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I want to prove that the power set of $\mathbb R$, which can be naturally identified as $\{0,1\}^{\mathbb R}$, is bijective to the set $\mathbb R^{\mathbb R}$. Can the following attempt be altered to get a bijection without invoking Cantor-Schröder-Bernstein ?

Here is my attempt:

Notation. For any $f:\mathbb R\to\mathbb R$, let $\operatorname{graph}f\overset{\text{Def.}}=\{(x, f(x))\in\mathbb R^2: x\in\mathbb R\}$. Let $g:\mathbb R\to\mathbb R^2$ be a bijection.${}^1$${}^2$${}^3$

Now I define the operator

\begin{split}B:\mathbb R^{\mathbb R}&\to\{0,1\}^{\mathbb R}, \\ f&\mapsto B(f),\end{split}

where

\begin{split} B(f): \mathbb R&\to\{0,1\}, \\ r&\mapsto \begin{cases} 1&\text{if } g(r)\in\operatorname{graph} f\\ 0&\text{if } g(r)\not\in\operatorname{graph} f \end{cases} \end{split}

Now, $B$ is injective, because if $f_1, f_2:\mathbb R\to\mathbb R$ satisfy $B(f_1)=B(f_2)$, then, since $g$ is a bijection, for any $(y^1, y^2)\in\mathbb R$, we have $(y^1,y^2)\in\operatorname{graph} f_1\iff (y^1, y^2)\in\operatorname{graph} f_2$, i.e. $f_1(y^1)=y^2\iff f_2(y^1)=y^2$ for any $(y^1,y^2)\in\mathbb R^2$, which means that $f_1=f_2$.

However, $B$ is sadly not surjective. Since the inclusion $\{0,1\}^{\mathbb R}\to\mathbb R^{\mathbb R}$ is trivially injective, Cantor-Schröder-Bernstein implies that the two sets are bijective.

Hanul Jeon
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Maximilian Janisch
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  • Let us examine the cardinal equality $2^{\mathfrak{c}}=2^{\aleph_0\cdot\mathfrak{c}}=(2^{\aleph_0})^{\mathfrak{c}} = \mathfrak{c^c}$. Thus, what we may need is the bijection between $\mathbb{R}$ and $\mathbb{N\times R}$. (We also need the bijection between $\mathbb{R}$ and the set of all infinite binary sequences, too.) – Hanul Jeon Apr 11 '21 at 09:27
  • @HanulJeon for that we can use the fact that $[0,+\infty)$ can be seen as a product of $\Bbb N \times [0,1)$ using fractional and integer parts, e.g. – Henno Brandsma Apr 11 '21 at 10:40

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