It is well known, that every statement involving only set-variables is provable in NBG set theory if and only if it is provable in ZFC. What confuses me however is that NBG has a global axiom of choice. Global choice implies that every set is ordinal definable (V=OD).

So the statement V=OD seems to be a counterexample: It only involves set-variables, it follows from NBG, but it is known to be independent of ZFC. Where am i wrong?


It is not true that global choice implies $V=OD$.

Global choice is the assertion that there is a class well-ordering of the universe. This is equivalent to the assertion that there is a global choice function, which selects from every non-empty set an element. It is not part of the axiom, however, that this class is definable. Rather, the class well-ordering of $V$ is simply one of the classes that is available in the second order part of the model, and there is no reason to suppose that all such classes are necessarily first-order definable, even from set parameters.

The axiom $V=HOD$, on the other hand, is equivalent to the assertion that there is a definable global choice function class, or equivalently, a definable well-ordering of the universe.

So the critical difference is in the question of whether the global well-ordering is definable or not. It might be interesting to note that there can be a global well-ordering that is definable from set parameters, and so still counts as a class in ZFC, without having $V=HOD$, simply because those parameter may be essential, and it may not be possible to define the order without them. This is what happens, for example, after adding a generic Cohen real over $L$, since in $L[c]$ we do not have $V=HOD$, and in fact $\text{HOD}^{L[c]}=L$ by the homogeneity of the forcing, but nevertheless one can use $c$ as a parameter to define in $L[c]$ a well-ordering of the universe, essentially the usual $L$-order relativized to $c$.

The proof that NGBC is conservative over ZFC is not difficult. Given any model $M$ of ZFC, one first forces global choice by class forcing: consider the class partial order consisting of well-ordered sequences of longer and longer set length. This forcing is $\kappa$-closed for every $\kappa$, and hence adds no new sets, but the generic filter $G$ provides a class well-ordering of $M$. Now, to form the GBC model $M[G]$, one considers as classes all the classes that are definable from set parameters and this new global well-order class. One can verify that the resulting model satisfies all the GBC axioms (and this amounts essentially to the verification that this kind of class forcing works as expected). Since the sets of $M[G]$ are the same as $M[G]$, it follows that any statement that is true in every GBC model is also true in every model of ZFC, and so the extension is conservative.

  • $\begingroup$ Good answer! Thank you for pointing me to L[c]. I now understand what's wrong with my "proof" of V=OD. $\endgroup$ Jun 21 '12 at 15:25

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