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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!

Asaf Karagila
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Mike F
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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.

Andrés E. Caicedo
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    You could say that Paul Cohen forced a new outlook onto the world of set theory. :-) – Asaf Karagila Feb 04 '11 at 09:54
  • Thanks for your answer! Though I'm afraid I'm not in a position to appreciate it any overly precise way... – Mike F Feb 04 '11 at 19:46
  • I guess this betrays that my accepting your answer was somewhat blind, but can I ask why the following is forbidden. The cardinality $c$ at least makes sense in ZFC so there is an ordinal $\alpha$ with $\aleph_\alpha = c$. Your first paragraph seems to suggest I should be able to find an extension of the universe of sets with the same cardinals where now $\aleph_\alpha > c$. You catch my drift? I'm not sure I understand what you mean by an extension with the same cardinals. – Mike F Feb 04 '11 at 19:57
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    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. – Steven Stadnicki Feb 04 '11 at 20:09
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    (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 $\aleph_1$. – Steven Stadnicki Feb 04 '11 at 20:14
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    @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) – Andrés E. Caicedo Feb 04 '11 at 20:14
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    (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.) – Andrés E. Caicedo Feb 04 '11 at 20:17
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    (cont.) The point is that I begin working on a model. Say, here ${\mathfrak c}=\omega_1$. I want to go to a model where ${\mathfrak c}\ge\omega_2$. For this, I add to $M$ a lot of reals that were not there before ($\omega_2$ of them), and form $M[f]$. If I don't do this carefully, in $M[f]$ I will have that ${\mathfrak c}\ge\omega_2$, for sure, but $\omega_2$ means here the "old one", the $\omega_2$ of $M$. Perhaps while adding reals by accident I added a bijection between the $\omega_2$ of $M$ and its $\omega_1$. Then perhaps in $M[f]$ we have ${\mathfrak c}=\omega_1$. (cont.) – Andrés E. Caicedo Feb 04 '11 at 20:20
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    (cont.) If, on the other hand, I am careful and make sure that "cardinals are the same", then in $M[f]$ I will have ${\mathfrak c}\ge\omega_2$, and $\omega_2$ is the "old one" but this is the same as the $\omega_2$ of $M[f]$, so really ${\mathfrak c}\ge\omega_2$, which is what we wanted. – Andrés E. Caicedo Feb 04 '11 at 20:22
  • (I agree this may be somewhat confusing at first. To understand forcing requires some understanding of models of set theory, and of what it means for a statement to be true in a model, as opposed to it being true "in the universe of sets".) – Andrés E. Caicedo Feb 04 '11 at 20:24
  • (And I see that Steven explained the same issue while I was typing. Sorry for the overlap.) – Andrés E. Caicedo Feb 04 '11 at 20:26
  • Andres: Not at all! I think the extra details on your answer are immensely useful, and I like it better than mine (which is a lot fuzzier on the specifics). – Steven Stadnicki Feb 04 '11 at 20:56
  • @Andres: Crash course in Cohen forcing over comments, eh? ;-) – Asaf Karagila Feb 06 '11 at 23:54
  • Andres, you write a sentence to the effect that: "If you fix an ordinal α, then it is consistent that $\beth_1>\aleph_α$." Is this literally true? I mean, what if $\alpha = \beth_1$? – goblin GONE Apr 03 '14 at 15:07
  • @user18921 Literally, what you have is this: Given any model $M$ of set theory, and any ordinal $\alpha$ of $M$, there is a forcing extension (essentially, another model of set theory) with the same ordinals and the same cardinals (meaning, if an ordinal $\beta$ was a cardinal of $M$, $\beta$ is still a cardinal of the extension, and any cardinal of the extension was one of $M$), and in this extension, $\beth_1>\aleph_\alpha$. If $\alpha=\beth_1$-of-$M$, all this means is that $\beth_1$-of-the-extension is strictly larger than $\beth_1$-of-$M$. – Andrés E. Caicedo Apr 03 '14 at 15:17
  • (@user18921 And it is not necessary to talk of models $M$ here, the discussion can be recast in a -- perhaps somewhat more cumbersome -- way that avoid this.) – Andrés E. Caicedo Apr 03 '14 at 15:19
  • @AndresCaicedo, any model, or should I be thinking of the standard ones? – goblin GONE Apr 03 '14 at 15:20
  • @user18921 Any model. Again, we can recast the whole discussion in a way that models are not even mentioned. The standard presentation of forcing uses countable transitive models, and this approach is perhaps the most intuitive, but the restriction is completely unnecessary. – Andrés E. Caicedo Apr 03 '14 at 15:21
  • Hmmm okay. I accept that you're correct, but it seems crazy though. Why doesn't this imply that $\beth_1 > \aleph_{\beth_1}$? – goblin GONE Apr 03 '14 at 15:25
  • Or rather, why doesn't it prove that $\beth_1 > \aleph_{\beth_1}$ is consistent with ZFC? – goblin GONE Apr 03 '14 at 15:43
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    @user18921 I do not see what is crazy about this. Anyway, the comment that starts "Literally" answers your question. The point is that what ordinals are cardinals is a matter of the existence of bijections between ordinals, and the ordinals do not change. The value of $\beth_1$ is a matter of the existence of a bijection between an ordinal and the set of reals (in the extension), and there may be new reals in the extension, so the set of reals in $M$ is not so relevant anymore. – Andrés E. Caicedo Apr 03 '14 at 16:04
  • @AndresCaicedo, okay I think I get it. Would be fair to say the following? Let $M$ denote the model we started with and $F$ the forcing extension. Then for any ordinal $\alpha$ of our model; $\aleph_\alpha^M = \aleph_\alpha^F,$ but there is no guarantee that $\beth_\alpha^M = \beth_\alpha^F$. Indeed, $\beth_\alpha^F$ could be strictly smaller then $\beth_\alpha^M$ (because we introduced a bunch of bijections) or it could be strictly larger (because we introduced a bunch of elements into $\mathcal{P}(\beth_0),$ so the old bijections aren't bijections any more.) – goblin GONE Apr 03 '14 at 16:53
  • Is that correct? I just want to make sure I'm not fooling myself here. – goblin GONE Apr 03 '14 at 16:55
  • Actually on second thought, I'm not sure that $\beth_\alpha^F$ can be strictly smaller than $\beth_\alpha^M,$ assuming the ordinals and cardinals of our models agree. – goblin GONE Apr 03 '14 at 17:20
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    @user18921 You are correct in saying that $\beth_\alpha^M$ and $\beth_\alpha^F$ could be different. The reason is that the former is computing the size of some power sets in $M$, and the latter is computing the size of the corresponding power sets in $F$. Since $F$ typically has more sets than $M$ did, what we expect is that these power sets are larger: If all ordinals and cardinals in $M$ and $F$ coincide, we necessarily have $\beth_\alpha^M\le\beth_\alpha^F$, and the inequality could be strict. – Andrés E. Caicedo Apr 03 '14 at 17:56
  • @AndresCaicedo, thanks, that makes perfect sense. – goblin GONE Apr 03 '14 at 17:58
  • @user18921 Now, there are forcing notions that for some $\alpha$, allow us find a $\beta<\beth_\alpha^M$ with a bijection between $\beth_\alpha^F$ and $\beta$, but when this happens, necessarily some cardinals of $M$ are no longer cardinals in $F$ (we added bijections witnessing this). – Andrés E. Caicedo Apr 03 '14 at 17:58