# Large cardinals

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?

• 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. – algori Jul 29 '10 at 23:34
• 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? – Stefan Geschke Jul 30 '10 at 21:04
• @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. – 16278263789 Jul 30 '10 at 23:17
• 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. – Stefan Geschke Aug 1 '10 at 10:14

• 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 $2^\kappa=\kappa^+$. – Stefan Geschke Aug 1 '10 at 10:04
• 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. – David Fernandez-Breton Dec 8 '13 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.