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 $S\in {\cal S}$?


2 Answers 2


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 $A_i=\cup_{k\in I_i} B_k$.

  • $\begingroup$ 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. $\endgroup$
    – Gro-Tsen
    Sep 6, 2017 at 9:40
  • $\begingroup$ Oh yes, just take $B_i=\{i!+1,\dots,(i+1)!\}$. $\endgroup$ Sep 6, 2017 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 $ud(S_n)=2^{-n}$.

  • 1
    $\begingroup$ 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. $\endgroup$
    – Dirk
    Sep 6, 2017 at 9:01

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