${\mathrm{\aleph }}_0$ 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

Let $\mathrm{cf}$, regular and singular be as in Cofinality $\mathrm{cf}(\alpha )$, and regular and singular cardinals, and let ${\mathrm{\aleph }}_{\alpha }$ be as in 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 ${\mathrm{\aleph }}_1={\omega }_1$. Then:

(a) In ZF. $\mathrm{cf}({\mathrm{\aleph }}_0)={\mathrm{\aleph }}_0$, so ${\mathrm{\aleph }}_0$ is regular.

(b) Assuming the Axiom of Choice (The Axiom of Choice). ${\mathrm{\aleph }}_{\alpha +1}$ is regular for every ordinal $\alpha$.

(c) In ZF. $\mathrm{cf}({\mathrm{\aleph }}_{\omega })={\mathrm{\aleph }}_0$, and ${\mathrm{\aleph }}_0<{\mathrm{\aleph }}_{\omega }$, so ${\mathrm{\aleph }}_{\omega }$ is singular.

(d) Assuming the Axiom of Choice. Every infinite cardinal below ${\mathrm{\aleph }}_{\omega }$ is regular, so ${\mathrm{\aleph }}_{\omega }$ is the least singular infinite cardinal.

Clause (b) is where the Axiom of Choice becomes indispensable, and the hypothesis is not decoration. The proof spends it once, to select an injection $g(\xi )\to {\mathrm{\aleph }}_{\alpha }$ for each $\xi$ below the cofinality, and there is no canonical such family to fall back on: the sets $g(\xi )$ are ordinals, but the injections are not determined by them. Clauses (a), (c) and the classification half of (d) are choice free.

Why regularity of ${\mathrm{\aleph }}_1$ is not a theorem of ZF. The proof of clause (b) selects one injection for each $\xi$ below the cofinality, and that selection is the entire content of the choice hypothesis: a union of countably many countable sets is not provably countable in ZF, and the same phenomenon is what would otherwise force $\mathrm{cf}({\mathrm{\aleph }}_1)={\mathrm{\aleph }}_1$. The choice ledger at the end of this page records how far this can fail.

Why ${\mathrm{\aleph }}_{\omega }$ is singular for a completely different reason. Nothing is chosen in clause (c): the map $n\mapsto {\mathrm{\aleph }}_n$ is definable, and it is short and cofinal simply because the index $\omega$ is a limit ordinal reached from below in $\omega$ steps. Singularity of ${\mathrm{\aleph }}_{\omega }$ is therefore a fact about the index, not about the size, and clause (c) holds in ZF.

What "least singular" means and what it does not. Clause (d) locates ${\mathrm{\aleph }}_{\omega }$ among the alephs: everything below it is regular, under choice. It says nothing about which cardinals above ${\mathrm{\aleph }}_{\omega }$ are singular, and it says nothing about $2^{{\mathrm{\aleph }}_0}$, whose position in the aleph hierarchy is not determined by anything on this page. What is determined is a constraint on that position, and it is Assuming the Axiom of Choice: $\kappa <{\kappa }^{\mathrm{cf}(\kappa )}$ for every infinite cardinal $\kappa$, and $\mathrm{cf}(2^{\kappa })>\kappa$; in particular $\mathrm{cf}(2^{{\mathrm{\aleph }}_0})>{\mathrm{\aleph }}_0$.