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Is there is limit to what $2^{\aleph_0}$ can be in a countable transitive model?

How large can be the value of the continuum in a countable transitive (standard) model of ZFC? For instance, if we wanted to consider the case where $2^{\aleph_0} = \aleph_{\omega_1}$, where would the $\aleph_1$ smaller cardinals come from, after all, the model is only countable?

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migrated from mathoverflow.net Oct 18 '15 at 17:33

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Assuming there exists any transitive model of ZFC, there exists a countable submodel by Lowenheim-Skolem. This countable submodel may not be transitive, but it is well-founded, so by Mostowski collapse it is isomorphic to a countable transitive model $M$.

Using forcing, you can then construct generic extensions of $M$ (which are also countable transitive models) in which $2^{\aleph_0}$ can be "arbitrarily large". More precisely, let us suppose for simplicity that $M$ satisfies GCH (this can be arranged by a class forcing, or by replacing $M$ with $L^M$). Then if $\kappa\in M$ is any ordinal such that $$M\vDash\text{"$\kappa$ is a cardinal of uncountable cofinality"},$$ there is a generic extension $M[G]$ of $M$ which has the same cardinals as $M$ and which satisfies $M[G]\vDash 2^{\aleph_0}=\kappa$. In particular, if (say) $M\vDash \kappa=\aleph_{\omega_1}$, then $M[G]$ is a countable transitive model in which $2^{\aleph_0}=\aleph_{\omega_1}$.

Note, however, that this does not mean that there actually are uncountably many cardinals in $M[G]$ below $2^{\aleph_0}$, any more than the fact that $M[G]\vDash 2^{\aleph_0}>\aleph_0$ means $M[G]$ actually contains uncountably many real numbers. What it means is that $M[G]$ thinks there are uncountably many cardinals less than what it thinks is $2^{\aleph_0}$. From the perspective of the outside universe, all of these "cardinals" below $2^{\aleph_0}$ are actually just countable ordinals (as is the cardinal $M$ thinks is $2^{\aleph_0}$), and there are only countably many of them. However, $M[G]$ doesn't know this, because no bijection from these sets to $\omega$ is an element of $M[G]$.

If you are still confused about this, you may find this Wikipedia page or the answers to this question helpful.

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  • $\begingroup$ Thks for the references. I "get" that M[G] lacks many bijections the outside world knows of, hence assigns the role of "cardinals" below $2^{\aleph_0}$ to various countable ordinals. Here's what still bothers me: Suppose outsiders examine those (countably many) countable ordinals (i.e the M[G] cardinals) and determine their sup (in the outside world), call it beta - (still a countable ordinal). Then we say to folks in M[G] world, "Point to $aleph_{beta}" and they respond with (say) aleph_{omega_1}. Does our "gotcha" question fail because we lack explicit simpler notation to express beta? $\endgroup$ – RedRover Oct 19 '15 at 4:04
  • $\begingroup$ Right, there can be no way to define $\beta$ such that the following two properties hold simultaneously: $M[G]$ thinks $\beta$ is the same thing as the full universe does (our definition is "absolute" for $M[G]$), and we can prove in ZFC that $\beta$ is countable. (For instance, $\omega_1$ is a definable ordinal which fails the first criterion, and $\omega_1^L$ is a definable ordinal which fails the second criterion.) If we could give such a definition of $\beta$, then since $M[G]$ is a model of ZFC, it would know that $\beta$ is countable. $\endgroup$ – Eric Wofsey Oct 19 '15 at 4:13
  • $\begingroup$ So, we might be able to "point out" $\beta$ to $M[G]$, but not in a way that is strong enough to let $M[G]$ tell that $\beta$ has to be countable. Or we might not even be able to point it out at all--there's no reason in general to expect that $\beta$ is even definable in the language of set theory without parameters. $\endgroup$ – Eric Wofsey Oct 19 '15 at 4:22
  • $\begingroup$ Yes, OK - I appreciate the follow-up comments, that clarifies things. $\endgroup$ – RedRover Oct 19 '15 at 13:38

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