$\operatorname{cf}(\aleph_\omega) = \aleph_0$, computed from the cofinal map $n \mapsto \aleph_n$

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

$C \;=\; \{\, \aleph_n : n \in \omega \,\} \;\subseteq\; \aleph_\omega$

(The successor cardinal $\kappa^{+}$, the alephs $\aleph_\alpha$, the beths $\beth_\alpha$, successor and limit cardinals, and the identifications $\aleph_0 = \omega$ and $\aleph_1 = \omega_1$) is cofinal in $\aleph_\omega$ (Cofinal subset of an ordinal) and satisfies $\lvert C\rvert = \aleph_0$, so

$\operatorname{cf}(\aleph_\omega) \;=\; \aleph_0 \;<\; \aleph_\omega$

(Cofinality $\operatorname{cf}(\alpha)$, and regular and singular cardinals) and $\aleph_\omega$ is singular.

Two things make the computation work, and both are visible in the display. The upper bound is the cofinal family itself: $\aleph_\omega$ is by definition the supremum of the $\aleph_n$, so a family indexed by $\omega$ already reaches it. The lower bound is structural: $\aleph_\omega$ is an infinite cardinal, hence a limit ordinal, so its cofinality is an infinite cardinal ($\operatorname{cf}(\alpha) \le \alpha$; $\operatorname{cf}(0) = 0$ and $\operatorname{cf}(\alpha + 1) = 1$; for a limit ordinal $\lambda$ the value $\operatorname{cf}(\lambda)$ is an infinite cardinal with $\operatorname{cf}(\operatorname{cf}(\lambda)) = \operatorname{cf}(\lambda)$, so it is regular; and every cofinal subset of $\lambda$ has cardinality at least $\operatorname{cf}(\lambda)$, a value that is attained) and cannot be smaller than $\aleph_0$.

Nothing is chosen, and that is the point. The cofinal family is the definable map $n \mapsto \aleph_n$, and Replacement makes its range a set. So singularity of $\aleph_\omega$ is a theorem of ZF, in contrast with the regularity of successor alephs, which is not ($\aleph_0$ is regular in ZF; assuming the Axiom of Choice every successor aleph $\aleph_{\alpha+1}$ is regular; $\operatorname{cf}(\aleph_\omega) = \aleph_0$, so $\aleph_\omega$ is singular, and under choice it is the least singular infinite cardinal).

The size of $\aleph_\omega$ 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 $\operatorname{cf}(\aleph_\lambda) \le \lvert \lambda \rvert$ 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, $\operatorname{cf}(\aleph_{\omega_1}) = \aleph_1$, so singular does not mean of countable cofinality computes a case where it is uncountable.

Why this is not a counterexample to anything about $2^{\aleph_0}$. It is the input to one: $\operatorname{cf}(2^{\aleph_0}) > \aleph_0$ (Assuming the Axiom of Choice: $\kappa < \kappa^{\operatorname{cf}(\kappa)}$ for every infinite cardinal $\kappa$, and $\operatorname{cf}(2^{\kappa}) > \kappa$; in particular $\operatorname{cf}(2^{\aleph_0}) > \aleph_0$) together with the value computed here is what refutes $2^{\aleph_0} = \aleph_\omega$ (FALSE: $2^{\aleph_0} = \aleph_\omega$).