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Suppose we don't believe the continuum hypothesis. Using Von Neumann cardinal assignment (so I guess we believe well-ordering?), is there any "familiar" ordinal number $\alpha$ such that, for non-tautological reasons, $\aleph_\alpha$ is provably larger than the cardinality of the continuum? I would hope not since it would seem pretty silly if something like $\alpha = \omega_0$ worked and we could say "well gee we can't prove that $c = \aleph_1$, but it's definitely one of $\aleph_1, \aleph_2, \ldots , \aleph_{73}, \ldots$". I (obviously) don't know jack squat about set theory, so this is really just idle curiosity. If a more precise question is desired I guess I would have to make it

For any countable ordinal $\alpha$ is the statement: $c < \aleph_\alpha$ independent of ZFC in the same sense as the continuum hypothesis?

assuming that even makes sense. Thanks!

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Mike: If you fix an ordinal $\alpha$, then it is consistent that ${\mathfrak c}>\aleph_\alpha$. More precisely, there is a (forcing) extension of the universe of sets with the same cardinals where the inequality holds.

If you begin with a model of GCH, then you can go to an extension where ${\mathfrak c}=\aleph_\alpha$ and no cardinals are changed, as long as $\alpha$ is not a limit ordinal of countable cofinality. For example, $\aleph_{\aleph_\omega}$ is not a valid size for the continuum. But it can be larger.

Here, the cofinality of the limit ordinal $\alpha$ is the smallest $\beta$ such that there is an unbounded function $f:\beta\to\alpha$. There is a result of König that says that $\kappa^\lambda>\kappa$ if $\lambda$ is the cofinality of $\kappa$. If $\kappa={\mathfrak c}$, this says that $\lambda>\omega=\aleph_0$, since ${\mathfrak c}=2^{\aleph_0}$ and $(2^{\aleph_0})^{\aleph_0}=2^{\aleph_0}$. Since $\aleph_{\aleph_\omega}$ has cofinality $\omega$, it cannot be ${\mathfrak c}$.

But this is the only restriction! The technique to prove this (forcing) was invented by Paul Cohen and literally transformed the field.

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    $\begingroup$ You could say that Paul Cohen forced a new outlook onto the world of set theory. :-) $\endgroup$
    – Asaf Karagila ♦
    Feb 4, 2011 at 9:54
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    $\begingroup$ Mike: The confusion here is that different 'versions' of ZFC can have different cardinals; for instance, you can't talk about 'the' $\omega_1$, but just the $\omega_1$ of your current model. How can this be? Well, think of $\omega_1$ as the set of all countable well-orderings; the word 'all' here is ill-defined, and it's possible for one model to know about countable well-orderings that another model doesn't. Saying that your extension has the same cardinals just means that you haven't introduced enough 'new' sets to change the cardinalities of old sets. $\endgroup$
    – Steven Stadnicki
    Feb 4, 2011 at 20:09
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    $\begingroup$ (cont) Likewise, the reason that $c$ can be equal to a given cardinal $\aleph_\alpha$ in one model and then larger (or smaller) than it in another is that the notion of 'the' set of all real numbers (or, if you prefer, all sets of integers) isn't necessarily well-defined, and again one model may know about sets of integers that another model doesn't. This is essentially how Cohen's proof works, by creating a whole lot of (necessarily non-constructible) new sets of integers, enough to make $c$ necessarily greater than $\aleph_1$. $\endgroup$
    – Steven Stadnicki
    Feb 4, 2011 at 20:14
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    $\begingroup$ @Mike: I am identifying cardinals with ordinals, i.e., a cardinal $\kappa$ is a special set, and any other set has size $\kappa$ if it is in bijection with the set $\kappa$. That $\kappa\ne\kappa'$ means that there is no bijection between $\kappa$ and $\kappa'$. It is perhaps easier now to argue in terms of models: Suppose that $M$ is a model of set theory, and that, in $M$, there is no bijection between $\kappa$ and $\kappa'$. It may still be the case that there is such a bijection $f$, but it does not belong to $M$. (Continues) $\endgroup$
    – Andrés E. Caicedo
    Feb 4, 2011 at 20:14
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    $\begingroup$ (cont.) Suppose we can extend $M$ to a model $M[f]$ that now has this bijection $f$. Then, in $M[f]$, at least one of $\kappa$ and $\kappa'$ is no longer a cardinal, since we now have a bijection between it and a smaller ordinal. The method of forcing allows us, precisely, to extend a model $M$ to a larger model $M[f]$ for some appropriate $f$. What I meant by "same cardinals" is that in $M[f]$ an ordinal is a cardinal iff it was a cardinal in $M$, i.e., I have not added any bijections as in the example above. I guess I should say why this is relevant. (Cont.) $\endgroup$
    – Andrés E. Caicedo
    Feb 4, 2011 at 20:17

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