## Intended interpretations of set theories

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In his Set Theory. An Introduction to Indepencence Proofs, Kunen develops $ZFC$ from a platonistic point of view because he believes that this is pedagogically easier. When he talks about the intended interpretation of set theory he says such things as, for example, that the domain of discourse $V$ is the collection of all (well-founded, when foundation is introduced) hereditary sets.

This point of view has always made me feel a bit uncomfortable. How can a variable in a first-order language run over the elements of a collection that is not a set? Only recently I realized that one thing is to be a platonist, and another thing is to believe such an odd thing.

A first-order theory of sets with a countable language can only prove the existence of countably many sets. Let me call them provable sets for short. Platonistically, we wish our intended interpretation of that theory to be one in which every provable set is actually the set the theory says it is. So we don't need our interpretation to contain every set, we just need that it contains at least the true provable sets. This collection is, really, a set, although it doesn't know it.

To be a bit more concrete, if one is a platonist and the cumulative hierarchy is what one has in mind as the real universe of sets, one can think that the $V$ of one's theory actually refers to a an initial segment of that hierarchy, hence variables run no more over the real $V$ but only over the elements of some $V_\alpha$.

There's a parallel to these ideas. For example, when we want to prove consistency with $ZFC$ of a given sentence, we do not directly look for a model of $ZFC$ where that sentence is true, but instead we take advantage of knowing that every finite fragment of $ZFC$ is consistent and that every proof involves only finitely many axioms.

My question is: then, is this position tenable or am I going awfully wrong? I apologize that this seems a philosophical issue rather than a mathematical one. I also apologize for stating things so simply (out of laziness).

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It's a pity only one answer can be accepted. Thanks a lot. – Marc Alcobé García Feb 15 2011 at 7:25

## 5 Answers

While Kunen takes for universe the collection of all hereditary sets, Marc proposes to restrict the universe to those hereditary sets which are first-order definable without parameters (Marc's "provable sets"). To make this more mathematical, let me rephrase the question as follows:

Suppose $M$ is a model of ZF. Does the subset $K$ of $M$ consisting of the first-order definable (without parameters) elements of $M$ form a model of ZF?

This turns out to be a delicate question which was answered here on MO some time ago. To summarize the answer there, Marc's proposed universe $K$ will be a model of ZF if and only if $M$ is a model of ZF + V = OD.

Returning to the more philosophical question, this says that Marc's proposal is tenable if and only if there is a reason to believe that the hereditary sets are all ordinal definable. Occam's Razor type arguments support that V = OD and, indeed, V = L is true. However, since it is much weaker than the rigid assumption V = L, the assumption V = OD is not incompatible with a much richer view of the universe...

There is another interpretation of Marc's "provable sets." Instead of merely requiring them to be first-order definable without parameters, one can require them to provably exist in any well-founded model of ZF. With this interpretation, one gets a potentially much smaller set: the minimal well-founded model of ZF. This minimal model is $L_\alpha$ where $\alpha$ is the smallest element of `$Ord\cup\{Ord\}$` such that $L_\alpha \models ZF$.

Note that it is entirely plausible that $\alpha = Ord$ (for example this will happen if we happen to live in the minimal model in question). The assumption that "$L_\alpha$ is a set" is equivalent to the existence of a well-founded model of ZF. From a philosophical standpoint, this is a rather strong assertion. For example, it implies that ZF is not only consistent but also $\omega$-consistent and much more...

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As far as I can see, if there is an ordinal $\alpha$ such that `$L_\alpha$` is a model of ZF, then the first such $\alpha$ is definable (since I've just defined it) and the corresponding `$L_\alpha$` is hereditarily definable. It seems that the collection of all hereditarily first-order definable (without parameters) sets will be considerably larger than this `$L_\alpha$`. Exactly where it lies seems to depend on lots of undecidable information about the ambient model. For example, cardinal exponentiation might encode some complicated reals, encoding complicated hereditarily countable sets. – Andreas Blass Feb 14 2011 at 18:16
You're completely right! I guess the correct adjective would be predicatively or something like that. – François G. Dorais♦ Feb 14 2011 at 19:02
Predicatively didn't seem right either, so I went with hierarchically... – François G. Dorais♦ Feb 14 2011 at 19:09
I wasn't happy with that either, so I rewrote the whole sentence. – François G. Dorais♦ Feb 15 2011 at 15:23

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If you assume that ZFC is consistent, then it follows from Gödel's completeness theorem that there is a set model of ZFC. You can then argue from a Platonistic point of view by taking the viewpoint of that set model. Note that Kunen also proves relative consistency results by moving to models that are not sets with respect to the viewpoint of your model. For example, he proves the relative consistency of the GCH with ZFC by moving to the inner model L. The reason he talks about fragments of ZFC for relative consistency results obtained via forcing is because CON(ZFC) does not guarantee the existence of a countable transitive model of ZFC.

Edit: Let me now put forth a more complete answer tying together philosophical and mathematical considerations. A strict finitist doesn't believe in the existence of the set of Natural numbers but that in itself does not prevent the individual from talking about properties shared by all of its elements. Even if you don't believe that it is a static collection of elements that can be put together into a single set, you can still syntactically prove theorems about Natural numbers (as discussed in more detail by Carl with reference to ZFC). But since $\mathbb{N}$ does not exist in this platonistic frame of thinking, your intuitive objection about quantifying over all well-founded sets is analogous to this problem where we try to quantify over all Natural numbers. Ultimately, some finitists will adopt a compromise where they accept the existence of infinite sets but only ones that can be constructed from the Natural numbers in a finitistic manner (e.g., expressible in a relatively weak theory such as primitive recursive arithmetic or PRA). A somewhat analogous resolution to your philosophical objection is to allow for the existence of definable classes as Kunen does by considering them as "abbreviations for expressions not involving them" or to treat classes as separate formal objects in a conservative extension of ZFC such as NBG. Of course, if you assume the existence of an inaccessible cardinal $\kappa$, then you have a nice set model of NBG, mainly the collection of $\Delta_0$-definable subsets of $V_{\kappa}$.

You can also take a semantic point of view but with a syntactic twist. Specifically, you can pretend that you live in a model of ZFC, but you don't know which one. You therefore can assert the existence of sets provable from the axioms of ZFC while acknowledging that there are sets out of reach. Similarly, you can assert ZFC-provable properties of all of the sets in your unspecified universe while realizing that you are unaware of the boolean truth value of statements independent of ZFC. This line of thinking is indeed present with boolean-valued models of ZFC where you talk about the probabilities of certain statements being true or certain sets given by names coming to fruition in forcing extensions.

Finally, you can throw yourself in a constructed set model $M$ guaranteed by ZFC's believed consistency and restrict to some subclass of $M$. Specifically, you can externally take a sufficient cut of its universe $V_{\alpha}^M$ (which may be all of $M$) as you alluded to or restrict yourself to its parameter-free definable sets if it happens to be a model of $V = OD$ as François mentions to get a model of ZFC. However, in both cases, you will potentially be sacrificing some of the richness exhibited by the universe $M$.

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To begin, its currently 6:00 am here and I am not entirely alive mentally, so if my answer is missing the point, let me know. In addition, I just thought I might make some additional points or comments not already mentioned in the other answers.

This is a common point of contention, and honestly there really isn't a good answer that avoids the issues of playing fast and loose with the consistency of $ZF$. Everyone really just comes up with a meta-mathematical justification which sort of sits the best with them and it really is a matter of philosophy and personality.

Firstly, with respect to this comment:

This point of view has always made me feel a bit uncomfortable. How can a variable in a first-order language run over the elements of a collection that is not a set? Only recently I realized that one thing is to be a platonist, an another thing is to believe such an odd thing.

Formally within the confines of $ZFC$ a variable cannot range over anything that is not a set. However this does not stop us from viewing or speaking about things we cannot formalize within $ZFC$, and Kunen takes liberties with this point of view. The thing that is actually going on when he talks about variables which range over proper classes, is we have just stepped outside of $ZFC$ and are now conversing in the meta-theory.

Secondly the comment you make here:

To be a bit more concrete, if one is a platonist and the cumulative hierarchy is what one has in mind as the real universe of sets, one can think that the V of one's theory actually refers to a an initial segment of that hierarchy, hence variables run no more over the real V but only over the elements of some $V_{\alpha}$.

is a very keen observation, and foreshadows a nice understanding of inaccessible cardinals assumptions. You see by postulating the existence of a strongly inaccessible cardinal we are in fact assuming that we have an initial segment of the cumulative hierarchy which satisfies all of $ZFC$. The only problem with this is that in doing so we have stepped up the consistency strength of system we are working in.

There's a parallel to these ideas. For example, when we want to prove consistency with ZFC of a given sentence, we do not directly look for a model of ZFC where that sentence is true, but instead we take advantage of knowing that every finite fragment of ZFC is consistent and that every proof involves only finitely many axioms.

is quite dead on. Because applying Levy reflection, coupled with the downward Löwenheim–Skolem theorem we can produce a countable model of enough of $ZFC$ for whatever argument we care to be trying to formulate. This view provides a lot of comfort when trying to construct forcing arguments. However, if this view is not comforting, there is a differing view on the matter, and it makes an appearance in the way of Boolean-valued models (which Kunen's text is kinda lite on).

EDIT: fixed some dumb typos.

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I have a few comments that I hope are useful even if they don't clarify things completely. As Michael Blackmon says, different people have different ways of resolving things to their own satisfaction. In the end, attempts to resolve things by reasoning about $V$ in a natural-language set-based metatheory are always going to be complicated because of the set-theoretic paradoxes. If these didn't exist, we could define $V$ in the naive way and not worry about it. But the paradoxes tell us that the idea of "the collection of all pure well-founded sets" is not as simple as we might have hoped.

1) "How can a variable in a first-order language run over the elements of a collection that is not a set?" This question seems to relate to the fact that most set theory books do not work with a semantic metatheory in the way that model theory books do. The counter-question is: from what perspective are we handling an interpretation of the language?

From the viewpoint of the object theory, quantifiers are no problem. By analogy, in Peano arithmetic we can quantify over all natural numbers even though we have no sets at all.

The way that most set theory books treat things, you want to pretend you are working in a metatheory like PRA that performs only syntactic (uninterpreted) manipulations of formulas. From this metatheoretic point of view, the variables of a ZFC formula don't range over anything at all, because the metatheory does not attempt to interpret them. The advantage of this approach are that it side-steps many philosophical problems, and it gives extra oomph to relative consistency results. The disadvantage is that it is divorced from the semantic way that set theorists actually think about models of ZFC.

It would be perfectly possible to work instead with an actual semantic metatheory that can handle models of ZFC as objects. In that situation, though, it wouldn't be the case that interpreted quantifiers would range over something that isn't a set, because now the interpretations are all sets; the variables under a certain interpretation would range over the object-sets of that interpretation, not over meta-sets. There is no reason, strictly speaking, that this metatheory would even have to be a set theory. For example, you could use something based on the multiverse axioms of Gitman and Hamkins, which is vaguely analogous to a category-theoretic axiomatization of the category of models of ZFC. In that case, it might not even be possible to directly quantify over meta-sets in the metatheory.

2) It isn't necessary to view "V" as a meta-theoretic definition. Instead, you can simply think of it as an object-theory definition, which stratifies the universe of discourse of a particular model into the levels of the cumulative hierarchy. Each interpretation has its own idea what $V$ denotes. In other words, there's no harm done if you just pretend you have fixed a particular model of ZFC and are working inside it. This is essentially what most set theory books do, even if they don't come out and say it. This means you can ignore any other models that might or might not exist; you've committed to just one of them.

3) The idea that the $V$ from any one model is always an initial segment of $V$ in another model was proposed by Zermelo (1930) "On boundary numbers and domains of sets", translation in From Kant to Hilbert v. 2. This proposal has echoes of the notion of "absolutely infinite" from Cantor. This idea of extending $V$ reappears as one of the multiverse axioms.

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Not being able to post comments, this "answer" intends to be a comment to the question: you say "For example, when we want to prove consistency with ZFC of a given sentence, we do not directly look for a model of ZFC where that sentence is true, but instead we take advantage of knowing that every finite fragment of ZFC is consistent and that every proof involves only finitely many axioms." But as far as I know, if we knew that every finite fragment of ZFC is consistent, we would know that $all$ of ZFC is consistent, wouldn't we? I think this would follow by the compactness theorem, hence we cannot assure that every finite fragment of ZFC is consistent. Please let me know if there's something I'm not understanding well here.

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This should really be a question. However such a question would duplicate mathoverflow.net/questions/18787/… – François G. Dorais♦ Mar 3 2011 at 12:47
There are two ways to interpret "every finite fragment". You could do it with one sentence $F$, "for every $n$, the conjunction of the first $n$ axioms of ZFC is not contradictory". Or you could make an axiom scheme that contains, for every standard $n$, the sentence saying that the  conjunction of the first $n$ sentences of ZFC isn't contradictory. If you add the single sentence $F$ as an axiom to ZFC, the resulting theory does prove Con(ZFC). If you add the axiom scheme to ZFC there is no reason to think that theory would prove Con(ZFC); nonstandard models have nonstandard axioms.  – Carl Mummert Mar 3 2011 at 12:53
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