Without axiom of choice, can we prove that there are only countably many disjoint intervals in real line.

Question

With the axiom of choice, we can select a ration number $r_k$ in each interval $[a_k,b_k]$ and we can easily prove $Q$ is countable,so this complete the proof. However,without the axiom of choice, does each interval has a rational number which we can select?

Answer 1

You don't need the axiom of choice to pick a rational number in each interval. Since one can define an explicit bijection between $\mathbb Q$ and $\mathbb N$, you can decide once and for all to pick the rational in each interval that maps to the smallest natural. Then there's no choice (in the sense of the axiom of choice) going on.