https://i.stack.imgur.com/Lo092.jpg https://i.stack.imgur.com/qXeFe.jpg
Theorem 2.3 (Compactness theorem). *
Let $\Gamma$ be a (possibly infinite) set of formulas such that every finite subset of $\Gamma$ has a model.
Then $\Gamma$ has a model.not hold if $i\left(P_{n+1}\right)=0 .$
That is, there is a finite subset $\Gamma^{\prime}$ of $\Gamma$ which has no model in which $P_{1}, \ldots, P_{n}, P_{n+1}$ take the values $i\left(P_{1}\right), \ldots, i\left(P_{n}\right), 0 .$
Then we define $i\left(P_{n+1}\right)=1$ and show that $\Phi(n+1),$ i.e., every finite subset of $\Gamma$ has a model in which $P_{1}, \ldots, P_{n}, P_{n+1}$ take the values $i\left(P_{1}\right), \ldots, i\left(P_{n}\right),$
- For let $\Delta$ be a finite subset of $\Gamma .$
Then $\Delta \cup \Gamma^{\prime}$ is a finite subset of $\Gamma$ and hence, by the induction hypothesis, $\Delta \cup \Gamma^{\prime}$ has a model in which $P_{1}, \ldots, P_{n}$ take the values $i\left(P_{1}\right), \ldots, i\left(P_{n}\right) .$ since $i$ is a model of $\Gamma^{\prime}, i\left(P_{n+1}\right)=1$
Let $i\left(P_{1}\right)=0$ and suppose $\Phi(1)$ does not hold. That is, there is a finite subset $\Gamma^{\prime}$ of $\Gamma$ which has no model in which $P_{1}$ takes the value $i\left(P_{1}\right)=0.$
Then we define $i\left(P_{1}\right)=1$ and show that $\Phi(1),$ i.e., every finite subset of $\Gamma$ has a model in which $P_{1}$ takes the value $i\left(P_{1}\right)=1.$ For let $\Delta$ be a finite subset of $\Gamma .$ Then $\Delta \cup \Gamma^{\prime}$ is a finite subset of $\Gamma$ and hence has a model $i$ since $i$ is a model of $\Gamma^{\prime}, i\left(P_{1}\right)=1.$
Suppose we have defined $i\left(P_{1}\right), \ldots, i\left(P_{n}\right)$ such that $\Phi(n) .$ Then we can extend the definition of $i$ to $P_{n+1}$ such that $\Phi(n+1) .$ For suppose that $\Phi(n+1)$ does.
I need a help understanding the proof appearing in the above pictures. I got it that it uses mathematical induction, but I have little sense of the whole reasoning. Especially I don’t get it why introduce the set Δ into the proof. What is it for?
Regarding $\Delta,$ they are trying to show that $\Phi(n+1),$ which means that every finite subset of $\Gamma$ has a model that agrees with $i$ on the values of $P_1,\ldots, P_{i+1}.$ So they say "let $\Delta$ be some finite subset of $\Gamma$" and then argue that it has a model that agrees with $i$ on the values of $P_1,\ldots, P_{i+1}.$
I will try to give some motivation for, and a clearer account of this proof.
Here is the issue: At the beginning all we know is that every finite subset has a truth assignment that satisfies it. But this isn't much help cause what we want is a single truth assignment that works for every sentence. It could be that each of the finite subsets requires a wildly different truth assignment to satisfy it, so there's no way of isolating one that works for everything.
So we chip away at it gradually, trying to show that we can at least demand some finite number of variables can be fixed and we still can find assignments of the remaining variables satisfy every finite subset of $\Gamma$.
First we argue that we can fix $P_1.$ Perhaps setting $P_1$ true works, but if it doesn't then setting it false will work, by the following argument: if $P_1$ true doesn't work, this means there is some finite $\Gamma_0\subset \Gamma$ that can only be satisfied by assignments where $P_1$ is false. If it also didn't work for $P_1$ false, then there would also be a finite subset $\Gamma_1\subset \Gamma$ that can only be satisfied by assignments where $P_1$ is true. But then $\Gamma_0\cup\Gamma_1$ would be a finite subset of $\Gamma$ that can't be satisfied by any assignment, since some statements need $P_0$ true and some need it false. This contradicts our assumption that every finite subset can be satisfied.
Okay, so we have fixed a value for $P_1$... say the value happened to be true. So in other words, we know that for any finite subset of $\Gamma$ there is an assignment that satisfies it where $P_1$ is true. Now, we argue that we can fix $P_2$ as well. The argument is exactly the same as the one we just gave for fixing $P_1$: if $P_2$ true doesn't work then this means there is some finite $\Gamma_0$ such that every satisfying assignment where $P_1$ is true has $P_2$ false. If it were also the case that there were a $\Gamma_1$ such that every satisfying assignment where $P_1$ is true has $P_2$ true. And this means $\Gamma_0\cup\Gamma_1$ is a finite set that can't be satisfied by any assignment where $P_1$ is true, which contradicts what we already know.
So say that $P_2$ false wound up working (as a side note, I should emphasize that it's possible that both options could work and we just pick one arbitrarily). This means we know that for any finite subset of $\Gamma$ there is an assignment that satisfies it where $P_1$ is true and $P_2$ is false. Then we do $P_3,$ just the same as above.
So we can keep going, filling out an infinite sequence $P_1=T, P_2=F, P_3=?, P_4=?\ldots.$ We consider the variable assignment where we assign each $P_i$ to the value it gets in that sequence and claim this assignment satisfies $\Gamma.$ Consider any sentence $\sigma\in\Gamma.$ Since $\sigma$ only contains a finite number of propositional variables, it will have a largest-indexed propositional variable $P_n.$
Now, by how we constructed our sequence, we know that there is an assignment that satisfies $\sigma$ and agrees with ours $P_1,\ldots, P_n$ ($\{\sigma\}$ is certainly a finite subset of $\Gamma$). But $P_1,\ldots, P_n$ are the only propositional variables that appear in $\sigma,$ so this means our interpretation must also satisfy $\sigma,$ since they agree on the only variables that affect its truth value.