Bijection between $2^{\mathbb R}$ and $\mathbb{ R ^ R}$

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.

Solution 1:

Once you have the bijection $P:2^\mathbb{N}\cong\mathbb{R}$, you can build your desired bijection as follows.

First, note that $\mathbb{N}\times 2^\mathbb{N}\cong 2^\mathbb{N}$, essentially by the bijection that associates $(n,A)$ with $\{n\}\cup (n+1+A)$, except that this will miss the empty set on the right, but we can fix this by composing with a map witnessing $2^\mathbb{N}-\{\varnothing\}\cong 2^\mathbb{N}$, such as shifting on a fixed countable subset.

Now simply observe that $$\mathbb{R}^\mathbb{R}\cong (2^\mathbb{N})^{(2^\mathbb{N})}\cong 2^{(\mathbb{N}\times 2^\mathbb{N})} \cong 2^{(2^\mathbb{N})} \cong 2^\mathbb{R}, $$

which provides the desired bijection. The first map is conjugation by $P$, simply composing with $P^{-1}$ and $P$ before and after; the second map is an easy exercise in parenthesis rearranging; the third map applies the observation above; and the final map applies $P$.