Category argument template. Let XX be a and let U1,U2,XU_1,U_2,\dots\subseteq X be and . Then

n=1Un\bigcap_{n=1}^{\infty}U_n

is dense in XX, and it is a because its complement is a countable union of . If XX\ne\varnothing, the intersection is therefore nonempty.

How to use the method

Conceptually, this packages the into a reusable method: encode the nnth requirement of a property as membership in a dense open set UnU_n, and then conclude that there are points satisfying all requirements at once (finite versions rely only on ).