Is having models with ever increasing cardinality of the power set of $\omega$ is a theorem of ZFC

first-order-logicset-theory

I'll present this claim informally and try to write it formally as much as I can.

Statement: every model of $\sf ZFC$ that statisfies the statement that the power set of $\omega$ is equal to a specific $\aleph_{\alpha}$, there is another model of $\sf ZFC$ that satisfies the statement that the power set of $\omega$ is strictly bigger than $\aleph_{\alpha}$.

$\forall x \forall M [(M \models ZFC + |P(\omega)|=x) \to \\ \exists y > x \ \exists N (N \models ZFC + |P(\omega)|=y)] $

Question: is the above statement a theorem of ZFC?

Best Answer

Again, you're conflating theories and objects: if $\alpha$ is an arbitrary ordinal in some model $M$ there's no reason for "$2^\omega=\aleph_\alpha$" to be in any sense expressible by a first-order sentence in the language of set theory. So while we can ask whether that expression is satisfied in a structure containing $\alpha$, it doesn't make sense to ask whether ZFC proves it (or anything involving it).

I think the right way to ask the intuitive question you have in mind is:

Does ZFC prove that, whenever $M\models$ ZFC and $\alpha\in Ord^M$, there is some $N\models$ ZFC with $Ord^M=Ord^N$ and $N\models 2^\omega>\aleph_\alpha$?

If we add the assumption that $M$ is countable, this is easy: forcing over arbitrary countable models can be formalized inside ZFC, so just consider the forcing adding $\aleph_{\alpha+1}$-many Cohen reals.

For uncountable models, however, the statement is clearly false: we could have an $\omega$-model which literally contains every real, so we clearly can't push the continuum any higher.

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