It's well known that the existence of an inaccessible cardinal proves $Con(MK)$. Joel Hamkins has a nice blog post (http://jdh.hamkins.org/km-implies-conzfc/) that explains what you get out of $MK$, and in particular that $MK \vdash Con_n(ZFC)$ for every $n$ [here $Con_n(ZFC)$ is the iteration of the consistency sentence $n$-many times, e.g. $Con_4(ZFC) =_{df} Con(Con(Con(Con(ZFC))))$].

Question: Is the exact consistency strength of $MK$ known?


1 Answer 1


Thanks for your kind words about my blog post. Let me try to answer your question.

To describe the consistency strength of a theory or assertion, we should compare the consistency of that theory or assertion to that of other more familiar or landmark theories or assertions. For example, the consistency strength of ZFC plus the continuum hypothesis is ZFC itself; the consistency strength of ZF+DC+all sets are Lebesgue measurable is the same as ZFC + there is an inaccessible cardinal. We compare our given theory or statement to a landmark theory.

The issue with your question, then, is that KM is itself such a landmark theory. The exact consistency strength of KM is: KM itself. Your question is a little like asking, "What is the exact consistency strength of ZFC plus an inaccessible cardinal?" The answer would be: ZFC plus an inaccessible cardinal.

But naturally such an answer will not satisfy. Perhaps you could explain what kind of answer you were seeking?

Meanwhile, it is possible to explain how the strength of KM relates to other large cardinals. It doesn't line up exactly with any of the usual large cardinals. (Although I view KM itself as a kind of large cardinal axiom.)

Lower bounds. The post on my blog to which you link explains that KM is strictly stronger than ZFC in consistency strength, and the argument given there shows that KM implies that there is a proper class club of cardinals $\kappa$ with $V_\kappa\prec V$. Thus, these are all worldly cardinals, and by elementarity each of them will be a limit of worldly cardinals. So one can begin to climb the degrees of worldliness, in the style of hyperMahloness, and see that there is a stationary proper class of hyperworldly cardinals, hyper-hyper worldly cardinals, and so on.

Another lower bound is provided by my recent paper V. Gitman, J. D. Hamkins, Open determinacy for class games, in review. Namely, we prove that the principle of clopen determinacy for proper class games is equivalent over GBC to the principle ETR of elementary transfinite recursion, which allows transfinite recursion over proper class well-founded relations, which are not necessarily set theory. That principle also gives the truth predicate, which leads to the worldly cardinals as in the previous paragraph. These theories are strictly weaker than GBC + $\Pi^1_1$-comprehension, which is also strictly weaker than KM.

Upper bound. Meanwhile, KM is strictly weaker than ZFC + there is an inaccessible cardinal. This is simply because if $\kappa$ is an inaccessible cardinal, then $\langle V_\kappa,\in,V_{\kappa+1}\rangle$ is a model of KM, and so from an inaccessible cardinal we can deduce $\text{Con}(KM)$ and $\text{Con}(KM+\text{Con}(KM))$ and more, iterating many times.

Equiconsistencies. It turns out that KM is equiconsistent with a natural strengthening of KM denoted by $\text{KM}^+$, which includes the class-choice principle: $\forall x\exists X\ \varphi(x,X)\to\exists Y\forall x\ \varphi(x,Y_x)$. The axiom says that if for every set $x$ there is a class $X$ with a certain property, then you can find a class $Y\subset V\times V$, whose slices $Y_x$ serve as witnesses. Gitman, Johnstone and I have proved that this assertion is not provable in KM itself, but one can construct a model of the stronger theory from any model of KM, so they are equiconsistent.

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    $\begingroup$ The last result is really nice! Do you know if KM+ is first-order conservative over KM? $\endgroup$
    – user104955
    Apr 17, 2016 at 17:25
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    $\begingroup$ Yes, it is, since the argument shows that every model of KM has a KM+ realization with the same first-order part. $\endgroup$
    – JDH
    Apr 17, 2016 at 17:27
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    $\begingroup$ Lovely; thanks, Joel! $\endgroup$
    – user104955
    Apr 17, 2016 at 17:38
  • $\begingroup$ Thanks Joel, this is absolutely brilliant! I suppose the answer I was seeking was whether there's a pure `size' based notion (similar to inaccessible, but weaker whilst stronger than ZFC). The lower bounds an upper bounds you give are very useful. I suppose, failing that, nice equiconsistencies was what I was after, in particular ones that are not directly derivative on KM itself (e.g. KM+). If we're viewing KM as a large cardinal axiom, then something comparable to the case of PFA and a Supercompact, assuming that Con(PFA) does in fact imply Con(Sct). Plenty to chew over though, thanks! $\endgroup$ Apr 20, 2016 at 6:48

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