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In this answer, Asaf Karagila says that it is consistent with ZFC that $\frak c=\aleph_\alpha$ for any ordinal $\alpha$ without cofinality $\omega$. It is not clear to me what exactly this means.

If $\alpha$ were a real natural number, this would make sense because those are definable (in the sense of there being a formula defining them in every model of ZFC at the same time). Did they mean definable ordinals? Even for those, could it not happen that they are not of cofinality $\omega$ in the metatheory but are in any model of ZFC? (if ZFC is not omega-consistent)

spaceisdarkgreen
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Carla_
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1 Answers1

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This is indeed technically incorrect, although it is quite common. Here is one precise statement, which to allay foundational concerns I should clarify is provable in $\mathsf{ZFC}$ itself (or indeed much less):

Suppose $\mathcal{M}\models\mathsf{ZFC}$ is countable and $\alpha\in\mathit{Ord}^\mathcal{M}$ is an ordinal in the sense of $\mathcal{M}$ such that $\mathcal{M}\models \mathit{cf}(\alpha)>\omega\wedge \alpha\ge\mathfrak{c}$. Then there is an end-extension $\mathcal{N}$ of $\mathcal{M}$ such that

  • $\mathcal{M}$ and $\mathcal{N}$ have the same ordinals,

  • $\mathcal{M}$ and $\mathcal{N}$ have the same cofinalities, and

  • $\mathcal{N}\models\aleph_\alpha=\mathfrak{c}$.

(We can drop the $\alpha\ge\mathfrak{c}$ hypothesis at the cost of weakening the conclusion from "$\mathcal{N}$ is an end extension of $\mathcal{M}$" to "$\mathcal{N}$ is an end extension of an inner model of $\mathcal{M}$ (namely $L^\mathcal{M}$).")

Here, "end extension" means that not only is $\mathcal{M}$ a substructure of $\mathcal{N}$ but moreover for each $m\in\mathcal{M}$ we have $$\{a\in\mathcal{N}:\mathcal{N}\models a\in m\}=\{a\in\mathcal{M}: \mathcal{M}\models a\in m\}.$$ In particular, forcing extensions are end extensions, so the result above follows from the standard forcing arguments. Meanwhile, the restriction to countable structures is just to ensure that appropriate extensions exist in the first place; the downward Lowenheim-Skolem theorem for first-order logic says that this isn't too restrictive. That said, we can drop the countability restriction if we allow Boolean-valued models (basically, this lets us genuinely extend $V$). In particular, the following is a $\mathsf{ZFC}$-theorem:

For any uncountable-cofinality $\alpha\ge \mathfrak c$ there is a c.c.c. (hence cofinality-preserving) poset $\mathbb{P}$ such that $\Vdash_\mathbb{P} 2^{\aleph_0}=\aleph_\alpha$.

The expression $\Vdash_\mathbb{P} 2^{\aleph_0}=\aleph_\alpha$ can be thought of as a statement about poset combinatorics inside $V$, or as a statement about Boolean-valued extensions of $V$.

Noah Schweber
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