Numerical and topological density

Question

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 $(\mathbb{N},\tau)$?

Answer 1

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.

Answer 2

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.

Answer 3

I initially thought that the following provides a negative answer, and posted this for comments. Will Brian points out below that the last claim is wrong. It may still be useful as a starting point, so I do not delete this answer. In the notation below, one seems to need a family $\mathcal{A}$ that is closed under finite intersections and such that $\mathcal{A}\setminus\{\emptyset\}\subseteq\mathcal{N}$ and every element of $\mathcal{N}$ contains an element of $\mathcal{A}$.

A semifilter is a family $\mathcal{F}$ of subsets of $\mathbb{N}$ such that for each set $A\in\mathcal{F}$, all supersets of $A$ are in $\mathcal{F}$. For a semifilter $\mathcal{F}$, define $$ \mathcal{F}^+:=\{A\subseteq\mathbb{N} : A^c\notin \mathcal{F}\}. $$ Then $\mathcal{F}^+$ is also the family of all sets that intersect all members of the semifilter $\mathcal{F}$. We also have that $\mathcal{F}^{++}=\mathcal{F}$.

Let $\mathcal{D}$ be the (semi)filter of all density 1 sets, and $\mathcal{N}$ be the family of neighborhoods, i.e., sets containing open sets, in the expected topology. You request that $\mathcal{N}^+=\mathcal{D}$ or, equivalently, that $$ \mathcal{N}=\mathcal{D}^+=\{A\subseteq\mathbb{N} : A^c\notin \mathcal{D}\}, $$ the family of all sets of positive upper density. (And here comes the gap...) But then there are two elements of $\mathcal{N}$ whose intersection is nonempty but has zero density; a contradiction.