Looking at the chart of cardinals in Kanamori's book, one realizes that all large cardinals are implied by stronger ones and imply weaker ones. For instance measurable implies Jonsson which implies zero sharp which implies weakly compact which implies Mahlo which implies inaccessible. So it seems as if all these large cardinal assumptions are linearly ordered by consistency strength. Is there a some assumption above ZFC that is not implied by and does not imply any of the linearly ordered large cardinals?

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    $\begingroup$ alephomega -- the large cardinal axioms in that list do not always imply one another, they are just ordered in terms of the consistency strength. See for example the discussion mathoverflow.net/questions/12804/… showing how a non-logician (myself in that case) can get a bit confused about this. Some of those axioms however do imply some other axioms, as explained in Joel Hamkins's answer in that thread. $\endgroup$
    – algori
    Jul 29, 2010 at 23:34
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    $\begingroup$ In the light of algori's comment, the question should be: Is there some statement A that is consistent with ZFC (which we cannot prove, but believe in) such that Con(ZFC+A) implies Con(ZFC) and Con(ZFC+A) does not imply the consistency of one of the usual large cardinals and is not implied by the consistency of one of the usual large cardinals? $\endgroup$ Jul 30, 2010 at 21:04
  • $\begingroup$ @Stefan, thank you for the clarification, this is exactly what I had in mind, I used the wrong formulation. It is possible? Except CH, is there such principles? I don't know if MA is a candidate because I don't know if it follows from some other statement, I only know you can force its truth by iteration with finite support. $\endgroup$ Jul 30, 2010 at 23:17
  • $\begingroup$ Well, both MA and CH have no consistency strength over ZFC. The consistency of either statement follows from the consistency of ZFC. But you are right, CH and MA are both examples for the kind of statements that you are looking for. That is why I said that you should not look for actual implications but for implications of consistency. $\endgroup$ Aug 1, 2010 at 10:14

2 Answers 2


By the well-known Levy-Solovay theorem, large cardinal properties are preserved under "small" forcing. Therefore CH is an assumption above ZFC which is not settled by large cardinal axioms.

  • $\begingroup$ I am going to look up this theorem, this sounds great. So large cardinals are absolute for certain types of forcing and for generic extensions, if I understand your answer. $\endgroup$ Jul 30, 2010 at 0:48
  • $\begingroup$ What exactly do you mean by "above ZFC"? Does not follow from ZFC and is consistent with, or the statement in question is consistent with ZFC and this consistency does not follow from the consistency of CH? $\endgroup$ Jul 30, 2010 at 20:58
  • $\begingroup$ @Stefan, yes by "above ZFC", I mean is not provable from ZFC but can be consistent with, which actually prompts another question: are there non trivial statements of cardinal arithmetic which are "in between" CH and ZFC, that is cardinal arithmetic statements provable when you add CH to ZFC but not provable from ZFC alone? Because presumably, CH does not say anything about singular cardinals for instance or even about $\endgroup$ Jul 30, 2010 at 23:07
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    $\begingroup$ Well, I can't really think of a cardinal arithmetic statement that follows from CH but not from ZFC, except for CH itself, of course. There are interesting theorems like Silver's theorem, though: If $\kappa$ is a singular cardinal of uncountable cofinality and for all $\lambda<\kappa$, $2^\lambda=\lambda^+$ (GCH holds below $\kappa$), then $2^\kappa=\kappa^+$. $\endgroup$ Aug 1, 2010 at 10:04
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    $\begingroup$ Well, a trivial example of a statement "between CH and ZFC" would be what is known as "the weak CH", namely that $2^{\aleph_0}<2^{\aleph_1}$. This is a trivial consequence of CH, but of course cannot be proved from ZFC alone (for example, Martin's axiom plus the negation of CH disproves it) and is known to be "strong enough" to yield interesting consequences. $\endgroup$ Dec 8, 2013 at 16:50

Woodin's Ω-conjecture implies that all large cardinal axioms are well ordered under the relation ``its consistency implies the consistency of''. See his paper in the Notices 2001/7. For this, of course, he does define what a large cardinal is.


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