How to prove that a regular cardinal cannot be expressed as a union of sets with less cardinality?

Question

My question is about the following:

Using the Axiom of Choice show that:

If $\kappa\ge\omega$ is a regular cardinal, $\gamma\le\kappa$, and $\langle A_\alpha\mid\alpha\lt\gamma\rangle$ is a sequence of sets each of cardinality less than $\kappa$, then $|\bigcup_{\alpha \lt \gamma} A_\alpha| < \kappa$

I found an answer here: Regular cardinals and unions but I can't see where it uses the axiom of choice. Any help would be much appreciated!

Answer 1

First of all, the definition of regular cardinals does not depend on the axiom of choice, even in $\sf ZF$ regular cardinals are those which cannot be expressed as a "small union of small sets". The proof that every successor cardinal is regular, however, does use the axiom of choice, i.e. there are models of $\sf ZF+\lnot AC$ in which $\aleph_1$ is singular, despite being the successor of $\aleph_0$.

Secondly, you are using the axiom of choice when you choose a well-ordering for $A_\alpha$. The proof would usually go as follows:

Assume without loss of generality that $A_\alpha$'s are pairwise disjoint. Let $\beta_\alpha$ be an order type of a well-order of $A_\alpha$, then we can embed $\bigcup A_\alpha$ into $[0,\beta_0)\cup\bigcup[\beta_\alpha,\beta_{\alpha+1})=\delta$, where $\delta$ is some ordinal. By the regularity of $\kappa$ we have to have $\delta<\kappa$ and therefore $\bigcup A_\alpha$ has size $<\kappa$.

But we cannot always choose such sequence of well-orders. For example, the union of countably many sets of size $2$ might not be well-orderable at all. Or the countable union of countable sets might have cardinality $\aleph_1$, in which case $\aleph_1$ is not regular.