# Suprema of lower density of sums and products of sets with lower density 0

We define the lower density of a set $A\subseteq \mathbb{N}$ by $$\operatorname{ld}(A) \ := \ \liminf_{n\to\infty}\frac{|A\cap\{1,\ldots,n\}|}{n}.$$ For $A,B\subseteq \mathbb{N}$, we set $$A + B \ := \ \{a+b: a\in A, b\in B\},$$ and similarly $$A \cdot B \ := \ \{a\cdot b: a\in A, b\in B\}.$$ Further let ${\cal P}_0(\mathbb{N}) := \{A\in{\cal P}(\mathbb{N}): \operatorname{ld}(A) = 0\}$ be the set of all sets of positive integers whose lower density is $0$.

Question: What are

• $\text{sup}\{\text{ld}(A+B): A,B\in {\cal P}_0(\mathbb{N})\}$, and
• $\text{sup}\{\text{ld}(A\cdot B): A,B\in {\cal P}_0(\mathbb{N})\}$?
• The first one is certainly 1. For a trivial construction, let $n_1\ll n_2\ll n_3\ll\ldots$. Then if $A=[n_1,n_2]\cup [n_3,n_4]\cup [n_5,n_6]$ while $B=\{0\}\cup [n_2,n_3]\cup [n_4,n_5]\cup\ldots$, their union is everything, but both have lower density 0. – Anthony Quas Jun 4 '15 at 7:28
• Many thanks, Anthony - do you want to post your construction as a (partial) answer? – Dominic van der Zypen Jun 4 '15 at 7:48
• Your first question is answered by some of the answers to this old question of yours, for instance, this answer of mine. – bof Dec 11 '15 at 4:44

Anthony Quas has already answered the first part. The second supremum is $1$ as well -- just let $A$ be the set of positive integers all of whose prime divisors are $2$ or $\equiv 1 (\!\!\mod 4)$, and let $B$ be the set of positive integers all of whose prime divisors are $\equiv 3 (\!\!\mod 4)$. Then $A \cdot B = \mathbb{N}$.