# Cardinality of power set of reals is equal to cardinality of all functions from reals to reals? [duplicate]

I want to prove that $P(\mathbb{R}) = \mathbb{R}^\mathbb{R}$ ?!

I know that $\mathbb{R} = \{0,1\}^ \omega = P(\omega)$

Also that $(A^B)^ C = A^{B \times C}$

And that $A = B$ imply that $P(A) = P(B)$

Here $P$ is the power set and $=$ means that they have the same cardinality

And $A,B,C$ are sets.

• $f:P(\mathbb{R})\to\{0,1\}^\mathbb R \subseteq \mathbb R^\mathbb R, f(A)=\chi_A$ is injective, would be one direction
– SK19
May 11, 2018 at 16:40

$|\mathcal{P}(\mathbb{R})|=2^\mathfrak{c}=2^{\aleph_0\times\mathfrak{c}}=(2^{\aleph_0})^\mathfrak{c}=\mathfrak{c}^\mathfrak{c}$
Where $\mathfrak{c}:=|\mathbb{R}|=2^{\aleph_0}$.
$|\mathscr{P}(\mathbb{R})| = 2^{|\mathbb{R}|} \le |\mathbb{R}^{\mathbb{R}}| \le |(2^{\mathbb{R}})^{\mathbb{R}}| = |2^{\mathbb{R} \times \mathbb{R}}| = 2^{|\mathbb{R}|}$