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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.

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

2 Answers 2

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$|\mathcal{P}(\mathbb{R})|=2^\mathfrak{c}=2^{\aleph_0\times\mathfrak{c}}=(2^{\aleph_0})^\mathfrak{c}=\mathfrak{c}^\mathfrak{c}$

Where $\mathfrak{c}:=|\mathbb{R}|=2^{\aleph_0}$.

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$|\mathscr{P}(\mathbb{R})| = 2^{|\mathbb{R}|} \le |\mathbb{R}^{\mathbb{R}}| \le |(2^{\mathbb{R}})^{\mathbb{R}}| = |2^{\mathbb{R} \times \mathbb{R}}| = 2^{|\mathbb{R}|}$

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