# Pairwise disjoint subsets of $\mathbb{N}$ with positive upper density

For $A\subseteq\mathbb{N}$ we define the upper density to be $$\text{ud}(A) = \lim\sup_{n\to\infty}\frac{|A\cap\{1,\ldots,n\}|}{n}$$

Is there an infinite set ${\cal S}$ of pairwise disjoint subsets of $\mathbb{N}$ such that $\text{ud}(S) > 0$ for all $S\in {\cal S}$?

Yes. Take the segments of integers $B_i=\{2^i,\dots,2^{i+1}-1\}$. Partition naturals onto disjoint infinite sets $I_1,\dots$. Take the sets $A_i=\cup_{k\in I_i} B_k$.
• You should emphasize that this realizes not only a positive upper density, but upper density $1$ for each of the $A_i$, thus answering a stronger and more interesting question than OP asked. – Gro-Tsen Sep 6 '17 at 9:40
• Oh yes, just take $B_i=\{i!+1,\dots,(i+1)!\}$. – Fedor Petrov Sep 6 '17 at 10:07
Yes there is. For $n\in\mathbb N$ take $S_n=\{k\in\mathbb N|k\equiv 2^{n-1}\pmod {2^n}\}$.
That is $S_1$ are the odd numbers, $S_2$ their doubles etc. Obviously $\bigcup S_n=\mathbb N$ and $ud(S_n)=2^{-n}$.
• You can extend this idea by taking any discrete valuation $\nu_p$ on the integers for a prime $p$ and construct the sets $S_n = \{k \in \mathbb{N} \mid \nu_p(k) = n\}.$ This should also generalize to many other rings. – Dirk Sep 6 '17 at 9:01