Partition of $\mathbb{N}$ into $\aleph_0$ infinite sets Theorem - There is a division of $\mathbb{N}$ for $\aleph_0$ sets so every set has cardinal of $\aleph_0$.
I don't know even where to start in order to prove this theorem. 
 A: For example:
Let a natural number $a$ belong to the $n$th set of the partition if $a$ has exactly $n$ prime factors. 
(And put $0$ and $1$ to any of them, say to the first one.)
A: Put a number in $A_n$ if it's divisible to $2^n$ but not by $2^{n+1}.$ Then $A_0,A_1,A_2,\dots$ are pairwise disjoint infinite sets whose union is $\mathbb N.$
A: When partitioning $\mathbb N$, the largest cardinal you're able to produce is $\aleph_0$, since you can't create sets with any more elements, and even if you partition $\mathbb N$ into its single elements, there are still only $\aleph_0$ elements. So in this case, we can conceptually simplify the problem into partioning $\mathbb N$ into infinite partitions where each partition has infinite cardinality.
There are tons of ways to do this. One way: partition $\mathbb N$ into sets containing $\{p,p^2,p^3, \dots \}$ for every prime $p$. Put all remaining elements into one set. There are $\aleph_0$ primes, so you will have $\aleph_0$ partitions. And each partition contains $\aleph_0$ elements.
A: Since there are a few different ways here already, here is my personal preference:
Put $1$ in the first partition. Put $2$ in the second partition and $3$ in the first partition. Put $4$ in the third partition, $5$ in the second partition and $6$ in the first partition. And so on.
