Is it necessary that model of theory is a set?

Question

From Model Theory article from wikipedia : "A theory is satisfiable if it has a model $ M\models T$ i.e. a structure (of the appropriate signature) which satisfies all the sentences in the set $T$". In structure definition there is also requirement for "container of a structure" to be set.

As we assume then, every model have to lay inside of some set-container. This obviously give us in serious trouble, as for set theory, there is no model of this type, and even maybe cannot be. One of possible explanations why set theory cannot be closed inside set ( which will lead us to some well known paradoxes) is opinion that "there can be no end to the process of set formation" so we have "structure" which cannot be closed inside itself which is obviously rather well state.

As we know that not every theory may have a model (see set-theory) then some question arises:

• What are the coses (other than pure practical - if they are set we know how to work with them) of putting so strong requirements for model to be set?

• Is there any way to weaken this requirement?

• Are there any "explorations" of possible extension of model theory with fundamental objects other than sets?

I presume that some categorical point of view may be useful here, but is there any? I am aware about questions asked before, specially here 8731, but it was asked in context of category theory which is of course valid point of view but somehow too fine. I would like to ask in general.

And last one, philosophical question: is then justified, that condition for a theory to have model in set universe is some kind of anthropomorphic point of view - just because we cannot build any other structures in effective way we build what is accessible for us way but it has no objective nor universal meaning? Is true that model theory is only a "universal algebra+logic" in universe of the set, or it justifications may be extended to some broader point of view? If yes: which one?

I have hope that this question is good enough for mathoverflow: at least please try to deal wit as kindly request for references.

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Remark: Well formulated point from n-CathegoryCafe discussion: "In the centre of Model Theory there is " fundamental existence theorem says that the syntactic analysis of a theory [the existence or non-existence of a contradiction] is equivalent to the semantic analysis of a theory [the existence or non-existence of a model]."

In fact the most important point is: may it be extended on non "set container" universes?

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I would like to thank everyone who put here some comments or answers. In is the most interesting that in a light of answer of Joel David Hamkins it seems that for first order theories (FOT) consistency is equivalent to having set model. It is nontrivial, because it is no matter of somehow arbitrary definition of "having model" but it is related to constructive proof of Completeness Theorem of Gödel. From ontological point of view it then states that for FOT there is no weaker type of consistency than arising from model theory based on sets, and in some way it is maximal form of consistency simultaneously. Then there is no way to extend for FOT equivalence to non-set containers, which is nontrivial part - the only theories which are consistent in FOT are those which has a set models and this statement is proved not using set theoretical constructions in nonconstructive ways. So it was important to me, and I learn a lot from this even if for specialist it is somehow maybe obvious. I have hope that I understand it right;-)

@Tran Chieu Minh: thank You for pointing to interesting discussion, I will try to understand the meaning of Your remarks here and there.

Answer 1

You seem to believe that it is somehow contradictory to have a set model of ZFC inside another model of ZFC. But this belief is mistaken.

As Gerald Edgar correctly points out, the Completeness Theorem of first order logic asserts that if a theory is consistent (i.e. proves no contradiction), then it has a countable model. To be sure, the proof of the Completeness Theorem is fairly constructive, for the model is built directly out of the syntactic objects (Henkin constants) in an expanded language. To re-iterate, since you have mentioned several times that you find something problematic with it:

• Completeness Theorem. (Goedel 1929) If a theory is consistent, then it has a model that is a set.

The converse is much easier, so in fact we know that a theory is consistent if and only if it has a set model. This is the answer to your question.

More generally, if a theory is consistent, then the upward Lowenheim-Skolem theorem shows that it has models of every larger cardinality, all of which are sets. In particular, since the language of set theory is countable, it follows that if ZFC is consistent, then it has models of any given (set) cardinality.

The heart of your confusion appears to be the mistaken belief that somehow there cannot be a model M of ZFC inside another structure V satisfying ZFC. Perhaps you believe that if M is a model of ZFC, then it must be closed under all the set-building operations that exist in V. For example, consider a set X inside M. For M to satisfy the Power Set axiom, perhaps you might think that M must have the full power set P(X). But this is not so. All that is required is that M have a set P, which contains as members all the subsets of X that exist in M. Thus, M's version of the power set of X may be much smaller than the power set of X as computed in V. In other words, the concept of being the power set of X is not absolute between M and V.

Different models of set theory can disagree about the power set of a set they have in common, and about many other things, such as whether a given set is countable, whether the Continuum Hypothesis holds, and so on. Some of the most interesting arguments in set theory work by analyzing and taking advantage of such absoluteness and non-absoluteness phenomenon.

Perhaps your confusion about models-in-models arises from the belief that if there is a model of ZFC inside another model of ZFC, then this would somehow mean that we've proved that ZFC is consistent. But this also is not right. Perhaps some models of ZFC have models of ZFC inside them, and others think that there is no model of ZFC. If ZFC is consistent, then these latter type of models definitely exist.

• Incompleteness Theorem. (Goedel 1931) If a (representable) theory T is consistent (and sufficiently strong to interpret basic arithmetic), then T does not prove the assertion "T is consistent". Thus, there is a model of T which believes T is inconsistent.

In particular, if ZFC is consistent, then there will be models M of ZFC that believe that there is no model of ZFC. In the case that ZFC+Con(ZFC) is consistent, then some models of ZFC will have models of ZFC inside them, and some will believe that there are no such models.

A final subtle point, which I hesitate to mention because it can be confusing even to experts, is that it is a theorem that every model M of ZFC has an object m inside it which M believes to be a first order structure in the language of set theory, and if we look at m from outside M, and view it as a structure of its own, then m is a model of full ZFC. This was recently observed by Brice Halmi, but related observations are quite classical. The proof is that if M is an ω-model, then it will have the same ZFC as we do in the meta-theory and the same proofs, and so it will think ZFC is consistent (since we do), and so it will have a model. If M is not an ω-model, then we may look at the fragments of the (nonstandard) ZFC inside M that are true in some V_α of M. Every standard fragment is true in some such set in M by the Reflection Theorem. Thus, by overspill (since M cannot see the standard cut of its ω) there must be some V_α in M that satisfies a nonstandard fragment of its ZFC. Such a model m = V_α^M will therefore satisfy all of the standard ZFC. But M may not look upon it as a model of ZFC, since M has nonstandard axioms which it thinks may fail in m.

Answer 2

The fundamental reason why models in model theory are required to be sets is that for such models $M$ there exists the satisfaction relation $M\models\phi[e]$ between formulas $\phi$ and evaluations $e$, obeying Tarski’s definition. This is not possible in general for class models in ZFC (or NBG)—for example, the existence of a satisfaction relation for $(V,\in)$ would imply the consistency of ZFC, which is not provable in ZFC.

There are other reasons for sticking to set models (for example, various constructions of models, in particular those using the compactness theorem, tend not to work for proper classes), but this is the most important one. On the other hand, in some parts of model theory it is convenient to work inside a “monster model” which may be a proper class (at least metaphorically, i.e., large enough so that it behaves as if it were a proper class for all arguments we want to employ it in), but one has to be careful when working in such a setup.

Answer 3

Gerald's answer is quite correct. This began as a comment justifying it, but because of length considerations I'm leaving it as an answer instead.

A model of ZFC is not a perfect replica of the category of sets crammed into a single set (in fact, to the extent that such a statement makes any sense at all, there is no such thing). Rather it is a set $M$ together with a binary relation, $\in$, such that the axioms of ZFC set theory -- i.e., a certain family of first order statement in the (countable) language of sets -- holds in $(M,\in)$.

There are a lot of things that such a model $M$ will not tell you about the category of sets. However, assuming -- as we generally do -- that there really is a category of sets satisfying each of the axioms of ZFC, then it follows from Godel's Completeness Theorem that it must have a model $M$ as above. This is a nontrivial result. Moreover, since the language of sets is countable it follows from Skolem-Lowenheim that models of ZFC exist of all infinite cardinalities.

To be fair, Skolem himself found this consequence -- the existence of a countable model of set theory -- to be somehow problematic ("Skolem's Paradox"). But modern set theorists and logicians simply don't feel this way: they have gotten used to the statement, which is not paradoxical in the strict logical sense (there is no contradiction) but rather merely sounds strange when you first hear it.

In much the same way, Tarski's motivation for the Banach-Tarski paradox was to exhibit the absurdity of the Axiom of Choice (AC). But nowadays 99.9% of all mathematicians are happy with AC, and the study of "paradoxical decompositions" is a flourishing subfield of geometric group theory.

As Joel David Hamkins explained to me previously here on MO, you could consider class-valued models of a theory and, depending upon taste, it may sometimes be convenient to do so. But again it is not necessary to do so because of Godel Completeness.

Answer 4

In some cases one takes a proper class as the model. But in fact if a theory (with finitely many symbols) is consistent then it has a countable model, and thus a model which is a set.

Answer 5

I think that this question was down-rated too quickly. It appears to me that modulo the confusion which was pointed out by previous posts there are some valid point that need to be addressed.

I just post my opinion. I don't think I have enough knowledge to give justification for them. I hope this will give some more fuel for the discussion.

1) I don't think there is truly much essential point to require a model to be a set in the sense of the Platonic concept of set i.e. to require that $ a \in b$ means $a$ is an element of $b$ in the real sense. An arbitrary interpretation of "is an element" will work. I think the more important requirement here is the requirement of "closed system" i.e. relations are defined everywhere, functions defined everywhere and won't give you an outside element. So being a set is a sufficient condition to capture this notion of "closed system" (in a philosophical sense). But it is not necessary.

2) That points out some way to weaken the requirement.

3) I don't know. I gave you the link to my previous question. I think Angus Macyntyre did suggest something like that. I have not find time to sit down and read all of them properly. My belief is that at least a category theory language is possible, but I don't know whether it can be extended to other kind of objects or not.