Let $\mathbb{N}$ denote the set of positive integers, and let's say that $A\subseteq \mathbb{N}$ is *numerically dense* if $$\text{lim inf}\_{n\to\infty}\frac{|A\cap\{1,\ldots,n\}|}{n} = 1.$$
Is there a topology $\tau$ on $\mathbb{N}$ such that the collection of numerically dense subsets of $\mathbb{N}$ equals the collection of topologically dense subsets of $(\mathbb{N},\tau)$?



Numerical and topological density
Suppose we have such a topology on $\mathbb{N}$. Let $\{A\_\alpha : \alpha < \mathfrak{c}\}$ be an uncountable almost disjoint family of subsets of $\mathbb{N}$. For each $\alpha$, let
$$B\_\alpha = \bigcup\{[2^n,2^{n+1}) : n \in A\_\alpha\}.$$
Notice that $\{B\_\alpha : \alpha < \mathfrak{c}\}$ is still an almost disjoint family, but it has the additional property that each $B\_\alpha$ meets every dense set of our alleged topology. In other words, each $B\_\alpha$ has nonempty interior. These interiors must be disjoint, since if $\alpha \neq \beta$ then the complement of $B\_\alpha \cap B\_\beta$ is dense. Thus, taking interiors, we get an uncountable family of disjoint nonempty subsets of $\mathbb{N}$, a contradiction. 


