$\Gamma\subseteq \text{Sent}(\mathcal{L})$ is maximal consistent iff it has models, and any two models are elementarily equivalent.

This note comes in my lecture notes after the Löwenheim-Skolem Theorem and a remark on the equivalence of "$$\Gamma \subseteq \text{Sent}(\mathcal{L})$$ is strongly maximal consistent (i.e. for any $$\psi\in\text{Sent}(\mathcal{L})$$, then $$\psi\in\Gamma$$ or $$\neg\psi\in\Gamma$$)" and "$$\Gamma = \text{Th}(\mathcal{A})$$ for some $$\mathcal{L}$$-structure $$\mathcal{A}$$". Where $$\mathcal{L}$$ is a first order predicate language and $$\text{Th}(\mathcal{A}):=\{\varphi \in \text{Sent}(\mathcal{L})|\mathcal{A}\models\varphi\}$$.

I define $$\Gamma$$ to be maximal consistent if it is consistent and for any $$\psi\in\text{Sent}(\mathcal{L})$$, we have $$\Gamma\vdash\psi$$ or $$\Gamma\vdash\neg\psi$$.

Now, to come back to the question, I see the forward direction, using the fact that $$\text{Th}(\mathcal{A})$$ is maximal consistent, so if $$\mathcal{A}$$ and $$\mathcal{B}$$ are models for $$\Gamma$$, then $$\Gamma\subseteq \text{Th}(\mathcal{A})\cap\text{Th}(\mathcal{B})$$ implies equality.

For the backwards direction I already have the maximal consistency of $$\text{Th}(\mathcal{A})$$, which is a set of sentences independent from the choice of model $$\mathcal{A}$$ for $$\Gamma$$, but $$\Gamma\subseteq\text{Th}(\mathcal{A})$$, and here I stop.

• What does "strongly" maximal consistent mean? – Alex Kruckman Apr 6 at 14:47
• Hi Alex, I thought it was a standard definition, just added. – Davide Apr 7 at 10:05

Call $$\Gamma\subseteq\mathrm{Sent}(\mathcal L)$$ complete if for every $$\phi\in\mathrm{Sent}(\mathcal L)$$: $$\phi\in\Gamma$$ or $$\neg\phi\in\Gamma$$.

We can then show that if $$\mathfrak A\models\Gamma$$ for a complete set of sentences $$\Gamma$$, then $$\Gamma=\mathrm{Th}(\mathfrak A)$$. For this, note first that as $$\mathfrak A\models\Gamma$$, we have $$\Gamma\subseteq\mathrm{Th}(\mathfrak A)$$. Now, let $$\phi\not\in\Gamma$$. Then, as $$\Gamma$$ is complete, we have $$\neg\phi\in\Gamma$$. Thus, $$\mathfrak{A}\models\neg\phi$$ and thus $$\phi\not\in\mathrm{Th}(\mathfrak A)$$.

Trivially, if $$\Gamma=\mathrm{Th}(\mathfrak A)$$, then $$\Gamma$$ is a complete theory as $$\mathrm{Th}(\mathfrak A)$$ is.

We can use this to now show another nice result about completeness of theories. For this, call $$\Gamma\subseteq\mathrm{Sent}(\mathcal L)$$ a theory if $$\Gamma\models\phi$$ implies $$\phi\in\Gamma$$ for any $$\phi\in\mathrm{Sent}(\mathcal L)$$.

Let $$\Gamma$$ be a satisfiable theory, then

$$\Gamma\text{ is complete iff all models are elementary equivalent}$$

For the direction from left to right, assume that $$\Gamma$$ is complete and let $$\mathfrak A,\mathfrak B$$ be two models of $$\Gamma$$. By the first paragraph, $$\mathrm{Th}(\mathfrak A)=\Gamma=\mathrm{Th}(\mathfrak B)$$ and thus $$\mathfrak A\equiv\mathfrak B$$.

For the direction from right to left, assume that all models of $$\Gamma$$ are elementary equivalent. Suppose that $$\phi\not\in\Gamma$$ for some $$\phi\in\mathrm{Sent}(\mathcal L)$$. Then, as $$\Gamma$$ is a theory, $$\Gamma\not\models\phi$$ and thus there is a structure $$\mathfrak A$$ such that $$\mathfrak A\models\Gamma$$ and $$\mathfrak A\not\models\phi$$. But now, for any $$\mathfrak B\models\Gamma$$, as $$\mathfrak A\equiv\mathfrak B$$, we have also $$\mathfrak B\models\neg\phi$$. Thus, $$\Gamma\models\neg\phi$$ and as $$\Gamma$$ is a theory, we have $$\neg\phi\in\Gamma$$.

• Thanks @blub, that's a very clear answer. But I was in particular trying to prove the statement in your 2nd paragraph but with $\Gamma$ being maximal consistent (i.e. $\Gamma$ is consistent and for any $\psi\in\text{Sent}(\mathcal{L})$, then $\Gamma\vdash\psi$ or $\Gamma\vdash\neg\psi$). But I think the completeness you define is a stronger condition? (indeed, that's what I define as strong maximal consistency) – Davide Apr 7 at 10:00
• @Davide If you want to consider a set $\Gamma$ with $\Gamma\vdash\psi$ or $\Gamma\vdash\neg\psi$ for any sentence $\psi$, you can take my argument up there applied to the deductive closure of $\Gamma$, that is $\Gamma^\vdash:=\{\phi\in\mathrm{Sent}(\mathcal L)\mid\Gamma\vdash\phi\}$. – blub Apr 7 at 10:06
• I am not sure I see your point. The condition for $\Gamma$ to be both deductively closed and complete is still stronger than being maximal. So your proof would show that if all models are elementarily equivalent, then $\Gamma$ is maximal consistent, but I'm not sure about the other way around. – Davide Apr 7 at 10:16
• @Davide I think you are using maximal in a non-standard sense. Normally, you refer to a set of formulae as maximal if there is no proper superset with some property. – blub Apr 7 at 10:17