If a sentence doesn't belong to any complete and closed theory...

Question

the question is:
If $\varphi$ doesn't belong to any $T$ complete and closed under $\vdash$ then $\left \{ \varphi \right \}\vdash \neg\varphi$.
I initially thought about solving it through a reciprocal proof ($A$ implies $B$ iff not $B$ implies not $A$), but because the general statement has the form "not $A$ and $B$ implies $C$", then I got into problem while attempting to find the real "not $A$", then I proceeded with a contradiction argument, and I proved that for a formula $\varphi$ that would belong to a complete theory and with this deduction closeness cannot satisfy such condition '$C$', but Im not sure if this is right.
Thank you in advance.

Answer 1

Lemma: If $\Sigma$ is a consistent set of $\mathcal{L}$-sentences, then there is a complete $\mathcal{L}$-theory $T$ containing $\Sigma$. (Proof below, but try proving it yourself first! The main idea is to use Zorn's lemma and the compactness theorem.)

Proof of Lemma: We will apply Zorn's lemma to the set of all consistent $\mathcal{L}$-theories containing $\Sigma$, partially ordered by inclusion. Since $\Sigma$ is consistent, it is in particular a member of this set, which is hence non-empty.

Now, suppose $(T_i)_{i\in I}$ is a chain of consistent theories, each of which contains $\Sigma$. Then let $T=\bigcup_{i\in I} T_i$. Certainly $T$ is a set of $\mathcal{L}$-sentences containing $\Sigma$, so we just need to show that it is consistent. By the compactness theorem, it suffices to show that $T$ is finitely consistent. However, if $\Delta\subseteq T$ is a finite subset of $T$, then there is some $i\in I$ such that $\Delta\subseteq T_i$, and hence (as a subset of a consistent set of sentences) $\Delta$ is consistent.

So $T$ is finitely consistent, hence consistent, and we can apply Zorn's lemma to obtain a maximal (with respect to inclusion) consistent $\mathcal{L}$-theory $T$ such that $\Sigma\subseteq T$. Then we just need to show that $T$ is complete; but indeed, if $\phi$ is an $\mathcal{L}$-sentence such that $T$ is consistent with $\phi$ (resp. $\neg\phi$), then by maximality of $T$ we have $\phi\in T$ (resp. $\neg\phi\in T$), as desired.

Given this lemma we may prove the result you want. Indeed, if $\varphi$ doesn't belong to any complete theory $T$, then applying the above lemma to the set $\Sigma=\{\varphi\}$ shows that $\varphi$ must be inconsistent. This means by definition that $\emptyset\models\neg\varphi$, and hence, by the completeness theorem, that $\emptyset\vdash\neg\varphi$, so in particular we have $\{\varphi\}\vdash\neg\varphi$, as desired.