I am reading an introductory logic lecture notes, and found this theorem. This is not proven in the lecture note I am reading, so I am wondering if my proof is correct. I am not very comfortable with what I will say, but $L$ is first order language with the usual sentential connectives.
A set $\Sigma$ of $L$-formulas is Satisfiable iff there exists an evaluation $v$ of sentential letters such that $\overline{v}(\sigma)=1$ for all $\sigma\in \Sigma$, where $\overline{v}$ is the unique extension of $v$ on $\Sigma$.
Compactness Theorem: If every finite $\Delta\subset \Sigma$ is satisfiable, then $\Sigma$ is satisfiable.
Proof: We will use the following lemma to prove the theorem: If $\sigma$ is an $L$-formula and $\Sigma \models \sigma$, then there exists some finite set $\Delta\subset\Sigma$ such that $\Delta\models \sigma$.
Assume for contradiction that every finite $\Delta\subset \Sigma$ is satisfiable and $\Sigma$ is not satisfiable. Let $w$ be a sentential letter of $L$. Then $\Sigma\models (w\wedge \neg w)$ because there is no evaluation $v$ such that $\overline{v}(\sigma)=1$ for all $\sigma\in \Sigma$. Then by the lemma, there is some finite set $\Delta\subset \Sigma$ such that $\Delta\models (w\wedge\neg w)$. As we assumed every finite subset of $\Sigma$ is satisfiable, let $v$ be an evaluation such that $\overline{v}(\delta)=1$ for all $\delta\in \Delta$. Then we reach a contradiction as $\overline{v}(w\wedge\neg w)=0$ so $\Delta\models (w\wedge \neg w)$ is not true.
Please let me know if anything is incorrect. Any comments will be appreciated. Thank you!