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I am reading proof of compactness theorem for sentential logic in Enderton's book, A mathematical introduction to logic.

(Compactness theorem) A set of wffs is satisfiable iff every finite subset is satisfiable.

He wrote a strategy before the proof:

  1. we take our given finitely satisfiable set $\Sigma$ and extend it to a maximal such set $\Delta$,

  2. we utilize $\Delta$ to make a truth assignment that satisfies $\Delta$.

I am struggling to prove that the set $\Delta$ is finitely satisfiable. In this vein, the textbook says:

Furthermore, (3) $\Delta$ is finitely satisfiable. For any finite subset is already a finite subset of some $\Delta$n and hence is satisfiable.

And why did we pick maximal $\Delta$ ? would it be valid in the sense that if some wff Ω has the same two properties that ∆ has and ∆ ⊆ Ω, then Ω = ∆?

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I am struggling to prove that the set $\Delta$ is finitely satisfiable.

You haven't explained the construction of $\Delta$, but I'm assuming $\Delta = \bigcup_{n\in \mathbb{N}}\Delta_n$ for some increasing chain of sets of formulas $\Delta_n$, each of which is finitely satisfiable.

What Enderton wrote is a fine explanation, but I'll fill in the details for you. Let $\delta = \{\varphi_1,\dots,\varphi_k\}$ be a finite subset of $\Delta$. Then for each $1\leq i \leq k$, since $\varphi_i\in \Delta$, $\varphi_i\in \Delta_{n_i}$ for some $n_i\in \mathbb{N}$. Let $N = \max(n_1,\dots,n_k)$. Then $\varphi_i\in \Delta_{n_i}\subseteq \Delta_N$ for all $1\leq i \leq k$, so $\delta\subseteq \Delta_N$. Now since $\Delta_N$ is finitely satisfiable, $\delta$ is satisfiable.

And why did we pick maximal $\Delta$?

Maximality of $\Delta$ is used in making the truth assignment that satisfies $\Delta$. You should read the proof carefully to see why we need $\Delta$ to be maximal. But the basic idea is this: we want $\Delta$ to decide the truth value of all atomic formulas, so that we don't have to make any "guesses" when we specify the truth assignment.

would it be valid in the sense that if some wff $\Omega$ has the same two properties that $\Delta$ has and $\Delta\subseteq \Omega$, then $\Omega = \Delta$?

I'm not sure what you're asking. This is the definition of maximality...

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