I'm not an experienced person in proof.
This is a proposition in Amann/Escher's Analysis.
Let $X$ be a nonempty set and $\mathcal{a}\in X$. For each $n \in\mathbb{N}^{\times}$, let $V_{n}:X^{n} \to X$ be a function. Then there is a unique function $f:\mathbb{N} \to X$ with the following properties:
(i) $f(0)=a$
(ii) $f(n+1)=V_{n+1}(f(0),f(1),...,f(n))$, $\forall n\in \mathbb{N}.$
The proof of uniqueness is trivial. My question is about the existence.
In Amann's book, he first claims that, for each $n \in \mathbb{N}$, there is a function $f_{n}:\{0,1,...,n\}\to X$ such that
$(0\le k<n)$
$f_{n}(0)=a$
$f_{n}(k)=f_{k}(k)$
$f_{n}(k+1)=V_{k+1}(f_{n}(0),\dots,f_{n}(k))$
and then he uses induction to prove that $f_{n}$ exists. $\textbf{But why construct this function?}$
I try to find the answer in requirements for function $f$.
In my opinion, in order to prove the function $f$ with (i),(ii) is existence. It is necessary to connect $f$ and $V_{n+1}$. So I guess the role of $f_{n}$ as a bridge. But I have no idea how to make this work.
And in my knowledge stoage, to prove a function $f:\mathbb{N}\to X$ is existence. It is equal to prove $\forall n\in \mathbb N$, specifies exactly one element of $X$ or use the relation language. $\textbf{Is there something I'm missing here?}$
From $f(0)=a,f(n+1)=V_{n+1}(f(0),f(1),...,f(n))$,$n\in \mathbb{N}$
I find that $f(1)=V_{1}(f(0))=V_{1}(a),f(2)=V_{2}(f(0),f(1))=V_{2}(a,V_{1}(a))...$
But I don't know if this works for my question.
I also notice that in next part, $f$ was defined: $f:\mathbb{N}\to X, f(n):=\begin{cases}a,n=0 \\ \\f_{n}(n),n\in \mathbb{N}^{\times } \end{cases} $
But I still can't figure out what works.
What should I do now? I can not find a road to finish the proposition. My wording may be incorrect as well.