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This answer hints that certain kinds of extensions are only guaranteed to exist for countable models of ZFC. Why?

One intuitive reason i can think of is that the metatheory might not have enough new sets to give to the extension. For example, suppose you have a model of ZFC whose $\omega$ and $2^\omega$ agree with the metatheory. Morally* it feels like it should be possible to extend this model, keeping $\omega$ the same but adding new elements to $2^\omega$. But this is not possible just because the metatheory already gave all its elements of $2^\omega$ to the original model and doesn't have any more to give it. If it had more subsets of $\omega$, maybe we could extend out model then.

Is this one of the reasons for it? If yes, if we restrict to extensions that are small enough, does it still work well? (For example can we (end-)extend any $\omega$-model whose $2^\omega$ is smaller than the meta $2^\omega$ to a $\omega$-model with larger $2^\omega$?). Are there other reasons for it?

*Some people might disagree if they believe in a platonic $2^\omega$ or something like that, but i am a finitist/formalist, so my morals might be different.

Carla_
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    Yes, that's one of the reasons for it. If we're talking specifically about forcing extensions, the "necessary and sufficient condition" for the extension to exist is the existence of a generic filter for the forcing, which can only be proven in general for countable models. – spaceisdarkgreen Jul 07 '24 at 15:33
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    More specfically, this is the Rasiowa-Sikorski lemma. – tomasz Jul 07 '24 at 18:11

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This is more of a comment, but it's too long:

You mention morals - or more precisely, philosophical frameworks. However, the distinction between countable and uncountable models exists even for the formalist ... since the relevant claims are made inside $\mathsf{ZFC}$. As is standard, we just don't say "The following is provable in $\mathsf{ZFC}$" before every claim of fact. In particular, $\mathsf{ZFC}$ proves the statement in the linked answer, but does not prove the analogous statement with the countability hypothesis dropped. This is a bare fact about the formalism, and doesn't depend on philosophical stances.

Where philosophy enters the picture is in the choice of theory within which to prove results. E.g. one might reasonably argue for a "multiverse"-style theory, within which an appropriate version of the statement in the linked answer is provable even without the countability assumption. But again, this provability will be acceptable by all philosophical stripes.

Noah Schweber
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    I didn't say i reject the provability of that statement. I just find the reason i gave for it silly and it feels like there should be a natural way to change the question so that the answer of the new question is positive. Of course "natural" is subjective tho. I mentioned morals/philosophical standpoints because i feel like a platonist wouldn't find that reason silly. But i didn't meant anything rigorous by morals. – Carla_ Jul 07 '24 at 20:45