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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?

  • Do you mean $a_{n+1}$ instead of $a_{++}$? – Matemáticos Chibchas Oct 31 '15 at 03:31
  • @MatemáticosChibchas My apologies. Tao defines $n++$ to indicate $n+1$. –  Oct 31 '15 at 03:33
  • N to N indicates the domain and codomain respectively. In layman's terms it means $f_n$ takes in a natural number and outputs a natural number. Note that codomain is different from range. – yoyostein Oct 31 '15 at 03:33
  • His claim is simply that sequentially composing all of the $f$ yields a function; that is, there is a function $a(n)$ (changing notation slightly) such that $a(1)=f_1(c)$, $a(2)=f_2(f_1(c))$, $a(3)=f_3(f_2(f_1(c)))$, etc. – Steven Stadnicki Oct 31 '15 at 03:36
  • @StevenStadnicki Oh, I see what he is getting at now. Thank you, that clears up my confusion. I would accept that as the answer if you were to reproduce it below. –  Oct 31 '15 at 03:42
  • @ChantryCargill Gladly! – Steven Stadnicki Oct 31 '15 at 04:04

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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.

Steven Stadnicki
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