Clarifying the metalanguage of models of ZFC

Question

Note: This is a follow-up to this earlier question (spun-off as a separate question to avoid "scope-creep" in the comments).

Suppose we have a ZFC universe $\mathcal{V}$, axiomatized using some metalanguage. Now, within $\mathcal{V}$ we may want to define a model of ZFC, call it $\mathcal{M}$. This raises the question of what metalanguage will be used to define the axioms as they pertain to $\mathcal{M}$.

Per my understanding, it seems like there are (at least) two possible choices:

• Option 1: The same metalanguage as was used to axiomatize $\mathcal{V}$.
• Option 2: The "internal metalanguage" within $\mathcal{V}$, whereby we encode sequences $\omega \mapsto \omega$ as "formulas" (in the manner described in the prior question).

My question then is:

1. Is my understanding (as just described above) in fact correct?
2. If so, how does one make sense of cases where it's not explicitly stated what metalanguage is being used to talk about axioms in relation to $\mathcal{M}$? Perhaps the "default" is to assume Option #1?

Update (an example):

Consider height-potentialist multiversism, according to which any universe $\mathcal{V}$ is a set (a level of the hierarchy) in some “taller” universe $\mathcal{W}$. Presumably we use the same metalanguage to talk about satisfaction of ZFC for all universes in the multiverse. But also, from the perspective of $\mathcal{W}$, the structure $\langle \mathcal{V}, \epsilon \rangle$ is a model. Thus (it seems) we have a case of Option #1, where we’re using the same metalanguage to talk about satisfaction of ZFC axioms for both a universe and for a model within it. Or am I missing something here?

Update 2 (another example):

To take a perhaps more straightforward “textbook” example: We start with a ZFC universe $V$, which requires that we have some (perhaps not explicit) metalanguage for talking about logic and formulas. Then we discuss the $L$ of that universe also satisfying the ZFC axioms. Finally, we may also assert the possibility that $L_{\alpha}$ (for some countable ordinal $\alpha$) is a countable transitive model of ZFC. When we speak of “satisfying the ZFC axioms” in the three cases of $V$, $L$, and $L_{\alpha}$ respectively, are we assuming the same metalanguage in all three cases?

Answer 1

There are a couple of ways to look at the situation

1. We are simply working in $\sf ZFC$, and there is no literal (or at least no specific) $V$. It is an implicit domain of discourse. Then we are talking within $\sf ZFC$ about models of $\sf ZFC$.
2. We are considering a model $V$ of $\sf ZFC$ (either have one literally, or within some background paradigm as in 1), and we are considering the prospect of models of $\sf ZFC$ inside it.

In the first case, when we reason inside the theory, there is not a way to form a sentence saying "$M$ satisfies all the metalanguage $\sf ZFC$ axioms". The default is, when we say a set is a model of $\sf ZFC$, we mean this in the sense of the internal language... that we can write down in a sentence.

However, we often also consider "proper class models" such as the constructible universe $L$ (or $V$ itself, as considered informally as the class of all sets), where there may be no way to define an internal satisfaction relation. In that case, when we say "$L$ is a model of $\sf ZFC + GCH$" or what-have-you, what we mean is that $\sf ZFC$ proves each instance of the schema that all metalanguage axioms of $\sf ZFC + GCH$ hold relative to $L$. This usage is sometimes considered sloppy, and of course many of the usual model theoretical principles fail if applied naively here, e.g. working in $\sf ZFC$, $V$ is always a "proper class model" of $\sf ZFC$, but this by no means indicates that $\sf ZFC$ has proved its own consistency.

In the second case, we still usually mean relative to $V$'s internal language for models that are sets in $V$, especially if we are, as you say, “working in $V$”, which is operationally equivalent to the first case. And If course this is always what is meant when we say $V\models (M \models \sf ZFC).$ But in the case where $V$ is not an $\omega$-model, it does make sense to consider (externally to $V$) $M\in V$ that are models just of external $\sf ZFC$ (or equivalently the standard fragment of $V$'s internal $\sf ZFC$). Amusingly, these always exist when $V$ is a non-$\omega$-model, even if $V\models \lnot \sf Con(ZFC).$

And as for "proper class models" in the second case, it makes sense, for instance, to say we define $V$'s constructible universe $L^V$ in $V$ (but crucially, in the sense of a definable subset, not a definable element as for set models), and when we say it's a model of $\sf ZFC + GCH,$ it's in the same sense as how $V$ is a model of $\sf ZFC$... it's not an internal statement in $V.$

Answer 2

Option 1 is not the default - in fact, I would argue it doesn't even make sense. Working within $\mathcal{V}$ (i.e., doing mathematics), you want to give a definition of what it means for $\mathcal{M}$ to be a model of ZFC. Well, you want to write something like "$\mathcal{M}$ satisfies every axiom of ZFC". When doing mathematics, we can't quantify over statements in the metalanguage. We can only quantify over mathematical objects (such as the axioms of ZFC as constructed "inside $\mathcal{V}$" as in Option 2).

Note that to give a definition of "$\mathcal{M}$ is a model of ZFC", we also need $\mathcal{M}$ to be a mathematical object, i.e., a set, not a proper class. This is also why mathematical discussions do not begin as you do, with "Suppose we have a ZFC universe $\mathcal{V}$...". $\mathcal{V}$ is not actually an object we can talk directly about in ZFC! So (responding to your second edit), what do people mean when they say that the constructible universe $L$, which is also a proper class, is a model of ZFC?

Well, it does make sense to approximate Option 1 in a way that makes sense for proper classes, not just sets - by a "metatheorem" or "theorem schema". Given a class $U$ (proper or not), defined by a metalanguage formula, we can take any metalanugage sentence $\varphi$ and relativize its quantifiers to $U$, obtaining a new sentence $\varphi^U$ ($\forall x$ becomes $\forall x\in U$ and $\exists x$ becomes $\exists x\in U$). Then if we say "$U$ is a class model of ZFC" we really mean that for each axiom $\varphi$ of ZFC (in the metalanguage), there is a proof of $\varphi^U$. Note that this is not - cannot be - a theorem, since we cannot quantify over sentences of the metalanguage. Instead, what we mean is that if you give me an axiom $\varphi$ of ZFC, I can produce for you a proof of the theorem $\varphi^U$. But it's a different theorem (with a different proof) for each axiom. Let's call this Option 1*.

Ok, so if we want "$\mathcal{M}$ is a model of ZFC" to be a real theorem, we need to go with Option 2. On the other hand, if $\mathcal{M}$ is a proper class, we are forced into Option 1*.

What if we're in a situation where Option 1* or Option 2 would both make sense, i.e., if $\mathcal{M}$ is a transitive set definable by a metalanguage formula, and we're happy to accept a theorem schema rather than a theorem? In this situation, Option 2 is strictly stronger than Option 1*. Indeed, suppose we can prove $\mathcal{M}\models \mathsf{ZFC}$ in the sense of Option 2. Now for any metalanguage axiom $\varphi$, there is an internal version $\varphi'$, and by our theorem, $\varphi'$ is true in $M$. Unpacking the definition of satisfaction, we find that we can prove $\varphi^M$. So $M$ is a model of ZFC according to Option 1*. The fact that Option 2 is stronger and less unweildy also makes it more useful - e.g., having a model of ZFC according to Option 2 has the consequence that ZFC is consistent, while the same is not true for Option 1*.

For this reason, Option 2 is the default (and by default, "model" means "set model"). For example, when people talk about the question of whether the set $L_\alpha$ is a model of ZFC, they mean Option 2.

Answer 3

Only option 2 makes sense. The axioms and metalanguage mentioned in your first sentence are not objects you have access to when you "work inside $\mathcal V$"; they're just the rules you follow, i.e. your system for mathematical reasoning. Following those rules, you establish definitions for objects inside $\mathcal V$ that represent syntactic formulas, models, the satisfaction relation, etc..

It's tempting to always want to examine "the metalanguage" as a mathematical object, but this doesn't always add anything useful - you're just adding more layers of the same questions. We know how to work in ZFC because we know how to follow the rules of first-order logic. We don't need to conceive a mathematical universe in which those rules are objects. (Doing so would require a system of mathematical reasoning...)

In other words, the reason that logic textbooks don't dwell on the metalanguage is the same reason algebra or analysis textbooks don't: it is not the thing we are studying.