0
$\begingroup$

Suppose $\mathbb{N}=\bigsqcup_{i\in\mathbb{N}}E_i$ with $\#E_i=\infty$ for each $i$.

  1. Is it possible that $\limsup_{N\to\infty}\frac{1}{N}\#(E_i\cap\{1,\ldots,N\})=0$ for all $i$, which would mean $\lim_{N\to\infty}\frac{1}{N}\#(E_i\cap\{1,\ldots,N\})=0$ for all $i$?
  2. Is it possible that $\liminf_{N\to\infty}\frac{1}{N}\#(E_i\cap\{1,\ldots,N\})=0$ for all $i$?
Improve this question
$\endgroup$
1
  • $\begingroup$ Let $k \geq 1$, and let $p_1, \dots, p_k$ be a finite set of primes. Denote by $S_{p_1, \dots, p_k}$ the set of natural numbers whose prime divisors are exactly $p_1, \dots, p_k$. These infinite sets partition $\mathbb{N} \setminus \{1\}$ disjointly and have $0$ density. $\endgroup$
    – Pablo
    Commented Nov 25, 2014 at 12:01

1 Answer 1

Reset to default
1
$\begingroup$

Yes, it is possible. We can construct such sets for example as follows: $E_0$ is going to be the set of all numberss of the form $n^2$. $E_1$ is going to be the set of numbers of the form $n^2+1$, unless it already appeared in $E_0$. $E_2$ is set of numbers of the form $n^2+2$, unless it already appeared in any of previous sets, and so on, $E_k$ is set of numbers of the form $n^2+k$, excluding numbers appearing in earlier sets. Feel free to show by yourself that this is indeed a partition into infinite sets of zero density.

Improve this answer
$\endgroup$

Not the answer you're looking for? Browse other questions tagged .