Solution
= Solution
The inclusion $f(\bigcap_nK_n)\subseteq\bigcap_nf(K_n)$ holds for every map. For the reverse choose $y\in\bigcap_nf(K_n)$ 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 $x\in\bigcap_nL_n$. Then $x\in\bigcap_nK_n$ and $f(x)=y$. Hence
$$
\boxed{f\left(\bigcap_nK_n\right)=\bigcap_nf(K_n).}
$$