TITLE: Numerical and topological density
QUESTION [7 upvotes]: 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 $(\mathbb{N},\tau)$?

REPLY [4 votes]: 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.