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

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?

• Yes, this is consistent. – Andrés E. Caicedo Jul 22 '16 at 12:21

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,

1. 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.

• For the OP: note that if $\beth_\alpha$ is not well-orderable, then $\beth_\beta$ is not well-orderable for any $\beta>\alpha$. So in fact $\neg AC$ is equivalent to "'most' $\beth$-numbers aren't well-orderable." – Noah Schweber Jul 22 '16 at 13:39
• @NoahSchweber Ahh thank you that is a very good point. I must say I didn't realize that though it's actually fairly obvious thanks to canonical inclusions and sub orders of well orders being well orders. – DRF Jul 22 '16 at 14:03
• @JacobWakem That's not true. Every aleph number is a generalized beth number, via $\beth_0(\kappa)=\kappa$, but "beth number" usually refers to numbers of the form $\beth_\alpha=\beth_\alpha(\omega)$. See en.wikipedia.org/wiki/Beth_number. – Noah Schweber Jul 22 '16 at 14:32
• @JacobWakem That is not true - in the absence of the axiom of choice, you need to make a distinction between cardinalities, or general cardinals, and cardinals in the sense of aleph numbers, or well-orderable cardinals. The "cardinality" of a non-well-orderable set will never be an aleph-number, and in the absence of the axiom of choice such sets may exist. Indeed the point of Asaf's answer is that every beth is an aleph IFF choice holds. I suggest you do some reading into set theory without the axiom of choice. – Noah Schweber Jul 22 '16 at 14:45
• You might start by looking at en.wikipedia.org/wiki/Aleph_number#Role_of_axiom_of_choice or math.stackexchange.com/questions/258510/…. – Noah Schweber Jul 22 '16 at 14:46