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

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.

• Shouldn't it be "infinite" ? – Max Jul 7 '18 at 16:17
• This seems very garbled, and isn't true as written. I suspect that the actual claim is: "If $\Sigma$ is a theory with no finite models which is $\kappa$-categorical for some infinite $\kappa$, then $\Sigma$ is complete" (which is true). – Noah Schweber Jul 7 '18 at 16:22
• What does it mean for a model to be $\kappa$-categorical? – bof Jul 7 '18 at 16:34
• What does it mean for a theorem to have a model? – Andrés E. Caicedo Jul 7 '18 at 17:49
• @Berci I know what it means for a theory to be $\kappa$-categorical. I'm wondering what the OP means by saying that the finite model $M$ is $\kappa$-categorical. – bof Jul 7 '18 at 22:03

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 $T$.

• I assume the first part of your answer is only relevant for a finite $\kappa$ (though of course as written the theorem stays true for infinite $\kappa$) – Max Jul 15 '18 at 20:07
• @Max Yes, it's only relevant for finite $\kappa$ but true as written for all $\kappa$. Incidentally, it's vacuous for infinite $\kappa$: by the compactness theorem, any theory with an infinite model has models of unboundedly large infinite cardinalities - so the hypothesis can't be satisfied. (And this is further refined by the Lowenheim-Skolem theorems, which together show that if a theory $T$ has an infinite model of cardinality at least that of $T$, then it has models in every infinite cardinality $\ge\vert T\vert$.) – Noah Schweber Jul 15 '18 at 20:09