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Partition of N into infinite number of infinite disjoint sets?

One answer for the above question about "Partition of N into infinite number of infinite disjoint sets" is the following (from Shai Covo):

Let $$ A_0 = \lbrace 1,3,5,7,9,\ldots \rbrace $$ and $$ A_1 = \lbrace 2^n 1 : n \in \mathbb{N} \rbrace, $$ $$ A_2 = \lbrace 2^n 3 : n \in \mathbb{N} \rbrace, $$ $$ A_3 = \lbrace 2^n 5 : n \in \mathbb{N} \rbrace, $$ $$ A_4 = \lbrace 2^n 7 : n \in \mathbb{N} \rbrace, $$ $$ A_5 = \lbrace 2^n 9 : n \in \mathbb{N} \rbrace, $$ $$ \cdots. $$ Noting that for any two distinct elements $r_1$ and $r_2$ of $A_0$ it holds $2^{n_1}r_1 \neq 2^{n_2}r_2$ $\forall n_1,n_2 \in \mathbb{N}$, we have that the $A_i$ are disjoint. On the other hand, let $2k$, with $k \in \mathbb{N}$, be an arbitrary even natural number. Considering its prime factorization, it is necessarily of the form $2k = 2^n r$, where $n \in \mathbb{N}$ and $r \in A_0$. Hence $2k \in \cup _{i = 1}^\infty A_i$, from which it follows that $\cup _{i = 1}^\infty A_i = \lbrace 2,4,6,8,10,\ldots \rbrace$, and so $\mathbb{N} = \cup _{i = 0}^\infty A_i$, with all the $A_i$ disjoint and countably infinite.

I am trying to figure out firstly in the above answer, how to formally prove that for any two distinct elements $r_1$ and $r_2$ of $A_0$ it holds $2^{n_1}r_1 \neq 2^{n_2}r_2$ $\forall n_1,n_2 \in \mathbb{N}$? And secondly, the above answer only proves that the set of even natural numbers is a subset of $\cup _{i = 1}^\infty A_{i}$ but it did not prove the other direction, may I ask how to prove that direction?

Thanks so much.

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1) if $r_1$ and $r_2$ are odd and $2^{n_1}r_1=2^{n_2}r_2$, unique factorization proves that $n_1=n_2$ and $r_1=r_2$.

2) For every odd number $r$, all the numbers in $A_r$ are even, since they are of the form $2^nr$ with $n>0$. Thus, $\bigcup\limits_{r\text{ odd}}A_r$ is contained in the set of even numbers.

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