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.


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 $\kappa$.

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    $\begingroup$ Forcing extensions of $V$ are not models of ZFC, for the ordinary meaning of "model". A model is a set. $\endgroup$ Sep 20, 2017 at 5:37
  • $\begingroup$ But L is a model of ZFC, proved by Godel $\endgroup$
    – user477899
    Sep 20, 2017 at 5:40
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    $\begingroup$ No. Godel proved that given an axiom of ZFC, its relativization to L is provable from ZF. Now you have a meta-theorem that L satisfies ZFC. $\endgroup$
    – Asaf Karagila
    Sep 20, 2017 at 11:43
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    $\begingroup$ You might want to take a look at Stability Theory, while not quite the same question, you can use it to count the number of non-isomorphic models of a theory. $\endgroup$
    – user185596
    Sep 20, 2017 at 12:16
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    $\begingroup$ I downvoted. In his answer to your previous question, Noah Schweber suggested that you learn some basic model theory before asking questions about the model theory of set theory, and he mentioned the Löwenheim-Skolem theorem as an explicit example. I'm going to repeat this recommendation, since Löwenheim-Skolem answers your question immediately. $\endgroup$ Sep 20, 2017 at 18:29

1 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.

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    $\begingroup$ It might be more interesting to ask how many models of each cardinality the theory has -- in other words how far it is from being categorical. On the other hand, I suspect that most theories of any foundational interest will have exactly $2^{\kappa}$ models of cardinality $\kappa$ for every infinite $\kappa$, so that would bring us no further. $\endgroup$ Oct 30, 2017 at 18:39
  • $\begingroup$ @HenningMakholm: Every unstable countable theory has $2^\kappa$ (isomorphism classes of) models of cardinality $\kappa$. I don't suppose any theory of foundational interest would be unable to define reals or natural numbers... $\endgroup$
    – tomasz
    Jun 28, 2018 at 23:25

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