What sort of large cardinal can $\aleph_1$ be without the axiom of choice?

Question

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?

Answer 1

-----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.

Answer 2

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.

Answer 3

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

Answer 4

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.