# Models of set theory

How can one talk of a semantics or model of set theory (lets say ZF or ZFC) when the definition of a structure (and potential model) needs a carrier set in the first place (by its definition)?

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Models are something you deal with inside some other theory. You can't reasonably ask for a model of every theory prior to any use of a theory... – Malice Vidrine Aug 28 '14 at 19:33
A similar question was asked on MathOverflow, which I answered at length: mathoverflow.net/questions/23060/set-theory-and-model-theory/… I recommend the answer :-) – Andrej Bauer Aug 28 '14 at 20:33
Perhaps the gist of my answer was the last sentence: if you ask "why are model theorists justified in using sets?" then I ask back "why are number theorists justified in using numbers?" – Andrej Bauer Aug 28 '14 at 20:34
@Malive that sounds reasonable - is there a common sense what is ingredient to any definition of model of a theory then? In beginner logic class this is clearly defined the usual(?) way with a carrier set and the usual symbols and the interpretation of all symbols in that structure that makes all sentences of a theory true. Now and then that definition is changed.. but in the end a theorem stating that a theory is consisten if and only if it has a model is valid for all these understandings of models? – YonedaLemma Aug 28 '14 at 21:00
@Andrej Thanks that is a very good article! – YonedaLemma Aug 28 '14 at 21:00

One can talk about $\sf ZFC$ in much weaker theories, like $\sf PRA$ or other fragments of arithmetic. These are sufficient to develop internally the basics rules of logic, and talk about axiom schemas and so on.

Then talking about models of $\sf ZFC$ is the same as talking about models of any other theory. Since $\sf ZFC$ is just a bunch of axioms, now defined internally as sets with certain properties, and a model of $\sf ZFC$ is just a set satisfying these axioms.

(And these don't exist in the meta-level of $\sf PRA$ or whatever, but models of group theory, ring theory or vector spaces don't exist in $\sf PRA$ either.

So if all that bothers you, ask yourself how we can talk about natural numbers before we have $\sf PRA$ and how we can talk about $\sf PRA$ before we have the natural numbers.)

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There is no reason one cannot talk about ZFC in ZFC, as far as I know? But one has to be careful to keep the model and the universe in which the model is studied separate. – Harald Hanche-Olsen Aug 28 '14 at 18:45
Yes, of course. – Asaf Karagila Aug 28 '14 at 18:46
@HaraldHanche-Olsen, just as one needs to be careful not to eat the wrapping of candies! – Mariano Suárez-Alvarez Aug 28 '14 at 19:13
@Mariano: But that's the best part. :( – Asaf Karagila Aug 28 '14 at 19:15
Mi little nephew seems to agree with you! :P – Mariano Suárez-Alvarez Aug 28 '14 at 19:27

I already answered a similar question at length on MathOverflow so let me try a short answer, suitable for this forum.

You mistakenly assume that it is the business of model theory or logic to somehow "produce" set theory out of nothing. Logic and model theory are just two branches of ordinary mathematics -- they do not precede them, although they are a bit peculiar because their object of study is mathematics itself (its methods, its possiblities, its limitations). Therefore, model theorists and logicians are "allowed" to use all the usual tools of mathematics (numbers, sets, topological spaces, and so on).

When logicians speak of "foundations" of mathematics, they may give the impression that they are "building the cathedral" starting from its foundation. But it is much better to view what they are doing as a study of how the cathedral is built and how we can improve it. For instance, logicians have observed the fact that almost all of modern mathematics can be expressed in the language of set theory, but this does not mean that we need to "secure" set theory before the rest of mathematics can be done. History is my witness: geometry, algebra, and analysis existed before set theory and logic came along.

P.S. I will not deny that historically a primary objective of logic was in fact to secure foundations, especially the kind of logic that Bertrand Russell did in his Principia mathematica. However, at least since Gödel we have known that such an endeavor must fail. In any case, I am expressing here my personal view that attempts to secure absolute foundations of mathematics are a bit like attempts to prove there is god. Ultimately it comes down to an act of faith.

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thanks - yes you are right, I am not so much keen on asking for foundations but to sort my understanding of models and consistency, and understanding the essence of syntax and semantics as logicians see it. – YonedaLemma Aug 28 '14 at 22:01
(and same to you - unfortunately one can only choose one official answer) – YonedaLemma Aug 28 '14 at 22:07

As you may be aware, by the Godel Incompleteness Theorem, it can not be proved in, for example, $\mathsf{ZFC}$ that there are models of $\mathsf{ZFC}$, one usually assumes $\mathsf{ZFC}$ is consistent and result are usually states as "If there exists a model of $\mathsf{ZFC}$, then ...". Another way to think of this is (again assuming the consistency of $\mathsf{ZFC}$, one assumes that we are working in a single fixed model $V$ and studying all the models that that $V$ has.

Another possible approach for studying models of $\mathsf{ZFC}$ is work in an axiom system that implies the existence of models of $\sf{ZFC}$. The common approach to this is augmenting $\sf{ZFC}$ with large cardinal axioms such as the inaccessible, measurable, etc. Then there actually exists a set model of $\sf{ZFC}$ and possible much more.

Finally, another common approach is rather than study model of the full $\sf{ZFC}$, one can instead study $\{\in\}$-structures satisfying certain axioms and having certain properties (like transitivity). By the usual compactness argument in logic, one can get nice models for finite amount of $\text{ZFC}$. For those who are doing independence results in set theory, models of finite amount of $\sf{ZFC}$ is often sufficent (essentially due to the fact that proofs of theorems are finite). This is a common approach to models used in forcing.

It should also be noted that sometimes in proving independence result, one works with "models" that are not even sets at all. These are really proper classes (or you could think of them as just formulas). The common examples are the well-founded $\text{WF}$ and Constructible $L$. Although these are not sets, they are used to prove independence results through relativization.

I suggest reading the appropriate sections of Kunen's Set Theory for the various mathematical formal approach to handling models of set theory.

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@YonedaLemma Godel Incompleteness implies that $\sf{ZFC}$ can not prove there is a model of $\sf{ZFC}$. – William Aug 28 '14 at 19:55
@YonedaLemma Sorry typo missing "not" in the first sentence. – William Aug 28 '14 at 19:56
Thanks very much! I am just thinking of a theorem like a theory is consisten if and only if it has a model - is this true for all the slightly different definitions of models? (At least I understand that $V$ assumes in the beginning the consistency of ZFC and therefore all models one constructs on this background do not proof the consistency of ZFC.) – YonedaLemma Aug 28 '14 at 21:09
(..and I would also accept your answer as the official answer but just because I can only choose one and the others where faster.) – YonedaLemma Aug 28 '14 at 21:58