The existence of the sequence corresponding to some asymptotic sequence

Question

The following proof of the axiom of choice by induction is obviously false:
Let $(\Lambda)_{i=1, 2, \ldots}$ be an infinite sequence of nonempty sets. When $i=1$, self-evident. We will assume this statement holds for $n$, then for $n+1$, we can choose element $a_{n+1}$ from $\Lambda_{n+1}$ because $\Lambda_{n+1}\neq \emptyset$. Therefore, using the axiom of induction, we have proved the above statement holds for every $n \in \mathbf{N}$.
However, from this argument the axiom of choice doesn't follow because we have proved for only every finite $n \in \mathbf{N}$, though we have to choose from infinite sequence of sets $(\Lambda_i)$.

Next, we will prove by induction. the asymptotic formula, say, $a_1=1$ and $a_{n+1} = a_n+1$, actually defines a sequence. $n=1$ is OK and if it holds for $n$, $n+1$ is OK. Then, we can say that for all $n\in\mathbf{N}$, we can get the finite sequence $a_1, a_2, \ldots, a_n$. However, does this argument proves the existence of the infinite sequence corresponding to the asymptotic formula? I think the wrong proof of axiom of choice and the latter proof of infinite sequence are parallel, but if the latter proof is wrong, I have no idea how to prove the existence of the corresponding sequence. Someone please help me!

Answer 1

When you give a recursive definition of a sequence, then you define a sequence uniquely.

You're not making arbitrary choices, since $a_{n+1}$ is picked uniquely once you know the values of $a_0,\ldots,a_n$. So giving a recursive formula and a starting condition makes no appeals to the axiom of choice.

This is different from the fake proof you presented, since there you have to choose an arbitrary element from each set. So the axiom of choice had to be involved in order to glue these choices together into an infinite sequence.