# Complete first order theories

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!

• 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 $A$." – Nagase Aug 23 '20 at 19:45