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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!

DiscipleOfKant
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  • That you feel that something must happen is not exactly an argument. – Mariano Suárez-Álvarez Mar 21 '18 at 21:08
  • For that statement to be correct, the theory must be consistent as well as complete. In fact, completeness is often defined to include consistency. However, according to the definition you gave, an inconsistent theory would be complete. – bof Mar 21 '18 at 21:35

2 Answers2

1

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?

Noah Schweber
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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 $Th(M)=T$.

YCB
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