Can cardinality $\kappa$ exist where $\forall n\in\mathbb{N} \beth_n<\kappa$,$\kappa<|\bigcup_{n\in\mathbb{N}}\mathbb{S}_n|$,$|\mathbb{S}_n|=\beth_n$

Question

The Wikipedia article on Beth numbers defines $\beth_\alpha$ such that $\beth_{\alpha} =\begin{cases} |\mathbb{N}| & \text{if } \alpha=0 \\ 2^{\beth_{\alpha-1}} & \text{if } \alpha \text{ is a successor ordinal} \\ \sup\{\beth_\beta: \beta<\alpha\} & \text{if } \alpha \text{ is a limit ordinal} \end{cases}$

As it happens, for any ordinal $\alpha$, the statement $\exists \mathbb{S} | \beth_{\alpha}<|\mathbb{S}|<\beth_{\alpha+1}$ is independent of ZFC. In fact, according to an answer I received elsewhere, any statement or negation of that form is independent of any other under ZFC.

Now, to define what I'm calling the union-power sequence $\mathcal{UP}_n$ on a set $\mathbb{S}$ as

$$\mathcal{UP}_n(\mathbb{S}) = \begin{cases} \mathbb{S} & \text{if } n=0 \\ \mathcal{P}(\mathcal{UP}_{n-1}(\mathbb{S})) \cup \mathcal{UP}_{n-1}(\mathbb{S}) & \text{if } n \ge 1\end{cases}$$

With that definition, I can trivially prove that $\beth_n = |\mathcal{UP}_{n}(\mathbb{N})|$.

I would expect that behavior to continue when extending the sequence into infinite ordinals. The most logical way to handle $\omega$ is, in most cases, the limit behavior as $n \to \infty$. In this case, that's

$$\mathcal{UP}_{\omega}(\mathbb{S})=\bigcup_{n\in\mathbb{N}}\mathcal{UP}_{n}(\mathbb{S})$$

And if I extend to all ordinals, it looks like:

$$\mathcal{UP}_\alpha(\mathbb{S}) = \begin{cases} \mathbb{S} & \text{if } \alpha=0 \\ \mathcal{P}(\mathcal{UP}_{\alpha-1}(\mathbb{S})) \cup \mathcal{UP}_{\alpha-1}(\mathbb{S}) & \text{if } \alpha \text{ is a successor ordinal} \\ \bigcup_{\beta < \alpha}\mathcal{UP}_{\beta}(\mathbb{S}) & \text{if } \alpha \text{ is a limit ordinal}\end{cases}$$

Now, here's what throws me:

It would seem intuitively that $\beth_\omega=|\mathcal{UP}_{\omega}(\mathbb{S})|$. But according to the above definition, $\beth_\omega$ is just the smallest cardinality larger than any $\beth_n$ for any natural number $n$.

The definitions are equivalent, if and only if $\nexists \mathbb{S}\forall n \in \mathbb{N}:|\mathcal{UP}_n(\mathbb{N})|<|\mathbb{S}|<|\mathcal{UP}_\omega(\mathbb{N})|$.

That is, if there's no cardinality strictly greater than any $\beth_n$ for $n\in\mathbb{N}$ but strictly smaller than $|\mathcal{UP}_{\omega}(\mathbb{N})|$. Since if any intermediate cardinality does exist, then the supremum of $\{\beth_n: n\in\mathbb{N}\}$ would be the smallest such cardinality.

In fact, for $\beth_\alpha=|\mathcal{UP}_\alpha(\mathbb{N})|$ to hold in general for ordinal $\alpha$, you'd have to have $\nexists \mathbb{S}\forall \beta < \lambda:|\mathcal{UP}_\beta(\mathbb{N})|<|\mathbb{S}|<|\mathcal{UP}_\lambda(\mathbb{N})|$ for any limit ordinal $\lambda$.

This seems intuitively true, but it looks enough like the Continuum Hypothesis to give me pause.

So, to recap, the statement $$\nexists \mathbb{S}\forall \beta < \alpha:|\mathcal{UP}_\beta(\mathbb{N})|<|\mathbb{S}|<|\mathcal{UP}_\alpha(\mathbb{N})|$$ is $$\begin{cases} \alpha=0 & \text{False but not interesting; since } \nexists \beta<0 \text{(}\beta\text{ being an ordinal), it amounts to }\nexists\mathbb{S}\forall\emptyset:\phi\\ \alpha \text{ is a successor ordinal} & \text{Undecidable in ZFC, and the whole point of the Generalized Continuum Hypothesis} \\ \alpha \text{ is a limit ordinal} & \text{Unknown to me, and what I'm asking about}\end{cases}$$

Is that statement provable, disprovable, or undecidable for limit ordinals in ZFC? And if it's not provable, is there a more compact way to write cardinalities of the form $|\mathcal{UP}_\alpha(\mathbb{N})|$ that holds for infinite ordinal values of $\alpha$?

Or does Wikipedia have the definition wrong? The definition for limit ordinals listed there reads more like what I'd expect for a $\aleph_\alpha$ definition.

Answer 1

Consolidating comments into an answer to my own question:

Yes, that statement is true. Both the specific

$$\nexists\mathbb{S}\forall n\in\mathbb{N}:|\mathcal{UP}_n(\mathbb{N})|<|\mathbb{S}|<|\mathcal{UP}_\omega(\mathbb{N})|$$

and the more general

$$\forall \text { limit ordinal }\lambda\nexists\mathbb{S}\forall \alpha<\lambda:|\mathcal{UP}_\alpha(\mathbb{N})|<|\mathbb{S}|<|\mathcal{UP}_\lambda(\mathbb{N})|$$

To see this, note that

• Given a set $\mathbb{S}$, $\sup\bigl\{|s|:s\in\mathbb{S}\bigr\}\le|\bigcup\mathbb{S}|\le\Big(|\mathbb{S}|\times \sup\bigl\{|s|:s\in\mathbb{S}\bigr\}\Big)$
• Given 2 cardinalities $\kappa_a$ and $\kappa_b$, if one or both of them is infinite and neither is 0, then their product will be equal to whichever is larger. That is, $\kappa_a \times \kappa_b=\max\{\kappa_a, \kappa_b\}$

From there, we can derive the value of $|\mathcal{UP}_\omega(\mathbb{N})|$

• Note that $\mathcal{UP}_\omega(\mathbb{N})$ is the union set of the union-power sequence in the naturals, $\bigcup\bigl\{\mathcal{UP}_n(\mathbb{N}):n\in\mathbb{N}\bigr\}$
  • I'll be shortening $\bigl\{\mathcal{UP}_n(\mathbb{N}):n\in\mathbb{N}\bigr\}$ to $\mathbb{U}$
• $\mathbb{U}$ is constructed with schema replacement from $\mathbb{N}$, so $|\mathbb{U}|=\beth_0$
• $\Bigl\{|s|:s\in\mathbb{U}\Bigr\}$ is $\bigl\{\beth_n:n\in\mathbb{N}\bigr\}$. So by definition of supremum, $\sup\Bigl\{|s|:s\in\mathbb{U}\Bigr\}$ is the smallest cardinality that's greater than or equal to any $\beth_{n\in\mathbb{N}}$
• As both $|\mathbb{U}|$ and $\sup\Bigl\{|s|:s\in\mathbb{U}\Bigr\}$ are both infinite cardinalities, their product is whichever is larger. In this case, it has to be $\sup\Bigl\{|s|:s\in\mathbb{U}\Bigr\}$, since it's the supremum of a set that happens to contain $|\mathbb{U}|$
• Therefore, $\sup\Bigl\{|s|:s\in\mathbb{U}\Bigr\}\le|\bigcup\mathbb{U}|\le\sup\Bigl\{|s|:s\in\mathbb{U}\Bigr\}$
• Therefore, $\sup\Bigl\{|s|:s\in\mathbb{U}\Bigr\}=|\bigcup\mathbb{U}|=|\mathcal{UP}_\omega(\mathbb{N})| $
• By definition of supremum, nothing can be both larger than every member of a given set and smaller than that set's supremum. Consequently, $\nexists\mathbb{S}\forall n\in\mathbb{N}:|\mathcal{UP}_n(\mathbb{N})|<|\mathbb{S}|<|\mathcal{UP}_\omega(\mathbb{N})|$
• A similar proof can be constructed for the more general $\forall \text { limit ordinal }\lambda\nexists\mathbb{S}\forall \alpha<\lambda:|\mathcal{UP}_\alpha(\mathbb{N})|<|\mathbb{S}|<|\mathcal{UP}_\lambda(\mathbb{N})|$

The same line of reasoning does not apply to successor ordinals. So at any ordinal value of $\alpha$, the statement $\exists\mathbb{S}:|\mathcal{UP}_\alpha(\mathbb{N})|<|\mathbb{S}|<|\mathcal{UP}_{\alpha+1}|$ remains safely undecidable without additional axioms.