TITLE: Is the compactness theorem (from mathematical logic) equivalent to the Axiom of Choice?
QUESTION [21 upvotes]: 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)?

REPLY [5 votes]: To supplement the other answers with some references at a more introductory level:

[1, p.109]: “In ZF, the Compactness Theorem (CTh) is a weakening of AC. ZF does not prove CTh, since there are models of ZF in which $\mathcal{P}(\mathbb{R})$ cannot be totally ordered. Also, ZF+CTh does not prove AC, or even that $\mathbb{R}$ can be well-ordered. Working in ZF, one can prove that CTh is equivalent to the Propositional Compactness Theorem, and to the Completeness Theorem. Also, in ZF, one can prove the Compactness and Completeness Theorems in the case that $\mathcal{L}$ is well-ordered;”
[2, Thm. I.15.13 on p.90] is the completeness theorem proved from AC+ (Def. I.3.3, p.19: “every set can be well-ordered”). Below Thm, I.15.13, a comment reads “It is known that the Completeness Theorem is not provable in ZF [...]. However, the Completeness Theorem is provable in ZF$^-$-P in the case that $\mathcal{L}$ can be well-ordered (in particular, when $\mathcal{L}$ is countable).”
Note that, in ZF, AC+ is equivalent to the axiom of choice [2, Thm. I.12.1 on p.68].
[3, p.144]: “We may then state the Gödel Completeness Theorem as
$$ \forall\, X\; (CON(X) \leftrightarrow \exists\, \mathfrak{M}\; (\mathfrak{M} \models X)) $$
and prove it within ZF (AC can be avoided here since the language is countable).”
[4, p.24]: In the course of proving results leading to the compactness theorem: “Note: If the vocabulary of $\mathcal{L}$ is finite or denumerable, then Zorn's lemma (and the axiom of choice) can be avoided in the usual way by enumerating the formulae in $F_{\mathcal{L} }$.”

[1] Kenneth Kunen, The foundations of mathematics, College Publications, 2009 (revised ed. 2012)
[2] Kenneth Kunen, Set theory, College Publications, 2011 (revised ed.2013)
[3] Kenneth Kunen, Set theory: An introduction to independence proofs, North Holland, 1980
[4] Albert C. Leisenring, Mathematical logic and Hilbert's $\varepsilon$-symbol, MacDonald Technical & Scientific, 1969