Finitely Satisfiable Theory has Maximal Finitely Satisfiable Extension/Proof 2

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Theorem

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

There is 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

Let $\mathcal A$ be the set of finitely satisfiable extensions of $T$.

By the lemma, for each element $S$ of $\mathcal A$ and each $\mathcal L$-sentence $\phi$, either $S \cup \left\{ {\phi}\right\} \in \mathcal A$ or $S \cup \left\{ {\neg \phi}\right\} \in \mathcal A$.

$\mathcal A$ has finite character, by the following argument:

Let $S \in \mathcal A$.

Let $F$ be a finite subset of $S$.

Then $S$ is satisfiable and hence finitely satisfiable.

Thus in $\mathcal A$.


Let $S$ be a theory on $\mathcal L$.

Let every finite subset of $S$ be finitely satisfiable.

Then every finite subset of $S$ is satisfiable.

Therefore $S$ is finitely satisfiable.

Thus $\mathcal A$ has finite character.

By the Restricted Tukey-Teichmüller Theorem, $\mathcal A$ has an element $T'$ such that:

for each $\mathcal L$-sentence $\phi$, either $\phi \in T'$ or $\neg \phi \in T'$.

$\blacksquare$


Boolean Prime Ideal Theorem

This theorem depends on the Boolean Prime Ideal Theorem (BPI), by way of Restricted Tukey-Teichmüller Theorem.

Although not as strong as the Axiom of Choice, the BPI is similarly independent of the Zermelo-Fraenkel axioms.

As such, mathematicians are generally convinced of its truth and believe that it should be generally accepted.