Bijection from $\mathbb R$ to $\mathbb {R^N}$

The nicest trick in the book is to find a bijection between $\mathbb R$ and $\mathbb{N^N}$, in this case we are practically done. Why?

$$\mathbb{(N^N)^N\sim N^{N\times N}\sim N^N}$$

And the bijections above are easy to calculate (I will leave those to you, the first bijection is a simple Currying, and for the second you can use Cantor's pairing function).

So if we can find a nice bijection between the real numbers the infinite sequences of natural numbers we are about done. Now, we know that $\mathbb{N^N}$ can be identified with the real numbers, in fact continued fractions form a bijection between the irrationals and $\mathbb{N^N}$.

We first need to handle the rational numbers, but that much is not very difficult. Take an enumeration of the rationals (e.g. Calkin-Wilf tree) in $(0,1)$, suppose $q_i$ is the $i$-th rational in the enumeration; now we take a sequence of irrationals, e.g. $r_n = \frac1{\sqrt{n^2+1}}$, and we define the following function:

$$h(x)=\begin{cases} r_{2n} & \exists n: x=r_n\\ r_{2n+1} & \exists n: x=q_n \\ x &\text{otherwise}\end{cases}$$

Now we can finally describe a list of bijections which, when composed, give us a bijection between $\mathbb R$ and $\mathbb{R^N}$.

1. $\mathbb{R^N\to (0,1)^N}$ by any bijection of this sort.
2. $\mathbb{(0,1)^N\to \left((0,1)\setminus Q\right)^N}$ by the encoding given by $h$.
3. $\mathbb{\left((0,1)\setminus Q\right)^N\to \left(N^N\right)^N}$ by continued fractions.
4. $\mathbb{\left(N^N\right)^N\to N^{N\times N}}$ by Currying.
5. $\mathbb{N^{N\times N}\to N^N}$ by a pairing function.
6. $\mathbb{N^N\to (0,1)\setminus Q}$ by decoding the continued fractions.
7. $\mathbb{(0,1)\setminus Q\to (0,1)}$ by the decoding of $h$, i.e. $h^{-1}$.
8. $\mathbb{(0,1)\to R}$ by any bijection of this sort, e.g. the inverse of the bijection used for the first step.

NOTE: This doesn't work

First, map all the $\mathbb R$s to $(0,1)\backslash \mathbb Q$. Then, for each sequence of irrationals from 0 to 1, set it up in a grid with as such:

$$s_1 = 0.d_{11}d_{12}d_{13}d_{14}d_{15}... \\ s_2 = 0.d_{21}d_{22}d_{23}d_{24}d_{25}... \\ s_3 = 0.d_{31}d_{32}d_{33}d_{34}d_{35}... \\...$$

And take the new irrational number by taking each diagonal similarly to how you create a bijection from the rationals to the naturals. That is:

$$r = 0.d_{11} \; d_{21}d_{12}\; d_{31}d_{22}d_{13} \; d_{41}d_{32}d_{23}d_{14}...$$

Now, does this map to every irrational? Not sure. Does this map to any rationals, I am pretty sure not. If r was repeating, I think that that would make the top row repeating. Not sure how to prove this though.

Why this doesn't work: Consider $r= 0.101001000100001....$. This is irrational and is only mapped to by $s_1 = 0.111111$ and $s_n = 0.0000...$ for all other n (and 0 isn't even in our set to begin with...).