Assuming countable choice, $\mathrm{cf}({\mathrm{\aleph }}_{{\omega }_1})={\mathrm{\aleph }}_1$, so singular does not mean of countable cofinality

Assume the Axiom of Countable Choice ${\mathrm{A}\mathrm{C}}_{\omega }$ (The Axiom of Countable Choice (${\mathrm{A}\mathrm{C}}_{\omega }$)). Then

$\mathrm{cf}({\mathrm{\aleph }}_{{\omega }_1})={\mathrm{\aleph }}_1<{\mathrm{\aleph }}_{{\omega }_1},$

so ${\mathrm{\aleph }}_{{\omega }_1}$ is singular (Cofinality $\mathrm{cf}(\alpha )$, and regular and singular cardinals) and its cofinality is uncountable (Finite, countably infinite, countable, uncountable).

This separates two conditions that the first singular example runs together. A singular cardinal is one reachable from below by a strictly shorter family; it need not be reachable by a countable one. Here the reaching family has length ${\omega }_1$ and no shorter one will do, and what rules out a shorter one is the boundedness theorem for ${\omega }_1$ (Assuming countable choice: every at most countable subset of ${\omega }_1$ is bounded below ${\omega }_1$, so no at most countable subset of ${\omega }_1$ is cofinal in it, and a supremum of at most countably many at most countable ordinals is at most countable), which is where ${\mathrm{A}\mathrm{C}}_{\omega }$ is spent.

Why countable choice appears, and where exactly. It is used once, at [L4]: without it, ${\omega }_1$ can consistently be the supremum of an $\omega$-indexed family of countable ordinals, and then the argument of step 1.2 collapses. That dependence is inherited, not introduced here — the published boundedness theorem carries the same hypothesis, and states so in its own title.

What "singular" does and does not mean. Singular says only $\mathrm{cf}(\kappa )\mathrm{\ne }\kappa$. The singular cardinal computed in $\mathrm{cf}({\mathrm{\aleph }}_{\omega })={\mathrm{\aleph }}_0$, computed from the cofinal map $n\mapsto {\mathrm{\aleph }}_n$ has countable cofinality, and a reader who meets only that example may take the two conditions to be the same. They are not: here the cofinality is ${\mathrm{\aleph }}_1$, uncountable, while the cardinal is still singular because ${\mathrm{\aleph }}_1$ is far below ${\mathrm{\aleph }}_{{\omega }_1}$.

The pattern behind both computations. For a limit ordinal $\lambda$ the family $\alpha \mapsto {\mathrm{\aleph }}_{\alpha }$ restricted to $\lambda$ is cofinal in ${\mathrm{\aleph }}_{\lambda }$, so $\mathrm{cf}({\mathrm{\aleph }}_{\lambda })\le \mid \lambda \mid$ always. The work is entirely in the lower bound, and it is a statement about $\lambda$ rather than about ${\mathrm{\aleph }}_{\lambda }$: it asks how short a family can be and still reach $\lambda$.