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
* $\text{sup}\{\text{ld}(A\cdot B): A,B\in {\cal P}\_0(\mathbb{N})\}$?



Suprema of lower density of sums and products of sets with lower density 0
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 $A \cdot B = \mathbb{N}$.


