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?

  • $\begingroup$ What if $\phi$ is $x=|P(\omega)|$? $\endgroup$ – Eric Wofsey Jun 2 '19 at 20:18
  • $\begingroup$ @EricWofsey, I've made an edit. Thanks $\endgroup$ – Zuhair Jun 3 '19 at 5:04
  • 2
    $\begingroup$ What does $M\models |P(\omega)|=x$ mean, if $x$ is an arbitrary set? $\endgroup$ – Eric Wofsey Jun 3 '19 at 5:14
  • $\begingroup$ @EricWofsey, I had the impression that any particular "substitution" of the variable $x$ would be seen by the model substituting the variable symbol $M$ as a constant. For example when $x$ is $\omega_1$, then the model satisfying $|P(\omega)|=\omega_1$ would so and so..... $\endgroup$ – Zuhair Jun 3 '19 at 11:06

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.

  • $\begingroup$ Of course, if one replaces model with "Boolean-valued model", then uncountability is no longer a problem. $\endgroup$ – Asaf Karagila Jun 3 '19 at 9:10
  • $\begingroup$ @AsafKaragila Quite right of course. $\endgroup$ – Noah Schweber Jun 3 '19 at 9:26
  • $\begingroup$ Also we need to add both assertions of countability of $M$ and of $N$. $\endgroup$ – Zuhair Jun 3 '19 at 11:00
  • $\begingroup$ I don't see $M \models |P(\omega)|=\aleph_{\alpha}$ mentioned in your statement? I think it should be added before the second comma in your statement $\endgroup$ – Zuhair Jun 3 '19 at 11:08
  • $\begingroup$ @Zuhair It doesn't need to be: $\alpha$ is an arbitrary element of $Ord^M$, whether $\aleph_\alpha=2^\omega$ in $M$ or not. So in particular we could add that hypothesis, but that would actually weaken the statement. $\endgroup$ – Noah Schweber Jun 3 '19 at 11:12

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