Prove with induction, without using the axiom of countable choice that a sequence exists

81 Views Asked by At

The question I was given is as follows:

Assume a set $|A|=|\mathbb{N}|$ and that $h:A\rightarrow\mathbb{N}$ is a bijection.

Prove that exists a sequence $\left\langle h_{k}|0<k\in\mathbb{N}\right\rangle $ such that $h_{k}:A^{k}\rightarrow\mathbb{N}$ is a bijection.

This seems pretty straight forward but if i assume I accomplished this, then I think I can write a proof that doesn't use the axiom of countable choice, for the theorem, "The set of all finite-length sequences of natural numbers is countable." which I thought required the axiom of countable choice.

Am iI wrong by thinking that I can prove the first part without the axiom of countable choice?

Am I wrong by thinking that I can prove the second theorem without the axiom of countable choice using what I proved in the first part?

Or am I wrong by thinking you need the axiom of countable choice to prove the second part?