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Am a beginner here. I would like to clarify the notation $\models$ used in set theory / model theory. I've read that given a structure $\mathfrak{M}$ and a set of sentences $\Sigma$ under a first-order language $\mathcal{L}$, we have $\mathfrak{M} \models \Sigma$ if every $\varphi \in \Sigma$ is "true" in $\mathfrak{M}$, or that $\mathfrak{M}$ is a model for $\Sigma$.

But how would we interpret the notation $\models$ ? I am confused with these two ways to understand $\mathfrak{M} \models \Sigma$:

(1) For each $\varphi \in \Sigma$, show that $\varphi$ is true in $\mathfrak{M}$ without thinking of $\models$ as a relation in the form of a truth definition

(2) The notation $\models$ refers to a truth definition relation in the metatheory where for all sentences $\phi$, we have $(\phi,T) \in \models$ iff $\phi$ is true in $\mathfrak{M}$ and $(\phi,F)$ otherwise. Then $\mathfrak{M} \models \Sigma$ iff for all $\varphi \in \Sigma$, we have $(\varphi,T) \in \models$. (but this definition has to be in the metatheory right ? since we are not assuming any Axioms here). If the domain of $\mathfrak{M}$ is a set, this can be done by recursion on the length of all formulas $\phi$ to come up with a truth definition (where the recursion on all possible $\phi$ is done as well in the metatheory) ...

So for example, if $\mathfrak{M}$ is a model for $\Sigma$, should it be interpreted as (1) or (2) ?

And if (2), am I correct that the recursion (for set models) is done in the metatheory without assuming any axioms, i.e. say replacement ?

EDIT (part 1):

As an example of what I am trying to ask here. If say, $\mathcal{L} = \{\epsilon\}$ and $\Sigma=\{\text{an infinite set exists}\}$, and say $\mathfrak{M}=\{V,\epsilon\}$, i.e. the proper class of all sets, then if we take (1) as an interpretation, we have $\mathfrak{M} \models \Sigma$, since we know that $\omega \in V$ by some finitistic proof in the metatheory. However, if we take (2) as the definition for $\models$, then it's not possible to have a truth definition in $\mathfrak{M}$ (even in the metatheory) by Tarski's undefinability theorem ...

EDIT (part 2):

I saw this link, and the accepted answer uses $\models$ for logical consequence, i.e. $A \models B$, where $A$ and $B$ are sentences of $\mathcal{L}$. But I am interested in $\mathfrak{M} \models A$, where $\mathfrak{M}$ is a structure and $A$ is a set of sentences of $\mathcal{L}$. One of the answers though said that $\mathfrak{M} \models A$ stands for $\models$ being the satisfaction relation or truth definition. So is that always the case ?

spacewindow
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    What difference do you see between (1) and (2)? My point is: when you unpack the meaning of "$\mathfrak{M}$ satisfies $\varphi$", don't you get (2)? If not, what's the distinction? – Alex Kruckman Jan 28 '25 at 23:30
  • @AlexKruckman, well I guess with (2) we have to evaluate all possible formulas, even those that are not in $\Sigma$, but for (1) we can evaluate only formulas in $\Sigma$ ... also in (2), we can "capture" the recursion by some formula ... – spacewindow Jan 28 '25 at 23:34
  • I don't really follow the question either. A structure can be thought of as a "mathematical universe" in which all the formal terms and sentences of the language $\mathcal L$ obtain meanings and truth values. The assignment of meanings (including the satisfaction relation between models and sentences, denoted by $\models$) is defined inductively. All of this takes place in the metatheory (as "$\models$" is not part of $\mathcal L$). It might help to look at some concrete examples with simple languages and small models. – Karl Jan 29 '25 at 00:09
  • @Karl okay I edited my question to hopefully make it clearer – spacewindow Jan 29 '25 at 00:30
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    Hmm, the example you added seems to be complicating/confusing things by adding another level of "meta". The domain of a structure is always a set, because the whole topic of model theory is developed using a set-theoretic framework (which we call "the metatheory" if we have to talk about it, but we prefer to just use it). Just because we're talking about models of a formal language that also looks set-theoretic doesn't mean we can start referring to our own proper classes. – Karl Jan 29 '25 at 00:47
  • @Karl, oh okay ... thanks for your comment, it's just that I thought of this question while reading a set theory book that has a subsection in model theory (i.e. Kunen set theory)... so some examples in the book also used proper class models – spacewindow Jan 29 '25 at 00:49
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    Ah, maybe you could include an example from that book. Using a proper class as a model certainly raises questions about how satisfaction is defined! (It's possible that some class function approach is implied, but that seems pretty sloppy to me.) – Karl Jan 29 '25 at 01:06

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