# Proving compactness theorem by an unusual way [closed]

The problem goes like this:

Consider the following statement; "Given a set $$\Sigma$$ of wffs, and a wff $$\tau$$, $$\Sigma\vDash\tau$$ iff there is a finite $$\Sigma_0\subset\Sigma$$ such that $$\Sigma_0\vDash\tau$$. By using the above statement, prove the compactness theorem; "A set of wffs is satisfiable iff it is finitely satisfiable."

Because of possible usefulness, we note that $$\Sigma\not\vDash\tau$$ iff $$\Sigma\cup\{\neg\tau\}$$ is satisfiable.

My input is the following:

Given $$\Sigma$$, take $$\sigma\in\Sigma$$.

$$\Sigma$$ is satisfiable iff

$$\Sigma\cup\{\sigma\}$$ is satisfiable iff

$$\Sigma\not\vDash\neg\sigma$$ iff

for all finite $$\Sigma_0\subset\Sigma$$, $$\Sigma_0\not\vDash\neg\sigma$$ iff

for all finite $$\Sigma_0\subset\Sigma$$, $$\Sigma_0\cup\{\sigma\}$$ is satisfiable then $$(*)$$

for all finite $$\Sigma_0\subset\Sigma$$, $$\Sigma_0$$ is satisfiable $$(*)$$

Now, the final step $$(*)$$ is difficult. How can I show "if" part?

• @spaceisdarkgreen I updated my input. Actually, the problem is not trivial as shown from my input. Please reconsider your thought.
– kkkk
Oct 22, 2020 at 6:16

Keep in mind that making a set of sentences smaller only makes it easier to satisfy: if $$\Gamma_0\subseteq\Gamma_1$$ and $$\Gamma_1$$ is satisfiable, then $$\Gamma_0$$ is satisfiable. In particular, if $$\Sigma_0\cup\{\sigma\}$$ is satisfiable then so is $$\Sigma_0$$.
(Maybe it's the other direction you're concerned about, that is going from the last line to the second-to-last line? In that case we just use the fact that $$\sigma\in\Sigma$$: if $$\Sigma_0$$ is a finite subset of $$\Sigma$$ then so is $$\Sigma_0\cup\{\sigma\}$$, so if every finite subset of $$\Sigma$$ is satisfiable then so is every set of the form $$\Sigma_0\cup\{\sigma\}$$ for $$\Sigma_0$$ a finite subset of $$\Sigma$$.)
All of this suggests that the particular $$\sigma$$ is really playing no role. In fact, there's a variant of the argument which ignores it (and I suspect is the intended solution) which is based off of the following equivalence $$(*)$$: for every set $$\Gamma$$ we have that that $$\Gamma$$ is satisfiable iff $$\Gamma\not\models\perp$$.
Using $$\perp$$ this way simplifies the argument: if $$\Gamma$$ is finitely satisfiable then by $$(*)$$ we know that for each finite $$\Gamma_0\subseteq\Gamma$$ we have $$\Gamma_0\not\models\perp$$, so by the quoted fact we have $$\Gamma\not\models\perp$$ which in turn means that $$\Gamma$$ is satisfiable (apply $$(*)$$ again).
(And the other direction, "If $$\Gamma$$ is satisfiable then $$\Gamma$$ is finitely satisfiable, is trivial: any $$M\models\Gamma$$ simultaneously witnesses the satifsiability of all subsets of $$\Gamma$$.)