Is the compactness theorem (from mathematical logic) equivalent to the Axiom of Choice? Or more importantly, is it independent of the axiom of choice. The compactness theorem states the given a set of sentences $T$ in a first order Language $L, T$ has a model iff every finite subset of $T$ has a model. So for any natural number $n, T(n)$ is a finite subset of $n$ sentences. Now if every finite subset has a model than adding a sentence $r$ to $T(n)$ gives us $T(n+1)$ which also has a model so there is some transfinite induction going on here when $T$ is countable. This seems to cry out for the application of Zorn's Lemma which is equivalent to the axiom of choice. So to summarize, is the compactness theorem consistent with ~AC (the negation of the axiom of choice which is independent of ZF set theory)? 

In general, the compactness theorem is equivalent to the ultrafilter lemma, which is known to be strictly weaker than the axiom of choice (so is consistent with its negation) but independent of ZF. The models the compactness theorem asserts exist can be constructed using ultraproducts.
Over a countable alphabet, the compactness theorem is provable in ZF. This is because it can be proven from the completeness theorem, which over a countable alphabet is also provable in ZF. 
