Today I have been presented the Compatness Theorem from mathematical logic
In the proof we use the following lemma:
Lemma: Let $T$ be a theory and $A$ a formula. If $T$ is finitely satisfiable, then either $T\cup \{A\}$ is finitely satisfiable or $T\cup\{\neg A\}$ is finitely satisfiable.
Proof If both $T\cup\{A\}$ and $T\cup\{\neg A\}$ are not finitely satisfiable, then there exist two finite subsets $F_1\subseteq T, F_2\subseteq T$, s.t. $F_1\cup \{A\}$ and $F_2\cup\{\neg A\}$ are both non-satisfiable. If this was not the case, it would follow that there is truth assignment function $v$ s.t. all formulas from $F_1\cup F_2$ are true. That would follow that $v(A)=1$ thus $F_1\cup\{A\}$ is satisfiable. Or $v(\neg A)=1$ and so $F_2\cup\{\neg A\}$ is satisfiable. But that's impossible, so the lemma holds.
My question: Where do we get those sets $F_1,F_2$ and why can we say that $F_1\cup\{A\}$ and $F_1\cup \{\neg A\}$ are not satisfiable?
Is this idea good?:
Denote $F'_1:=F_1\cup\{A\},F_2':=F_2\cup\{\neg A\}$. And because $F_1'\cup F_2'=F_1\cup F_2\cup \{A,\neg A\}$, thus $F_1'\cup F_2'$ are both not satisfiable, because we are never going to get truth assignment such that both $A,\neg A$ are true, right?
Now, the first part, by definition, we have that $T$ is finitely satisfiable if every finite subset of $T$ is satisfiable. We assume $T\cup \{A\}$ is not finitely satisfiable and neither is $T\cup \{\neg A\}$. That means we have in fact two theories, so for each, we find that one particular finite subset which is not satisfiable, so let's say, we find a subset $G_1\subseteq T\cup \{A\}$ also $G_2\subseteq T\cup \{\neg A\}$ and now, because the $A$ is "complementary", we say that this is some $F_1\cup \{A\}=G_1$ and $F_2\cup \{\neg A\}=G_2$. And it can also be the case that $F_1=F_2$, right? I would just like to know if my argumentation is correct.
Best Answer
$F$ is finitely satisfiable iff every finite subset of it is satisfiable.
We assume that $T ∪ \{ A \}$ is not.
This means that we have a finite subset of it that is not satisfiable. It cannot be $\{ A \}$, because $A$ is clearly satisfiable.
Thus we must have a subset $F_1$ of $T$ plus $\{ A \}$, because being $T$ finitely satisfiable, every finite sub of $T$ is satisfiable, and thus $F_1$ alone is so.
The same reasoning applies to $F_2$ which is satisfiable but $F_2 ∪ \{ ¬ A \}$ is not.
Now, if $F_1$ and $F_2$ are finite subsets of $T$, also $F_1 ∪ F_2$ is, and thus by finitely satisfiability of $T$, also $F_1 ∪ F_2$ is satisfiable.
This means that we have a $v$ such that $v(φ)= \text T$ for every $φ ∈ F_1 ∪ F_2$. But then, either $v(A)= \text T$ or $v(¬ A)= \text T$.