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I'm currently reading Terence Tao's book Analysis I, and I got stuck on his proposition 2.1.16 about recursive definitions ; actually, I don't understand his (informal) proof of it.

First, here are the five Peano axioms he's using :

Axiom 2.1. $0$ is a natural number.

Axiom 2.2. If $n$ is a natural number, then $n{++}$ is also a natural number.

Axiom 2.3. $0$ is not the successor of any natural number ; i.e., we have $n{++} \ne 0$ for every natural number $n$.

Axiom 2.4. Different natural numbers must have different successors ; i.e., if $n, m$ are natural numbers and $n \ne m$, then $n{++} \ne m{++}$. Equivalently, if $n{++} = m{++}$, then we must have $n=m$.

Axiom 2.5. (Principle of mathematical induction). Let $P(n)$ be any property pertaining to a natural number $n$. Suppose that $P(0)$ is true, and suppose that whenever $P(n)$ is true, $P(n{++})$ is also true. Then $P(n)$ is true for every natural number $n$.

And here is the proposition, and the proof I'm stuck on :

Proposition 2.1.16 (Recursive definitions). Suppose for each natural number $n$, we have some function $f_{n} : \mathbb{N} \rightarrow \mathbb{N}$ 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 $n$.

Proof. (Informal) We use induction. We first observe that this procedure gives a single value to $a_0$, namely $c$. (None of the other definitions $a_{n{++}}:= f_{n} (a_n)$ will redefine the value of $a_0$, because of Axiom 2.3.) Now suppose inductively that the procedure gives a single value to $a_n$. Then it gives a single value to $a_{n{++}}$, namely $a_{n{++}}:= f_{n} (a_n)$. (None of the other definitions $a_{m{++}}:= f_{m} (a_{m})$ will redefine the value of $a_{n{++}}$, because of Axiome 2.4.) This completes the induction, and so $a_n$ is defined for each natural number $n$, with a single value assigned to each $a_n$.

I don't understand how does that prove anything. I feel like all he's saying is that « all natural numbers are different from each others » ; I don't understand why we couldn't have, for example, $a_0 = a_1 = c$ (for instance, if we consider the Fibonacci sequence : $0, 1, 1, 2, ...$, we have $a_1 = a_2 = 1$).

I actually found another proof of uniqueness on Wikipedia (regarding the Recursion theorem) and I think I understood it, but I really would like to understand Tao's point of view as well.

Finally, shouldn't we prove that such a function exists as well, instead of just supposing it exists ? I found this document which gives a proof for the existence of such a function (at pages 2-3), but I'm a little bit confused about the way they do it: can we really "imagine" or "create" such sets as $S$ and $g$, like they do, and draw conclusions from them? Another thing that disturbs me regarding their approach is that Tao doesn't talk about sets before chapter 3, and the proposition above comes from chapter 2... so the question here is: how could we prove the existence of such functions as the ones described in the recursion theorem without knowing anything about sets and set theory?

So, as you can see... I'm pretty confused about a lot of things. Any help would be greatly appreciated!

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justdoit
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    We could have $a_0 = a_1 = c$ just as long as $a_1=f_0(a_0)=f_0(c)=c$. Nothing in his definition prevents that. – Arby Apr 13 '17 at 06:43
  • @Arby Thanks for your comment. But then what does the proposition mean? – justdoit Apr 13 '17 at 08:02
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    He is manufacturing the sequence ${ a_n }$. He does not say that all $a_n$ are distinct; what he is saying is that the "procedure" never loops. i.e. that it never assign two different values to $a_n$ ensuring that the result is a function $F : \mathbb N \to \mathbb N$ such that $F(n)=a_n$. – Mauro ALLEGRANZA Apr 13 '17 at 08:36
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    I can't explain any simpler than Tao does in the paragraph before the proposition. – Arby Apr 13 '17 at 08:43
  • @MauroALLEGRANZA Thanks a lot, I think I finally got it! But there's one more thing I'm not sure about: why is this informal proof a thing since $a_{n{++}}$ is defined as $f_{n}(a_n)$? What I mean is that we suppose, as an hypothesis of the proposition, that $f_n$ is a function, so shouldn't it be "obvious" (I hate that word, but still) that $a_{n{++}}$ has a unique value for all $n$? In other words: even if I now understand what he did in his proof, I just wonder if that proof still makes sense, considering the hypothesis he made. – justdoit Apr 13 '17 at 09:10
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    NO; consider the "tricky" case of the inductive step: we have that $a_{n++} := f_n(a_n)$. This is a definition, and thus we have nothing to say "against" it; the issue is: "it may happens that for a "later" $m$ the corresponding def $a_{m++} := f_m(a_m)$ makes a loop ?" i.e. that $m++=n++$ (and not that $a_{m++} = a_{n++}$ that is perfectly acceptable...) ? The answer is NO, because if $m++=n++$, then - by Ax.2.4 - $m=n$ and thus $f_m(a_m)=f_n(a_n)$ and we have no risk of "multiple values". – Mauro ALLEGRANZA Apr 13 '17 at 09:20
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    You can apply it to the recursive def of sum: $+$. Basis : $x+0 :=0$ here we have $c=0$ and Ind.step: $x+(n++)=(x+n)++$ here we have $a_{n++}=x+(n++)$ and $a_n=(x+n)$ and $f_n=s$ (the successor function). The Th about recursive defs licenses us to assert that we have a "single" function $+$ such that $+(x,n)$ is a well-defined value for every $n$. – Mauro ALLEGRANZA Apr 13 '17 at 09:25
  • @MauroALLEGRANZA I'm getting confused again. If the issue is that $m{++}$ could be equal to $n{++}$, then souldn't we have $f_{m}(a_m) = a_{m{++}}=a_{n{++}}$ (because $m{++}=n{++}$) $= f_{n}(a_n)$, and therefore $f_{m} (a_m) = f_{n}(a_n)$? And what was disturbing me in my last comment was that a function is "a relation that associates an input to a single output according to some rule" (source: wikipedia), and therefore each $a_{n++}$, defined as the (single) output $f_{n}(a_n)$, had to be unique by the hypothesis made in the proposition. – justdoit Apr 13 '17 at 09:44
  • Not enough place to write everything in one single comment: So, I was sceptical about the validity of the proof because we accept that $a_{n{++}}$ is defined as the output of a function (the $f_n$ function), which has to be unique by definition. So now I feel like we're not proving anything at all with this proof. – justdoit Apr 13 '17 at 09:47
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    Last try :-) assume that $m++=n++$ but $m \ne n$. When we arrive at $m++$ after having alteady defined $a_{n++}$ we define $a_{m++}$; but by assumption we are looping on the "index" and thus we get a new value $a^_{n++}$ for the "index" that we have already used. But under our assumption $m\ne n$ and thus $f_m(a_m)$ can be different from $f_n(a_n)$ and thus we have that $a^{n++} \ne a{n++}$, getting the undesired multi-value. – Mauro ALLEGRANZA Apr 13 '17 at 10:26
  • @MauroALLEGRANZA I think I finally got what I was not understanding: I did not consider the fact that we have a different function for each $n$, thus allowing the possibility to have two different values for a same index. Thanks a lot again for your patience and for your explanations, that helped a lot (and sorry for being so slow to understand things...!) :-) – justdoit Apr 14 '17 at 15:58
  • can you repost the document you have in the question I'm in exactly the same place as you with the same questions – Moe Elsharif Jan 17 '23 at 13:33

1 Answers1

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The proposition states that if we have a countably family of functions

$$\{ f_n \mid f_n : \mathbb{N} \rightarrow \mathbb{N}, n \in \mathbb{N}\}$$

then we can always, by choosing a seed $c \in \mathbb{N}$, use the family to define a unique sequence of natural numbers $\{ a_k \}_{k \in \mathbb{N}}$ given by

  • $a_0 = c$
  • Otherwise, $a_{n++} = f_n(a_n)$

For an example of such a sequence, consider the family where $f_n(x) = x^2 + n$. Let $c = 2$; then we get the sequence

$$ \{ 2, 2^1+1, (2^1+1)^2+2, \ldots \} = \{ 2, 3, 11, 124, \ldots \}$$

The proof of uniqueness relies on Axiom 2.5; the property $P(n)$ is

$P(n)$: For all $n \in \mathbb{N}$, $a_n$ is uniquely defined.

$P(0)$ holds, since there is precisely one clause defining $a_0$.

Now we show that if $P(n)$ is true, then so is $P(n{++})$.

So we must show that $a_{n++}$ is uniquely defined. Axiom 2.3 tells us that we cannot have used the first clause, $a_0 = c$, to define $a_{n++}$, so we must have used the second clause, $a_{n++} = f_n(a_n)$, to define $a_{n++}$. But since $a_n$ is uniquely defined and $f_n$ is a function, the value $a_{n++}$ is also unique. This completes the proof.

Weishi Z
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Hans Hüttel
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