What is the value of $\beth_{\omega_1}^{\aleph_0}$?

Question

It is well known that $\beth_\omega^{\aleph_0} = \beth_{\omega+1}$. This follows since for strong limit $\kappa$, we have $\kappa^\kappa = \kappa^{\mathrm{cf}(\kappa)}.$

Question. To the extent that ZFC can pin it down, what is the value of $\beth_{\omega_1}^{\aleph_0}$?

Clearly its between $\beth_{\omega_1}$ and $\beth_{\omega_1+1}$ (again using $\kappa^\kappa = \kappa^{\mathrm{cf}(\kappa)}$ for the upper bound), but that is all I have been able to puzzle out.

I've actually been thinking about this for a while. At first I was thinking that perhaps since $\aleph_0 < \mathrm{cf}(\beth_{\omega_1}),$ we can safely deduce $\beth_{\omega_1}^{\aleph_0} = \beth_{\omega_1}.$ That is, I thought it might be a general principle that if $\nu < \mathrm{cf}(\kappa)$, then $\kappa^\nu = \kappa.$ Hence this question. Unfortunately, however, the conjecture is false (see Asaf's answer in the link), and I cannot think of anything else to try.

Answer 1

Here's another way to see that $\beth^\omega_{\omega_1}\leq \beth_{\omega_1}$.

First, note that by a simple induction we can show that for uncountable ordinals $\alpha$, $\beth_\alpha = |V_\alpha|$. In particular $\beth_{\omega_1} = |V_{\omega_1}| $. So every function from $\omega$ to $\beth_{\omega_1}$ can be coded as a function from $\omega$ to $V_{\omega_1}$. Since any such function will be a subset of $V_{\omega_1}$ of size $\omega$, it will have to exist at some $V_\alpha$ for $\alpha<\omega_1$. Thus, $\beth^\omega_{\omega_1}\leq \beth_{\omega_1}$.

Answer 2

I think I got it.

Definition. Whenever $\kappa$ is a cardinal number, write $\mathcal{B}(\kappa)$ for the boundful powerset of $\kappa$ i.e. the set of all bounded subsets of $\kappa$.

Proposition. If $\kappa$ is a strong limit cardinal, then $|\mathcal{B}(\kappa)| = \kappa$.

We need to show that $|\mathcal{B}(\kappa)| \leq \kappa.$ Now I think it is a general principle that $|\mathcal{B}(\kappa)| = 2^{<\kappa}.$ But for strong limit $\kappa$, we have $2^{<\kappa} = \kappa.$ Hence the result follows.

Corollary. If $ν<\mathrm{cf}(\kappa)$ and $\kappa$ is a strong limit cardinal, then $κ^ν=κ$.

This follows since $κ^ν$ is the cardinality of $[\kappa]^\nu$ (the set of all subsets of $\kappa$ with cardinality precisely $\nu$). By hypothesis, every element of $[\kappa]^\nu$ is bounded, hence $[\kappa]^\nu \subseteq \mathcal{B}(\kappa).$

Corollary to that. $$\beth_{\omega_1}^{\aleph_0} = \beth_{\omega_1}$$