1\}_{j \geq 1}$ converges, does the set $\{n_j : j \geq 1\}$ have density zero in $\mathbb{N}$?

Answer 1

Suppose $n_j\alpha\bmod1$ converges to $\beta$ and $\{n_j\}$ has positive lower density, i.e., there exists a $d>0$ such that $\{n_j\}\cap [1,N]$ has at least $dN$ elements if $N\gg0$. Fix $M\in \Bbb N$. Then there must exist infinitely many intervals $[k+1,k+M]$ such that $\{n_j\}\cap [k+1,k+M]$ has at least $dM$ elements. So if we pick our $M$ such that $dM>2$, there are infinitely many pairs of numbers $n,m\in\{n_j\}$ with $n<m<n+M$. By pigeon-hole, there exists a $1\le \Delta<M$ such that there are infinitely many $n$ with $n,n+\Delta\in\{n_j\}$. For large enough such $n$, the difference $(n+\Delta)\alpha\bmod 1-n\alpha\bmod 1=\Delta\alpha\bmod 1\pm1$ must become arbitrarily close to $0$. We conclude that $\Delta\alpha\equiv 0\pmod 1$, i.e., $\alpha\in\Bbb Q$.