On page 24, one finds the claim that the compactness theorem follows from Tychonoff's theorem. Here is how.
Start with the two-point "space" {F, T}, with the discrete topology (every set is open). This is trivially compact.
Now take the product space
{F, T} x {F, T} x {F, T} x ...
consisting of all the truth assignments. That is, a point in the
product space has an A_1 component (either F or T), an A_2 component, and
so forth. (This would make sense even if the set of sentence symbols were
some arbitrary set, instead of looking like the positive integers.)
This space is assigned the product topology. Under this topology,
an open set consists of all truth assignments meeting some restriction
on finitely many components, but with no restriction on
the other components. (Check: The intersection of two open sets is
open. The union of lots of open sets is open.) By Tychonoff's theorem,
the space of truth assignments is compact.
Now suppose that some set of formulas
{sigma1, sigma2, sigma3, ...}
is unsatisfiable. We will proceed to make an open cover for the
space. (Notation: "Mod Phi" denotes the set of all truth assignments
that satisfy Phi.) The sets
Mod{not-sigma1},
Mod{(not-sigma1) v (not-sigma2)},
Mod{(not-sigma1) v (not-sigma2) v (not-sigma3)}, ...
form an monotonically increasing sequence of open sets. In fact, they form an
open cover, because any point outside all of the sets would satisfy
the entire set
{sigma1, sigma2, sigma3, ...}, which by supposition is unsatisfiable.
By compactness, this open cover has a finite subcover.
By the monotonicity, some one set
Mod{(not-sigma1) v (not-sigma2) v ... v (not-sigma_k)}
covers the space. Which is to say that
{sigma1, sigma2, ... , sigma_k}
is unsatisfiable.
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