# Are there axioms for $\mathrm{ZFC}$ that imply that $\aleph_1$ is very large?

A bit of philosophy: under the usual definition of the aleph numbers, ZFC proves the sentence "$\aleph_1$ is an ordinal." However, in some sense $\aleph_1$ isn't really an ordinal (in my opinion), because its position in the ordinals varies greatly between different models of $\mathrm{ZFC}$. Its not "fixed." (Of course, if you believe in a true set-theoretic universe, you will assert that its $\aleph_1$ is the true $\aleph_1$ and hence that there is a fixed, true ordinal corresponding to $\aleph_1$.)

Anyway, philosophy aside, I was wondering if there are axioms for $\mathrm{ZFC}$ that either directly assert, or otherwise imply that $\aleph_1$ is very large in the ordinals. Is this even possible? (I don't think the continuum hypothesis could reasonably be construed to be such an axiom, but please comment if you think otherwise.)

• ????? $\aleph_1$ is always the minimum uncountable ordinal. $2^{\aleph_1}$ can vary greatly. Dec 17 '14 at 14:16
• @Martín-BlasPérezPinilla note that $\beth_1$ can vary greatly on the scale of aleph numbers, while $\aleph_1$ can vary greatly on the scale of ordinal numbers. They're different scales. Dec 17 '14 at 14:18
• The aleph numbers are simply the cardinals. Dec 17 '14 at 14:28
• Yes, assuming the axiom of choice, they're precisely the infinite cardinals. (I'm not sure how I feel about this. A non-empty part of me wishes that the aleph numbers started at the cardinal $0$, so that $\aleph_n=n$ for all finite ordinals $n$.) Dec 17 '14 at 14:30
• I don't think $\mathsf{CH}$ is really much of a restriction. There are natural axioms that assert the existence of generic elementary embeddings with $\aleph_1$ as a critical point. Via (standard) reflection arguments, this translates into "largeness" properties of $\aleph_1$. Some of these axioms contradict $\mathsf{CH}$, while others imply it. Dec 17 '14 at 15:30

As Andres points out in the comments, $\sf CH$ is not the actual player here.

Consider the case that there is an inaccessible cardinal $\kappa$ which is itself a limit of inaccessible cardinals. By forcing we can arrange $\kappa$ to be $\omega_1$. This will certainly say that $\omega_1$ is large in the sense that it is the limit of inaccessible cardinals in an inner model. Moreover if $X$ is real number, then either we can "compute" the entire collapse from $X$, or that $\kappa$ is inaccessible in the model generated by $X$.

But that is not a "natural axiom". Here is one, however, $0^\#$ exists. It means that $\omega_1$ is a very large cardinal in $L$ and it is the limit of equally large large cardinals.

Similar "sharps" imply the same thing. Consequentially, any large cardinal axiom which imply the existence of sharps will say that $\omega_1$ is "quite" large. For example a measurable cardinal implies the existence of $0^\#$, and since these things are upwards absolute, then we can say that having "an inner model with a measurable" (which is a reasonable large cardinal axiom) implies that as well.

• Asaf, a silly question: but in the "sharp" case $\omega_1$ isn't the smallest uncountable cardinal in the model, right? Dec 17 '14 at 17:40
• Martín, if $V$ is the universe of set theory, there might be a class $L\subseteq V$ which contains all the ordinals and behaves like a universe of set theory itself. In fact, there is always one called $L$ which is Goedel's constructible universe. The ordinals which are cardinals in $V$ will remain cardinals in $L$, but it might be the case that some countable ordinal $\alpha$ was actually a cardinal in $L$. This means that $L$ didn't know that we can enumerate this ordinal. The assertion that $0^\#$ exists means that $V$ is "very far" from $L$, and it follows from that that $\omega_1$ of $V$ Dec 17 '14 at 17:54
• [...] is not the same as $\omega_1$ of $L$. But of course the existence of $0^\#$ is much much stronger, and it states all sort of crazy things, for example it means that $\aleph_\omega$ is a regular cardinal in $L$, so in particular the set $\{\omega_n\mid n\in\omega\}$ is not a set in $L$. Remember, though, that these cardinals are "calculated" in $V$, not in $L$. This would be similar to saying that $\sqrt[3]2$ is the unique solution of $x^3-2=0$, but in a smaller field it might not exist, and in a larger field it might not be unique anymore. Being a cardinal is relative to the universe. Dec 17 '14 at 17:56
• Yes, I know that being a cardinal isn't absolute. My point is that the variation of size of $\omega_1$ is totally "external". Dec 17 '14 at 18:03
• Well, here it's really the opposite. It's the fact that $\omega_1$ is really much larger (and in fact a large cardinal, in the technical sense of the word) in smaller universes. Dec 17 '14 at 18:11

By the downward Löwenheim - Skolem theorem there is a countable transiive model of every theory that holds in any set-sized transitive model and large cardinal axioms generally imply that there is a transitive set that satisfies all weaker large cardinal axioms. Thus the stronger the large cardinal axioms that hold in $$V$$ (or in sets) the larger $$\omega_1$$ since every ordinal in a countable transitive model is a countable ordinal, and thus less than $$\omega_1$$.