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Assuming the axiom of choice it is very easy to see that $\aleph_1$ is a regular Joe of a successor cardinal. It is not very large in any way except the fact that it is the first uncountable cardinal.

If however we begin with a model of ZFC+Inaccessible, we can construct models of ZF in which $\aleph_1$ is somewhat inaccessible in the sense that $\aleph_1\nleq 2^{\aleph_0}$ If, on the other hand, we start with a model of ZF whic has this property then there exists an inner model with an inaccessible cardinal.

It can be that $\aleph_1$ is a measurable cardinal, you can even have that every subset of $\omega_1$ contains a club, or is non-stationary; and it is possible for $\aleph_1$ to have the tree property (I only know of models by Apter in which all successor cardinals have the tree property; but that would require a proper class of very large cardinals).

In general we say that $\aleph_1$ is P-large for a large cardinal property P, if it is consistent with ZF that $\aleph_1$ has property P, and from such model we can produce a model of ZFC+$\kappa>\aleph_0$ has property P.

Question: Is there a limit on how P-large can $\aleph_1$ be? (e.g. P can be tree property/$\kappa$-complete ultrafilter/supercompact measures/etc.) and are there properties P such that for $\aleph_1$ to have them we require more than ZFC+P?

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What do you mean by $\aleph_1 \not\leq 2^{\aleph_0}$ ? Assuming choice that's just wrong and inaccessibles don't change that. –  Johannes Hahn Feb 3 '12 at 16:37
    
@Johannes: Asaf is thinking about the Solovay model - en.wikipedia.org/wiki/Solovay_model –  François G. Dorais Feb 3 '12 at 16:47
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Is $\aleph_1$ a regular Joe or a regular John? Sorry bad joke. Under AD $\aleph_1$ has the strong partition property. –  Carlo Von Schnitzel Feb 4 '12 at 6:18
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If $\kappa = \aleph_1$ is measurable as witnessed by a measure $\mu$, then $\kappa$ is measurable in $L[\mu]$, which is a model of AC. –  Trevor Wilson Feb 28 '12 at 23:49
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Concerning "It can be that $\aleph_1$ is a measurable cardnal, namely that every subset of $\omega_1$ contains a club or is nonstationary": I disagree with "namely". Although both statements are consistent relative to large cardinals, they are not equivalent. I believe that $\aleph_1$ can be measurable without the club filter being ultra. In fact, I believe that this is what happens in Jech's model. –  Andreas Blass Aug 20 '13 at 19:14
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3 Answers 3

up vote 10 down vote accepted

-----edited to include corrections, thanks Joel and Tanmay-----

From a model of "ZFC + $large(\kappa)$" you can get a model of "ZF + $large(\kappa)$ + $\kappa=\omega_1$" if $large(\ )$ is a large cardinal property that is preserved under small forcing and can be written in the form

"for every set of ordinals $X$, there is a set $Y$ such that $\phi(X,Y)$ holds", for some upwards absolute formula $\phi$,

so for example weakly compact, Ramsey, measurable. The model you'd use is basically Jech's model for making $\omega_1$ measurable in "$\omega_1$ can be measurable".

For details see the comments below, Tanmay's answer and my thesis "Symmetric Models, Singular Cardinal Patterns, and Indiscernibles" Chapter 1, section 3.3. These large cardinals are there called "preserved under symmetric forcing".

For the other way around (a limit on how large can $large(\ )$ be), as Trevor Wilson said, measurable in ZF gives measurable in ZFC, and I guess similar arguments would work for large cardinal properties that are "preserved under symmetric forcing". I guess this is not a very good limit though and I can't come up with/remember something better right now. If I do I'll come back to answer.

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Don't you need to say something about the complexity of $\phi$? Do you mean that $\phi$ is $\Delta_0$? –  Joel David Hamkins Aug 20 '13 at 12:48
    
I have your thesis in my Dropbox quick access. But I will have to wait until I'm home this one time. Thanks for the reference! –  Asaf Karagila Aug 20 '13 at 14:51
    
Joel, right, this holds for $\phi$ a downwards absolute formula with two free variables, and I also forgot to include that $large(\ )$ has to be preserved under small forcing. The proof that this works uses that this sort of symmetric model satisfies the "approximation lemma", which basically means that all sets of ordinals of the symmetric model $V(G)$, for some generic $G$, are also in some $V[G']\subseteq V(G)$, where $G'$ is an initial part of $G$ (so it's small forcing). Asaf I'm really happy to hear someone uses it! –  Ioanna Aug 20 '13 at 19:11
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Could you edit your answer to include the appropriate restrictions on $\phi$? –  François G. Dorais Aug 21 '13 at 14:48
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It is also possible to make $\omega_1$ supercompact, using the same Jech construction. The only extra thing required here is to prove that fine measures generate fine measures in "small" forcing extensions. I did a small project (under the supervision of Benedikt Loewe) on this a few months back, and the write-up can be found here. The relevant result being Lemma 26.

That $\omega_1$ can be supercompact was first shown by Takeuti in 1970 in "A relativisation of axioms of strong infinity to $\omega_1$", where he also (independently of Jech) showed that $\omega_1$ can be measurable. I should add that his terminology is a bit different, but I do not have access to the paper right now, so I cannot tell you what he calls (what we now call) supercompact cardinals. I should also add that all of my terminology comes from Ioanna's thesis.

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Thanks for the link! I just remembered the conversation we had back then, and that you already then spotted the mistake with the "downwards" absoluteness. I'm so glad this notation/terminology was helpful to you. –  Ioanna Aug 23 '13 at 8:57
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This got too long for a comment: Ioanna, can't $\phi$ even be $\Sigma_1$? As I understand it, the main trick is that using the 'Approximation Lemma' in your thesis, you can show that any set of ordinals $X$ in the ($\mathsf{ZF+ \neg AC}$) model obtained by the Jech construction actually exists in some intermediate ($\mathsf{ZFC}$) submodel. These models are typically also forcing extensions by "small" partial orders, and so you can use the Levy-Solovay theorem for the large cardinal property to tell you that the cardinal is still `large', and so there is a set $Y$ such that $\phi(X,Y)$ holds, and then by upwards absoluteness you have that $\phi(X,Y)$ holds in the $\mathsf{ZF+\neg AC}$ model too.

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This is exactly the correct answer. Upwards absoluteness is, of course, what is needed here. –  Ioanna Aug 23 '13 at 8:42
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But since you have an existential quantifier "there is $Y$" right before $\phi$, does this change really matter? That is, if you insisted $\phi$ is $\Delta_0$ you could simply put the existential quantifier into $Y$ and have exactly the same properties. That is, I don't think you get any new large cardinal properties by going to $\Sigma_1$ rather than $\Delta_0$. –  Joel David Hamkins Aug 23 '13 at 11:47
    
I just used the term "upwards absolute" because this is what the proof needs and I'm unsure if all upwards absolute formulas are $\Sigma_1$. Do you know if they are? –  Ioanna Aug 27 '13 at 8:19
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