Following this question and that one illustrating how induction in NBG can be tricky, I realized I'm also confused about the notion of “model” of NBG. The goal of this confusion is to hopefully lift this confusion (for me and, perhaps, for other people).

A model of NBG inside NBG is a collection $V$ of (internal) “sets”, a collection $W$ of (internal) “classes” and a collection $E$ of belonging relations, with $V \subseteq W$ and $E \subseteq V\times W$, which satisfies the axioms of NBG. Depending on how “collection” is interpreted in this definition, this seems to give three different possible concepts:

  • a “small model” of NBG, or (set,set)-sized model, is one in which $V$, $W$ and $E$ are all sets (in the sense of the “external” NBG, of course),

  • a “medium model” of NBG, or (set,class)-sized model, is one in which $V$ is a set but $W$ and $E$ are classes,

  • a “large model” of NBG, or (class,class)-sized model, is one in which $V$, $W$ and $E$ are all classes,

Actually, a medium model is the same as a small model because Extensionality for classes tells us we can identify $W$ with a subset of $\mathscr{P}(V)$, so if $V$ is a set then so is $W$. So we can forget about medium models.

Now small models are the ones Gödel's completeness theorem tells us something about: it is a theorem of ZFC that there exists a model of NBG iff NBG is consistent, and since theorems of ZFC are the same as theorems of NBG that talk only about sets, it is a theorem of NBG that there is a small model of NBG iff NBG is consistent (and this is also equivalent to the existence of a small model of ZFC and to the consistency of ZFC).

If we look at models of ZFC inside NBG (which have just $V$ and $E$), the existence of a small (i.e., set-sized) model is tantamount to an arithmetic statement, as pointed out in the previous paragraph; on the other hand the existence of a large (i.e., class-sized) model should be obvious: just take the class of all sets. Except I'm not sure we can even express in NBG the fact that a class is a model of ZFC because ZFC has infinitely many axioms (and as I learned in the questions mentioned at the start we can't even perform mathematical induction on classes), so I'm treading on thin ice here.

But for models of NBG, which has finitely many axioms, this difficulty should not arise, so it appears the notion of “large model” of NBG inside NBG is well-defined, and I don't see an obvious way to get one, nor an argument why it should imply the consistency of NBG. Hence:

Question: What does the existence of a “large model” of NBG, as a statement of NBG, tell us? Is it equivalent to an arithmetical statement? Is it equivalent to a statement purely about sets?

More general plea: Please help me clear up the confusion which is probably obvious in this question!

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    $\begingroup$ You cannot prove the existence of a large model of NBG in NBG. The reason is that for every model $M$ of ZFC, the collection of definable classes of $M$ is a model of NBG. If a large model of NBG existed in every such model, then using compactness, there would exist a single formula that defines a larger model of NBG provably in ZFC, i.e., there would be an interpretation of NBG in ZFC. (A parametric interpretation, to be precise.) This is impossible: NBG is finitely axiomatized, hence you would get an interpretation of NBG (hence of ZFC) in a finite fragment of ZFC. But ZFC proves the ... $\endgroup$ Apr 13 at 14:08
  • $\begingroup$ ... consistency of each of its finite subtheories, hence it would prove its own consistency, contradicting Gödel’s theorem. So I’m assuming the consistency of ZFC all along. (If you take NBG to include the axiom of global choice, the above is not quite true; I’d have to work with ZFC expanded with an extra function symbol for a global choice function. But this still proves the consistency of each of its finite subtheories, and is a conservative extension of ZFC, thus the argument continues to work.) $\endgroup$ Apr 13 at 14:13
  • $\begingroup$ Ah, never mind, the existence of a large model of NBG does imply the consistency of NBG, making the argument above redundant. See my answer below. $\endgroup$ Apr 13 at 14:35

1 Answer 1


In NBG, the existence of a large model of NBG implies (and therefore is equivalent to) the consistency of NBG.

Since NBG is finitely axiomatized, all its axioms have quantifier rank bounded by a constant $k$. Thus, the cut-elimination theorem (which is provable in $I\Delta_0+\mathrm{SUPEXP}$, and a fortiori in NBG) ensures that if NBG is inconsistent, then there is a proof of inconsistency in NBG in which all formulas have quantifier rank at most $k$ (or $k+1$ or so, depending on the exact details of the proof system). But if $M=(V,W,E)$ is a large model of NBG, you can easily define (as a class) a satisfaction predicate for $M$ for formulas of fixed quantifier rank. Then you can prove by induction on the length of a proof that all formulas in a proof from NBG using only formulas of rank at most $k$ are true, and in particular, no such proof can be a proof of inconsistency.

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    $\begingroup$ Nice argument. So more generally NBG can be replaced by any finitely axiomatizable fragment of Kelley-Morse extending NBG. Also the result applies to the context of second order arithmetic. $\endgroup$
    – Ali Enayat
    Apr 13 at 16:14
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    $\begingroup$ The argument shows that the existence of a large model of $T$ is equivalent to the consistency of $T$ for any standard finitely axiomatized theory $T$; it does not need to be a fragment of KM. $\endgroup$ Apr 13 at 17:18
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    $\begingroup$ Yes Emil, my reference to KM was just to emphasize that the result applies to much stronger finitely axiomatized class theories than just NBG. $\endgroup$
    – Ali Enayat
    Apr 13 at 17:48

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