Product of two sets with density zero has density zero? Let $A$ and $B$ be two subsets of $\mathbb N$ which have asymptotic density zero. Define $A\times B$ as the set of integers of the form $ab$ with $a\in A$ and $b\in B$. Must $A \times B$ also have asymptotic density zero?
I conjecture that it must , but I am not sure.
 A: Separate the set of primes $P$ into two sets $A_p,B_p$ by considering the partial products of $(1-1/2)(1-1/3)(1-1/5)\cdots$ with each factor of the form $1-1/p$ where the primes $p$ run through the sequence $2,3,5,\cdots$ of primes. We alternately go for partial products each at most $1/2,$ so that $A_p$ begins with $2$ [since $1-1/2\le 1/2$] and then $B_p$ continues with $3,5,7$ since $(1-1/3)(1-1/5)(1-1/7)=16/35=0.45...$ but using only $3,5$ is not enough for the first $B_p$ group [$(1-1/3)(1-1/5)=8/15=0.53...$ is not at most $1/2$] 
We are assured that we never reach a point where we cannot make the next $A_p$ or $B_p$ group of primes have their partial product of $1-1/p$ terms at most $1/2,$ since the product of this expression over all primes $p$ diverges to $0$. If my calculation is right, then the next group of primes in list $A_p$ begins with $11$ and ends with $83$
Now we define the subsets $A$ and $B$ of $\mathbb{N}$ so that $A$ consists of all positive integers which may be formed using only the prime factors (to whatever powers) from $A_p$, and similarly $B$ is the integers formable using only powers of primes from $B_p$
It then is clear at least that $AB$ is all of $\mathbb{N},$ while arguments similar to that showing the natural density of primes is zero using the full product of the factors $1-1/p$ seem to me to show that each of the above $A,B$ have density zero. (I intend to return to this last issue with more details, but thought I'd put this admittedly partial answer up for now, maybe someone could help finish it, or show the $A,B$ had positive densities after all, then I'd delete.)
Here at any rate is a heuristic attempt to show e.g. that the natural density of $A$ is zero. A positive integer is in $A$ iff none of its prime divisors lie in the set $B_p$ of primes. The latter were chosen in such a way that the product of terms $(1-1/p)$ for $p \in B_p$ is zero, since we have partial products bounded by $1/2,$ then further by $(1/2)^2,$ etc. And in a probability sense, $(1-1/p)$ is the likelihood that a "random" integer is not divisible by $p.$ So if we believe such probability arguments, we can say the probability of a positive integer not being divisible by a prime in the set $B_p$ is zero, which is to say the probability of a positive integer to lie in set $A$ is also zero.
