= Solution
For nonempty <compact sets>, define the <distance from a point to a closed set> by $d(x,B)=\min_{b\in B}|x-b|$ and the <Hausdorff distance> by
$$
d_H(A,B)=\max\{\sup_{a\in A}d(a,B),\sup_{b\in B}d(b,A)\}.
$$
The minima and suprema exist: the sets are nonempty and <compact>, and the distance function is <Lipschitz continuous>. The <Hausdorff distance> is finite, nonnegative and symmetric. If it is zero, every point of $A$ belongs to the <closed set> $B$ and conversely, so $A=B$; the converse is immediate.
For $a\in A$, choose $b\in B$ attaining $d(a,B)$ and then $c\in C$ attaining $d(b,C)$. The <triangle inequality> gives $d(a,C)\leq d(a,B)+d_H(B,C)$. Taking the supremum over $a$, and doing the same with $A,C$ exchanged, proves $d_H(A,C)\leq d_H(A,B)+d_H(B,C)$. Hence this is a <metric>.
The nested <compact sets> have the <finite intersection property> inside $K_1$, so $K=\bigcap K_n$ is nonempty. It is a <closed subset> of $K_1$ and therefore <compact>. Because $K\subseteq K_n$, only $\sup_{x\in K_n}d(x,K)$ contributes to the <Hausdorff distance>. If this did not tend to zero, some increasing indices $n_j$ and $x_j\in K_{n_j}$ would satisfy $d(x_j,K)\geq\epsilon>0$. A <convergent subsequence> in $K_1$ has limit $x$. For each fixed $m$, its tail belongs to the <closed set> $K_m$, so $x\in K$. But then $d(x_j,K)\leq|x_j-x|\to0$, a contradiction. Thus $\boxed{d_H(K_n,K)\to0}$.
Back to article page
