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In Shulman's 2008 paper 'Set Theory for Category Theory', he includes amongst the axioms of $\sf NBG$ the axiom of limitation of size. Being well known to imply the axiom of global choice, it seems to me that since $\sf ZFC+I$ doesn't imply global choice, this should be inconsistent with his later remark that $\sf ZFC+I$ has a model of $\sf NBG$ given by $V_{\kappa}$ for the sets, and $\operatorname{Def}(V_{\kappa})$ for the proper classes. I'm sure there is something here I don't quite understand, I would greatly appreciate some clarification.

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    $\begingroup$ That later remark does not say that ZFC+I implies global choice; it says that it provides a model for global choice. A well-order of $V_\kappa$ in the universe will feel like a global well-order if you think $V_\kappa$ is the whole universe. $\endgroup$
    – KP Hart
    Commented Dec 8 at 13:41

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Care is needed here.

It is true that $\sf ZFC+I$ implies the consistency of $\sf NBG$. And it is true that $\sf ZFC+I$ does not prove Global Choice.

However, when we talk about consistency statements, namely "Theory $T$ proves that theory $S$ has a model", it is a weaker condition. Similarly, $\sf ZF+DC+LM$ (here $\sf LM$ stands for "every set of reals is Lebesgue measurable") proves the consistency of $\sf ZFC$, even if it refutes the Axiom of Choice.

So, what we are really saying here is that there is a model of $\sf NBG$ in which there is a Global Choice function. And since the model is a set in our universe of $\sf ZFC+I$, a choice function on the model can, potentially, be a good candidate for being a Global Choice function.

Nevertheless, the statement you seem to refer to is itself wrong. It is certainly consistent that inside $\operatorname{Def}(V_\kappa)$ there are no well-orders of $V_\kappa$. But it is true that we can find other ways around this. We can take $V_{\kappa+1}$ or we can restrict to $L_\kappa$ and $L_{\kappa+1}=\operatorname{Def}(L_\kappa)$, for our model of $\sf NBG$.

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    $\begingroup$ Depending on your preexisting knowledge, Jech might be good. Or Enderton. Or Just & Weese. You can take a look on my website, there are lecture notes in set theory which briefly touch on $L$. After reading those, Jech would probably be a suitable option. $\endgroup$
    – Asaf Karagila
    Commented Dec 8 at 18:35
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    $\begingroup$ Thanks for pointing out that mistake. If and when I revise the note I'll fix it. Is there a natural subset of $V_{\kappa+1}$ that we could take as the classes to obtain a model of NBG that isn't a model of Morse-Kelley? E.g. could we use $\mathrm{Def}(V_\kappa \cup \{x\})$ where $x$ is some fixed well-ordering of $V_\kappa$? $\endgroup$ Commented Dec 9 at 8:05
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    $\begingroup$ Coincidentally, here's a blog post today by @JoelDavidHamkins proving that whenever $V_\kappa$ is a model of ZFC (a much weaker condition than being inaccessible), there is a subset of $V_{\kappa+1}$ that extends it to a model of NBG. If I understand correctly, he indeed uses $\mathrm{Def}(V_\kappa \cup \{x\})$ for some well-ordering $x$ of $V_\kappa$, but he uses class forcing to construct a particularly good such $x$. (I don't quite follow why the good properties are needed.) $\endgroup$ Commented Dec 14 at 1:49
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    $\begingroup$ @Mike: Suppose that I picked a well order whose first $\omega$ segment is actually a cofinal sequence of ordinals. This would break Replacement on sets, so you can't have that. The idea to add a generic well-order is to ensure this sort of problem doesn't happen. $\endgroup$
    – Asaf Karagila
    Commented Dec 14 at 11:27
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    $\begingroup$ Mike and Asaf, yes, that is right. Except that the argument isn't "adding" the generic well order $\leq$ of $V_\kappa$ by forcing over $V$ to add it in the usual forcing manner, but rather proving that there is already in $V$ such an order that is $\text{Def}(V_\kappa)$-generic, and this is enough to get NBG in $\langle V_\kappa,\in,\leq\rangle$, just in in the usual conservativity argument for NBG over ZFC. So we can freely add a global choice order to any worldly cardinal and preserve NBG. $\endgroup$ Commented 2 days ago

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