For predicative formula I mean a formula where all the quantifiers are restricted to sets.

In Zermelo-Fraenkel set theory the axiom of comprehension is replaced by the axiom of separation that doesn't lead to inconsistencies like Russel paradox.

In NGB the idea to reduce the power of axiom of comprehension is accomplished considering predicative formulas only. Otherwise, I think it might be some formulas that allows a class to belong to another class, that is absurd.

Can someone express a clearer reason for this choice?

Thanks in advance :)

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    $\begingroup$ Please don't use math mode to produce italics. Math font is not italic font. $\endgroup$ Feb 10, 2021 at 17:26
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    $\begingroup$ About capitalization of "von Neumann", see here. Since he chose not to capitalize the "von" when writing his full name, it should remain uncapitalized unless it starts a sentence. $\endgroup$ Feb 10, 2021 at 18:11
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    $\begingroup$ Not sure about the historical reason, but metamathematically, NBG is a conservative extension of ZFC, and so it can be considered as just ZFC with syntactic sugar. In particular the two are equiconsistent. Once impredicative comprehension is allowed, we get a stronger theory which proves the consistency of ZFC and much more $\endgroup$ Feb 10, 2021 at 18:30
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    $\begingroup$ @JasonZeshengChen I've folded your reference into my answer. $\endgroup$ Feb 10, 2021 at 18:49

1 Answer 1


Class-quantifier-comprehension does not in fact force (proper) classes to be elements of classes: remember that, regardless of the type of variable being quantified over, the object produced by a comprehension axiom is a priori a collection of sets. So you won't run into this problem here.

In fact, allowing class quantifiers in the comprehension axioms is relatively safe. Working in $\mathsf{ZFC}$, suppose $\kappa$ is inaccessible and consider the two-sorted structure $\mathfrak{M}=(V_\kappa, V_{\kappa+1})$ (here I'm following the Bernays approach where each set "appears twice," once as a set and again as a class). It's easy to check that $\mathfrak{M}$ satisfies all the NGB axioms together with the stronger comprehension principle allowing quantification over "classes" (= elements of $V_{\kappa+1}$), the point being that every collection of $\mathfrak{M}$-sets (= every subset of $V_\kappa$) is an $\mathfrak{M}$-class (= element of $V_{\kappa+1}$) due to the nature of the $V$-hierarchy.

Put another way, $\mathsf{ZFC}$ + "There is an inaccessible cardinal" proves the consistency of this strengthening of $\mathsf{NGB}$. And as usual, this can be optimized by being a bit more careful about the details. So allowing quantification over classes in the comprehension axioms for $\mathsf{NGB}$ will not actually result in an inconsistency, barring a gigantic surprise.

So why does $\mathsf{NGB}$ not allow quantification over classes in the comprehension axiom? Well, I think this comes down to the philosophical motivation for $\mathsf{NGB}$: a certain amount of predicativity is desired for its own sake, not merely to avoid contradictions. Not being an historian I don't have much to say here, but that's my understanding of the situation.

One important point here is that the quantifier restriction lets us prove that $\mathsf{NBG}$ is, appropriately construed, a conservative extension of $\mathsf{ZFC}$. However, I don't think this motivated that axiomatic choice, since I believe the conservativity result was only observed after $\mathsf{NBG}$ was codified. But I could be wrong here, too - the early set-theoretic literature is sometimes a bit out of whack in terms of what was known when vs. what was published when.

EDIT: And let me add to that observation the dual, which is that strengthening comprehension to allow quantification over classes does increase the consistency strength. See this summary of Hamkins, with the caveat that Hamkins' $\mathsf{GBC}$ is what we've been calling $\mathsf{NGB}$ (and thanks to Jason Zesheng Chen for reminding me of it in a comment above).

  • $\begingroup$ @HanulJeon How could it fail to satisfy full separation? Every subset of $V_\kappa$, definable in $\mathfrak{M}$ or not, is an element of $V_{\kappa+1}$. The structure $\mathfrak{M}$ will fail to satisfy set replacement (or limitation of size if you prefer) if $\kappa$ is singular, though, which is why it fails to satisfy $\mathsf{MK}$. $\endgroup$ Feb 10, 2021 at 23:16
  • $\begingroup$ That said, you're right, if $\kappa$ is singular then $\mathfrak{M}$ doesn't satisfy $\mathsf{NGB}$ after all since it fails to satisfy limitation of size - I briefly forgot that was part of $\mathsf{NGB}$! Fixing ... $\endgroup$ Feb 10, 2021 at 23:19

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