Number of different partitions of N What is the number of different partitions of the set N of integers? My reasoning thus far goes as follows: the number must be at least as big as ω because any given integer can be matched with any other integer in a single partition cell. The number must be greater than ω (the first infinite cardinal), because diagonalization kicks in as soon as you try to enumerate the set. But it can be no bigger than ω(2), because each partition is a collection of subsets of N and thus a subset of P(N) (the power set of N). Hence, the set of all partitions is a subset of P(P(N)). Therefore, given the continuum hypothesis, the number I am looking is either the first or the second uncountable ordinal. Any ideas? 
 A: It seems that generally you have the right idea, but you are also very very confused.
It's not clear what $\omega(2)$ is. Is it $^\omega2$ or $2^\omega$? Is it something else? Assuming that it is the set of all binary sequences, then indeed this is the cardinality of $\mathcal P(\omega)$, which is also the cardinality of the real numbers.
Later on, however, you claim that this is a subset of $\mathcal{P(P(}\omega))$, and therefore, if the continuum hypothesis holds then this is either the cardinality of the real numbers, or its power set.
Well...

*

*First you say that this cannot be larger than the cardinality of the real numbers, then you say that it is either the real numbers of its power set, from these two you can easily conclude that the set of all partitions of $\omega$ has the same cardinality as the real numbers and $\mathcal P(\omega)$.


*The continuum hypothesis merely says that $2^{\aleph_0}=\aleph_1$, or that there is no set whose cardinality is strictly between $\Bbb N$ and $\Bbb R$. It says nothing, absolutely nothing about the cardinality of $\mathcal P(\Bbb R)$, or the possible intermediate cardinalities between $\Bbb R$ and its power set.


*The correct way of approaching this is to note that every partition cannot have more than $\aleph_0$ parts, therefore every partition is completely determined by a function from $\omega$ to itself.
Since there are only $2^{\aleph_0}$ of these functions, we have an easy upper bound.


*Finally, it is not very hard to show that the set of partitions of $\omega$ into two parts already has at least $2^{\aleph_0}$ elements, which establishes a lower bound.
"And that's all I have to say about that."
