I read through the answers to previous questions regarding Beth numbers and was unable to find the answer to my question, so I hope this isn't a duplicate.
I am studying the definition of Beth numbers, specifically:
$\beth_0:=\aleph_0$
$\beth_{\alpha+1}:=2^{\beth_\alpha}$
$\beth_\lambda:=\displaystyle\sup_{\alpha<\lambda}\beth_\alpha$ for limit ordinals $\lambda$
How is the third line of the definition justified? How do we know that the power set operation can be applied an infinite number of times? Is there a way to show rigorously that $\beth_\omega$ for example, exists? Would I need a version of the Axiom of Choice?
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$\begingroup$To expand on my comment, recall the following version of the Recursion Theorem reads, where $\phi$ is a formula in the language of set theory:
Suppose that $\forall x \exists! y\phi(s,y)$, and define $G(s)$ to be the unique $y$ such that $\phi(s,y)$ (note the use Replacement). Then we can define a formula $\psi$ for which the following are provable.
$\forall x \exists! y \psi(x,y)$, so $\psi$ defines a functions $F$, where $F(x)$ is the $y$ such that $\psi(x,y)$.
$\forall\xi\in ON [F(\xi)=G(F\upharpoonright\xi)]$.
where $F\upharpoonright\xi$ means $F$ restricted to $\xi$. For a proof and some comments on the statement of the theorem, see Kunen's Foundations of Mathematics, page 45.
To apply this theorem, we need only specify $G$. We let:
$$ G(s)= \begin{cases} \aleph_0 & \text{if $s=0$}\\ 2^{s(\eta)} & \text{if $s$ is a function with dom$(s)=\eta+1$ a successor ordinal}\\ \sup_{\alpha<\lambda} s(\alpha) & \text{if $s$ is a function with dom$(s)=\lambda$ a limit ordinal}\\ 0 & \text{otherwise} \end{cases} $$
You can now apply (2) in the theorem to this $G$ to see that $F$ is the desired $\beth$ function.
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