# Proof that every consistent theory has a maximally consistent extension via the Compactness Theorem

I am trying to prove that it follows from the Compactness Theorem that every consistent first-order theory has a maximally consistent extension (I hope that this will allow me to turn the usual Henkin proof of the Completeness Theorem into a proof that does not rely on the Axiom of Choice; recall that the Compactness Theorem is strictly weaker than the Axiom of Choice). However, I am not really sure about the details of my proof; for some reason, it feels a bit fishy. The proof goes like this:

Let $$L$$ be a language, let $$T$$ be a (syntactically) consistent $$L$$-theory and denote by $$F(L)$$ the set of all $$L$$-formulas. We will prove that it follows from the Compactness Theorem that $$L$$ can be extended to a maximally (syntactically) consistent theory. Let $$\mathscr{L}$$ be the language that has one constant symbol $$c_\phi$$ for each $$\phi\in F(L)$$, one unary predicate symbol $$\mathcal{T}$$ and one $$n$$-ary predicate symbol $$\operatorname{Con}^n$$ for each $$n\in\mathbb{N}$$. Next, let $$\Sigma$$ be the $$\mathscr{L}$$-theory consisting of the sentences $$\mathcal{T}(c_\phi)$$ for all $$\phi\in T$$, $$\mathcal{T}(c_\phi)\lor\mathcal{T}(c_{\neg\phi})$$ for all $$\phi\in F(L)$$, $$\forall x_1\ldots x_n(\mathcal{T}(x_1)\land\dots\land\mathcal{T}(x_n)\rightarrow\operatorname{Con}^n(x_1,\ldots,x_n))$$ for all $$n\in\mathbb{N}$$ and $$\operatorname{Con}^n(c_{\phi_1},\ldots,c_{\phi_n})$$ for all $$\phi_1,\ldots,\phi_n\in F(L)$$ such that $$\{\phi_1,\ldots,\phi_n\}$$ is syntactically consistent and all $$n\in\mathbb{N}$$. We will use the Compactness Theorem to prove that $$\Sigma$$ has a model. Let $$\Sigma_0\subseteq\Sigma$$ be finite, let $$\{\phi_1,\ldots,\phi_n\}$$ be the set of all $$L$$-formulas $$\phi$$ such that $$c_\phi$$ occurs somewhere in $$\Sigma_0$$ and let $$X=\{\phi_1,\ldots,\phi_n\}\cup\{\neg\phi_1,\ldots,\neg\phi_n\}.$$ We prove by induction on $$n$$ that $$X$$ can be turned into a model of $$\Sigma_0$$. It suffices to find a consistent extension $$T'\subseteq X$$ of $$T\cap X$$ such that either $$\phi_i\in T'$$ or $$\neg\phi_i\in T'$$ for all $$i$$. If $$n=1$$, this is trivial. Suppose that the claim holds for $$k$$ and let $$n=k+1$$. Let $$T''\subseteq\{\phi_1,\ldots,\phi_k\}\cup\{\neg\phi_1,\ldots,\neg\phi_k\}$$ be the extension given by our induction hypothesis. Then, either $$T''\cup\{\phi_n\}$$ or $$T''\cup\{\neg\phi_n\}$$ is consistent; set $$T'=T''\cup\{\phi_n\}$$ if $$T''\cup\{\phi_n\}$$ is consistent and $$T'=T''\cup\{\neg\phi_n\}$$ otherwise. Hence, by the Compactness Theorem, $$\Sigma$$ has some model $$\mathcal{M}$$. It is clear that the theory $$T^*=\{\phi\in F(L):c_\phi^\mathcal{M}\in\mathcal{T}^\mathcal{M}\}$$ is a maximally consistent extension of $$T$$. QED

Does this proof work? For some reason, the step where I add the sentence $$\operatorname{Con}^n(\phi_1,\ldots,\phi_n)$$ for all $$n$$-tuples $$(\phi_1,\ldots,\phi_n)$$ such that $$\{\phi_1,\ldots,\phi_n\}$$ is consistent feels particularly fishy to me but I can't really explain why. Given a finite set of $$L$$-formulas, do we always know (can we always find out) whether it is consistent or not?

I am very grateful for any comments/help!

• It seems just like routine non-constructive reasoning to add $\mathrm Con^n(\ldots)$ for all the consistent sets without necessarily having a way to decide what they are. But I think what you actually need is $\lnot \mathrm Con^n$ for the inconsistent sets, or I don't see how it's clear that $T^*$ is consistent. Apr 11 at 15:12
• This can be streamlined a bit to use propositional compactness (taking the first order sentences to be propositional variables). Which is similar to constructing the Lindenbaum algebra and using the ultrafilter lemma directly. Apr 11 at 15:28
• (The issue I might raise with using first order compactness is how are you weakening choice in the proof of first order compactness to the ultrafilter lemma? I can’t see a way of doing it that doesn’t feel like we’re doing the same work twice.) Apr 11 at 15:46
• @spaceisdarkgreen Thanks for the comments! I agree that I should have added $\neg\operatorname{Con}(\ldots)$ for inconsistent sets instead. I'm not really sure why I was confused about that part so I guess you are right. I will look into what you said about propositional compactness/Lindenbaum algebras. For your final comment: There is a proof that the ultrafilter lemma implies the compactness in "The Axiom of Choice" by Jech. However, I agree that Jech's proof of this almost looks like a proof of the completeness theorem.
– Jon
Apr 11 at 17:04