cardinality problem Let $\mathbb{R}^\mathbb{R}$ be the set of all functions $f: \mathbb{R} \to \mathbb{R}$ and $P(\mathbb{R})$ be the power set of $\mathbb{R}$. How to show that they have the same cardinality?
 A: Note that $\mathbb R$ has the same cardinality as $\mathbb R\times\mathbb R$. Therefore their power sets have the same size.
In particular $f\colon\mathbb R\to\mathbb R$ is an element in $\mathcal P(\mathbb{R\times R})$, and therefore $\mathbb{R^R}$ has cardinality of at most $2^{|\mathbb R|}$, and the other direction is trivial, because: $$2<|\mathbb R|\implies |2^\mathbb R|\leq|\mathbb{R^R}|\leq|2^\mathbb R|$$
Thus equality holds.
A: We know that $|\mathbb R|=\mathfrak c=2^{\aleph_0}$
$|\mathcal P(\mathbb R)|=2^{\mathfrak c}$
$|\mathbb R^{\mathbb R}|=\mathfrak c^{\mathfrak c}$
So we only need to show that $2^{\mathfrak c}=\mathfrak c^{\mathfrak c}$.
We have
$$\mathfrak c^{\mathfrak c} = (2^{\aleph_0})^{\mathfrak c} = 2^{\aleph_0\mathfrak c}=2^\mathfrak c,$$
since $\aleph_0\mathfrak c=\mathfrak c$.
We get the last equality from 
$$\mathfrak c \le \aleph_0 \mathfrak c \le \mathfrak c \mathfrak c = 2^{\aleph_0} 2^{\aleph_0} = 2^{\aleph_0+\aleph_0} = 2^{\aleph_0} = \mathfrak c$$
and from Cantor-Benstein theorem.
A: Denote $A=\{0,1\}$. You can use the fact that $\mathbb{R} \simeq A^{\mathbb{N}}$, where $\simeq$ means "have the same cardinality". So we have
$$
\mathbb{R}^{\mathbb{R}} \simeq \left(A^{\mathbb{N}}\right)^{\mathbb{R}} \simeq A^{\mathbb{N} \times \mathbb{R}}.
$$
Now we use the fact that $\mathbb{N}\times\mathbb{R} \simeq \mathbb{R}$, so
$$
\mathbb{R}^{\mathbb{R}} \simeq A^{\mathbb{R}} \simeq P(\mathbb{R}),
$$
QED.
