Past exam of the options II course of the University of Oxford / 2008 / E / E4

In the usual real line, take $\boxed{F_n=[n,\infty )}$ for $n\ge 1$. The sets are closed, nonempty and nested, but no real number belongs to all of them.

Every $K_n$ is closed in the Hausdorff ambient space, because it is compact. Inside the compact space $K_1$, these sets have the finite intersection property: a finite intersection equals the member with largest index and is nonempty. If the total intersection were empty, their open complements would cover $K_1$; a finite subcover would contradict that property. Thus

If no $K_n$ were contained in $U$, the sets $L_n=K_n\setminus U$ would be nonempty, nested and compact, since each is closed in $K_n$. Part (i) would supply a point in ${\bigcap }_n{L}_n$, belonging to every $K_n$ but not to $U$, a contradiction. Thus some $K_{n_0}\subseteq U$, and nesting gives the same containment for every later index.

The inclusion $f({\bigcap }_n{K}_n)\subseteq {\bigcap }_{n}f(K_n)$ holds for every map. Conversely, let $y$ belong to the right side. The sets $L_n=K_n\cap f^{-1}(\{y\})$ are nonempty and nested. They are compact because singletons are closed in a Hausdorff space and $f$ is continuous. By part (b)(i), some $x$ lies in every $L_n$. Then $x\in {\bigcap }_n{K}_n$ and $f(x)=y$, proving equality of the two sets.