If a theorem is $\kappa$-categorical and has a model of order $\kappa$ when $\kappa<\aleph_0$ , then it's complete [FALSE CLAIM]

first-order-logiclogicmodel-theorypredicate-logic

For a theory $\Sigma$, if it has a finite model $M$ of order $\kappa$ which is $\kappa$-categorical, then $\Sigma$ is complete.

It was mentioned in a class, but I wasn't able to find a proof, neither here nor in Enderton.
Would be thankful for a reference or some guidance.

Edit:
This has turned out to be a false claim, the correct claim is for $\kappa$-categorical with an infinite $\kappa$, with no finite models. See the answer by @Noah Schweber.

Best Answer

As written this is not true. E.g. consider the theory $T$ of groups, and let $\kappa=7$; there is a unique group up to isomorphism of order $7$, but $T$ is certainly not complete.

What is true is the following (until the second part of this answer, $\kappa$ is finite; the statement below is true for infinite $\kappa$ too, but vacuous in that case since by the compactness theorem no theory will satisfy the hypothesis if $\kappa$ is infinite):

Suppose $T$ is a theory with exactly one model of cardinality $\kappa$ (up to isomorphism), and no models of cardinality $\not=\kappa$. Then $T$ is complete.

The proof of this is straightforward. Suppose there were some sentence $\varphi$ such that $T\not\vdash\varphi$ and $T\not\vdash\neg\varphi$. Then both $S_1=T\cup\{\varphi\}$ and $S_2=T\cup\{\neg\varphi\}$ are consistent. By the completeness theorem, there are models $\mathcal{M}_1$ of $S_1$ and $\mathcal{M}_2$ of $S_2$. But then:

  • We can't have $\mathcal{M}_1\cong\mathcal{M}_2$, since they disagree about $\varphi$.

  • We have $\mathcal{M}_1,\mathcal{M}_2\models T$, since $S_1, S_2\supseteq T$.

  • But then $\mathcal{M}_1$ and $\mathcal{M}_2$ are nonisomorphic models of $T$ of cardinality $\kappa$ (since $T$ has no models of cardinality $\not=\kappa$), contradicting the $\kappa$-categoricity of $T$.

Incidentally, a similar argument shows:

Let $\lambda$ be an infinite cardinal. Suppose $T$ is a theory with no finite models and exactly one model up to isomorphism of cardinality $\lambda$ (that is, $T$ is $\lambda$-categorical). Then $T$ is complete.

This is actually a bit trickier, since we don't have the assumption that every model of $T$ has cardinality $\lambda$, merely that $T$ has no finite models. We have to use the Lowenheim-Skolem theorem (together with the fact that $T$ has no finite models) to show that if there are two non-elementarily-equivalent models of $T$, then there are two non-elementarily-equivalent models of $T$ of cardinality $\lambda$, and since non-elementary-equivalence implies non-isomorphism this would contradict the $\lambda$-categoricity of $T$.

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