Without AC is there a relationship between $\beth$ and $\aleph$ numbers?

Question

Assuming AC we know that all $\beth_\alpha$'s will be $\aleph_\beta$ for some $\beta$ since they can be well ordered.

Can anything interesting be said about their relationship without AC? Is it possibly consistent that with the exception of $\beth_0$ none of the $\beth$'s can be well-ordered?

Answer 1

In $\sf ZF$ the two are equivalent:

1. The axiom of choice.
2. Every $\beth$ number is an $\aleph$ number.

You can prove this by going through the following equivalent statement,

3. The power set of an ordinal can be well-ordered.

So if the axiom of choice fails, we know that there is some $\alpha$ such that $\beth_\alpha$ cannot be well-ordered. Interestingly, the least such $\alpha$ can be a limit cardinal, and in fact it can be $\omega$.

Other than that we cannot say much. We know it is consistent that $\aleph_1$ is incomparable with $\beth_1$, for example, in which case there is no $\beth$ number (other than $\beth_0$, that is) which can be well-ordered.