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I encountered the following problem in L. Gasínski, N.S. Papageorgiou - Exercises in Analysis - Part 1: (2.63)

"Suppose that $X$ and $Y$ are two topological spaces and $f : X \longrightarrow Y$ is a closed function such that for every $y \in Y$ , the set $f^{−1}(y) \subseteq X$ is compact. Show that for every compact set $K \subseteq Y$ , the set $f^{−1}(K)\subseteq X$ is compact."

The proof given uses nets. Is there an alternative, nice proof that uses either ultrafilters, open covers or the finite intersection property characterization of compactness? I'm stuck where I should use the closedness of the map.

Werner
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1 Answers1

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First: we may assume that $Y$ is compact, then we need to show that $X$ is compact. I’m dropping the Hausdorff assumptions because they’re probably not the interesting one.

Let $(U_i)_{i \in I}$ be an open cover of $X$. For each $J \subset I$, $F_J=X \backslash \bigcup_{j \in J}{U_j}$ is a closed subset of $X$ so $V_J=Y \backslash f(F_J)$ is open. $V_J$ is the set of $y \in Y$ such that every pre-image of $Y$ is in some $U_j$ for $j \in J$.

For every $y \in Y$, $(U_i)_{i \in I}$ is an open cover of $f^{-1}(y)$ so there exists a finite $J_y \subset I$ such that $y \in V_{J_y}$. Thus, the set $(V_{J_y})_y$ is an open cover of $Y$, and there are thus $y_1,\ldots,y_r$ such that $Y$ is covered by the $V_{J_{y_i}}$. Then $X$ is covered by the $U_t$ for $t$ in the (finite) reunion of the $V_{J_{y_i}}$.

Aphelli
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