TITLE: Is the completeness theorem for first-order logic relative to one's choice of set theory?
QUESTION [9 upvotes]: By the completeness theorem for first-order logic, every consistent theory has a model. However, to even make sense of the word "model," I believe we're assuming a set theory. So is there a set theory which yields a notion of "model" such that the completeness theorem fails?

REPLY [10 votes]: If one assumes $\mathsf{ZF}$ the completeness theorem is not provable. Since the only difference from the "usual" set theory is the fact the axiom of choice is missing, there is no real difference in how we define a language, structure, and so on. This means that the notion of "model" is the same.
In fact assuming $\mathsf{ZF}$ the following theorems are equivalent:

The completeness theorem for first-order logic.
The compactness theorem for first-order logic.
Every filter can be extended to an ultrafilter.

Note that if $\mathsf{ZF}$ is consistent then $\mathsf{ZFC}$ is consistent, so we cannot prove the negation of the completeness theorem either. However it was proved that it is also consistent that the completeness theorem fails. In universe you can construct a counterexample (from an object whose existence is assured by additional axioms, of course).