Continuous images of nested compact intersections ($f({\bigcap }_n{K}_n)={\bigcap }_{n}f(K_n)$)

Let $X$ be a Hausdorff space, let $K_n$ be a nested sequence of nonempty compact subsets of $X$, and let $f:X\to Y$ be a continuous map into a T1 space. Then the displayed equality holds. The forward inclusion is true for every map. For the reverse, fix $y\in {\bigcap }_{n}f(K_n)$. Each $F_n=K_n\cap f^{-1}(\{y\})$ is nonempty and closed in the compact set $K_1$: the $K_n$ are closed by the Hausdorff assumption, while the fiber is closed because $Y$ is T1 and $f$ is continuous. The nested sets $F_n$ have the finite intersection property, so compactness gives a point in their intersection. Its image is $y$.

The proof only needs the relevant fibers to be closed in $K_1$, rather than all of $Y$ to be T1. A Hausdorff space as codomain is a familiar sufficient special case. Mere continuity into an arbitrary space does not suffice.

Take $X=[0,1]$ with its usual topology and $Y$ the Sierpiński space with open singleton $\{1\}$. The map $f(0)=0$, $f(x)=1$ for $x>0$ is continuous, since $f^{-1}(\{1\})=(0,1]$ is relatively open in $X$. The sets $K_n=[0,1/n]$ are nonempty, nested and compact, butThe fiber of $1$ is not closed, which is exactly where the compact-intersection proof fails. This distinguishes the valid Hausdorff-codomain theorem from a false extension to arbitrary codomains.