Past exam of the mathematics course of the University of Cambridge / 2005 / iii / Paper 6 / 1 / i

The Baire category theorem states that in a complete metric space, a countable intersection of open dense sets is dense. Equivalently, a nonempty complete metric space cannot be a countable union of nowhere dense sets, where a set is nowhere dense if its closure has empty interior.

Here is the proof. Let $G_1,G_2,\dots$ be open dense sets and let $U$ be any nonempty open set. Choose a point $a_1\in U\cap G_1$ and a radius $0<r_1<1/2$ such that the closed ball $B_1=\overline{B}(a_1,r_1)$ lies inside $U\cap G_1$. Inductively choose $a_{m+1}$ in the nonempty open set $B(a_m,r_m)\cap G_{m+1}$, and then a radius $0<r_{m+1}<2^{-m-1}$ such thatThus the closed balls are nested and their radii tend to zero. For $j,k\ge m$, both centers lie in $B_m$, so $d(a_j,a_k)\le 2r_m$. They form a Cauchy sequence. Completeness gives a limit $a$, and closedness puts it in every $B_m$. Consequently $a\in U\cap {\bigcap }_m{G}_m$. As $U$ was arbitrary, the intersection is dense.

To obtain the equivalent formulation, take complements of the closures of the nowhere dense sets; those complements are open dense. Conversely, complements of open dense sets are closed with empty interior. In particular, a countable closed cover of a nonempty complete metric space must contain a set with nonempty interior.