# Proof of the Compactness Theorem for propositional logic by topological compactness

I'm reading Russ's proof of the compactness theorem, wherein we suppose $\Gamma = \{\phi_1, \phi_2,...\}$ is a set of propositional sentences with all of the propositional variables chosen from $V = \{p_1,p_2,...\}$. We identify $2^V$ as the set of all truth-assignments--I take it this means that although $2^V$ is the power-set of $V$, an element like $\{p_1,p_3,p_5,...\}$ would be identified with the truth-assignment $\tau$ in which $\tau(p_1)=T$ and $\tau(p_2)=F$ and so on. I believe he also says to identify these with $X\times X\times ...$ where $X=\{T,F\}$ so that in my example this truth-assignment is identified with the tuple $(T,F,T,F,T,...)$, and to give this the product topology. He notes that, by the Tychonoff theorem $2^V$ is compact. From here he defines $D_\phi = \{\tau | \tau \vDash \phi\}$ and notes that this is always both open and closed.

Getting to the proof of compactness, we want to show that if every finite subset of $\Gamma$ has a model then so does $\Gamma$. He points out that $\Gamma$ has a model if and only if $\cap_{\phi\in\Gamma}D_\phi\ne\emptyset$. So far I'm good with all this.

Next he says that, due to the compactness of $2^V$, this is equivalent to $\cap_{\phi\in\Gamma_0}D_\phi\ne\emptyset$ for every finite $\Gamma_0\subset\Gamma$. Here's where I get lost. How does compactness get us this, where are we using open covers?

I have a suspicion about what the open covers are but I'm not seeing exactly the move, still. I suspect that the open sets in the cover are, for any given $\Gamma_0$ and $\phi\in\Gamma_0$, the set of all associated $D_\phi$ and $2^V-D_\phi$, hence why he remarked that $D_\phi$ is both open and closed. How does the fact that every open cover has a finite subcover connect to these particular finite subcovers?

• Google for the Finite Intersection Property, a well´known and extremely useful characterization of compactness. It should be explained in every textbook on general topology. Jan 13, 2017 at 6:10

One of the key theorems that characterize compactness says that a topological space $X$ is compact if and only if every collection $\mathcal{C}$ of closed subsets of $X$ such that the intersection of every finite subset of $\mathcal{C}$ is nonempty, has nonempty intersection.