The choice principle $\text{AC}\_{\text{WO}}$ proves a large amount of cardinal arithmetic. It's well-known to imply DC, that successor cardinals are regular, and that for all $X$, there is $\lambda$ such that $\aleph(X)=\aleph^\*(X)=\lambda^+.$ Furthermore, we have the following:



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> ($\text{ZF + AC}\_{\text{WO}}$) If $\aleph(^{\text{cf}(\kappa)} \kappa) = \lambda^+,$ then $\text{cf}(\lambda)>\text{cf}(\kappa)$ or $\text{cf}(\lambda)=\omega.$
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Pf: Fix $\lambda < \aleph(^{\text{cf}(\kappa)} \kappa)$ such that $\omega\_1 \le \text{cf}(\lambda) \le \text{cf}(\kappa)$ and a cofinal sequence $\langle \alpha\_{\xi}: \xi<\text{cf}(\lambda) \rangle \subset \lambda.$ Notice that the domination ordering $\le^\*$ on $^{\text{cf}(\lambda)} \lambda$ is well-founded, since it has no descending sequences and DC holds. Any well-ordering $\prec$ of $\lambda$ embeds into $(^{\text{cf}(\lambda)} \lambda, \le^\*)$ by $\gamma \mapsto (\xi \mapsto \text{ot}({\gamma \downarrow\_{\prec}} \cap \alpha\_{\xi}, \prec)),$ so $\text{rk}(^{\text{cf}(\lambda)} \lambda, \le^\*) \ge \lambda^+.$


We thus have $\lambda^+ < \aleph^\*(^{\text{cf}(\lambda)} \lambda) = \aleph(^{\text{cf}(\lambda)} \lambda) \le \aleph(^{\text{cf}(\lambda) \cdot \text{cf}(\kappa)} \kappa) = \aleph(^{\text{cf}(\kappa)} \kappa).$ $\square$


Of course, AC proves this result without the uncountable cofinality hypothesis, which raises the question,



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> Does $\text{ZF + AC}\_{\text{WO}}$ prove that $\aleph(^{\text{cf}(\kappa)} \kappa) = \lambda^+$ for some $\lambda$ with $\text{cf}(\lambda) > \text{cf}(\kappa)?$
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Note that a failure of this implication has consistency strength at least a Woodin cardinal, because then $\lambda$ is a singular cardinal with countable cofinal sequence $A,$ and $\text{HOD}(A)$ computes $\lambda^+$ incorrectly.


Of particular interest is the simplest non-trivial case:



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> Does $\text{ZF + AC}\_{\text{WO}}$ prove $\aleph(\mathbb{R})=\Theta \neq \aleph\_{\omega+1}?$
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A proof from a stronger partition principle like PP or WPP would also be of interest.



Does $\text{AC}_{\text{WO}}$ prove $\Theta \neq \aleph_{\omega+1}?$
($\text{ZF + AC}\_{\text{WO}}$) For any cardinals $\kappa\_1, \kappa\_2,$ there is $\lambda$ such that $\aleph(^{\kappa\_2}\kappa\_1)=\lambda^+$ and $\text{cf}(\lambda)>\kappa\_2.$


Pf: Let $\lambda$ be such that $\aleph(^{\kappa\_2}\kappa\_1)=\lambda^+,$ and fix a cofinal sequence $\langle\gamma\_{\xi}: \xi<\text{cf}(\lambda) \rangle \subset \lambda.$ Choose injections $f\_{\alpha}: \alpha \rightarrow \lambda$ for $\alpha<\lambda^+.$ For such $\alpha,$ we recursively define $g\_{\alpha}: \text{cf}(\lambda) \rightarrow \lambda$ by setting $g\_{\alpha}(\xi) = \min(\lambda \setminus \{g\_{\beta}(\xi): \beta \in f\_{\alpha}^{-1}(\gamma\_{\xi})\}).$


Notice that $\alpha \mapsto g\_{\alpha}$ injects $\lambda^+$ into $^{\text{cf}(\lambda)} \lambda,$ so $\aleph(^{\kappa\_2} \kappa\_1)=\lambda^+ < \aleph(^{\text{cf}(\lambda)} \lambda) \le \aleph(^{\text{cf}(\lambda) \cdot \kappa\_2} \kappa\_1).$ Clearly $\text{cf}(\lambda)> \kappa\_2.$ $\square$


Corollary: $\text{AC}\_{\text{WO}}$ does prove $\Theta \neq \aleph\_{\omega+1}.$


