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Working in $\textsf{ZF} + \text {there is a strongly inaccessible cardinal}$.

Let $\kappa$ be the first strongly inaccessible cardinal, and let $\lvert V_\kappa\rvert= \kappa$, then $(V_{\kappa+1}, \in)$ would be a model of $\textsf{MK}$.

Now, is it consistent to add that $V_{\kappa+1}$ is non-well-orderable?

If yes, then what's the benefit of having $V_{\kappa+1}$ well-orderable, on the theory $\textsf{MK}$? I mean what additional axioms in the language of $\textsf{MK}$ would this confer?

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    $\begingroup$ TeX note: \sf applies to all following text, so $\sf ZF + \text{there is a strongly accessible cardinal}$ \sf ZF + \text{there is a strongly accessible cardinal} has everything in a sans-serif font, which presumably is not what was intended. You can use {\sf ZF} + \text{…} or, in the modern LaTeX style where formatting commands take arguments, \textsf{ZF} + \text{…}. I have edited accordingly. $\endgroup$
    – LSpice
    Commented Oct 29, 2023 at 2:53
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    $\begingroup$ @LSpice, Thanks! $\endgroup$ Commented Oct 29, 2023 at 3:00

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Since the well-ordering of the classes would not be a class, but a meta-class, you'd have to say more about what you mean exactly. For example, do you want to go to third-order set theory? Or do you want to augment the structure with an additional relation that is a well order? Or perhaps one is simply interested in the case that there is a definable well ordering in this model of its classes.

But let me answer for the case of a definable well order of the classes.

First, it is consistent that indeed there is a definable well ordering of the classes in this model. For example, if you had started in a model of V=L, then $V_\kappa=L_\kappa$, and the subsets are those that arise in the constructible hierarchy before $L_{\kappa^+}$. But all those sets be interpreted using class codes (much like reals can interpret hereditarily countable sets). In this case, you can define the $L$-order for subsets of $V_\kappa$, and there will be a (second-order) definable well ordering.

It isn't always true in KM that you have a definable well-ordering. The reason is that KM is known not to prove the class choice principle CC, which asserts that whenever every $x$ has a class $A$ with $\varphi(x,A)$, then there is a class $U\subset V\times V$ such that $\varphi(x,U_x)$ for every $x$, where $U_x$ is the section above $x$. If there were a definable well order of the classes, then we could define $U$ to use the least class in each section, and so we'd get CC as a consequence of class comprehension.

The CC principle is extremely useful in class theory, enabling many bi-interpretation results.

Finally, let me point out that your models $V_{\kappa+1}$ do not always have a definable well ordering of the classes. To see this, start with $\kappa$ inaccessible and then force to add $\kappa^+$ many Cohen subsets to $\kappa$, which preserves inaccessibility. Now, with any parameter, there will be many mutually generic sets over those parameters, but in a symmetric-model type argument, there can be no condition forcing that a particular one of them is least. So there will be no definable well order there.

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  • $\begingroup$ Nice! This answer greatly enrich the subject. What I had in mind is the second option, that is we add a new binary relation symbol $\prec$ to the signature of $\sf MK$, then add the axiom schema that $\prec$ is a well order over classes, i.e. an areflexic, transitive, connected and well founded relation over all classes. $\prec$ need not be definable. From this answer a definable well order would be beneficial, but would an indefinable one be too? $\endgroup$ Commented Oct 29, 2023 at 19:34
  • $\begingroup$ If the expanded language was allowed to appear in the comprehension scheme, then such a well order relation would enable the proof of CC. So the theory would be strictly stronger than KM and not conservative over it. $\endgroup$ Commented Oct 29, 2023 at 19:37
  • $\begingroup$ Would it prove the consistency of KM? $\endgroup$ Commented Oct 29, 2023 at 19:39
  • $\begingroup$ No, because every model of KM can be reduced to the L of the model, which has a definable well order for its classes. $\endgroup$ Commented Oct 29, 2023 at 19:40
  • $\begingroup$ For the record, of course I wanted $\prec$ to be allowed to appear in comprehension. So, here there is a clear benefit gained in that it would enable CC. And you said it enables many bi-interpretation results. $\endgroup$ Commented Oct 29, 2023 at 19:47

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