Why does every complete theory have a model?

Question

I have to prove that $T$ is complete iff $T=Th(\frak{A})$ for a $\tau$-structure $\mathfrak{A}$.

If $Th(\mathfrak{A})=T$ and $\psi$ is a $FO(\tau)$ proposition (I don't know the english expression but I want to say that $\psi$ has no free variables). Then either $\mathfrak{A}\models\psi$ or $A\not\models\psi$. Thus $T$ is complete.

Now for the other direction assuming $T$ is complete, if I would know that there is a structure $\mathfrak{A}$ such that $\mathfrak{A}\models T$. Then I could say $ T\subseteq Th{\mathfrak{(A)}}$. Then looking at $\psi\in FO(\tau)$ in the case of $\mathfrak{A}\models \psi$, then because of completeness we have also either $T\models \psi$ or $T\not\models \psi$ , the second option is not possible. Because of the subsetrelation if $\mathfrak{A}\not\models \psi$ then also $T\not\models \psi$.

The question is how can I prove the existence of a model for $T$?

Answer 1

This is highly nontrivial - it's the Completeness Theorem CT ("completeness" in its name refers to the proof system's completeness, not the completeness of any particular first-order theory.). I've given a quick summary of its proof in the first part of this answer of mine.

Specifically, CT says that every consistent theory has a model. A theory $T$ is complete iff $T$ is consistent and for each $\varphi$ exactly one of $\varphi$ and $\neg\varphi$ is in $T$. Since complete theories are consistent, CT says that complete theories in particular have models.

(OK, that may not be the definition of "complete theory" you're using; the other common one is that $T$ is complete iff $T$ is consistent and for each $\varphi$ exactly one of $\varphi$ or $\neg\varphi$ is in $T$. But that doesn't change the argument above.)

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In case you're already familiar with the completeness theorem for propositional logic, it's worth noting that despite their similarities they are really fundamentally different - the propositional version is comparatively trivial. Similarly, the compactness theorem for propositional logic has a quick topological proof, whose idea breaks down for the first-order case.

Basically, the difference is that in propositional logic the semantics is already very close to the syntax: a "model" in the sense of propositional logic is simply an assignment of truth values to the sentences which satisfies some basic rules (e.g. it makes $A\wedge B$ true iff it makes $A$ true and it makes $B$ true). By contrast, the semantics for first-order logic is extremely complicated: a structure is much more than just its theory (whereas in the propositional case, a structure literally is its theory).

One technical way to make this precise is to observe that we can whip up a single first-order sentence with no computable model (e.g. take $I\Sigma_1+\neg Con(I\Sigma_1)$ and apply Tennenbaum's theorem).