# Does $\text{AC}_{\text{WO}}$ prove $\Theta \neq \aleph_{\omega+1}?$

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:

($$\text{ZF + AC}_{\text{WO}}$$) If $$\aleph(^{\text{cf}(\kappa)} \kappa) = \lambda^+,$$ then $$\text{cf}(\lambda)>\text{cf}(\kappa)$$ or $$\text{cf}(\lambda)=\omega.$$

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,

Does $$\text{ZF + AC}_{\text{WO}}$$ prove that $$\aleph(^{\text{cf}(\kappa)} \kappa) = \lambda^+$$ for some $$\lambda$$ with $$\text{cf}(\lambda) > \text{cf}(\kappa)?$$

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:

Does $$\text{ZF + AC}_{\text{WO}}$$ prove $$\aleph(\mathbb{R})=\Theta \neq \aleph_{\omega+1}?$$

A proof from a stronger partition principle like PP or WPP would also be of interest.

($$\text{ZF + AC}_{\text{WO}}$$) 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 $$\Theta \neq \aleph_{\omega+1}.$$