# Some questions about *Set Theory* by Frank R. Drake

Hello everyone,

I am currently studying set theory on my own on the book Set Theory, an Introduction to Large Cardinals by Frank R. Drake and I have a couple of serious doubts.

Drake first introduces the usual (not formal) definition (due to Tarski) of satisfaction on a given collection A and then the usual definition of a model of some theory $S$ ($M \models S$). Then he gives (more or less) the the following definition:

Given a theory $S$ and a formula $\phi$ we write $S \vdash \phi$ iff for all structures $M$ we have $$M \models S \text{ iff } M \models \phi.$$

Of course this is not the usual definition of $S \vdash \phi$ (there exists a finite sequence of formulas, etc..) but (I think) the two are equivalent because of the Completeness Theorem (even for classes in the sense below). This is a minor point but I feel it could be somewhat related to the major problems below.

After the presentation of some of the axiomatic development of ZF, Drake starts to discuss (now informally) some model theory and the notion of absoluteness between structures (again he uses the word "collection" referring to the domain of the structures). Then he returns to formality and explains how we can define the notion of satisfaction within ZF, i.e. how we can formally write the statement

$ZF \vdash x \models \phi$, for some $\phi$,

where of course $x$ is a set.

Then he says that the whole properties about absoluteness could be easily formalized within ZF so that we should have, e.g.

$ZF \vdash x \models \phi \leftrightarrow y \models \phi$, if $x$ and $y$ are both transitive models of ZF and $\phi$ is a $\Delta_1^{ZF}$ formula.

Drake defines the universe of the constructible sets, $L$, and, in order to prove consistency results, uses strongly (for example) the following fact:

$V \models \phi$ iff $L \models \phi$, if $\phi$ is a $\Delta_1^{ZF}$ formula.

I've been thinking about this for days and I arrived to the conclusion that this is only a short form for

$ZF \vdash \phi$ iff $ZF \vdash \phi^L$, if $\phi$ is a $\Delta_1^{ZF}$ formula.

i.e., from the Completeness Theorem and the definition of satisfaction,

$ZF \vdash \phi \leftrightarrow \phi^L$, if $\phi$ is a $\Delta_1^{ZF}$ formula.

So, if you have had the kindness to read up to this point, my questions are the following:

a) Is my reasoning correct?

b) If my reasoning is correct, is $ZF \vdash \phi^M$ a "good" definition of $M \models \phi$ (M proper class)? "Good" meaning "consistent with Tarski's definition of satisfaction.

c) If my reasoning is correct, is there a easy way to show that the absoluteness properties remain the same even for a model which is a proper class?

d) How can we speak about such a thing as $V$? I am really uncomfortable about it.

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For your main question, you are right that there is a subtle issue with the claim that $$V\models\phi\iff L\models\phi\text{ for }\Delta_1^{ZF}\text{ assertions }\phi.\qquad\qquad (\star)$$ But your interpretation is not as strong as one can give here.

First, supporting your worries, let's point out that we cannot expect to prove the theorem as a single assertion in ZF. This is just because if we happen to be living in a world where $\neg\text{Con}(ZF)$, then technically every formula is $\Delta_1^{ZF}$, because for a formula $\phi$ to be $\Delta_1^{ZF}$ means that it is provably equivalent to a $\Sigma_1$ assertion and also provably equivalent to a $\Pi_1$ assertion. In a world where ZF proves anything, then every formula is $\Delta_1^{ZF}$. But we can live in such a world where ZF is inconsistent but $V$ and $L$ satisfy different formulas. Thus, Drake cannot be claiming that $(\star)$ is a theorem of ZF.

Rather, on the positive side, Drake is likely claiming $(\star)$ as a meta-theoretic assertion, a theorem scheme. (And most uses of $\Delta_1^{ZF}$ amount to such theorem schemes or claims in the meta-theory.) Specifically, what $(\star)$ is asserting is that if in the meta-theory we happen to observe that $\phi$ has complexity $\Delta_1^{ZF}$, which means that in the meta-theory we have a proof in ZF that $\phi$ is equivalent to a $\Sigma_1$ assertion and also another proof that it is equivalent to a $\Pi_1$ assertion, then for this particular $\phi$ we also have a proof that $\phi\iff\phi^L$, or in other words, that $V\models\phi$ if and only if $L\models \phi$. This is easy, since the $\Sigma_1$ variation is upwards absolute and the $\Pi_1$ variation is downwards absolute.

Your interpretation of the claim is not about truth in $V$ for such assertions $\phi$, but merely about provability of $\phi$ in all models of ZF, and so it fails to apply in many instances where the scheme version of the claim does apply.

Thus, I would say that your proposal to interpret $M\models \phi$ by $\text{ZF}\vdash\phi^M$ is flawed. If $M$ is any definable class, then for any particular $\phi$, we may run the Tarski definitiion of truth sufficiently to define what it means for $M\models\phi$. Slightly more generally, for any meta-theoretic natural number $n$, we can write down the Tarskian truth definition for truth in $M$ of any given assertions of complexity $\Sigma_n$.

What we can't do, and this is likely the source of your worries, is give ourselves a full account of truth-in-$M$ for all formulas. This is precisely forbidden by Tarski's theorem on the non-definability of truth. But when you restrict the complexity of the assertions, as you do here to $\Delta_1$, then we have a fully robust $\Sigma_n$ theory of truth applicable to any class to which we can refer.

Finally, to assert that $V\models\phi$ is the same as to assert $\phi$. But it is sometimes convenient to have a notation for the class of "everything", when one is comparing this universe to inner models such as $L$.

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Thanks a lot. So is it correct to say that, if $\phi$ is a $\Delta_1^{ZF}$ formula then $$V \models \phi \Leftrightarrow L \models \phi$$ implies for this particular $\phi$ $$ZF \vdash \phi \Leftrightarrow ZF \vdash \phi^L?$$ For, if this is not true I cannot see how Drake can use the metatheoretic claim in order to prove finitistic consistency results (which, for example, are based on the fact that $ZF \vdash \phi^L$ for all $\phi$ axiom of ZF). –  Archbishop Jul 4 '12 at 15:53
Yes, and even more: if $\phi$ is a particular formula known to be provably $\Delta_1$, then $ZF\vdash \phi\iff\phi^L$, and from this it follows easily that $ZF\vdash\phi\iff ZF\vdash\phi^L$. –  Joel David Hamkins Jul 4 '12 at 16:02

Joel has covered most of this topic very well, but let me first amplify one of the points he made. As he says, the proposal to interpret $M\models\phi$ as $\text{ZF}\vdash\phi^M$ is flawed. There is, however, a correct interpretation of $M\models\phi$ that looks rather similar to this proposal, namely just $\phi^M$ (i.e., delete "$\text{ZF}\vdash$" from the flawed proposal). In other words, truth of one sentence (or satisfaction of one formula) in a definable class is extremely easy to define. This subsumes the (currently) next-to-last sentence in Joel's answer, because $\phi^V$ is simply $\phi$.

As Joel also explains, one can define truth in proper classes for far more than a single formula, namely for the class of all $\Sigma_n$ formulas for any fixed $n$, but this is considerably more complicated than what I wrote above about a single $\phi$.

Finally, let me also comment on the line between the last two shaded boxes in the question, "i.e., from the Completeness Theorem and the definition of satisfaction". You seem to believe that "$T\vdash\alpha$ iff $T\vdash\beta$" says the same thing as "$T\vdash(\alpha\iff\beta)$". (You only state this belief for a particular $T$, $\alpha$ and $\beta$, so I apologize if I've incorrectly interpreted the generality of your belief.) In fact, these two do not say the same thing. The second is stronger than the first. The first only means (via the Completeness Theorem) that if all models of $T$ satisfy $\alpha$ then they all satisfy $\beta$ and vice versa; it would allow for $\alpha$ and $\beta$ to be true for entirely different collections of models of $T$ as long as each fails in at least one model. The second, in contrast, means that exactly the same models of $T$ satisfy $\alpha$ as satisfy $\beta$.

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