cardinality of the set of countable partitions of $\mathbb{R}$ What is the cardinality of the set 
$$A=\{ P| P\  \text{is a countable partition of the reals} \}$$ ?
I am searching on this for a while. I think the cardinality is $2^w$ where $w$ is the cardinality of $\mathbb{N}$. Clearly $|A|\geq |\mathbb{R}|=2^w$, but I cannot prove the other direction. I tried (but did not succeed)to construct an injection $A \rightarrow C$ for some set $C$ with  cardinality equal to $2^w$. 
 A: I think it's not very hard. A countable partition of reals can be seen as just a function from the reals into natural numbers, up to a permutation of the naturals. So we have
$$\lvert \mathbf N^{\mathbf R}\rvert=\aleph_0^{2^{\aleph_0}}=2^{2^{\aleph_0}}$$
(because $\aleph_0\leq 2^{\aleph_0}$ we can do the last substitution)
On the other hand, there are only $\aleph_0^{\aleph_0}=\mathfrak c<2^{\mathfrak c}$ permutations of natural numbers, so there are $2^{2^{\aleph_0}}$ equivalence classes of functions from reals to naturals under the relation of equivalence up to a permutation of naturals (because each has at most $\mathfrak c$ elements), which are precisely correspondent to the countable partitions of reals.
A: What is a countable partition? It means that we have some function from $\mathbb R$ into $\mathbb N$ and every part in the partition is the fiber of such function. However counting like this we count many partitions several times, for example take a partition into two parts, then we can see it as a function whose range is $\{0,1\}$ or $\{2,3\}$ or any other pair of numbers.
However note that there are exactly $2^{\aleph_0}$ many ways to re-order the natural numbers therefore at most continuum many functions generate the same partition.
So we have $\aleph_0^{2^{\aleph_0}}=2^{2^{\aleph_0}}$ many functions and each is equivalent to at most $2^{\aleph_0}$ many of them. Therefore there are $2^{2^{\aleph_0}}$ many partitions.
Why can't there be another number? Well, we denote by $\frak p$ the cardinality of all countable partitions, then we have that $2^{2^{\aleph_0}}=2^{\aleph_0}\cdot\frak p$, by the general argument appearing in Does $k+\aleph_0=\mathfrak{c}$ imply $k=\mathfrak{c}$ without the Axiom of Choice? we have that $2^{2^{\aleph_0}}=\frak p$.
A: Hint:  how many subsets does $\mathbb R$ have?  Can't (almost) all of them be part of a partition?
