FALSE: ${\mathrm{\aleph }}_{\alpha }$ is regular for every ordinal $\alpha$

FALSE. Every aleph is regular: $\mathrm{cf}({\mathrm{\aleph }}_{\alpha })={\mathrm{\aleph }}_{\alpha }$ for every ordinal $\alpha$ (Cofinality $\mathrm{cf}(\alpha )$, and regular and singular cardinals, The successor cardinal ${\kappa }^{+}$, the alephs ${\mathrm{\aleph }}_{\alpha }$, the beths ${\mathrm{\beth }}_{\alpha }$, successor and limit cardinals, and the identifications ${\mathrm{\aleph }}_0=\omega$ and ${\mathrm{\aleph }}_1={\omega }_1$).

The claim is plausible because the alephs a reader meets first are regular: ${\mathrm{\aleph }}_0$ is regular in ZF, and assuming the Axiom of Choice every successor aleph is regular (${\mathrm{\aleph }}_0$ is regular in ZF; assuming the Axiom of Choice every successor aleph ${\mathrm{\aleph }}_{\alpha +1}$ is regular; $\mathrm{cf}({\mathrm{\aleph }}_{\omega })={\mathrm{\aleph }}_0$, so ${\mathrm{\aleph }}_{\omega }$ is singular, and under choice it is the least singular infinite cardinal). It fails at the first aleph whose index is a limit ordinal, and the failure is a theorem of ZF requiring no choice principle at all.

Which alephs the theorem does certify. ${\mathrm{\aleph }}_0$, in ZF; and every ${\mathrm{\aleph }}_{\alpha +1}$, assuming the Axiom of Choice (${\mathrm{\aleph }}_0$ is regular in ZF; assuming the Axiom of Choice every successor aleph ${\mathrm{\aleph }}_{\alpha +1}$ is regular; $\mathrm{cf}({\mathrm{\aleph }}_{\omega })={\mathrm{\aleph }}_0$, so ${\mathrm{\aleph }}_{\omega }$ is singular, and under choice it is the least singular infinite cardinal). So the false claim is not wrong everywhere — it is wrong exactly where the index is a limit ordinal reached from below by a short family, and $\omega$ is the smallest such index.

Singularity here is about the index, not about the size. The cofinal family that witnesses $\mathrm{cf}({\mathrm{\aleph }}_{\omega })={\mathrm{\aleph }}_0$ is $n\mapsto {\mathrm{\aleph }}_n$, indexed by $\omega$ because the subscript $\omega$ is a limit of an $\omega$-sequence. Nothing about how large ${\mathrm{\aleph }}_{\omega }$ is enters the argument, and nothing is chosen, which is why clause (c) of ${\mathrm{\aleph }}_0$ is regular in ZF; assuming the Axiom of Choice every successor aleph ${\mathrm{\aleph }}_{\alpha +1}$ is regular; $\mathrm{cf}({\mathrm{\aleph }}_{\omega })={\mathrm{\aleph }}_0$, so ${\mathrm{\aleph }}_{\omega }$ is singular, and under choice it is the least singular infinite cardinal is choice free while clause (b) is not.

The claim is not repaired by assuming choice. Adding the Axiom of Choice certifies more alephs as regular, but it does not touch the witness above: the refutation is a theorem of ZF and remains one in ZFC.