SOLVED
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].
