# Is every class that does not add sets necessarily added by forcing?

We know there are many situations in which we can force over a model $M$ of GBC to add a class $G$ without adding any sets. That is, the extension $M[G]$ satisfies GBC and has the same sets as $M$. This technique is used, for example, in the proof that GBC is a conservative extension of ZFC, by forcing to add a universal choice function to a model of ZFC.

I'd like to know whether every class that can be 'safely' added to a model (preserving GBC and adding no sets) arises through forcing:

If $M$ is a model of GBC and $G$ is a class (that is, a subcollection of the sets of $M$, not a class member of $M$) such that $M[G]$ satisfies GBC and has the same sets as $M$, then is $G$ necessarily generic for some partial order $P \in M$?

I feel 'morally certain' that the answer must be no. Certainly the analogous question about sets, 'Can every set be added by forcing?' has a negative answer - we cannot, for example, force over $L$ to add a measure to a cardinal $\kappa$ (or to add $0^\\#$, or any other set that might increase the consistency strength). However, I don't see how to adapt this kind of argument to classes, if we are not allowed to add sets. I'd love to see a counterexample (a proof of a positive answer would also be welcome!).

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A $\Pi^1_1$-singleton class could not be added by forcing. I haven't worked out how to make this work, but here is the plan... Find a formula $\forall Z \phi(X,Z)$ where $\phi$ only has set quantifiers but $Z$ ranges over classes which provably has at most one solution $X$ but no solution definable by comprehension using only set quantifiers. The unique solution cannot be added by forcing over the model of NBG where all classes are defined by comprehension using only quantification over sets. –  François G. Dorais Jan 9 '13 at 22:58
@Jonas: I deleted my answer since it did not take your requirement that $G$ not be a member of the GB-model. On the other hand, I have an argument that shows your hunch is right if we also insist that the notion of forcing $P$ be Ord-closed in the sense of $M$; and I will try to see if this condition can be removed. –  Ali Enayat Jan 10 '13 at 5:29
François, do you expect that we should be able to add a $\Pi^1_1$ singleton class over any model of ZFC? My answer seems to illustrate your plan, for models that admit a satisfaction class, since "being a satisfaction class" is even first-order expressible in the language with a predicate for the class. –  Joel David Hamkins Jan 10 '13 at 16:39
I expected that this would be possible but Ali pointed out that it might not be. Satisfaction classes are examples of $\Pi^1_1$ singletons but I wanted to avoid these since they can add new sets. It would be interesting to know which models have $\Pi^1_1$ singletons. –  François G. Dorais Jan 10 '13 at 18:22

It is a fantastic question, Jonas! I've spent hours with it now, going back and forth several times about which way it might go.

But finally, I've got a negative answer, at least for some models $M$. My idea is that some models of ZFC admit what are called satisfaction classes, but these can never be added by class forcing.

For any transitive model $M$ of ZFC, let $S$ be the set of pairs $\langle{}^\ulcorner\varphi{}^\urcorner,a\rangle$, for which $\varphi[a]$ holds in $M$. This class $S$ is the (unique) class of pairs $\langle n,a\rangle$ satisfying the Tarskian inductive definition of truth. Further, we may sometimes add $S$ as a class to $M$ and still have GBC in $(M,S)$. This is true, for example, when $M=V_\kappa$ for an inaccessible cardinal $\kappa$; but also it is true for many other models.

So let us suppose that $M$ admits such a unique satisfaction class.

Can $S$ be added by class forcing over $M$ without adding sets? If we regard $M$ as a GB model by endowing it with only its definable classes, then the answer is no.

To see this, suppose that $S$ was added in the class forcing extension $M[H]$, where $H\subset\mathbb{P}$ is $M$-generic for the class forcing $\mathbb{P}$, which adds no sets. So $S=\dot{A}_H$ for some class $\mathbb{P}$-name $\dot{A}$, and in particular, both $\dot{A}$ and $\mathbb{P}$ are definable in $M$.

Now, the key step is that although truth itself is not definable, by Tarski's theorem, the property of $S$ that it satisfies the inductive definition of truth has complexity merely $\Delta_1(S)$. Thus, one of the assertions about $S$ that is true in $M[H]$ is that it satisfies the Tarskian definition of truth. Thus, there must be a condition forcing that $\dot{A}$ is a satisfaction class, obeying Tarski's inductive definition.

But since the satisfaction class of $M$ is unique, this means that there is no choice for the generic filter whether or not to place a pair into or out of $\dot{A}$. Thus, inside $M$ by simply looking at which pairs can be added at all to the class named by $\dot{A}$, we can define $S$ in $M$. But this contradicts Tarski's theorem on the non-definability of truth. QED

Let me argue next that we don't actually need the satisfaction class to be unique, and the same argument will work whenever $M$ admits a satisfaction class at all, with GBC in $(M,S)$. This can conceivably happen in non-standard models $M$, with non-standard Gödel codes. Nevertheless, the standard part of the satisfaction class, that is, for standard Gödel codes, will be unique, and so we can still get that stable part of the satisfaction class from the name $\dot{A}$---the pairs that are forced into $\dot{A}$ by every condition. This will still be a satisfaction predicate, contrary to Tarski's theory of truth, even if $M$ is not an $\omega$-model and possibly admits several satisfaction classes.

Finally, let me point out that not every model of ZFC admits a satisfaction class. For example, if $M$ is pointwise definable, as in our joint paper on Pointwise definable models of set theory, then this property would be revealed by the satisfaction class, and so we cannot add this class without revealing to $M$ that it is countable.

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What a great example, Joel! The fact that the satisfaction class is compatible with GBC but is not definable is a very nice combination. Does existence of a satisfaction class increase the consistency strength of GBC? –  jonasreitz Jan 10 '13 at 16:23
I guess not, because you can make a compactness argument that there must be a ZFC model which admits a satisfaction class with GBC, since for any standard $n$ we can have a $\Sigma_n$-satisfaction class. That is, we can write down the theory of what it means to be a satisfaction class with GBC still holding, and apply compactness. –  Joel David Hamkins Jan 10 '13 at 16:33
Incidentally, since not every model $M$ admits a satisfaction class with GB, there is still room here for a better answer, which either proves it negatively for all models $M$, or which finds some $M$ to which one cannot add classes except generically. –  Joel David Hamkins Jan 10 '13 at 16:35
Meanwhile, it seems that the compactness argument shows that every model $M$ can be elementarily embedded into a model that admits a satisfaction class (just add the elementary diagram of $M$ to the theory before applying compactness). Thus, the existence of a satisfaction class with GBC is conservative over ZFC, and there can be no obstacle coming just from the theory of $M$. –  Joel David Hamkins Jan 10 '13 at 16:48
@Joel: Nice answer; the question you pose in your comment is intriguing. Of course, we want to concentrate on countable $M$ because there are such things as "rather classless models" of $ZF$, all of whose classes are definable. –  Ali Enayat Jan 10 '13 at 17:19