Show that if $f$ is a continuous map from a compact space into a Hausdorff space is a closed map.

Question

Show that if $f$ is a continuous map from a compact space into a Hausdorff space is a closed map.

Answer 1

Let $F$ be a closed subset of the domain. Then $F$ is compact and therefore $f(F)$ is compact too. But compact subspaces of Hausdorff spaces are closed.

Answer 2

Let $C$ be closed in the domain. It is therefore also compact. The image of a compact set under a continuous map is compact. So $f(C)$ is a compact subset of a Hausdorff space, which is closed. Hence $f$ is a closed map.

Answer 3

Suppose $F \subseteq X$ is closed. Let $y \notin f[F]$. For each $z \in f[F]$ we have $z \neq y$, so there are disjoint open sets $U_z$ and $V_z$ such that $z \in U_z$, $y \in V_z$. This uses Hausdorffness of $Y$. Then for all $z \in f[F]$, $f^{-1}[U_z]$ is open in $X$ by continuity of $f$, and $F \subseteq \cup \{f^{-1}[U_z]: z \in f[F]\}$ (suppose $x \in F$, then $f(x) \in U_z$ for $z = f(x) \in f[F]$, and so $x \in f^{-1}[U_z]$).

So $\{f^{-1}[U_z]l z \in f[F]\} \cup \{ X\setminus F\}$ is an open (as $F$ is closed) cover of the compact space $X$, so there are finitely many $z_1,\ldots z_n \in f[F]$ such that $X= (\bigcup_{i=1}^n f^{-1}[U_{z_i}]) \cup (X \setminus F)$. Now define $V := \cap_{i=1}^n V_{z_i}$, which is an open neighbourhood of $y$ as a finite intersection of such neighbourhoods.

Suppose $z \in V \cap f[F]$, then $z = f(x)$ for some $x \in F$. Then $x \in f^{-1}[U_{z_j}]$ for some $j \in \{1,\dots, n\}$. So $z =f(x) \in U_{z_j}$, but also $z \in V \subseteq V_{z_j}$, contradicting the disjointness of $U_{z_j}$ and $V_{z_j}$. So $V \cap f[F] = \emptyset$,and as this holds for all $y \notin f[F]$, we have that $f[F]$ is closed in $Y$.