cardinality of the set of all partitions of $\mathbb N$ Calculate $|A|$ with $A=\{ F \subset \mathcal P(\mathbb N): F \text{ is a partition of } \mathbb N\}$. A partition of $\mathbb N$ is a family of subsets $F=\{P_i\}_{i \in I} \subset \mathcal P(\mathbb N)$, non-empty such that $\bigcup_{i \in I} P_i= \mathbb N$ and $P_i \cap P_j= \emptyset$
My attempt at a solution:
I think that $|A|=c$
I could prove that $c\leq |A|$
I've defined a surjective function $f$, with $f:A \to\{0,1\}^{\mathbb N}$ as follows:
$f(F)=\{a_n\}_{n \in \mathbb N}$ such that $a_n=1$ if $\{n\} \in F$ or $a_n=0$ if $\{n\} \not \in F$.
Lets check for surjectivity:
If $\{a_n\}_{n \in \mathbb N} \in  \{0,1\}^{\mathbb N}$, choose $F \in A$ as $F=(\bigcup_{n \in \mathbb N : a_n=1} \{\{n\}\}) \cup (\{\mathbb N \setminus \bigcup_{n \in \mathbb N : a_n=1} \{n\}\})$, then, clearly $f(F)=\{a_n\}_{n \in \mathbb N}$ by how I've constructed the partition $F$.
This proves $c=|\{0,1\}^{\mathbb N}|\leq |A|$
I need help to show the other inequality, and also, I would like to know if this part of the proof is ok. Thanks in advance. 
 A: Yes, what you’ve done works fine. Another approach that is perhaps just a little simpler is to let $E=\{2n:n\in\Bbb N\}$, and for each $A\subseteq E$ let $\mathscr{P}(A)=\{A\cup\{1\},\Bbb N\setminus(A\cup\{1\})\}$; then $\mathscr{P}$ is an injection from $\wp(E)$ to the family of partitions of $\Bbb N$.
To get the other inequality, notice that every partition of $\Bbb N$ is a countable subset of $\wp(\Bbb N)$. If $\mathscr{P}$ is a partition of $\Bbb N$, we can list the members of $\mathscr{P}$ in increasing order of their smallest elements; this uniquely associates to $\mathscr{P}$ a specific sequence, possibly finite, of subsets of $\Bbb N$. If the sequence is finite, we can extend it to an infinite sequence by repeating the last term of the finite sequence.
Thus, it suffices to show that there are at most $\mathfrak{c}$ infinite sequences of subsets of $\Bbb N$. This is equivalent to showing that there are at most $\mathfrak{c}$ infinite sequences of functions from $\Bbb N$ to $\{0,1\}$. If ${}^XY$ denotes the set of functions from $X$ to $Y$, find a bijection between ${}^{\Bbb N\times\Bbb N}\{0,1\}$ and ${}^{\Bbb N}\{0,1\}$ to complete the proof.
A: You've done the hard part.
$A$ has as elements countable sequences from $\mathcal{P}(\mathbb{N})$, i.e. it's a subset of $\mathcal{P}(\mathbb{N})^{\mathbb{N}}$, so all you need to do is exhibit bijections
$$\left(\{0,1\}^{\mathbb{N}}\right)^{\mathbb{N}} \cong \{ 0, 1 \}^{\mathbb{N} \times \mathbb{N}} \cong \{0,1\}^{\mathbb{N}}$$
to get that
$$|A| \le |\mathcal{P}(\mathbb{N})^{\mathbb{N}}| = \mathfrak{c}$$
A: Your proof is correct.
For the second part: one way to do it is the following.  First show that for such a partition $F = \{P_i\}_{i\in I}$, $I$ is at most countable.  Then, identify each such family $F = \{P_i\}_{i\in \mathbb{N}}$  (or $F = \{P_i\}_{0\leq i \leq m}$) with the function $f:\mathbb{N}\rightarrow \mathbb{N}$ given by $f(n) = i$ if $n \in P_i$.  This gives an injection from $A$ to $\mathbb{N}^\mathbb{N}$. Now you just need to prove (or recall, if you already know it) that the cardinality of the set of functions from $\mathbb{N}$ to $\mathbb{N}$ is $\mathfrak{c}$.
