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The answer is no. Mahloness is much stronger than this. Every Mahlo cardinal $\kappa$ is a limit of such cardinals. One can see this, because there is a club of $\gamma<\kappa$ with $V_\gamma\prec V …

If all we want is that for every measure $U$ there is a strongness embedding with a seed for $U$, then the answer is yes. To see this, suppose $\kappa$ is a strong cardinal, and let $U$ be any $\kappa …

I don't agree that it is difficult to formalize the local/global distinction, and indeed, I think that there is a largely agreed-upon technical meaning for these notions. Specifically, a property is …

On the one hand, we cannot easily describe any such list of axioms, since if $\Lambda$ was computably enumerable, then by your third bullet point, the arithmetic consequences of ZFC+(V=L)+$\Lambda$ wo …

This is a very nice question (and indeed, I remember asking myself this question when I was a graduate student). The answer in general is that no, you do not get any extra strength from having these …

Yes, this situation can occur. One should simply undertake the dual of the construction you had suggested with iterated ultrapowers. Specifically, suppose that $\mu$ is a normal measure on a measura …

The answer is yes. Theorem. The direct limit ultrapower you describe is well-founded if and only if $\bigcap_n A_n\neq \emptyset$ whenever $A_n\in U$ for all $n$. Proof. You've already noted the con …

I'm glad to hear you're reading our paper, which can be found here: The foundation axiom and elementary self-embeddings of the universe. Click through to the arxiv for a pdf — and I note that the titl …

I assume that by $M_{\kappa,\mu}$ you intend the ultrapower of $V$ by $\mu$. Both of the statements are inconsistent, even in the case of only two measurable cardinals, let alone a proper class of t …

If $\kappa$ is weakly inaccessible, then it is a limit cardinal and hence $\kappa=\aleph_\lambda$ for some limit ordinal $\lambda$. Since the cofinality of $\aleph_\lambda$ is the same as the cofinali …

I view these kinds of hypotheses as on a continuum stretching from weak compactness up through all the levels of the indescribability hierarchy, second order, third order and so on. This hierarchy of …

My perspective on this issue is that there are a variety of ways to take the claim of the Kunen inconsistency, and we needn't pick a particular one as the only right one. Rather, we gain a fuller pers …

No, this is not possible. If κ is supercompact, then in particular it is measurable, and there is a normal measure μ on κ that concentrates on non-measurable cardinals. If j:V to M is the correspond …

In general, $\lambda$-supercompactness, if consistent, does not imply $\lambda$-strongness. One can see this by observing that the smallest cardinal $\kappa$ that is $\kappa^+$-supercompact is never $ …

I would answer question 1 by saying that $M=\text{Ult}(V,U)$ if and only if $j$ is isomorphic to an ultrapower by some normal measure. This is another way of saying that normal measures are minimal wi …