Basic question about natural density

Question

Suppose that we have a sequence of finite sets $A_1, A_2, \ldots$, which partition $\mathbb{N}$. I am making no other assumptions on the $A_n$ - i.e. there could be any amount of interleaving between them. Now suppose we have $S\subset\mathbb{N}$. If $\lim_{n\rightarrow\infty} \frac{|S\cap A_n|}{|A_n|}=0$, does it follow that $S$ has a natural density of 0?

(And if so, while I'm at it, can 0 be replaced by other numbers? Natural density be replaced by upper, lower density? I mostly just care about the density 0 (equivalently density 1) case, though.)

EDIT: And if the statement is false, is there some sufficiently-little-interleaving condition on the $A_n$ I could assume that would make it true?

Answer 1

Fix an increasing sequence of positive integers $(k(n))_n$ and let $A_n=[k(n),k(n+1))$.

Fix a sequence of positive integers $(\ell(n))_n$ such that $\ell(n)\le k(n+1)-k(n)$ for every $n$ and define $S$ as the union of the intervals $[k(n),k(n)+\ell(n))$.

Assume that $\ell(n)\ll k(n+1)-k(n)$. Then $|S\cap A_n|\ll|A_n|$ hence your hypothesis is met.

Assume furthermore that $\ell(n)\ge uk(n)$ for a given positive $u$, for every $n$. Then, for every $N=k(n)+\ell(n)$, $$ \frac1N|S\cap\{1,2,\ldots,N\}|\ge\frac{\ell(n)}{k(n)+\ell(n)}\ge\frac{u}{1+u} $$ hence the natural density of $S$ cannot be $0$ and your conclusion does not hold.

Every condition above on the sequences $(k(n))$ and $(\ell(n))$ is met if, for example, $k(n)=2^{n^2}$ and $\ell(n)=2^{n^2+n}$.

In this specific example, the sequence of densities of general term $N^{-1}|S\cap\{1,2,\ldots,N\}|$ converges to $0$ when restricted to the integers $N=k(n)$ and converges to $1$ when restricted to the integers $N=k(n)+\ell(n)$, hence the set of limit points of the whole sequence of densities is the interval $[0,1]$.

Answer 2

This is false. Let $S$ be any infinite coinfinite subset of $\mathbb{N}$, and let $A_n$ contain $1$ element of $S$ and $n$ elements of $\mathbb{N} \setminus S$ (it's clear we can always find such a partition, for example by recursively letting $A_n$ contain the smallest suitable numbers not already in $A_i$ for some $i < n$). Then $\lim_{n \to \infty} \frac{|S \cap A_n|}{|A_n|}=\lim_{n \to \infty} \frac{1}{n+1}=0$, but the density of $S$ can be anything in $[0,1]$, or not exist at all.

In other words: your condition on $S$ and the $A_n$s makes no use of the order structure on $\mathbb{N}$, and just treats $\mathbb{N}$ as an unstructured set. It thus can't imply anything about $S$ that isn't already implied by its cardinality and cocardinality.