# Finitely Satisfiable Theory has Maximal Finitely Satisfiable Extension/Proof 1

## Theorem

Let $T$ be a finitely satisfiable $\mathcal L$-theory.

Then there exists a finitely satisfiable $\mathcal L$-theory $T'$ which contains $T$ as a subset such that:

for all $\mathcal L$-sentences $\phi$, either $\phi \in T'$ or $\neg \phi \in T'$.

### Lemma

Let $T$ be a finitely satisfiable $\mathcal L$-theory.

Let $\phi$ be an $\mathcal L$-sentence.

Then either $T \cup \left\{ {\phi}\right\}$ or $T \cup \left\{ {\neg \phi}\right\}$ is finitely satisfiable.

## Proof

The set of all finitely satisfiable $\mathcal L$-theories containing $T$ forms an ordered set using subset inclusion as the ordering.

Let $C$ be a nonempty chain in this ordered set.

Let $\displaystyle T_C = \bigcup_{\Sigma \mathop \in C} \Sigma$.

Let $\Delta$ be a finite subset of $T_C$.

Then there exists a single $\Sigma$ in $C$ which contains $\Delta$.

Since this $\Sigma$ is finitely satisfiable by definition, this means that $\Delta$ is satisfiable.

Hence $T_C$ is finitely satisfiable.

Since each $\Sigma \in C$ is contained in $T_C$, this means that $T_C$ is an upper bound for $C$ in the ordered set.

Thus, by Zorn's Lemma, there is a finitely satisfiable $\mathcal L$-theory $T'$ containing $T$ such that $T'$ contains all other such theories.

Let $\phi$ be an $\mathcal L$-sentence.

Let $\phi \nsubseteq T'$.

Aiming for a contradiction, suppose $T' \cup \left\{ {\phi}\right\}$ were finitely satisfiable.

Then by definition of $T'$, $T'$ would contain $T' \cup \left\{ {\phi}\right\}$ as a subset.

This would mean that $T'$ contains $\phi$, which contradicts the assumption.

Thus $T' \cup \left\{ {\phi}\right\}$ is not finitely satisfiable.

By the lemma, $T' \cup \left\{ {\neg \phi}\right\}$ is finitely satisfiable.

Thus, by definition of $T'$, $T'$ contains $T' \cup \left\{{\neg \phi}\right\}$ as a subset.

Hence $T'$ contains $\neg \phi$.

$\blacksquare$

#### Axiom of Choice

This proof depends on the Axiom of Choice, by way of Zorn's Lemma.

Because of some of its bewilderingly paradoxical implications, the Axiom of Choice is considered in some mathematical circles to be controversial.

Most mathematicians are convinced of its truth and insist that it should nowadays be generally accepted.

However, others consider its implications so counter-intuitive and nonsensical that they adopt the philosophical position that it cannot be true.