Solution
= Solution
The inclusion $f(\bigcap_nK_n)\subseteq\bigcap_nf(K_n)$ 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 $x\in\bigcap_nK_n$ and $f(x)=y$, proving \b[equality of the two sets].