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Recursion Theorem:

Let $A$ be a set, $a\in A$, and $f \colon A\to A$ a mapping. Then there exists a unique mapping $g \colon \Bbb N\to A$ such that

1. $g(0)=a$

2. $g(n+1)=f(g(n))$

I formalize a proof based on the comments of Noah Schweber at here and here .

Fix $a\in A$. Let $Y=\{(n,x)\in\mathbb{N}\times A\mid \text{ there exist a sequence of length}$ $n$, in which first term is $a$, last term is $x$, and each term after the first is $f$ of the previous term$\}$.

Now we prove by induction: for each $n\in \mathbb{N}$, there exists a unique $y\in A$ such that $(n,y)\in\mathbb{N}\times A$.

For $n=0\implies$ only $y=a$ such that $(0,a)\in Y$. Assume that for $n=k$ there exists a unique $y\in A$ such that $(k,y)\in Y\implies$ for $n=k+1$, there exits a unique $f(y)$ such that $(k+1,f(y))\in Y$.

$\implies$ For all $n\in \mathbb{N}$, there exists a unique $y\in A$ such that $(n,y)\in Y$

$\implies$ Set $Y$ defines a function $g \colon \Bbb N\to A$ such that

1. $g(0)=a$

2. $g(n+1)=f(g(n))$

Please check my proof above! Many thanks!

Akira
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  • Seems perfectly fine to me. You may note that not only does $Y$ define a function $g$ -- it is exactly this function (or the graph thereof -- depending on your definition of 'function'). – Stefan Mesken Feb 24 '18 at 10:08
  • Hi @StefanMesken, how do we know that "there exist a sequence of length $n$, in which first term is $a$, last term is $x$, and each term after the first is $f$ of the previous term" is a well-defined statement under Zermelo–Fraenkel set theory? – Akira Feb 24 '18 at 10:17
  • You can formalize it as follows: $\exists s \subseteq \mathbb N \times A \wedge \forall n \in \mathrm{dom}(s) \exists ! a \in A \colon (n,s) \in s \wedge 0 \in \mathrm{dom}(s) \wedge (0,a) \in s \wedge \forall n \in \mathrm{dom}(s) \forall m < n \colon m \in \mathrm{dom}(s) \wedge n + 1 \in \mathrm{dom}(s) \implies (n+1, f(s(n)) \in s \wedge \neg \exists n < m (m \in \mathrm{dom}(s) \implies s(n) = x)$. – Stefan Mesken Feb 24 '18 at 10:27
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    Mind you that I don't encourage you to use the formalization. You should aim to become sufficiently comfortable with these notions that you know how to formalize it without the need of actually doing it. Overly formalized mathematics is almost always a disservice to both the author and the reader. – Stefan Mesken Feb 24 '18 at 10:28
  • Your formalization seems to confuse me: for all $(n,x)$, $n$ is the length of the sequence $s\implies n$ is the fixed number. But the way you used $n$ in the expressions makes me feel like $n$ is a free variable. Please explain the meaning of $\neg \exists n < m (m \in \mathrm{dom}(s) \implies s(n) = x$ ! Furthermore, I think you meant $(n,a)\in s$, not $(n,s)\in s$. – Akira Feb 26 '18 at 03:17
  • Yes, I meant $(n,a) \in s$ and $n$. The above formula says $s$ is a subset of $\mathbb N \times A$, $s$ is a function, $s(0) = a$, $s$ satisfies $f(s(n)) = s(n+1)$ whenever $n+1 \in \mathrm{dom}(s)$ and if $m$ is the maximal element of $\mathrm{dom}(s)$ then $s(m) = x$. (I now realize that you need to add that there is a maximal element in the domain of $s$ -- otherwise $s$ might be an infinite function.) – Stefan Mesken Feb 26 '18 at 07:54

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