Every finite subset of $\Gamma$ is consistent implies $\Gamma$ is consistent

Question

Thm: If every finite subset of $\Gamma$ is consistent then $\Gamma$ is consistent.

My notes claims that it can be implied from compactness of $\vdash$. Meaning: If $\Gamma \vdash A$ then there's a finite subset $\Delta\in\Gamma$ such that $\Delta\vdash A$.

I don't see how it explains the theorem above.

I'd be glad for an explanation.

Answer 1

Suppose to the contrary that $\Gamma$ is not consistent. Then there is a sentence $\varphi$ such that $(\varphi \land \lnot\varphi)$ is a theorem of $\Gamma$. Then by Compactness $(\varphi \land \lnot\varphi)$ is provable from some finite subset $\Delta$ of $\Gamma$. It follows that some finite subset of $\Gamma$ is not consistent.

Answer 2

Hint: Prove the contrapositive. Suppose that $\Gamma$ is inconsistent: in particular, then for some $\alpha$, $\Gamma\vdash \alpha$ and $\Gamma\vdash \neg\alpha$. Use compactness to find finite $\Delta_1, \Delta_2$ such that $\Delta_1\vdash \alpha$ and $\Delta_2\vdash \neg\alpha$. Can you find an inconsistent finite subset in terms of $\Delta_1$ and $\Delta_2$?