I'm well aware of the standard proof based on cardinality arithmetic to show that these two sets have the same cardinality and the question of defining a bijection between the two sets came up. I worked on it a little while and was wondering if there were any gaps, or a simpler method of coming up with such a bijection.
The main idea I had was to take a family of sequences indexed by the real numbers. So that each one of these sequences corresponds to a unique real number. Then I can map each one of these sequences to another number, by applying the binary function from $\mathbb R$ to $\{0,1\}$ and mapping the infinite binary sequence to a real number, use one of the standard bijections.
First let $E$ be a family of sequences indexed by the real numbers. Such that $E_{xi}\neq E_{yj}$ for all $x,y \in \mathbb R$ and $i,j \in \mathbb N$ also with the property that
$\mathbb R= \bigcup_{x \in \mathbb R, i \in \mathbb N} E_{xi}$.
What sort of theorems would I need to prove such a family existed, is it enough to note that the reals can be represented as a uncountable union of countably infinite sets?
Also let $P: 2^{\mathbb N} \rightarrow \mathbb R$ be you favorite bijection.
Now consider an element of $2^{\mathbb R}$ as a function $f: \mathbb R \rightarrow \{0,1\}$ and $g \in \mathbb{ R^R}$ in a similar manner. Then our bijection between the two sets will be the function $\Phi$ which maps $f$ to a function $g$, such that $g(x)=P(F(E_x))$ where $F$ applies $f$ to each element of the sequence $E_x$.
I'm fairly sure this is an onto and one-to-one function. The one-to-one property should follow because the sequences are disjoint, cover $\mathbb R$ and $P$ is a bijection. The onto argument is a little hazier, but it should be possible to recreate the function $f$, by first inverting each element with $P$ and then defining $f$ based on these sequences.