Krdytkyu

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...

If $mathcal AvDashSigma$ 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 $Sigmasubseteq T$, so $mathcal AvDashSigma$, as desired

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 $Sigma subseteq Th(mathcal A)$.