equivalent statement of compactness theorem

Question

My book on logic says that there are two equivalent compactness theorems:

1. Let $\Gamma$ be a set of propositional formulas, and let $\phi$ be a formula. If $\Gamma\models \phi$, then there is a finite subset $\Gamma_0\subset\Gamma$ such that $\Gamma_0\models\phi$.

2. If for each finite $\Gamma_0\subset\Gamma$ we have a model, then for $\Gamma$ we have a model.

Now for $2\implies 1$, my book writes the following: Suppose 2. is true. Let $\Gamma$ be a set of propositional formulas, and let $\phi$ be a formula such that $\Gamma\models\phi$. Then there is no model for $\Gamma\cup\{\neg\phi\}$, and because of our assumption, there exists a finite $\Gamma_0\subset\Gamma$ such that there is no model for $\Gamma_0\cup\{\neg\phi\}$. But then we have $\Gamma_0\models\phi$.

I don't understand how our assumption leads to the existence of $\Gamma_0\subset\Gamma$ such that $\Gamma_0\cup\{\neg\phi\}\models\perp$. I'm confused, because $\Gamma_0\cup\{\neg\phi\}$ has no model, so it seems like we can't apply the assumption, except for $\Gamma$ separately, but I don't see how.

Answer 1

Since your $\Gamma_0 \cup \{\neg \phi\}$ has no model we have $\Gamma_0 \cup \{\neg \phi\} \models \bot$ (remind that $\Theta\models\varphi$ if every model of $\Theta$ is a model of $\varphi$ and that $\bot$ has no model by definition).

Since $\Gamma_0 \cup \{\neg \phi\}$ has no model it follows that every model of $\Gamma_0$ is not a model of $\neg \phi$, it makes $\phi$ false. So every such model must make $\phi$ true, hence $\Gamma_0 \models \phi$.

Answer 2

Suppose 2. is true.

Let $\Gamma$ be a set of propositional formulas, and let $\phi$ be a formula such that $\Gamma\models\phi$.

Then there is no model for $\Gamma\cup\{\neg\phi\}$, and because of our assumption, there exists a finite $\Gamma_0\subset\Gamma$ such that there is no model for $\Gamma_0\cup\{\neg\phi\}$.

Otherwise, consider the finite subsets of $\Gamma \cup \{\phi\}$.

If it is $\Gamma_0 \subset \Gamma$, if it does not have a model, then $\Gamma_0 \vDash \phi$ vacuously.

If it is in the form of $\Gamma_0 \cup \{\neg \phi\}$, then by "otherwise", it has a model.

So by 2, $\Gamma \cup \{\neg \phi\}$ would have a model, contradiction.

But then we have $\Gamma_0\models\phi$.