Model Theory – Is It Necessary That Model of Theory is a Set?

big-picturemodel-theoryreference-request

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.


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?


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.

Best Answer

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 Vα 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 Vα in M that satisfies a nonstandard fragment of its ZFC. Such a model m = VαM 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.

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