# 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?

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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.

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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.

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In case some steps are still unclear, this overview might be useful. –  Martin Sleziak Nov 2 '12 at 16:12

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.

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