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In model theory, if L is a first order language, by the definition of a L-structure $\mathcal{M}$ it is partly given by a non-empty set $M$ called the universe or domain of $\mathcal{M}$.

From where did we get this set? I know how I could define a first order set theory by giving the language L and then axioms of ZFC for example. Do we assume the existence of the set theory when reasoning about the models, interpretation, embeddings, etc.? What kind of set theory is it? What statements about this set theory are true and which or not?

In short: Is there any starting point, if yes, what is it?

Asaf Karagila
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Dávid Natingga
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    You may find useful my comment here. – Andrés E. Caicedo Jul 21 '13 at 17:37
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    In model theory, just like in commutative algebra, or partial differential equations, we are doing ordinary mathematics. Nowadays, most mathematics is done using set-theoretic language. And most proofs could in principle be written as formal ZFC proofs, though for obvious reasons one does not do this (too messy, and it would hide the intuition). – André Nicolas Jul 21 '13 at 17:39

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Yes. Usually we do model theory in $\sf ZFC$, but sometimes we need to assume further assumptions (e.g. $\sf CH$ or large cardinals of some sort, etc.).

Thinking about it deeper, if the language is infinite, where does it live? If it is uncountable then we can't even fully express it in "plain logic" without appealing to some set theory.

I'm not sure what statements are "true about this set theory", because that would probably solve a lot of open questions in mathematics.

Asaf Karagila
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  • Well, basic model theory can be developed with only some rudimentary or naive set theory. (Like most branches of mathematics.) That said, advanced model theory is almost exclusively done with ZFC as the metatheory (like most branches of mathematics). – tomasz Jul 22 '13 at 20:07
  • @tomasz: I'd say $\sf ZFC$ and then some. To assure saturated models, if I recall correctly, you need some $\kappa$ such that $\kappa^{<\kappa}=\kappa$. This is independent of $\sf ZFC$ so you need to assume that something like that exists. It can also be beneficial to use large cardinals from time to time. So perhaps it's better to be more careful and say that we work in an extension of $\sf ZFC$ as a metatheory. – Asaf Karagila Jul 22 '13 at 20:16
  • You don't really need saturated models. They're easier to think about, but all you really need is $\kappa$-saturated and $\kappa$-strongly homogeneous for sufficiently large $\kappa$, and to make those ZFC is enough. But sure, sometimes extensions are used. For instance, when we're dealing with absolute things we can assume CH without any reduction of generality. – tomasz Jul 22 '13 at 20:38
  • In any case, I'm pretty sure most things done using the monster model can be done without it, and it's really there for sake of simplicity, like CTMs in set theory. – tomasz Jul 22 '13 at 20:44