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

Question

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.

Answer 1

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.

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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

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

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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 $\neg\phi\in\Gamma$.