21

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?

Asaf Karagila
  • 38,140
  • 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 03 '12 at 16:37
  • @Johannes: Asaf is thinking about the Solovay model - http://en.wikipedia.org/wiki/Solovay_model – François G. Dorais Feb 03 '12 at 16:47
  • Since $\aleph_1 \leq 2^{2^{\aleph_0}}$ in ZF, I have a hard time thinking of $\aleph_1 \nleq 2^{\aleph_0}$ as an inaccessibility statement. – François G. Dorais Feb 03 '12 at 16:52
  • Johannes: Thanks for the comment. Of course I mean that without the axiom of choice, as Francois said. I have edited the question to reflect that. – Asaf Karagila Feb 03 '12 at 18:45
  • Francois: However $\forall\mathfrak p(\mathfrak p<\kappa\Rightarrow\kappa\nleq2^\mathfrak p)$ is a strong limit property in ZFC which reflects in $\aleph_1$ in Solovay's model. – Asaf Karagila Feb 03 '12 at 18:47
  • 1
    Is $\aleph_1$ a regular Joe or a regular John? Sorry bad joke. Under AD $\aleph_1$ has the strong partition property. – Rachid Atmai Feb 04 '12 at 06:18
  • "It follows from ZF + axiom of determinacy that ω1 is measurable, and that every subset of ω1 contains or is disjoint from a closed and unbounded subset." - http://en.wikipedia.org/wiki/Measurable_cardinal#Measurable – Amit Kumar Gupta Feb 10 '12 at 08:37
  • @Amit: AD implies infinitely many measurable cardinals. In Jech's Set Theory he shows how from one measurable $\aleph_1$ can become measurable in a symmetric extension; however I have no idea about the converse. Does the measurability of $\aleph_1$ implies an inner model with a measurable cardinal? – Asaf Karagila Feb 10 '12 at 09:44
  • 6
    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
  • 4
    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
  • @Andreas, I think that I agree with you (although I didn't know that back then). Isn't it true that if the club filter is a measure, then it is normal? In that case Gitik constructed a model in which there is a measurable without a normal measure, collapsing it to be $\aleph_1$ should give a model of this sort as well. – Asaf Karagila Aug 21 '13 at 12:20
  • @AsafKaragila One has to be careful about normality of the club filter on $\aleph_1$ (whether it's ultra or not). It's provable without choice that any diagonal intersection of clubs is club, but we need to say the same for any diagonal intersection of sets from the club filter. So it seems as if we need choice in order to pick, for all the given sets in the club filter, subsets that are clubs. If I remmeber correctly, Kleinberg had a trick for getting around this, but I think it depended on a strong partition relation. – Andreas Blass Aug 21 '13 at 15:06
  • @Andreas: Interesting. I'll try to find out more (maybe the Bilinsky-Gitik paper is a good start) – Asaf Karagila Aug 21 '13 at 15:13
  • @AndreasBlass my knowledge of this is restricted to the 1970's, but Kleinberg showed that if $\gamma\rightarrow(\gamma)^\omega_\lambda$ for every $\gamma<\lambda$, then the $\omega$-club filter on $\gamma$ is a measure on $\gamma$. From there, he goes on to show that, Assuming the $\omega$-club filter on $\gamma$ is $\gamma$-additive, then $\gamma\rightarrow(\gamma)^\gamma$ implies that the measure on said filter is in fact normal. So, in particular, the strong partition relation on $\omega_1$ gives us that the $\omega$-club filter on $\omega_1$ is a normal measure on $\omega_1$. – Shehzad Ahmed Aug 21 '13 at 18:48
  • @AndreasBlass : The trick that Klenberg uses involves the fact that if $x$ is any unbounded subset of a regular uncountable cardinal $\gamma$, then ${\cup p : p\in [x]^\omega}$ is $\omega$-club in $\gamma$.

    Fix a $g$ regressive on the $\omega$-club filter for $\gamma$, let $A$ be the $\omega$-club set witnessing this. Let $G: [A]^\omega\rightarrow 2$ be given by $G(p)=0$ if and only if $g(\cup p)$ is less than the least member of $p$. Note then that there is a size $\gamma$ homogenous set $C$ for this partition, and note that $G''[C]^\omega = 0$...

     – Shehzad Ahmed Aug 21 '13 at 19:25
  • ... Kleinberg then uses ${\cup p : p\in [C]^\omega}$ to find a $\beta$ such that ${\alpha<\gamma : g(\alpha)=\beta}$. – Shehzad Ahmed Aug 21 '13 at 19:27

4 Answers4

21

-----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.

Ioanna
  • 1,252
  • 13
  • 15
  • 1
    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
  • 1
    Could you edit your answer to include the appropriate restrictions on $\phi$? – François G. Dorais Aug 21 '13 at 14:48
8

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.

tci
  • 662
  • 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 08:57
8

In the paper

" The relative consistency of a "large cardinal'' property for $\omega_1$, Rocky Mountain J. Math. 20 (1990), no. 1, 209–213."

it is shown than a model of ZFC with a huge cardinal can extend to a model of ZF in which $\omega_1$ is huge.

The link to the paper: http://projecteuclid.org/euclid.rmjm/1181073173

Asaf Karagila
  • 38,140
Mohammad Golshani
  • 31,966
  • Thank you Mohammad. I changed the link to the stable link (which you can find in the "Links and Identifiers" section), as it's usually better to use stable links, rather than copying the address from the address bar. – Asaf Karagila Oct 31 '13 at 09:42
2

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.

tci
  • 662
  • 1
    This is exactly the correct answer. Upwards absoluteness is, of course, what is needed here. – Ioanna Aug 23 '13 at 08:42
  • 2
    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 08:19