In ZF, does there exist an ordinal of provably uncountable cofinality?

Question

Question is in the title. In ZFC, one can prove that $\aleph_{\alpha+1}$ is regular, so there is a large source of cardinals with uncountable cofinality, but in ZF, it is consistent that ${\rm cf}(\aleph_1)=\aleph_0$, and most conceivable limit alephs also have cofinality $\aleph_0$. I recall reading in a book that it is "unknown" if there are any ordinals that are provably of uncountable cofinality in ZF, so this is really a reference request for progress on this problem. Is this a "provably unprovable" problem? Are there any large cardinal hypotheses (other than AC) that shed some light on the problem?

Answer 1

No. As Miha quotes, it is consistent (relative to some very very large cardinals) that no initial ordinal (read: $\aleph$ number) has an uncountable cofinality. Since the cofinality of an ordinal is always an initial ordinal, this finishes the proof.

Note that very large cardinals are necessary. If $\operatorname{cf}(\omega_1)=\operatorname{cf}(\omega_2)=\omega$ then there is an inner model with a Woodin cardinal. So to have all the cardinals with countable cofinality you have to expect some proper class of very large cardinals.

Gitik proved this from a proper class of strongly compact cardinals, which is quite a large assumption. (Note, however that "proper class of ..." is quite scary, but still weaker from something like "inaccessible cardinal which is a limit of ...")