Why is it that the complete L-theories containing $\Sigma$ are exactly those of models of $\Sigma$?

Question

Consider a set of L-sentences $\Sigma$ and a set of complete L-theories $T_i$ containing $\Sigma$ i.e. $\Sigma \subseteq T_i$.

Why is it that for each complete L-theory $T_i$ containing $\Sigma$ are exactly the the theories $Th(\mathcal A)$ of models of $\Sigma$?

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I also want to make explicit what I believe the question is asking because I got confused after reading the answer. So I believe the question/natural language means:

If ( $T_i$ is complete & $\Sigma \subseteq T_i$) THEN ($T_i = Th(\mathcal A_i$) for some $\mathcal A \models \Sigma$)

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Assume:

• $ \mathcal A $ is an L-structure in some language L
• I'm assuming $Th(\mathcal A) = \{ \sigma : \mathcal A \models \sigma \}$
• recall an L-theory is closed under provability i.e. $T \vdash \sigma \implies \sigma \in T$

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Honestly my thoughts on this are very limited except that I guess completeness theorem should play a role?

$$ \Sigma \vdash \sigma \iff \forall \mathcal A \models \Sigma, \mathcal A \models \sigma$$

I'm not even sure how $Th(\mathcal A)$ relates to $T_i$. This makes it seem all the L-theories are the same when they are not the same by assumption...

Answer 1

If $\mathcal A\vDash\Sigma$ then $\mathrm{Th}(\mathcal A)$ clearly contains $\Sigma$ so this proves one inclusion.

For the reverse inclusion suppose we have a complete theory $T$ containing $\Sigma$, we want to show that it is of the form $\mathrm{Th}(\mathcal A)$ where $\mathcal A$ is a model of $\Sigma$. Since $T$ is complete any two models of $T$ are elementarily equivalent so they all have the same complete theory and $T=\mathrm{Th}(\mathcal A)$ where $\mathcal A$ is any model of $T$. Note that $\Sigma\subseteq T$, so $\mathcal A\vDash\Sigma$, as desired

Answer 2

Let me try to answer it, as an exercise and to make sure I understood it.

Theorem: Let $\Sigma$ be a set of L-sentences and $T$ be an L-theory. Then $T$ is a complete theory extending $\Sigma$ if and only if there is some model $\mathcal A$ of $\Sigma $ s.t. $T = Th(\mathcal A)$.

$(\Rightarrow)$ (this is the hard direction) Suppose $\Sigma \subseteq T$ and $T$ is complete (so $T \vdash \sigma $ or $T \vdash \neg \sigma$ and the definition of complete I know and like includes consistency automatically). Consider any two models of $T$ call them $\mathcal B_1, \mathcal B_2$ (Note such a model satisfies $\Sigma$ because $T$ extends $\Sigma$). Recall the completeness theorem:

For any consistent $\Sigma$ set of L-sentences, for each $\sigma$ L-sentence, $$ \Sigma \vdash \sigma \iff \forall \mathcal A \models \Sigma, \mathcal A \models \sigma $$

Thus because of completeness of $T$ we have for all L-sentences $\sigma$:

1. $T \vdash \sigma \implies $ all its models $\mathcal B \models \sigma$
2. $T \vdash \neg \sigma \implies $ all its models $\mathcal B \models \neg \sigma$

we have a decision from $T$ which implies a decision on every model of it. Since this holds for all sentences then any two models satisfy the same set of L-sentences and thus are elementarily equivalent. So pick any model of $T$ and define $Th(\mathcal B) = \{ \sigma : \mathcal B \models \sigma \}$. Since 1) and 2) are actually IFF's then we have the things $T$ proves are the same as the things true in it's models. Since $T$ is an L-theory then all it proves it contains. Thus, these two facts together means that $T = Th(\mathcal B)$.

($\Leftarrow$) Suppose $\mathcal A \models \Sigma$ and $T = Th(\mathcal A)$. We want to show $T=Th(\mathcal A)$ extends $\Sigma$ and is complete. Since $T$ is defined in terms of a model and models always yield a truth value either $\mathcal A \models \sigma$ (true) or $\mathcal A \models \neg \sigma$ (false), then we have have $T=Th(\mathcal A)$ is complete.

Now we need to show $\Sigma \subseteq T = Th(\mathcal A)$ (extends $\Sigma$). Since $\mathcal A \models \Sigma$ then by definition $\forall \sigma \in \Sigma, \mathcal A \models \sigma$. The condition $\mathcal A \models \sigma $ is the condition we need for being in $Th(\mathcal A)$. Thus, we have $\Sigma \subseteq Th(\mathcal A)$.