The first infinite cardinal, $\aleph_0$, has many large cardinal properties (or would have many large cardinal properties if not deliberately excluded). For example, if you do not impose uncountability as part of the definition, then $\aleph_0$ would be the first inaccessible cardinal, the first weakly compact cardinal, the first measureable cardinal, and the first strongly compact cardinal. This is not universally true ($\aleph_0$ is not a Mahlo cardinal), so I am wondering how widespread of a phenomenon is this. Which large cardinal properties are satisfied by $\aleph_0$, and which are not?

There is a philosophical position I have seen argued, that the set-theoretic universe should be uniform, in that if something happens at $\aleph_0$, then it should happen again. I have seen it specifically used to argue for the existence of an inaccessible cardinal, for example. The same argument can be made to work for weakly compact, measurable, and strongly compact cardinals. Are these the only large cardinal notions where it can be made to work? (Trivially, the same argument shows that there's a second inaccessible, a second measurable, etc., but when does the argument lead to more substantial jump?)

EDIT: Amit Kumar Gupta has given a terrific summary of what holds for individual large cardinals. Taking the philosophical argument seriously, this means that there's a kind of break in the large cardinal hierarchy. If you believe this argument for large cardinals, then it will lead you to believe in stuff like Ramsey cardinals, ineffable cardinals, etc. (since measurable cardinals have all those properties), but this argument seems to peter out after a countable number of strongly compact cardinals. This doesn't seem to be of interest in current set-theoretical research, but I still find it pretty interesting.

  • 3
    $\begingroup$ Even measurability and strong compactness are out if defined in terms of elementary embeddings. The point is that definitions that are equivalent for uncountable cardinals may be inequivalent for $\aleph_0$. $\endgroup$ Nov 9, 2010 at 16:17
  • $\begingroup$ But if I knew the answer, I wouldn't be asking the question, would I? Many of the large cardinal axioms have alternate characterizations -- honestly I thought that was part of the case for their mathematical interest. The language characterization of weakly compact and strongly compact cardinals applies to $\aleph_0$, as does the ultrafilter characterization of measurable and strongly compact. I don't know enough about large cardinals to know if the cardinals I mentioned are unusual that regard, or typical. $\endgroup$
    – arsmath
    Nov 9, 2010 at 16:45
  • $\begingroup$ So for example, supercompactness has a characterization in terms of normal measures. Being a Vopenka cardinal has a characterization in terms of families of binary relations. $\endgroup$
    – arsmath
    Nov 9, 2010 at 16:52
  • 9
    $\begingroup$ It's slightly amusing that 0, 1, and 2 have some large cardinal properties that 3, 4, 5, ... lack. 2 is not a sum of fewer than 2 numbers, each of which is smaller than 2. Similarly for 0 and 1. Maybe the paradox dissolves when you realize that "large" means in effect large by comparison to smaller cardinals. $\endgroup$ Nov 9, 2010 at 17:25
  • 1
    $\begingroup$ I just mean to use an ultrapower $j:V\to M$ by an ultrafilter on $\omega$. The critical point is $\omega$, since the predecessors of $\omega$ are not isomorphic to those of $j(\omega)$ by $j$, and this is sufficient for measurability: $\kappa$ is measurable if ($\kappa$ is uncountable and) there is an embedding $j:V\to M$ such that $\kappa$ is least such that the respective predecessors of $\kappa$ and $j(\kappa)$ are not isomorphic by $j$. You don't need a $\kappa$-th member in $M$ to get an induced measure. (My remark about well-foundedness below $\omega$ was a trivial observation...) $\endgroup$ Nov 23, 2010 at 16:33

2 Answers 2


This probably isn't the kind of thing anyone just knows off hand, so anyone who's going to answer the question is just going to look at a list of large cardinal axioms and their definitions, and try to see which ones are satisfied by $\aleph _0$ and which aren't. You could've probably done this just as well as I could have, but I decided I'd do it just for the heck of it.

First of all, this doesn't cover all large cardinal axioms. Second, many large cardinal axioms have different formulations which end up being equivalent for uncountable cardinals, or perhaps inaccessible cardinals, but may end up inequivalent in the context of $\aleph _0$. So even for the large cardinals that I'll look at, I might not look at all possible formulations.

  1. weakly inaccessible - yes, obviously
  2. inaccessible - yes, obviously
  3. Mahlo - no, since the only finite "inaccessibles" are 0, 1, and 2 as noted by Michael Hardy in the comments, and the only stationary subsets of $\omega$ are the cofinite ones
  4. weakly compact:
    • in the sense of the Weak Compactness Theorem - yes, by the Compactness Theorem
    • in the sense of being inaccessible and having the tree property - yes, by Konig's Lemma
    • in the sense of $\Pi ^1 _1$-indescribability - no, it's not even $\Pi ^0 _2$-indescribable as witnessed by the sentence $\forall x \exists y (x \in y)$
  5. indescribable - no, since it's not even $\Pi ^0 _2$-indescribable
  6. Jonsson - no, the algebra $(\omega, n \mapsto n \dot{-} 1)$ has no proper infinite subalgebra
  7. Ramsey - no, the function $F : [\omega ]^{< \omega} \to \omega$ defined by $F(x) = 1$ if $|x| \in x$ and $0$ otherwise has no infinite homogeneous set
  8. measurable:
    • in the sense of ultrafilters - yes, by Zorn's Lemma, and because filters are $\omega$-complete by definition, i.e. closed under finite intersections
    • in the sense of elementary embeddings - no, obviously
  9. strong - no, obviously (taking the elementary embedding definition)
  10. Woodin - ditto
  11. strongly compact:
    • in the sense of the Compactness Theorem - yes, by the Compactness Theorem
    • in the sense of complete ultrafilters - yes, as in the case of measurables
    • in the sense of fine measures - yes, by Zorn's Lemma, and because filters are $\omega$-complete by definition
  12. supercompact:
    • in the sense of normal measures - no, if $x \subset \lambda$ is finite and $X$ is the collection of all finite subsets of $\lambda$ which contain $x$, then the function $f : X \to \lambda$ defined by $f(y) = \max (y)$ is regressive, but for any $Y \subset X$, if $f$ is constant on $Y$ with value $\alpha$, then Y avoids the collection of finite subsets of $\lambda$ which contain $\{ \alpha + 1\}$ and hence Y cannot belong to any normal measure on $P_{\omega }(\lambda)$
    • in the sense of elementary embeddings - no, obviously
  13. Vopenka - no, take models of the empty language of different (finite) sizes
  14. huge - no, obviously (taking the elementary embedding definition)
  • 1
    $\begingroup$ Thanks. I half-expected that there was a succinct characterization in terms of some general property somewhere in the literature. "Everything that happens for $\aleph_0$ happens again" has the flavor of a reflection principle, and there's a big literature on reflection principles, so I found it plausible that it had a tidy answer. $\endgroup$
    – arsmath
    Nov 10, 2010 at 8:13
  • $\begingroup$ (+1) Interesting summary, Amit! I enjoyed reading your answer! :-) $\endgroup$ Jun 4, 2018 at 8:07

There is a sense in which rank-into-rank axioms could be considered generilzations of $\omega$. Woodin discussed this in his paper Suitable Extender Models $I$, given the following conjecture, which I denote $(^*)$.

$(^*)$ Assume $V=\text{Ultimate}-L$. Then there is some $j: V_{\lambda+1}\prec V_{\lambda+1}$ if and only if $L(P(\lambda))\nvDash AC$.

In the same vein is the following conjecture, described by Vincenzo Dimonte in $I0$ and rank-into-rank axioms.

Question 11.2: Is it true that $\text{Ultimate}-L\vDash I0(\lambda)$ if and only if $\text{Ultimate}-L\vDash L(V_{\lambda+1})\nvDash AC$?

If either one of these were true, then it would be a shocking affirmation of the existence of $I1$ and $I0$ cardinals respectively (Assuming the proof does not show that there are none). The reason for this is that if $V=\text{Ultimate}-L$, then there is a proper class of Woodin cardinals and so in particular $L(\mathbb R)=L(P(\omega))=L(V_{\omega+1})\vDash AD$ and so $L(P(\omega))=L(V_{\omega+1})\nvDash AC$.

This corresponds with the sense in which it is not the critical point $\kappa$ of $j: V_{\lambda+1}\prec V_{\lambda+1}$ which is important, but the $\lambda$ itself. Note that $\lambda$ is a strong limit of cofinality $\omega$, and also $\omega$ is a strong limit of cofinality $\omega$, and finally a theorem of Shelah is that if $\lambda$ is a strong limit of uncountable cofinality, then $L(P(\lambda))\vDash AC$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.