Let $d$ denote the lower density on $\mathbb{N}$, $a>0, $ $\mathbb{N}_{a}:=\left\{ B\subset\mathbb{N}:{d}(B)\geq1-a\right\} $ and $A\subset\mathbb{N}$.
If $A\cap B\neq\emptyset$ for every $B\in\mathbb{N}_{a}$ then clearly $d$$(A)\geq a$.
Does there exist a countable family $\mathcal{F\subset}\mathbb{N}_{a}$ such that if $A\cap B\neq\emptyset$ for every $B\in\mathcal{F}$ then $d(A)\geq a?$
This is not true. Actually for any countable family of infinite subsets you can construct a subset of the naturals that intersects every member and that has lower density as low as you want.