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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 $$\text{lim sup}_{n\to\infty}\frac{|A\cap\{1,\ldots,n\}|}{n} = 1?$$

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  • $\begingroup$ It might help to notice that $(A\cap B)\subset A$. $\endgroup$ – Olivier Oloa Feb 7 '16 at 19:49
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    $\begingroup$ @OlivierOloa How? $\endgroup$ – Did Feb 7 '16 at 20:06
  • $\begingroup$ This can be stated equivalently as: Set with density zero form an idal. Sets with asymptotic density equal to 1 form a filter. $\endgroup$ – Martin Sleziak Apr 14 '16 at 5:20
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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*}

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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.

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