# Is this a bijection between $\mathcal{P}(\mathbb{R})$ and $\mathbb{R}^\mathbb{R}$

I wish to prove the existence of a bijection $\mathcal{P}(\mathbb{R}) \sim\mathbb{R}^\mathbb{R}$

So, I know (and have proven) that $\mathcal{P}(\mathbb{R}\times\mathbb{R}) \sim \mathcal{P}(\mathbb{R})$, so I thought I would use that. My strategy is to construct two injective maps \begin{align} \phi &: \mathbb{R}^\mathbb{R} \rightarrow \mathcal{P}(\mathbb{R}\times\mathbb{R}) \\ \psi &: \mathcal{P}(\mathbb{R}\times\mathbb{R}) \rightarrow \mathbb{R}^\mathbb{R} \end{align} in order to use the Schröder-Bernstein theorem to prove the existence of a bijection.

For $f\in \mathbb{R}^\mathbb{R}$, let $\phi(f) = \{(x,f(x)) : x \in \mathbb{R} \}$. Clearly, if $\phi(f)=\phi(g)$ then $f(x)=g(x)$ for all $x\in \mathbb{R}$ by the definition of $\phi$, so it is injective.

For $S\in \mathcal{P}(\mathbb{R}\times\mathbb{R})$, first map it bijectively to $S' \in \mathcal{P}(\mathbb{R})$ via the bijection I know exists between $\mathcal{P}(\mathbb{R}\times\mathbb{R})$ and $\mathcal{P}(\mathbb{R})$. Now, let $\psi(S')(x) = \begin{cases} 0 &\text{if } x \notin S' \\ 1 &\text{if } x \in S'\end{cases}$

Suppose that $\phi(S_1)=\psi(S_2)$, i.e. $\psi(S_1)(x) = \psi(S_2)(x)$ for all $x\in \mathbb{R}$. Then clearly $x\in S_1 \iff x\in S_2$ and $x \notin S_1 \iff x \notin S_2$, so $S_1 = S_2$ and the map is injective.

Now, with these injective maps, the bijection I know exists and the Schröder-Bernstein theorem, I have $\mathcal{P}(\mathbb{R}\times\mathbb{R}) \sim \mathcal{P}(\mathbb{R}) \sim \mathbb{R}^\mathbb{R}$.

Is this correct? Are there improvements I can make? Is there an easier way to go about this proof? Thank you in advance.

• That's how I'd do it myself. Good job! – Stefan Mesken May 7 '18 at 8:58