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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?

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There is a paper by Gitik, "All uncountable cardinals can be singular", Israel Joural of Mathematics, 35, pp 61--88, whose main theorem seems to be that ZF + "all alephs have countable cofinality" is consistent relative to a proper class of strongly compact cardinals. – Miha Habič Jun 10 '13 at 8:32
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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 ...")

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"So to have ..." this is not a proof, right? Anyway, the best known lower bound currently is: Daniel Busche and Ralf Schindler. The strength of choiceless patterns of singular and weakly compact cardinals, Ann. Pure Appl. Logic, 159 (2009), 198-248. They prove that if all uncountable $\aleph$-cardinals are singular, then there is forcing extension of an inner model, where $\mathsf{AD}$ holds in $L(\mathbb R)$. (In strength this is not too high: Just $\omega$ Woodin cardinals. We do expect much more to be true.) – Andrés E. Caicedo Jun 10 '13 at 14:19
@Andres: No, it's not a proof at all. It's based a bit about my intuition and the little I recall from "recent" papers that I was looking at some time ago. Later I might have some time to refine that "bound" (with references, of course). – Asaf Karagila Jun 10 '13 at 14:37
@Andres: After a bit of cheating and asking Gitik, he replied and said that he is unaware of any weaker principle which can prove this consistency result. – Asaf Karagila Jun 12 '13 at 12:59

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