A note on $\mathscr{B}$-free sets and the existence of natural density (2506.10218v1)

Abstract: Given $\mathscr{B}\subseteq \mathbb{N}$, let $\mathcal{M}\mathscr{B}=\bigcup{b\in\mathscr{B}}b\mathbb{Z}$ be the correspoding set of multiples. We say that $\mathscr{B}$ is taut if the logarithmic density of $\mathcal{M}\mathscr{B}$ decreases after removing any element from $\mathscr{B}$. We say that $\mathscr{B}$ is minimal if it is primitive (i.e.\ $b| b'$ for $b,b'\in\mathscr{B}$ implies $b=b'$) and the characteristic function $\eta$ of $\mathcal{M}\mathscr{B}$ is a Toeplitz sequence (i.e.\ for every $n\in \mathbb{N}$ there exists $s_n$ such that $\eta$ is constant along $n+s_n\mathbb{Z}$). With every $\mathscr{B}$ one associates the corresponding taut set $\mathscr{B}'$ (determined uniquely among all taut sets by the condition that the associated Mirsky measures agree) and the minimal set $\mathscr{B}^*$ (determined uniquely among all minimal sets by the condition that every configuration appearing on $\mathcal{M}{\mathscr{B}^*}$ appears on $\mathcal{M}\mathscr{B}$: for every $n\in \mathbb{N}$, there exists $k\in \mathbb{Z}$ such that $\mathcal{M}{\mathscr{B}^*}\cap [0,n]=\mathcal{M}\mathscr{B} \cap[k,k+n]-k$). Besicovitch [2] gave an example of $\mathscr{B}$ whose set of multiples does not have the natural density. It was proved in [7, Lemma 4.18] that if $\mathcal{M}{\mathscr{B}'}$ posses the natural density then so does $\mathcal{M}\mathscr{B}$. In this paper we show that this is the only obstruction: every configuration $ijk\in {0,1}^3$ (with $ij\neq 01$), encoding the information on the existence of the natural density for the triple $\mathcal{M}\mathscr{B},\mathcal{M}{\mathscr{B'}},\mathcal{M}{\mathscr{B}^*}$, can occur. Furthermore, we show that $\mathcal{M}\mathscr{B}$ and $\mathcal{M}_{\mathscr{B}'}$ can differ along a set of positive upper density.