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.

Here is an example of a topology $\tau$ on $\mathbb{N}$, such that

$$\{S \in 2^{\mathbb{N}} \colon S \,\, \mbox{is numerically dense}\} \subsetneq \{D \in 2^{\mathbb{N}} \colon D \,\, \mbox{is a dense subset of} \,\,(\mathbb{N},\tau)\}.$$

We say that a subset $c$ of $\mathbb{N}$ is closed if either the sum of the reciprocals of the elements of $C$ converges or $C = \mathbb{N}$. It can be shown that the family $\{\mathbb{N} \setminus C \colon C \,\, \mbox{is closed}\}$ determines a topology $\tau$ on $\mathbb{\mathbb{N}}$. It can also be proven that $D \in 2^{\mathbb{N}}$ is a dense subset of $(\mathbb{N},\tau)$ if and only if the series of the reciprocals of the elements of $D$ diverges: since the series of the reciprocals of any numerically dense subset of $\mathbb{N}$ diverges, it follows that any such set belongs to $$\{D \in 2^{\mathbb{N}} \colon D \,\, \mbox{is a dense subset of} \,\,(\mathbb{N},\tau)\}$$ and the purported inclusion follows.