How many nonisomorphic models of ZFC (or other theories) are there?

Question

Given a theory $T$, one could define it's "weakness" by the number of models it has (i.e. how satisfiable it is.) However, even though ZFC is particularly strong, every forcing extension of $V$ is a model of ZFC, which means that not only is the number of models it has a cardinal, but it is not even a proper class.

The reason this definition isn't too useful is one could "cheat the system" by having very few symbols in the theory, leaving room for other symbols to exist (like the symbols for names used in poset forcing.)

Instead, we could use this definition of "weakness": $$\mathrm{Iso}(\mathcal{M})=\{\mathcal{N}:\mathcal{M}\cong\mathcal{N}\}$$ $$St(T)=\{\mathrm{Iso}(\mathcal{M}):\mathcal{M}\models T\}$$ $$T\;\mathrm{is}\;\kappa-\mathrm{weak}\Leftrightarrow|St(T)|=\kappa$$ $$T\;\mathrm{is}\;\mathrm{totally}\;\mathrm{weak}\Leftrightarrow \neg\exists\kappa\in\mathrm{Card} (T\;\mathrm{is}\;\kappa-\mathrm{weak})$$ So, a theory is $\kappa$-weak if there are $\kappa$ nonisomorphic models of $T$, and it is totally weak if there are $\kappa$ nonisomorphic models of $T$ for every cardinal $\kappa$. So if $T$ is $\kappa$-categorical for $\lambda$ different $\kappa$, then $T$ is at least $\lambda$-weak (and therefore uncountably categorical theories are all totally weak). Also note that $0$-weak theories are exactly those which are inconsistent. Finally, note that if there are only $\kappa$ different models of $T$, then there are at most $\kappa$ different nonisomorphic models of $T$ (the only situation where this happens is when every model of $T$ is nonisomorphic to every other model of $T$), meaning that $T$ is at most $\kappa$-weak.

An example of a totally weak theory is the theory of an algebraically closed field with characteristic $0.$

An example of a $1$-weak theory is $\forall x(x=c)$, where $c$ is a constant symbol. Though this theory may not seem strong, it has only one isomorphism class of models, which is just those with universe $\{c^\mathcal{M}\}$. For it's size, this theory is relatively strong.

QUESTIONS:

Has this already been defined under a different name? If not, has the name been taken by some other definition?

Is this a good definition of weakness? Is it useful in any way?

Are there $\kappa$-weak theories for every cardinal $\kappa$? One can show (by using arguments similar to that of the example of a $1$-weak theory) that there are for finite $\kappa$.

Answer 1

Pretty much every interesting theory is "totally weak". In particular, if $T$ is any (first-order) theory which has a model of cardinality $\geq n$ for all $n\in\mathbb{N}$ (in particular, if $T$ has an infinite model), then $T$ has models of arbitrarily large cardinality (and thus a proper class of nonisomorphic models). Namely, if $\kappa$ is any infinite cardinal, adjoin $\kappa$ constant symbols to your language and add axioms saying all these constants are distinct. Any finite subset of the resulting theory has a model (since any finite subset requires only finitely many of the constants to be distinct and $T$ has a model with enough distinct elements), and hence by compactness the entire theory has a model, which is a model of $T$ with at least $\kappa$ distinct elements.

(More strongly, by the Löwenheim-Skolem theorem, $T$ has a model of cardinality exactly $\kappa$ if $\kappa$ is greater than or equal to the cardinality of the language of $T$.)

So a theory that is not totally weak can only have finite models of bounded cardinality. Over a language with only finitely many relation and function symbols, there can then be only finitely many isomorphism classes of models.