Understanding Analysis Exercise 1.2.4.
Produce an infinite collection of sets $A_1, A_2, A_3,\ldots$ with the property that every $A_i$ has an infinite number of elements, $A_i \cap A_j = \varnothing$ for all $i \not= j$, and $\bigcup_{i=1}^\infty A_i = \mathbb{N}$.
I understand that there is a very straightforward solution, i.e., arranging $\mathbb{N}$ into a square grid (which, in hindsight, looks pretty obvious), but I wanted to verify if what I came up with works.
Let $P=\{2,3,5,7,11,\ldots\}$ such that $P_n$ is the $n$th prime.
Let $A_1$ equal the set of all even numbers plus 1. Let $A_2$ equal the set of all multiples of $P_2$ excluding the set of even numbers. Let $A_3$ equal the set of all multiples of $P_3$ excluding the multiples of $P_1,P_2$, and so on. In other words, include only numbers whose smallest prime factor is $P_i$.
I can't see anything wrong with this logic, but I just started real analysis and did not see anything remotely similar to it in the various answers I found online. Also, regardless of its correctness, is there any way to concisely express the logic? My best try is below.
Let $A_1=\{1\} \cup \{2k : k \in \mathbb{N}\}$
For $i \geq 2$, $A_i=\{P_i \cdot k : k \in \mathbb{N}\} \backslash \bigcup_{j=1}^{i-1}\{P_j \cdot k : k \in \mathbb{N}\}$
Thanks in advance!