We say a set $A\subseteq\mathbb{Z}$ is *arithmetical* if there are integers $a>0,b\geq 0$ such that $A=\{ax+b:x\in\mathbb{Z}\}$.


Is there $S\subseteq\mathbb{Z}$ such that $$S\cap A\neq\varnothing \neq (\mathbb{Z}\setminus S)\cap A$$
for every arithmetical set $A\subseteq\mathbb{Z}$?



Set "crossing" all arithmetical integer sets
Yes. Enumerate all arithmetic sets $A\_1, A\_2, \ldots$. Choose $n\_k\in A\_k$ with $|n\_k|>2^k$. The set $\{n\_1, n\_2, \ldots\}$ intersects any arithmetic set by construction, but it has zero density, thus does not contain a whole arithmetic set.


Actually you may construct such $S$ for each infinite sequence $A\_1, A\_2, \ldots$ of infinite sets: on $k$-th step choose $n\_k\in A\_k\cap S$ and $m\_k\in A\_k\setminus S$ so that $n\_1, m\_1, n\_2, m\_2, \ldots$ remain distinct.


