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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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    This is a definition by recursion; you need the recursion theorem, which does not use Choice. – 2017-01-31
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    The supremum of any set of cardinals/ordinals exists. Recursion gives you that each of the $\beth_n$ exist, so their supremum $\beth_\omega$ exists. For larger limit ordinals, you would use Transfinite Induction or Recursion. – 2017-01-31
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    You do need AC (Axiom of Choice) to show that $\beth (x)$ is always a (cardinal ) ordinal. Without AC you cannot even show that $\mathbb R$ can be well-ordered. – 2017-02-01
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    @user254665 It depends exactly what you mean by "$\beth(x)$". There is a natural choiceless definition: the supremum of the cardinals onto which $\mathcal{P}(x)$ surjects. This definition requires no choice to analyze. – 2017-02-01

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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.

  1. $\forall x \exists! y \psi(x,y)$, so $\psi$ defines a functions $F$, where $F(x)$ is the $y$ such that $\psi(x,y)$.

  2. $\forall\xi\in ON [F(\xi)=G(F\upharpoonright\xi)]$.

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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    +1. Note that we can sidestep the Recursion Theorem in the specific case the OP asks about: it is is a direct consequence of Replacement. There is a formula $\varphi(x, y)$ in the language of set theory such that for all $n\in\omega$, we have $\varphi(n, a)\iff a=\beth_n$; by Replacement, the range of this function exists, and the union of the range is exactly $\beth_\omega$. – 2017-01-31
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    @NoahSchweber that's actually pretty neat, thanks for pointing it out! – 2017-01-31
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    @NoahSchweber using that same logic could I build upon $\beth_\omega$ by applying Replacement again (similar to the construction of the ordinal numbers)? Example: $\forall n\in\omega$, we have $\phi_(n,a) \Leftrightarrow a=\beth_{\omega +n}$, which gives me $\beth_{\omega+\omega}$... and so on and so on. So long as I only use cardinals I've previously constructed? – 2017-02-01