Let $D>0$ be the diameter of the bounded set $S$, and choose $x_0\in S$. Then $S\subseteq B(x_0,D)$. It is enough to prove a <volumetric covering bound> for the unit ball.

Choose a maximal $1/4$-separated set $N$ in $B(0,1)$. The balls of radius $1/8$ centred at points of $N$ are disjoint and lie in $B(0,9/8)$. Comparing <Lebesgue measure>[volumes] gives
$$
|N|\left(\frac18\right)^n\leq\left(\frac98\right)^n,
\qquad |N|\leq9^n.
$$
Maximality means that the balls of radius $1/4$ centred at $N$ cover the unit ball.

After translating and scaling, at most $9^n$ balls of radius $D/4$ cover $S$. Assign each point of $S$ to one covering ball containing it. This gives at most $9^n$ disjoint pieces, each of diameter at most $D/2<D$. Thus the claim holds with the absolute constant $C=9$.

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 109 / 4 / iv / Solution

The unit ball in $\mathbb R^n$ can be covered by at most $9^n$ balls of radius $1/4$. A maximal $1/4$-separated set supplies the centres, and disjoint radius-$1/8$ balls around them fit inside the ball of radius $9/8$.


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 Incoming links: Volumetric covering bound