0
$\begingroup$

Let $T$ be a set of sentences in some first-order language $L$, and assume $T$ is negation-complete, so that for any sentence of $\phi\in L$, either $T\vdash\phi$ or $T\vdash\lnot\phi$. Assume as well that $T$ has a categorical-in-cardinality-$\kappa$ axiomatization $\Gamma$, i.e. assume there exists a set of sentences $\Gamma$ expressed in $L$ such that (1) the deductive closure of $\Gamma$ is $T$, and (2) all models of $\Gamma$ with cardinality $\kappa$ are isomorphic.

Let $\Omega$ be an arbitrary axiomatization of $T$. Does it follow that $\Omega$ must also be categorical in $\kappa$?

I don't know much logic beyond the definitions, so I don't have a supply of examples of $\kappa$-categorical theories to test the conjecture on. And I don't see immediately why it would be impossible for one axiomatization to be $\kappa$-categorical without all of them being so.

$\endgroup$
3
$\begingroup$

If T and S are theories with the same deductive closure, then they have the same models. In particular if one is k-categorical then so is the other.

Note that completeness is irrelevant.

$\endgroup$
  • $\begingroup$ Ah, duh. Thanks. $\endgroup$ – symplectomorphic May 12 '17 at 23:32
2
$\begingroup$

If $\Gamma$ is an axiomatization of a theory $T$, then models of $\Gamma$ are exactly the same thing as models of $T$. Indeed, if a model $M$ satisfies every statement in $\Gamma$, then it satisfies every statement in $T$, since every statement in $T$ is a consequence of the statements in $\Gamma$. The converse is trivial, since $\Gamma\subseteq T$.

So what models a theory has doesn't depend on how you axiomatize it. As a result, neither does categoricity of a theory, since that's just a statement about what models the theory has.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.