Structure for a complete theory

Question

I am having troubles writing a proof for the following question.

Show that any complete theory is of the form ThU for some L-structure U.

Where a theory is defined as a set of L-sentences closed under deducibility and a complete theory just means that for any sentence, either that sentence or its negation is a member of the theory. ThU is the set of sentences that hold in U.

I don't know if it's because I am overthinking it or missing something. I feel like any complete theory (say T) must have an underlying structure (say U) so we can then just say that theory, T, is of the form ThU but I feel like that is trivial.

Thanks!

Answer 1

Let $T$ be a complete theory. If $T$ is consistent, by completeness theorem exists a model $M$ such that $M\models T$.

Then, you can to show that $Th(M)=T$.

Answer 2

You need to add the hypothesis that $T$ is consistent.

There are three pieces to this proof:

• First, show that $Th(M)$ is always consistent, for any structure $M$. This is the soundness theorem, which is easy to prove.

• Second, show that if $T_0\subseteq T_1$ and $T_0$ is complete and $T_1$ is consistent then $T_0=T_1$. Basically, you can't "add anything" to a complete theory without making it inconsistent. This will be immediate from the definition of "complete."

• The interesting part is showing that any consistent theory, complete or not, has a model. This is the completeness theorem, and is a deep result (and in my opinion the most surprising result of basic model theory).

Combining these three points, suppose that $T$ is a complete consistent theory. Then:

• By point (3), we have $M\models T$ for some $M$. We now want to show that $T=Th(M)$.

• We have $T\subseteq Th(M)$ by the above; by point (1), $Th(M)$ is consistent, and by assumption $T$ is complete. What can you conclude?