Past exam of the Probability, statistics and options II course of the University of Oxford / 2008 / E / 4

In the usual real line take $\boxed{F_n=[n,\infty )}$. Each set is nonempty and closed, and $F_{n+1}\subseteq F_n$. No real number lies in them all, because the positive integers are unbounded. Thus their intersection is empty. These sets are not compact.

Compact subsets of a Hausdorff space are closed. Regard every $K_n$ as a closed subset of the compact set $K_1$. Any finite intersection equals the member with largest index and is therefore nonempty. The finite intersection property for closed subsets of a compact space impliesEquivalently, an empty intersection would give an open cover of $K_1$ by their complements, with a finite subcover contradicting nested nonemptiness.

Suppose no such $n_0$ exists. Nestedness then makes every $K_n\setminus U$ nonempty. These are nested compact sets: $X\setminus U$ is closed and $K_n$ is compact. By (i) their intersection is nonempty. But that intersection is $({\bigcap }_n{K}_n)\setminus U$, contrary to the hypothesis. Thus $K_{n_0}\subseteq U$ for some $n_0$, and nestedness gives $\boxed{K_n\subseteq U\text{ for all }n\ge n_0}$ immediately.

The inclusion $f({\bigcap }_n{K}_n)\subseteq {\bigcap }_{n}f(K_n)$ holds for every map. For the reverse choose $y\in {\bigcap }_{n}f(K_n)$ and let $L_n=K_n\cap f^{-1}(\{y\})$. These sets are nonempty and nested. A Hausdorff space has closed singletons, and continuity makes the fibre closed, so each $L_n$ is compact. Part (b)(i) gives a point $x\in {\bigcap }_n{L}_n$. Then $x\in {\bigcap }_n{K}_n$ and $f(x)=y$. Hence