The natural numbers are a infinite union of disjoint countably infinite sets

Question

I know this question and similar questions are posted with many examples, but I can't find my solution anywhere. I thought my solution was pretty obvious and so I am thinking I have done something wrong. Can you help me find the error in my ways?

Let $A_i= \{2^{n_1}3^{n_2}5^{n_3}...p_i^{n_i} \ \mid n_1,n_2,n_3,...,n_{i-1}\in \mathbb{N}\cup\{0\} , n_i\in \mathbb{N}\}$. Obviously each $A_i$ is infinite. w.l.o.g suppose $a\in A_i\cap A_j$ where $i< j$. Since $a\in A_j$, it must contain a non zero power of $p_j$ in its prime factorization, but no element of $A_i$ has a non zero power of $p_j$ in its prime factorization. Every natural numbers shows up in some $A_i$ and we are done.

edit* as pointed out below, I will add $1$ to $A_1$

Answer 1

I think you're fine.

Simpler phrasing which defines essentially the same sets: let $A_i$ be the set of numbers which have exactly $i$ prime factors, and additionally put $1$ into $A_1$ (and, depending on your persuasion, put $0 \in A_1$ if $0$ is a natural number).

Another example: define the set $A_i$ to be the natural numbers with $i$ instances of $1$ in their binary expansion. (Here, $\mathbb{N}$ excludes $0$.)

Answer 2

As you were told by others, what you did is more or less fine.

A different solution would be: if $A_0$ is the set of all odd natural numbers and if, for $n\in\mathbb N$,$$A_n=\left\{k\in\mathbb{N}\,\middle|\,2^n\mid k\wedge2^{n+1}\nmid k\right\},$$then each $a_n$ is countable and their union is a disjoint union. Furthermore, the union of all $A_n$'s is $\mathbb N$.

Answer 3

Indirect proof: is know that there is a bijection $f:\Bbb N\times\Bbb N\longrightarrow\Bbb N$. Take $$A_n = \{f(m,n):m\in\Bbb N\}.$$