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, namely that every subset of $\omega_1$ contains a club, or is nonstationary; 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?
