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

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