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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
  • $\text{sup}\{\text{ld}(A\cdot B): A,B\in {\cal P}_0(\mathbb{N})\}$?
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    $\begingroup$ 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. $\endgroup$
    – Anthony Quas
    Commented Jun 4, 2015 at 7:28
  • $\begingroup$ Many thanks, Anthony - do you want to post your construction as a (partial) answer? $\endgroup$
    – Dominic van der Zypen
    Commented Jun 4, 2015 at 7:48
  • $\begingroup$ Your first question is answered by some of the answers to this old question of yours, for instance, this answer of mine. $\endgroup$
    – bof
    Commented Dec 11, 2015 at 4:44

1 Answer 1

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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 $A \cdot B = \mathbb{N}$.

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