Is there any way to avoid using Axiom of Choice in proving this theorem?

Question

I asked for proof verification of a proof about nest of intervals here, where I appeal to a theorem:

Theorem: Let $a,b \in \mathbb R$ such that $a <b$ and $X := \{p \in \mathbb Q \mid a<p<b\}$. Then $|X| = \aleph_0$.

Here is my attempt that appeals to Axiom of Choice:

Lemma: For all $a,b \in \mathbb R$ such that $a <b$, there exists $p \in \mathbb Q$ such that $a<p<b$.

Since $X \subseteq \mathbb Q$, $|X| \le |\mathbb Q| = \aleph_0$. Thus $|X| \le \aleph_0$.

Let $\mathcal I$ be the collection of all intervals in $\mathbb R$. By Axiom of Choice and our lemma, there is a function $f:\mathcal I \to \mathbb Q$ such that $f(I) \in I$ for all $I \in \mathcal I$.

We define a function $g:\mathbb N \to X$ recursively by $g(0) = f((a,b))$ and $g(n+1) = f((g(0),b))$. It is easy to verify that $g$ is injective and thus $\aleph_0 = |\mathbb N| \le |X|$.

As a result, $|X| = \aleph_0$.

My questions:

1. Does my attempt contain logical gaps/errors?

2. Is there any way to avoid using Axiom of Choice in proving my theorem?

Answer 1

Of course the axiom of choice is not needed.

First of all, $\Bbb Q$ is countable. Just enumerate it, and choose the least element in the enumeration from each interval.

Secondly, you're asking if $X$, a subset of $\Bbb Q$ is infinite, since we already know that $\Bbb Q$ is countably infinite. That is all. And since $\Bbb Q$ is dense, the axiom of choice is not needed at all.

Answer 2

I think your proof is correct. You can certainly avoid choice, though. Use the lemma twice to get two distinct rationals $q,r$ in $(a,b)$. Now let $q_0=q$, $q_i=\frac12(q_{i-1}+r)$ for $i\geq 1$. We have $(q_i)_{i\geq 0}$ is an infinite sequence of distinct rationals in $(a,b)$.