Prove that $\omega_{\alpha+1}$ is regular

Question

Here, under the section 'Regular and Singular Cardinal', there is this sentence 'Assuming the Axiom of Choice, $\omega_{\alpha+1}$ is regular for each $\alpha$' . May I know how to prove this? Also, if we don't assume AC, does the statement hold?

Answer 1

There are several ways to define regularity. You can do it using order preserving functions; or you can do it using partitions. I like working with the partitions, for this proof.

$\kappa$ is regular if every partition of $\kappa$ into less than $\kappa$ parts, there is at least one part of size $\kappa$.

For example $\aleph_\omega$ is not regular, since we can partition it into parts of size $\aleph_n$ for each $n\in\omega$, the partition is small and all its parts are small.

Now using the axiom of choice, the union of $\aleph_\alpha$ sets of size $\aleph_\alpha$ has size $\aleph_\alpha$ (we use the axiom of choice to choose an enumeration for each set, if the enumeration is given then we don't need the axiom of choice). Therefore, if we partition $\aleph_{\alpha+1}$ into less than $\aleph_{\alpha+1}$ parts, we can assume that we partitioned it into $\aleph_\alpha$ of them, and if each part has size $<\aleph_{\alpha+1}$ then we might as well assume that each part has size $\aleph_\alpha$.

But then now taking union of all these parts we only covered $\aleph_\alpha$ elements, not $\aleph_{\alpha+1}$.

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As for the axiom of choice, yes. It is consistent that $\omega_1$ is the countable union of countable sets. In particular this means that it might not be regular without the axiom of choice. Assuming additional axioms to be true, Moti Gitik shown that it is consistent that every infinite cardinal is the countable union of smaller cardinals. So $\omega$ is the only [infinite] cardinal that is regular.