Let $A=\{n_1<n_2<\cdots\}\subset \mathbb{N}$ be a lacunary sequence (so there exists some $\epsilon>0$ with $n_{k+1}\geq (1+\epsilon)n_k$ for all $k$). Must there exist an irrational $\theta$ such that $$\{ \|\theta n_k\| : k\geq 1\}$$ is not dense in $[0,1]$ (where $\| x\|$ is the distance to the nearest integer)?

Let $A=\{n_1<n_2<\cdots\}\subset \mathbb{N}$ be a lacunary sequence (so there exists some $\epsilon>0$ with $n_{k+1}\geq (1+\epsilon)n_k$ for all $k$). Must there exist an irrational $\theta$ such that $$\{ \|\theta n_k\| : k\geq 1\}$$ is not dense in $[0,1]$ (where $\| x\|$ is the distance to the nearest integer)?

Solved independently by de Mathan [dM80] and Pollington [Po79b], who showed that, given any such $A$, there exists such a $\theta$, with $$\inf_{k\geq 1}\| \theta n_k\| \gg \frac{\epsilon^4}{\log(1/\epsilon)}.$$ This bound was improved by Katznelson [Ka01], Akhunzhanov and Moshchevitin [AkMo04], and Dubickas [Du06], before Peres and Schlag [PeSc10] improved it to $$\inf_{k\geq 1}\| \theta n_k\| \gg \frac{\epsilon}{\log(1/\epsilon)},$$ and note that the best bound possible here would be $\gg \epsilon$.

This problem has consequences for 894.

The conclusion "$\{\|\theta n_k\|\}$ is not dense in $[0,1]$" is formalized as the sequence $(\theta n_k)$ not being dense modulo one; see the formalization notes above.