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What is the easiest way to show that $$ \lvert \mathbb{N}^\mathbb{R} \rvert = \mathcal{P}(\mathbb{R}) ? $$ I already know one cumbersome method, but I'd like to know if there is a simpler/cleverer way. If there's a technique that just uses basic cardinal arithmetic, that would be perfect.

Another question I have is:

Could using the heuristics \begin{align*} \lvert \mathbb{N}^{\mathbb{R}} \rvert \cong \lvert \mathbb{Z}^\mathbb{R} \rvert \cong \lvert 2^\mathbb{R} \rvert &= \mathcal{P}(\mathbb{R}) \\ \lvert \mathbb{N}^\mathbb{N} \rvert \cong \lvert \mathbb{N}^\mathbb{Z} \rvert \cong \lvert \mathbb{Z}^\mathbb{N} \rvert \cong \lvert \mathbb{Z}^\mathbb{Z} \rvert \cong \lvert 2^\mathbb{N} \rvert \cong \lvert 2^\mathbb{Z} \rvert &= \mathcal{P}(\mathbb{N}) \end{align*} get me into a lot of trouble?

Aaron Hendrickson
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Jeremy Lindsay
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1 Answers1

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We have an injective map $\mathcal P(\Bbb R)\to\Bbb N^{\Bbb R} $ given by $ A\mapsto \mathbf 1_A$, where $\mathbf 1_A(x)=\begin{cases}1&x\in A\\0&x\notin A\end{cases}$.

We obtain an injective map $\Bbb N^{\Bbb R}\to \mathcal P(\Bbb R)$ by mapping $f\colon \Bbb R\to\Bbb N$ to $\{\,g(x)+f(x)\mid x\in \Bbb R\,\}$, where $g\colon \Bbb R\to [0,1)$ is an injective map, e.g., $g(x)=\frac1{e^x+1}$.

Then $|\mathcal P(\Bbb R)|=|\Bbb N^{\Bbb R}|$ follows by Schröder-Bernstein.

Hagen von Eitzen
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