$\mathrm{cf}({\mathrm{\aleph }}_{\omega })={\mathrm{\aleph }}_0$, computed from the cofinal map $n\mapsto {\mathrm{\aleph }}_n$

Work in ZF; no choice principle is used. The set

$C=\{{\mathrm{\aleph }}_n:n\in \omega \}\subseteq {\mathrm{\aleph }}_{\omega }$

(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 ${\mathrm{\aleph }}_1={\omega }_1$) is cofinal in ${\mathrm{\aleph }}_{\omega }$ (Cofinal subset of an ordinal) and satisfies $\mid C\mid ={\mathrm{\aleph }}_0$, so

$\mathrm{cf}({\mathrm{\aleph }}_{\omega })={\mathrm{\aleph }}_0<{\mathrm{\aleph }}_{\omega }$

(Cofinality $\mathrm{cf}(\alpha )$, and regular and singular cardinals) and ${\mathrm{\aleph }}_{\omega }$ is singular.

Two things make the computation work, and both are visible in the display. The upper bound is the cofinal family itself: ${\mathrm{\aleph }}_{\omega }$ is by definition the supremum of the ${\mathrm{\aleph }}_n$, so a family indexed by $\omega$ already reaches it. The lower bound is structural: ${\mathrm{\aleph }}_{\omega }$ is an infinite cardinal, hence a limit ordinal, so its cofinality is an infinite cardinal ($\mathrm{cf}(\alpha )\le \alpha$; $\mathrm{cf}(0)=0$ and $\mathrm{cf}(\alpha +1)=1$; for a limit ordinal $\lambda$ the value $\mathrm{cf}(\lambda )$ is an infinite cardinal with $\mathrm{cf}(\mathrm{cf}(\lambda ))=\mathrm{cf}(\lambda )$, so it is regular; and every cofinal subset of $\lambda$ has cardinality at least $\mathrm{cf}(\lambda )$, a value that is attained) and cannot be smaller than ${\mathrm{\aleph }}_0$.

Nothing is chosen, and that is the point. The cofinal family is the definable map $n\mapsto {\mathrm{\aleph }}_n$, and Replacement makes its range a set. So singularity of ${\mathrm{\aleph }}_{\omega }$ is a theorem of ZF, in contrast with the regularity of successor alephs, which is not (${\mathrm{\aleph }}_0$ 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).

The size of ${\mathrm{\aleph }}_{\omega }$ plays no role. Only the index $\omega$ is used: it is a limit ordinal reached from below by an $\omega$-indexed family, and the aleph operation is continuous at limits, so the same computation gives $\mathrm{cf}({\mathrm{\aleph }}_{\lambda })\le \mid \lambda \mid$ for any limit $\lambda$ by exactly the argument of steps 1.2 and 1.3. What that bound is worth depends on $\lambda$, and Assuming countable choice, $\mathrm{cf}({\mathrm{\aleph }}_{{\omega }_1})={\mathrm{\aleph }}_1$, so singular does not mean of countable cofinality computes a case where it is uncountable.

Why this is not a counterexample to anything about $2^{{\mathrm{\aleph }}_0}$. It is the input to one: $\mathrm{cf}(2^{{\mathrm{\aleph }}_0})>{\mathrm{\aleph }}_0$ (Assuming the Axiom of Choice: $\kappa <{\kappa }^{\mathrm{cf}(\kappa )}$ for every infinite cardinal $\kappa$, and $\mathrm{cf}(2^{\kappa })>\kappa$; in particular $\mathrm{cf}(2^{{\mathrm{\aleph }}_0})>{\mathrm{\aleph }}_0$) together with the value computed here is what refutes $2^{{\mathrm{\aleph }}_0}={\mathrm{\aleph }}_{\omega }$ (FALSE: $2^{{\mathrm{\aleph }}_0}={\mathrm{\aleph }}_{\omega }$).