If set $A$ has asymptotic density $d_A$ and set $B$ has asymptotic density $d_B$, then does it follow that $d_{A \cap B} \leq {d_A}\cdot{d_B}$?

Question

Let $A, B \subset \mathbb{N}$.

If set $A$ has asymptotic density $d_A$ and set $B$ has asymptotic density $d_B$, then does it follow that $d_{A \cap B} \leq {d_A}\cdot{d_B}$?

I am currently unable to come up with a specific counterexample. Intuitively, I do know that if $d_A = d_B = 0$, then $d_{A \cap B} = 0$, but I have no proof.

Edit January 15 2017

In view of Rosie's counterexample, one could perhaps ask whether the reverse inequality $${d_A}\cdot{d_B} \leq d_{A \cap B}$$ always holds?

Answer 1

No. If $0<\delta<1$ and $A=B$ and $d_A=d_B=\delta$ then $0<\delta^2=d_A d_B

The reverse doesn't hold either. If $0<\delta<1$ and $B$ is the complement of $A$, and $d_A=\delta$, then $d_B=1−\delta$, so $0=d_{A\cap B}

I don't know what set your $A, B$ are subsets of. Let $S$ be some function so that $S(x)$ is some set where "the asymptotic density of $A$" means $\lim_{x\to\infty} \;d(A\cap S(x))$. So if you were thinking of subsets of $\mathbb{N}$, then $\{1,\dots,x\}$ would do, or if it were $\mathbb{R}$, then the interval $[-x, x]$. So then, the point of $d_{A\cap B}\;$ as I see it is $\lim_{x\to\infty}\; d_{C(x)}$ where $C(x)=\{(a,b)\mid a\in A\cap S(x), b\in B\cap S(x)\}$. The density among $C(x)$ of the pairs $\{(a,a)\mid a\in A\cap B\cap S(x)\}$ is something else entirely.