I was reading "Classical Recursion Theory" by Odifreddi and it starts with this phrase:
Recall that Classical Recursion Theory is the study of real numbers or, equivalently, functions over the natural numbers.
I come from Computer Science so this was puzzling to me at first. I understand he's referring to the fact that there is a bijection between $\mathbb{N}$ $\rightarrow$ $\mathbb{N}$ and $\mathbb{R}$. I know that |$\mathbb{N}$ $\rightarrow$ $\mathbb{N}$| is greater than |$\mathbb{N}$| (Cantor's diagonal), but how can you find the real $x$ corresponding to a certain function?
I know that given a real $x$ we can construct a function like this: $f(n) =$ the $n$-th digit of $x$. But what is the other way around? We can't use digits in this case, I think, because we have numbers with more than one digit; I don't think changing the digit system would help (I think every digit system has a finite alphabet?) and that way we would link more functions to the same real number.
I know that you can't compute reals in a strict sense (you would need $\infty$ digits) but I was wondering if there was at least some sketch-procedure of how to represent a function with a real number, to have at least logically a glimpse of what is the number $x$ corresponding to a generic function $f$: $\mathbb{N}$ $\rightarrow$ $\mathbb{N}$.
Here's an idea I didn't check in detail, but maybe it helps:
Let $x_0\ge 0$ be any real number and recursively define for $n\in\mathbb N=\{0,1,\dots\}$: \begin{align*} a_n &= \lfloor x_n \rfloor, \\ x_{n+1} &= g(x_n-a_n), \end{align*} for $g$ the bijection $[0,1)\to\mathbb R_{\ge 0}$ given by $g(x) = \frac{x}{1-x}$ with inverse given by $g^{-1}(y) = \frac{y}{1+y}$.
I claim that the map $\mathbb R_{\ge 0} \to \mathbb N^{\mathbb N}$ given by $x_0 \mapsto (a_n)_{n\in\mathbb N}$ is a bijection. Compose this with your favorite bijection between $\mathbb R_{\ge 0}$ and $\mathbb R$ to obtain what you asked for.