Let B be a Banach space, and let B* be its dual Banach space (the space of continuous linear functionals on B). Then the unit ball of B* is compact in the weak-* topology.
What?
In a finite dimensional vector space, the unit ball is compact. This is essentially the Heine-Borel theorem. But even for something as "simple" as an infinite dimensional Hilbert space, it is not compact (e.g. in L2, the sequence of unit basis vectors does not have any subsequence which converges in norm). WIBNI we still managed to have some convergence properties?
Alaoglu's theorem answers that need! If we have a closed bounded set in B* (and note that it's bounded in norm!), then it's compact in the weak-* topology.
Is it useful?
YES, very! Here's one way in which Alaoglu's theorem gets used.
We can often find the set of probability measures on some Ω within the elements of norm 1 in the dual of some vector space. If Ω is a countable set, then any probability measure P on Ω is just an assignment of a (nonnegative) probability pi to every element xi∈Ω, subject to the restriction that ∑pi=1. In other words, we may find P within l1(Ω), on the closed unit ball. This Banach space turns out to be the dual Banach space of the space of all sequences (a1,a2,...) which converge to 0. So Alaoglu's theorem guarantees that any sequence of probability measures P1,P2,... has some subsequence which converges in the weak-* topology.
What does this mean? Well, suppose Pnk->P in the weak-* topology. It turns out that this means that for any event E, Pnk(E)->P(E). In particular, taking E=Ω, we see that P(Ω)=1; taking E={xi} for any i, we see that P({xi})≥0. So P itself is a probability measure. Without any real work, we've shown a peculiar compactness property for probability measures!