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

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?

• What if $\phi$ is $x=|P(\omega)|$? – Eric Wofsey Jun 2 '19 at 20:18
• @EricWofsey, I've made an edit. Thanks – Zuhair Jun 3 '19 at 5:04
• What does $M\models |P(\omega)|=x$ mean, if $x$ is an arbitrary set? – Eric Wofsey Jun 3 '19 at 5:14
• @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..... – 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.

• Of course, if one replaces model with "Boolean-valued model", then uncountability is no longer a problem. – Asaf Karagila Jun 3 '19 at 9:10
• @AsafKaragila Quite right of course. – Noah Schweber Jun 3 '19 at 9:26
• Also we need to add both assertions of countability of $M$ and of $N$. – Zuhair Jun 3 '19 at 11:00
• 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 – Zuhair Jun 3 '19 at 11:08
• @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. – Noah Schweber Jun 3 '19 at 11:12