I have 2 questions about a passage from Tao's Analysis I in which he defines recursive sequences. Questions have been asked about this passage here, here, here, and here; but I have different questions than those in those posts. Here are my questions, below which I will paste the passage followed by Tao's later formalization of the proposition he proves here:
Definitively, what is the "procedure" of which Tao speaks (mentioned in the first sentence of his informal proof)? It seems to be the infinite procedure "first defining $a_0$ to be some base value, $a_0:=c$ for some number $c$, and then by letting $a_1$ be some function of $a_0, a_1:=f_0\left(a_0\right), a_2$ be some function of $a_1, a_2:=f_1\left(a_1\right)$, and so forth", but such is never clarified- perhaps the "procedure" could be making the assumption that for all $n$, if $n=0, a_n=c$, and $a_{n++}=f_n(a_n)$, from which we deduce that each $a_n$ is uniquely defined?
If it is indeed that infinite procedure, how does it make sense to talk about an infinite procedure complementing and referencing its temporal aspects, as Tao does by saying "constantly redefined" in the penultimate paragraph, which makes it seem as if the conflicting definitions are happening at different points in time?
One consequence of the axioms is that we can now define sequences recursively. Suppose we want to build a sequence $a_0, a_1, a_2, \ldots$ of numbers by first defining $a_0$ to be some base value, e.g., $a_0:=c$ for some number $c$, and then by letting $a_1$ be some function of $a_0, a_1:=f_0\left(a_0\right), a_2$ be some function of $a_1, a_2:=f_1\left(a_1\right)$, and so forth. In general, we set $a_{n++}:=f_n\left(a_n\right)$ for some function $f_n$ from $\mathbf{N}$ to $\mathbf{N}$. By using all the axioms together we will now conclude that this procedure will give a single value to the sequence element $a_n$ for each natural number $n$. More precisely :
Proposition 2.1.16 (Recursive definitions). Suppose for each natural number n, we have some function $f_n: \mathbf{N} \rightarrow \mathbf{N}$ 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 $a_{n++}=f_n\left(a_n\right)$ 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 $a_{n++}:=f_n\left(a_n\right)$ 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 $a_{n++}:=f_n\left(a_n\right)$. (None of the other definitions $a_{m++}:=f_m\left(a_m\right)$ will redefine the value of $a_{n+1}$, because of Axiom 2.4.) This completes the induction, and so $a_n$ is defined for each natural number $n$, with a single value assigned to each $a_n$.
Note how all of the axioms had to be used here. In a system which had some sort of wrap-around, recursive definitions would not work because some elements of the sequence would constantly be redefined.
For context, Tao later formalizes proposition 2.1.16:
Exercise 3.5.12 This exercise will establish a rigorous version of Proposition 2.1.16 that avoids circularity (in particular, avoiding the use of any object that required Proposition 2.1.16 to construct).(i) Let $X$ be a set, let $f: \mathbf{N} \times X \rightarrow X$ be a function, and let $c$ be an element of $X$. Show that there exists a function $a: X \rightarrow X$ such that$$a(0)=c$$and$$a(n++)=f(n, a(n)) \text { for all } n \in \mathbf{N},$$
However, I haven't yet gotten to the part of the book where this formalization occurs (it is in a later chapter), so functions and sets haven't been rigorously defined for me- so I would appreciate an answer to focus more on the concepts from the text I posted above this formalization.