By definition $\mathbb R^\mathbb R \subset \mathbf 2^{\mathbb R\times\mathbb R}$. So all we need is a surjection $\mathbb R \twoheadrightarrow \mathbb R \times \mathbb R$ of which there are plenty such as [space filling curves](https://en.wikipedia.org/wiki/Space-filling_curve). If you have a bijection between $\mathbf 2^\mathbb N$ and $\mathbb R$, then a lot of this stuff is easier to do in terms of $\mathbf 2^\mathbb N$. For example, $\mathbf 2^\mathbb N \times \mathbf 2^\mathbb N \cong \mathbf 2^{\mathbb N + \mathbb N} \cong \mathbf 2^\mathbb N$ with the last bijection via the straightforward one showing $\mathbb N \cong \mathbb N + \mathbb N$ (where $+$ represents the disjoint union/coproduct).
That said, you should be able to fairly directly extract a bijection from a proof using cardinal arithmetic. Of course, there's no real reason to apply Schröder-Bernstein at that point.
Derek Elkins
·
2023-01-14T15:46:25Z (over 1 year ago)