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 $i$?



Are the natural numbers a disjoint union of infinite sets of zero asymptotic density?
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.


