Set Theory request for knowledge: Need criticism of where my logic is flawed with respect to the intersection of infinite sets. Example in post. Greetings fellow problem solvers! I'll start off by thanking anyone who takes the time to look over what I've done and redirect me. Now I'm not looking for solutions to this problem. My approach to learning this is by working out why something I have done is incorrect then refining it until I come to a proper solution.
The exercise: Produce an infinite collection of sets $A_{1}, A_{2}, A_{3},...$ with the property that every $A_i$ has an infinite number of elements $A_{j}\ \cap\ A_i = \emptyset$ for all $i\neq j$ and $\bigcup_{i=1}^{\infty}A_{j} = N.$ Where N is the Natural Numbers
My solution to this is as follows.
$$A_1 = \{1, 2,..., n\}\\
A_2 = \{n+1, n+2, ..., k\}\\
A_3 = \{k+1, k+2, ..., m\}\\
A_i = \{m+1, m+2, ...\}$$
Where $n, k, m \in$ natural numbers.
Now to me, the above shows every $A_i$ has an infinite number of elements and  $A_{j}\ \cap\ A_i = \emptyset$ for all $i\neq j$ and $\bigcup_{i=1}^{\infty}A_{j} = N.$ Where N is the Natural Numbers.
I feel like what I've done here is a horrible abuse of notation, at best. Worst case the logic doesn't even follow. Regardless I hope it gives enough insight into how I'm thinking about the intersection of these sets, for someone to help redirect my thought process.
Again I appreciate any feedback,
 A: Your sets are pairwise disjoint, and their union is $\Bbb N$, but none of them has an infinite number of elements: all of them are finite. Your $A_1$ has $n$ elements for some positive integer $n$; that is a finite number. Your $A_2$ has $k-n$ elements, again a finite number, your $A_3$ has $m-k$ elements, and so on.
There is the additional problem that you’ve not really explained how to keep going, though since the idea doesn’t work, this is perhaps of secondary significance. Still, it might be useful in the future to know how to write up the idea. Here is one way:

Let $\langle n_k:k\in\Bbb Z^+\rangle$ be a strictly increasing sequence of non-negative integers such that $n_1=0$, and for each $k\in\Bbb Z^+$ let $$A_k=\{m\in\Bbb Z^+:n_k+1\le m\le n_{k+1}\}\,.$$

Now let’s get back to the problem. Here is a sketch of one of the nicer ways to solve it.

*

*Show that for each positive integer $n$ there are unique non-negative integers $m$ and $k$ such that $n=2^k(2m+1)$.

Then for each non-negative integer $m$ let
$$A_m=\left\{2^k(2m+1):k\ge 0\right\}$$
and show that $\{A_m:m\ge 0\}$ satisfies the requirements of the problem.
A: First of all, what exactly are $n,k,m$, etc. in the definitions of your sets? E.g. is $A_1=\{1,2,3\}$ ($n=3$), or $\{1,2,3,4,5\}$ ($n=5$) or ...?
More importantly, contra your claim, each $A_i$ you've described is finite - e.g. if $n=6$ and $k=17$, then $A_2$ has only $17-6=11$ elements.
Since the $A_i$s have to be infinite, nothing like $\{a, a+1,..., b\}$ is going to work. Your $A_i$s have to be more complicated - essentially, they'll have to "interleave" with each other.
A good first step is to solve the following simpler problem:

Find two sets $A_1,A_2$ such that $A_1\cap A_2=\emptyset$ but $A_1$ and $A_2$ each have infinitely many elements.

Don't try anything too fancy:

 For example, set $A_1=\{$evens$\}$ and $A_2=\{$odds$\}$.

Now how about three infinite, pairwise-disjoint sets - or more generally, $n$-many for $n$ finite?

 Try mod $3$, or mod $n$ more generally.

Getting infinitely many infinite, pairwise-disjoint sets is harder and the above idea doesn't immediately generalize. However, it should suggest some things to try, and there are various ways to get this to work. Here's one idea:

 Given a prime number $p$, let $Pow_p$ be the set of powers of $p$. How big is each $Pow_p$? If $p,q$ are distinct primes, what can you say about $Pow_p\cap Pow_q$? And finally, how many primes are there?

