TITLE: Tao's Analysis Proposition 2.1.16
QUESTION [5 upvotes]: I am not understanding what Tao is claiming in proposition 2.1.16. He writes the following.
Proposition 2.1.16 (Recursive definitions). Suppose for each
natural number $n$, we have some function $f_n : \mathbb{N} \to \mathbb{N}$ from
the natural numbers to the natural numbers. Let $c$ be a natural
number. Then we can assign a unique natural number $a_n$ to each
natural number $n$, such that $a_0 = c$ and $a_{n++} = f_n(a_n)$ for each
natural number $n$.
Does $f_n$ necessarily map to all of $\mathbb{N}$ or could it be something like $f_n = 1$? If not, what does he write that indicates this? Does $f_n : \mathbb{N} \to \mathbb{N}$ imply that?
If that's the case then is his claim that if $f_n$ uniquely maps each $n$ to separate values, then recursively applying it will never end up repeating a number?
If not, what is proposition 2.1.16 actually saying then?

REPLY [2 votes]: Tao's claim here is that sequentially composing all of the $f$ yields a function from $\mathbb{N}$ to $\mathbb{N}$; changing notation slightly from $a_i$ to $a(i)$, he's saying that there's a function $a(n)$ such that $a(1) = f_1(c)$, $a(2) = f_2(f_1(c))$, $a(3)=f_3(f_2(f_1(c)))$, etc.  This is a version of the  Recursion Theorem — there, there's a single function $f()$ rather than a collection of functions $\{f_i\}$, but the two can be shown to be equivalent.