# Contents

## Background

###### Definition

(locally compact topological space)

A topological space $X$ is called locally compact if for every point $x \in X$ and every open neighbourhood $U_x \supset \{x\}$ there exists a smaller open neighbourhood $V_x \subset U_x$ whose topological closure is compact and still contained in $U$:

$\{x\} \subset V_x \subset \underset{\text{compact}}{Cl(V_x)} \subset U_x \,.$

## Statement and proof

###### Proposition

(proper maps to locally compact spaces are closed)

Let

1. $(X,\tau_X)$ be a topological space,

2. $(Y,\tau_Y)$ a locally compact topological space according to def. 1,

3. $f \colon X \to Y$ a continuous function.

Then:

If $f$ is a proper map, then it is a closed map.

###### Proof

Let $C \subset X$ be a closed subset. We need to show that every $y \in Y \backslash f(C)$ has an open neighbourhood $U_y \supset \{y\}$ not intersecting $f(C)$ (by this prop.).

By local compactness of $(Y,\tau_Y)$ (def. 1), $y$ has an open neighbourhood $V_y$ whose topological closure $Cl(V_y)$ is compact. Hence since $f$ is proper, also $f^{-1}(Cl(V_y)) \subset X$ is compact. Then also the intersection $C \cap f^{-1}(Cl(V_y))$ is compact, and hence so is

$f(C \cap f^{-1}(Cl(V_y))) = f(C) \cap (Cl(V)) \; \subset Y \,.$

This is also a closed subset, since compact subspaces of Hausdorff spaces are closed. Therefore

$U_y \coloneqq V_y \backslash ( f(C) \cap (Cl(V_y)) ) = V_y \backslash f(C)$

is an open neighbourhod of $y$ not intersecting $f(C)$.

Last revised on May 16, 2017 at 13:49:24. See the history of this page for a list of all contributions to it.