Past exam of the mathematics course of the University of Cambridge / 2019 / ii / Paper 2 / 2H (Topics in Analysis)

The sum is nonempty because $A$ and $B$ are nonempty. It is a bounded set: if $|a|\le M_A$ and $|b|\le M_B$, then $|a+b|\le M_A+M_B$.

To prove that it is a closed set, let $x_j=a_j+b_j\in A+B$ be a convergent sequence with $x_j\to x$. By the Heine-Borel theorem, $A$ is a compact set, so $(a_j)$ has a convergent subsequence $a_{j_r}\to a\in A$. ThenSince $B$ is closed, $x-a\in B$, and hence $x=a+(x-a)\in A+B$. Therefore $A+B$ is nonempty, closed and bounded, so $A+B\in \mathcal{K}$.

Every $K_n$ is contained in the bounded set $K_1$, so the intersection $K={\bigcap }_{n\ge 1}{K}_n$ is bounded. It is closed because an arbitrary intersection of closed sets is closed.

It remains to prove that $K$ is nonempty. Choose $x_n\in K_n$. The set $K_1$ is a compact set, so some subsequence $x_{n_j}$ converges to a point $x\in K_1$. For each fixed $m$, all sufficiently late terms satisfy $n_j\ge m$ and hence $x_{n_j}\in K_{n_j}\subseteq K_m$. Since $K_m$ is closed, $x\in K_m$. This holds for every $m$, so $x\in K$. Thus $K\in \mathcal{K}$.

Because $K\subseteq K_n$, one of the two directed terms in the given Hausdorff distance vanishes, andSuppose this does not tend to zero. Then some $\epsilon >0$, a strictly increasing sequence $(n_j)$ and points $x_j\in K_{n_j}$ satisfyAll $x_j$ lie in the compact set $K_1$, so after passing to a subsequence, $x_j\to x$. For every fixed $m$, eventually $n_j\ge m$, so $x_j\in K_m$ and the closedness of $K_m$ gives $x\in K_m$. Hence $x\in K$, contradicting $|x_j-x|\to 0$. Therefore $\rho (K_n,K)\to 0$.