One of the unexplained phenomena about large cardinal axioms is that they tend to be in a linear order (in terms of consistency strength). So it brings to mind other natural questions about this "order".

So my question is - For any two large cardinal axioms $A1$, $A2$ such that $ZFC + A1$ proves $ZFC + A2$ is consistent, can we define a large cardinal axiom $A3$ such that $ZFC + A1$ proves $ZFC + A3$ is consistent, and $ZFC + A3$ proves $ZFC + A2$ is consistent?

Meaning, is the order known\thought to be densely ordered (assuming the ordering itself is not some random phenomena)? Alternatively, can we build an example (even a highly contrived one) of $A1$, $A2$, where any axiom we add to $ZFC$ is either stronger than $ZFC + A1$, weaker than $ZFC + A2$, or equiconsistent to one of them (meaning the ordering is not dense)?

P.S - I know there is no agreed upon definition of "large cardinal axiom", but if we discuss their linear order we should be able to ask questions like this as well.

  • $\begingroup$ We expect the consistency strength degrees of large cardinals to be pre-well-ordered. Essentially, because we ultimately expect to identify these degrees with certain "mice", and these should be orderable via comparison processes. $\endgroup$ Commented Mar 15, 2016 at 15:20
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    $\begingroup$ A good place to read about why this leads to a (pre-)well-ordering is MR1720574 (2000k:03101). Löwe, Benedikt(1-CA); Steel, John R.(1-CA) An introduction to core model theory. (English summary) Sets and proofs (Leeds, 1997), 103–157, London Math. Soc. Lecture Note Ser., 258, Cambridge Univ. Press, Cambridge, 1999. Particularly see Section 2.5 on the mouse order. (Note the link above is to a postscript file.) $\endgroup$ Commented Mar 15, 2016 at 15:20
  • $\begingroup$ @Andrés: Are these mice nice? And if there is just an iterable mouse, is it nouse? And if it's not, does that mean that the iterable mouse is a terrible mouse? $\endgroup$
    – Asaf Karagila
    Commented Mar 15, 2016 at 15:23

1 Answer 1


If by large cardinal axioms you mean axioms which directly allude to cardinals, then the answer is negative.

Namely, if a large cardinal axiom is something like "There exists a cardinal $\kappa$ with such and such properties" and these properties imply certain amount of inaccessibility, in which case we rule out things like "$\sf ZFC$ is consistent" or "$0^\#$ exists", then the answer is negative.

To see that, simply note that there will always be a smallest large cardinal. If you want it to be such that $V_\kappa\models\sf ZFC$, then the least worldly cardinal is your smallest large cardinal. Of course this is debatable, and we don't have a mathematical definition of what is a large cardinal axiom. But much like pornography, "we know it when we see it".

So if we accept that worldly cardinals are the smallest large cardinals, then "There are two worldly cardinals" proves the consistency of "There is a worldly cardinal", but there are no intermediate notions to be found.

Of course you might argue that $\sf ZFC$ with "the theory $\sf ZFC+\exists \kappa$ worldly is consistent" proves the consistency of the existence of worldly cardinals, but now this becomes a question of whether or not this is a large cardinal axiom. Some people might argue that it is, others might argue that it's not.

For what it's worth, statement about "this theory is consistent" are arithmetic, so if we only assume $\sf PA$ with "this theory is consistent" then we can prove the consistency of said theory. But this is really not what we mean when we talk about large cardinals, is it now?

  • $\begingroup$ Of course, I am strictly treating the case where the two extensions of ZFC have different consistency strength. One could argue that we cannot find anything intermediate between "There is a weakly compact cardinal" and "There is a successor of a regular cardinal with the tree property", because the two theories have been shown to be equiconsistent. $\endgroup$
    – Asaf Karagila
    Commented Mar 15, 2016 at 15:26
  • $\begingroup$ Interesting. Though the example with worldly cardinals makes me wonder what would happen if I'd amend the question to demand a large cardinal axiom that states the existence of an unbounded class of a certain type of cardinals (instead of a single cardinal). $\endgroup$
    – Alon Navon
    Commented Mar 15, 2016 at 20:59
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    $\begingroup$ That would usually be the the same. A worldly limit of inaccessible cardinals is the smallest large cardinal above "proper class of inaccessible cardinals" in terms of consistency strength. $\endgroup$
    – Asaf Karagila
    Commented Mar 15, 2016 at 21:32

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