Question related to ordinal number without using Axiom of Choice.

Question

Can we proof this result without using Axiom of Choice :- $$A\cap \alpha=\emptyset \,\,\,\, \mbox{and}\, \, \, A\times \alpha \sim A\cup \alpha$$ then there is an $A^{'} \subset A$ such that $\alpha \sim A^{'}$ or $B\subset \alpha$ such that $A\sim B$.

Answer 1

Yes, this is Tarski's lemma with which he proved that $\kappa^2=\kappa$ implies the axiom of choice.

Suppose that $f\colon A\cup\alpha\to A\times\alpha$ is a bijection. Either for every $a\in A$ there is some $\beta<\alpha$ such that $f(\beta)=(a,\gamma)$, in which case consider the map $a\mapsto\min\{\beta\mid\exists\gamma\ f(\beta)=(a,\gamma)\}$, then by the fact that $f$ is a bijection this map is injective from $A$ into $\alpha$.

If this is not the case, then there is some $a$ such that $f^{-1}\Big(\{a\}\times\alpha\Big)\subseteq A$ and this defines an injection from $\alpha$ into $A$.