Past exam of the mathematics course of the University of Cambridge / 2025 / ii / Paper 4 / 12I / Solution

A subset $A$ of a metric space $X$ is a nowhere dense set whenEquivalently, every nonempty open set contains a nonempty open subset disjoint from $A$.

The Baire category theorem says that if $X$ is a complete metric space and $G_1,G_2,\dots$ are open dense subsets of $X$, then ${\bigcap }_{n\ge 1}{G}_n$ is dense in $X$. Equivalently, no nonempty open subset of $X$ is a countable union of nowhere-dense sets.

To prove it, take a nonempty open set $V$. Since $G_1$ is open and dense, there is a closed ballwith $0<r_1<2^{-1}$. Inductively, openness and density of $G_{n+1}$ allow us to chooseThe balls are nested, and for $m>n$ their centres satisfyThus $(x_n)$ is Cauchy and has a limit $x\in X$. Every closed ball contains the tail of the sequence, so it contains $x$. HenceAs $V$ was arbitrary, the intersection is dense.