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I vaguely remember solving this problem before. I also vaguely remember the solution had something to do with prime numbers.It makes sense, since the primes are enumerable, and we can express any natural as a product of primes.

If we didn't have the nonintersecting condition, the solution would simply be that the sets are $$A_p = \{np: n\in\mathbb{N}\}$$ where $p$ is a fixed prime.

If I remember correctly, a slight modification on that solves the problem so I'm looking for a hint.

Or perhaps there's another solution not involving primes at all.

Hunter_Gatherer_n
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    You may want $n$ to not be anything already in an $A_q$ with $q <p$. There are other solutions too, for example looking at powers of $2$ but not other primes – Henry May 30 '22 at 16:19
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    Hint: Consider prime factorizations. – Karl May 30 '22 at 16:20
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    Another approach: divide $\Bbb N$ into just two countably infinite sets, and then recursively subdivide one of them in the same way. – Karl May 30 '22 at 16:22
  • I'll try both approaches. Soon I'll post an attempted solution. – Hunter_Gatherer_n May 30 '22 at 16:24
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    What if you look at the highest power of a fixed prime that divides each number? – JonathanZ May 30 '22 at 16:25
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    When you use the word "countable", do you mean "countably infinite"? (I ask, as many authors will use "countable" to mean "there exists an injection from the set to the natural numbers", hence a finite set is also considered countable). – Xander Henderson May 30 '22 at 16:47
  • I mean "countably infinite". The author from the book I'm currently reading distinguishes between finite and countable. – Hunter_Gatherer_n May 30 '22 at 16:51
  • another interesting answer: $0,2,4,6,..., 1,3,7,13,19,27,...,5,9,15,23,..., 11,17,25,...,\dots$ – C Squared May 30 '22 at 23:39

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$\textbf{Sketch of Solution:}$

Let $$A_1 = \{n\in\mathbb{N}: 2\mid n\},$$ $$A_2 = \{n\in\mathbb{N}: 3\mid n\text{ and } n\not\in A_1\}, $$ $$A_3 = \{n\in\mathbb{N}: 5\mid n\text{ and } n\not\in A_1,A_2\}.$$

In general, if $p_j$ is the $j$-th prime, then $$ A_j = \{n\in\mathbb{N}: p_j\mid n\text{ and } n\not\in A_1,A_1\cdots A_{j-1}\}. $$

There's an enumerable quantity of primes, so there's an enumerable quantity of such sets. Moreover, since $\mathbb{N}$ is enumerable, each $A_j$ is enumerable, for $p_j^n\in A_j$ for all $n\in\mathbb{N}$. By construction $A_i\cap A_j=\emptyset$ whenever $i\neq j$.

So $$\mathbb{N}=\bigcup_{j\in\mathbb{N}}A_j$$ is the desired representation of $\mathbb{N}.$

Hunter_Gatherer_n
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