$\omega_2$ is not a countable union of countable sets

Question

Without using axiom of choice, can we show that $\omega_2$ is not a countable union of countable sets? I know this cannot be done for $\omega_1$.

Answer 1

Let $\omega_2 = \bigcup_n A_n$ where $A_n$'s are countable and pairwise disjoint. Let $\alpha_n$ be the order type of $A_n$. Then $\bigcup_n A_n$ can be injected into $\alpha_1 + \alpha_2 + \dots$. Since each partial sum is $< \omega_1$, the whole sum is $\leq \omega_1$. Contradiction! Hopefully axiom of choice wasn't used here.

Answer 2

Yes, this can be proved without the axiom of choice.

Assume $\omega_2 = \cup_n E_n$ with $E_n$ countable. Then each $E_n$ is well-ordered, hence is isomorphic to a unique ordinal $\alpha_n < \omega_1$ via a unique isomorphism.

This makes it easy to construct a surjection $\omega \times \omega_1 \to \omega_2$, a contradiction since $|\omega \times \omega_1| = \aleph_1$.