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

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  • $\begingroup$ Do you need inf? $\endgroup$ – მამუკა ჯიბლაძე Feb 9 '16 at 8:20
  • $\begingroup$ Just a comment: Furstenberg topology does not work, and finite non-empty subsets cannot be open. Where does this question come from? $\endgroup$ – Paolo Leonetti Feb 9 '16 at 14:30
  • $\begingroup$ @Paolo I was goofing around with dense sets (in the sense of numerically dense above) and wondered whether they could be dense with respect not only to this "lim (inf)" measure, but with respect to a topology. $\endgroup$ – Dominic van der Zypen Feb 9 '16 at 16:14
  • $\begingroup$ @PaoloLeonetti We can say more - nonempty set $A$ with asymptotic density zero can't be open, because then $\mathbb N\setminus A$ would be numerically and hence topologically dense, yet it wouldn't intersect $A$. $\endgroup$ – Wojowu Feb 10 '16 at 9:32
  • $\begingroup$ @Wojowu yes, we can also add that if $X$ has upper asymptotic density $x>0$ then its interior $\mathrm{Int}(X)$ has upper asymptotic density $x$ too (and $X/\mathrm{Int}(X)$ has density $0$); also, if $X,Y$ are open sets then $X \cap Y$ is empty or has positive upper asymptotic density.. $\endgroup$ – Paolo Leonetti Feb 10 '16 at 9:38
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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.

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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.

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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.

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  • $\begingroup$ @SalvoTringali Can you give an example of two members of $F$ such that their intersection is not in $F$? Also see math.stackexchange.com/questions/1644932/… $\endgroup$ – Dominic van der Zypen Mar 6 '16 at 15:07
  • $\begingroup$ @Dominic. I jumped to hasty conclusions. I take my comment back, and agree that $F$ is a filter. On the other hand, what's the meaning of the superscript '+' appended to a subfamily of $\mathcal P(\mathbf N)$? $\endgroup$ – Salvo Tringali Mar 6 '16 at 16:24
  • $\begingroup$ @Boaz: I follow your argument up until the last line. If $\mathcal N$ is defined as the family of sets containing open sets (but they need not be open), then I see no contradiction in having two elements of $\mathcal N$ with nonempty, zero-density intersection. $\endgroup$ – Will Brian Mar 8 '16 at 14:39
  • $\begingroup$ @WillBrian: Thanks. I've updated the "answer" accordingly. $\endgroup$ – Boaz Tsaban Mar 8 '16 at 19:58

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