Intersection of dense sets in $\mathbb{N}$ Let's call $A\subseteq\mathbb{N}$ dense if $$\text{lim inf}_{n\to\infty}\frac{|A\cap\{1,\ldots,n\}|}{n} = 1.$$
Is the intersection of two dense sets dense again? Or does the intersection of two dense sets at least satisfy the weaker statement $$\text{lim sup}_{n\to\infty}\frac{|A\cap\{1,\ldots,n\}|}{n} = 1?$$
 A: Yes. Equivalently, $A$ is dense if its complement $A^c$ satisfies
$$
\limsup_{n\to\infty} \frac{\#(A^c \cap \{1,\dots,n\})}n = 0.
$$
And this property certainly respects finite addition, since
\begin{align*}
\#\big((A\cap B)^c \cap \{1,\dots,n\}\big) &= \#\big( (A^c \cap \{1,\dots,n\}) \cup (B^c \cap \{1,\dots,n\}) \big) \\
&\le \#(A^c \cap \{1,\dots,n\}) + \#(B^c \cap \{1,\dots,n\}).
\end{align*}
A: This answer is about the lim sup version of the definition. It is not the same for a set in this sense to be "dense" (maybe call it "lim sup dense") as it is for a set to be "lim inf dense" (using the lim inf version).
We can make $A,B$ disjoint and have each of them "lim sup dense". For each $n=1,2,...$: On step $2n-1$ put enough consecutive naturals in $A$ to bring its density so far up to $1-1/(n+1).$ Then on step $2n$ put enough of the so far unused consecutive naturals in $B$ to bring its density so far up to again $1-1/(n+1).$ Alternating in this way, there are always infinitely many left so one can bring the $A$ or $B$ density up to where one wants.
