9
$\begingroup$

While studying Hodges' A shorter model theory I came across this observation:

Given a first order language $L$, we say that an $L$-theory $T$ is complete if $T$ has models and any two of its models are elementary equivalent. [...] the compactness theorem implies that any complete theory in $L$ is equivalent (i.e. has the same models) to a theory of the form $\text{Th}(A)$ for some $L$-structure $A$.

Now, I don't see how the compactness theorem comes into the picture. Why do we need it? Given the definition of complete theories it is immediate to me that a complete theory is equivalent to the theory of one of its models. What am I missing?

Thanks!

$\endgroup$
1
  • 3
    $\begingroup$ I agree that this is a strange comment. For what is worth, he does not mention compactness in the corresponding passage from Model Theory (cf. p. 43). There, he just says: "Of course, if $A$ is any $L$-structure then $Th(A)$ is complete in this sense; and conversely any complete theory in $L$ is equivalent to a theory of the form $Th(A)$ for some $L$-structure $A$." $\endgroup$ – Nagase Aug 23 '20 at 19:45
7
$\begingroup$

This is indeed a mistake. As Nagase says, it's not present in the original ("big") model theory book. My suspicion is that Hodges added it after mixing up two notions of completeness: "satisfiable and all models are elementarily equivalent" versus "contains each sentence or its negation." Using the latter sense of completeness we do indeed need compactness to identify complete theories with theories of structures.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.