Bijection from $\mathbb R$ to $\mathbb {R^N}$ How does one create an explicit bijection from the reals to the set of all sequences of reals? I know how to make a  bijection from $\mathbb R$ to $\mathbb {R \times R}$.
I have an idea but I am not sure if it will work. I will post it as my own answer because I don't want to anchor your answers and I want to see what other possible ways of doing this are.

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...).
