Can't find the demonstration of a theorem about recursion

Question

Here's the theorem :

Let $E$ be a set, $g$ a function from $E$ into $E$, and $a$ an element of $E$. There exists a unique function $f$ defined from $\mathbb{N}$ into $E$ such that $f(0)=a$ and $f(n+1)=g(f(n))$ for every natural integer $n$.

I'm just trying to find a proof about that theorem. I'm just a student; all I know is that this theorem is true and I would like to have a proof about it, but I really don't know how to achieve it.

Answer 1

$\def\nn{\mathbb{N}}$Consider the following procedure $f$ that requires an input from $\nn$:

  $f$ on input $n$ does the following:

    If $n = 0$:

      Output $a$.

    Otherwise:

      Output $g(f(n-1))$.

Note that $f(0) = a$. Also note that, given any $n \in \nn$, if $f(n)$ produces an output then $f(n+1)$ also does and furthermore $f(n+1) = g(f(n))$. Therefore by induction $f(n)$ produces an output for every $n \in \nn$, and $f(n+1) = g(f(n))$ for every $n \in \nn$. Thus $f$ is a function on $\nn$ satisfying the desired properties. Finally, it is easy to prove (by induction again) that any two functions satisfying the desired properties are the same.

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The above proof is based on intuitive concepts, but these are not available in some formal systems such as ZF set theory. In ZF a function is defined as a set of ordered pairs (intended to encode (input,output) pairs) such that no two have the same first entry (each input maps to only one output). To build the desired $f$ we must build approximations.

The first step is to let $P(k)$ be the assertion that there is a function $h$ with domain $\{0..k\}$ such that $h(0) = a$ and $h(n+1) = g(h(n))$ for every $n \in \nn_{<k}$. Then clearly $P(0)$ and $P(k) \to P(k+1)$ for any $k \in \nn$, by explicit construction. Therefore by induction $P(k)$ for every $k \in \nn$.

The second step is to show that any function witnessing $P(k)$ is unique for the given $k \in \nn$. Again this is by induction.

The third step is to let $(h_k)_{k\in\nn}$ be a sequence of functions such that $h_k$ witnesses $P(k)$ for every $k \in \nn$, and show that $h_i$ and $h_j$ agree on $\{0..i\}$ for every $i,j \in \nn$ such that $i < j$. Again, this is by induction ('on $j$').

The fourth step is to let $f = \bigcup_{k\in\nn} h_k$, and to check that it is a function (which follows from the pairwise agreement) and that it satisfies the desired properties (which follows from the fact that $h_k$ witnesses $P(k)$ for every $k \in \nn$.

Longwinded? I agree. But that is just how it is. Essentially no shortcut.