Past exam of the mathematics course of the University of Cambridge / 2024 / ii / Paper 2 / 22G / Solution

The Baire category theorem states that a complete metric space is not a countable union of closed sets with empty interior. To prove it, let $G_1,G_2,\dots$ be open dense sets and start with any nonempty open set $V$. Inductively choose closed ballswith the first ball inside $V\cap G_1$. The centres are Cauchy. Completeness supplies a limit lying in every ball, hence in $V\cap {\bigcap }_n{G}_n$. Thus the intersection is dense, which is the equivalent form of the theorem.

Now each $nS$ is closed and ${\bigcup }_{n\ge 1}nS=X$. Baire givesfor some $x,r,n$. Symmetry gives $B(-x,r)\subset nS$. If $\Vert z\Vert <r$, then $x+z$ and $-x+z$ lie in these two balls; convexity makes their midpoint $z$ belong to $nS$. Hencewhich proves the closed convex absorbing set has an origin neighbourhood result.

Convexity cannot be dropped. In $X=\mathbb{R}$, setThis is closed and symmetric but has gaps arbitrarily close to zero. For any $x\ne 0$, choose $k$ so that the intervalhas length at least one, and choose an integer $n$ in it. Then $x/n\in S$, so ${\bigcup }_{n}nS=\mathbb{R}$, while $S$ is not a neighbourhood of zero. This is a closed symmetric absorbing set without an origin neighbourhood.

Solved by gpt-5.6-sol high.