# Does the Continuum Hypothesis hold for sets of a certain rank?

Hi I hope this question is accurate. Gödel and Cohen could show that the Continuum Hypothesis (CH) is independent from ZFC using models in which CH holds, and fails respectively. My question now is:

If we take a large enough part of the universe, say $V_{\omega_{\omega}}$ , does CH hold in it? And why can't we go on to argue like this:

If $V_{\omega_{\omega}} \models CH$ then there is a bijection between $\omega_1$ and $2^{\aleph_0}$, and as $V_{\omega_{\omega}}$ is transitive this is 'really' a bijection thus CH holds in ZFC.

If $V_{\omega_{\omega}} \models \lnot CH$ then, as any possible bijection between $\omega_1$ and $2^{\omega}$ is already an element of $V_{\omega_{\omega}}$, CH is inside ZFC refuteable?

This are maybe crude argumentations but I can't answer that by myself.

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You mean $\omega_1$ and $2^{\aleph_0}$ –  Asaf Karagila Sep 5 '10 at 17:54

The Continuum Hypothesis, viewed as the assertion that every subset of $P(\omega)$ is either countable or bijective with $P(\omega)$, is expressible already in $V_{\omega+2}$, since that structure has the full $P(\omega)$ and all subsets of it, as well as all functions between such subsets (one should use a flat pairing function for ordered pairs, which doesn't require one to increase rank for pairs). Thus, CH holds if and only if it holds in any (or all) $V_\alpha$ for $\alpha\geq\omega+2$.
Note also that CH is not expressible in $V_{\omega+1}$, since two models of set theory can have the same reals and hence the same $V_{\omega+1}$, but still disagree about CH. For example, if you start with a model of $\neg$CH, and then force to collapse the continuum to $\omega_1$, then you will recover CH while preserving $V_{\omega+1}$. –  Joel David Hamkins Sep 5 '10 at 18:37
But if CH holds iff it holds in a model (i.e. $V_{\omega +2}$), why can't we argue that CH cannot be independent from ZFC by this argument: $V_{\omega +2} models CH or non-CH - hence ZFC proves or refutes CH. – user8996 Sep 5 '10 at 19:05 Some models of ZFC have a$V_{\omega+2}$that satisfies CH, and others have their$V_{\omega+2}$satisfying$\neg$CH. Not all models of ZFC agree on$V_{\omega+2}$or even on$V_{\omega+1}$or even$V_\omega$, for that matter. – Joel David Hamkins Sep 5 '10 at 20:00 Ah now everything makes sense. Thank you – user8996 Sep 5 '10 at 20:08 Adding to the previous answers: Note that$V_{\aleph_\omega}$is not a model of ZFC. Hence, what holds or not in$V_{\aleph_\omega}$does not formally tell you what follows from ZFC and what doesn't. This issue aside, as mentioned above,$V_{\aleph_\omega}$knows whether or not CH holds in the real world. What Gödel did to show the consistency of CH was to construct a "narrow" class$L$inside a model of ZFC (resp. ZF if you also want to also prove the consistency of AC with ZF) such that$L$satisfies ZFC+CH. The point here is that even if CH fails in the large model of set theory,$L$contains so few subsets of$\omega$that CH actually holds (roughly). So, L is narrow, while$V_{\aleph_\omega}$is short and wide (it contains all set up to a certain rank). A short wide structure (as long it is long enough) is correct about CH, a narrow structure might not be. - The outcome of CH is already known at very low rank, before rank$\omega+\omega$. Indeed, the set bijections between subsets of$\mathcal{P}(\omega)$are already available in$V_{\omega+\omega}$, so the question whether there are sets of intermediate cardinality between$\aleph_0$and$2^{\aleph_0}$is already known there. (Also, note that there is a natural copy of$\omega_1$inside$V_{\omega+\omega}$, namely the isomorphism classes of wellorderings of$\omega\$.)