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My understanding is that large cardinals are ordered by "consistency strength", but how does this correlate with their size (cardinality)?

More specifically, are there any systematic results on the lines of: If A and B are two types of large cardinals such that Cons(ZFC + Type A exists) => Cons( ZFC + Type B exists) THEN Cardinality of smallest Type A cardinal >= Cardinality of smallest Type B cardinal

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  • $\begingroup$ I recall some result that the least supercompact is smaller than the least measurable if both exist, despite consistency strength going other way around. $\endgroup$
    – Wojowu
    Commented Sep 16, 2015 at 5:21
  • $\begingroup$ @Wojowu That's not true, as you wrote, if $\kappa$ is supercompact, then there are $\kappa$-many measurables below it. $\endgroup$
    – Mohammad Golshani
    Commented Sep 16, 2015 at 6:28
  • $\begingroup$ @Wojowu: You might have meant the result I cite in my answer. $\endgroup$
    – Asaf Karagila ♦
    Commented Sep 16, 2015 at 6:39
  • $\begingroup$ Note that the structure that is embodied in a large cardinal definition may be quite delicate, and the consistency of that structure is the issue, not how big the underlying cardinal is. $\endgroup$
    – David Roberts ♦
    Commented Sep 16, 2015 at 7:01
  • $\begingroup$ Here is a related MathOverflow question comparing consistency and implication strength orders with each other. It could be of your interest as well. $\endgroup$
    – Morteza Azad
    Commented Aug 29, 2017 at 10:18

4 Answers 4

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I may note that a cardinal of type $A$ may has more consistency strength of a cardinal of type $B$, while the smallest cardinal of type $A$ is smaller than the least cardinal of type $B$ (assuming cardinals of both types exist). For example:

The consistency of a huge cardinal implies the consistency of a supercompact cardinal, nevertheless, the least huge cardinal is smaller than the least supercompact cardinal (assuming both exist).

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Similar to what Mohammad writes, but slightly different, Magidor found the identity crisis of strongly compact cardinals:

It is consistent that the least strongly compact cardinal is the least measurable cardinal, and it is consistent that the least strongly compact cardinal is the least supercompact cardinal.

This is despite the fact strongly compact cardinals sit far above measurable cardinals in consistency strength; and as Mohammad points out in the comments, a supercompact has many measurable cardinals below it.

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Typically, large cardinals with stronger consistency strength are larger, but there is an important general exception:
If a large cardinal axiom is $Σ_{k+1}$, then typically the least example is larger than the least example for a $Σ_k$ axiom regardless of the consistency strength. This is because $Σ_{k+1}$ large cardinal axioms typically imply that the cardinal is a $Σ_k$ elementary substructure of $V$ (and thus larger than the least example satisfying a $Σ_k$ axiom). However, a consistency-wise stronger $Σ_k$ axiom typically implies existence of $V_κ$ that satisfies a consistency-wise weaker $Σ_{k+1}$ axiom.

For example, the least strong cardinal ($Σ^V_3$ existence axiom) is larger than the least Woodin cardinal ($Σ^V_2$ existence axiom), which in turn is larger than the least cardinal strong up to an inaccessible ($Σ^V_2$ existence axiom), which in turn implies existence of $V_κ$ satisfying "ZFC + there is a strong cardinal".

There are also ad hoc exceptions (such as, at least in some generic extensions, strongly compact cardinals), but their relative infrequency is a testament to the coherency of the large cardinal hierarchy.

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Since the relevant kind of large cardinal was just named on Joel Hamkins's blog, let me point to it as an interesting dual to some of the existing answer.

A cardinal $\kappa$ is otherworldly if there is some $\lambda$ such that $V_\kappa\prec V_\lambda$. (Exercise 4.1.7 in Drake's large cardinal book shows that otherworldly implies worldly, hence the name). We say $\kappa$ is totally otherworldly if there are arbitrarily large $\lambda$ such that $V_\kappa\prec V_\lambda$.

Now totally otherworldly cardinals are very low in consistency strength: a single inaccessible cardinal suffices to show the consistency of a proper class of totally otherworldlies, via the usual Löwenheim–Skolem argument. But every totally otherworldly cardinal is $\Sigma_2$-correct, which makes it at least larger than the least measurable, the least superstrong, and the least huge, if consistent.

So this is an example of very low consistency strength but very large size.

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    $\begingroup$ Actually totally otherworldly cardinals are even $\Sigma_3$-correct. Suppose $\psi = \exists x \forall y \phi(x, y)$ is a $\Sigma_3$ formula (where $\phi$ is $\Sigma_1$). If $\kappa$ is totally otherworldly, there is $\lambda$ such that $V_\kappa \prec V_\lambda$ and there is a witness $x \in V_\lambda$ of $\psi$. Then $\phi(x, y)$ is absolute between $V_\kappa$, $V_\lambda$, and $V$, as noted in this blog post, so $\forall y \phi(x, y)$ is downward absolute and $\psi$ is absolute between $V_\kappa$ and $V_\lambda$ by elementarity. $\endgroup$
    – Arvid Samuelsson
    Commented Sep 23, 2021 at 20:47

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