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

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 $$\Sigma$$?

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 $$\mathcal A \models \Sigma$$)

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. $$T \vdash \sigma \implies \sigma \in T$$

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...

• Does your concept of a "complete" theory include a hidden requirement that the theory is consistent? Otherwise the claim you want an explanation for is not true. – Henning Makholm Dec 25 '18 at 19:02
• Yes it does. The formalization Im familiar with assumes complete theories are built from consistent theories by lindenbaum. – Pinocchio Dec 25 '18 at 19:04

## 2 Answers

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

• Im confused are they all the same theory? – Pinocchio Dec 25 '18 at 18:58
• Not necessarily, in general given $\Sigma$ there are many complete theories extending it – Alessandro Codenotti Dec 25 '18 at 19:05
• so for each $T_i$ we have a corresponding $Th( \mathcal A_i )$? – Pinocchio Dec 25 '18 at 19:42
• @Pinocchio This is the precise statement I'm proving: Let $\Sigma$ be a set of $L$-sentences and let $T$ be an $L$-theory. Then $T$ is a complete theory extending $\Sigma$ if and only if $T=\mathrm{Th}(\mathcal A)$ for some $\mathcal A$ which is a model of $\Sigma$. Note that if $\Sigma=\varnothing$ we recover the usual result that a theory is complete iff it is the theory of some structure – Alessandro Codenotti Dec 25 '18 at 20:55
• @Pinocchio "Is it just because it's complete?" No, it's because it's consistent! "How do you know any two models of $T$ are elementarily equivalent?" Because it's complete! – Alex Kruckman Dec 26 '18 at 4:51

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 $$\Sigma \subseteq Th(\mathcal A)$$.