Contents

# Contents

## Statement

###### Lemma

Let $X$ be a topological space which is

1. locally compact

in the sense that for every open neighbourhood $U$ there exists a smaller open neighbourhood whose topological closure is compact and still contained in $U$;

Then there exists a countable open cover $\{U_i \subset X\}_{i \in \mathbb{N}}$ of $X$ such that for each $i \in I$

1. the topological closure $Cl(U_i)$ is a compact subspace

2. $Cl(U_i) \subset U_{i +1}$.

###### Proof

By sigma-compactness of $X$ there exists a countable cover $\{K_i \subset X\}_{i \in \mathbb{N}}$ of compact subspaces. We use these to construct the required cover by induction.

For $i = 0$ set

$U_0 \coloneqq \emptyset \,.$

Then assume that for $n \in \mathbb{N}$ we have constructed a set $\{U_i \subset X\}_{i \in \{1, \cdots, n\}}$ with the required properties.

In particular this implies that

$Q_n \coloneqq Cl(U_n) \cup K_{n-1} \;\subset X$

is a compact subspace. We now construct an open neighbourhood $U_{n+1}$ of this union as follows:

Let $\{U_x \subset X\}_{x \in Q_n}$ be a set of open neighbourhood around each of the points in $Q_n$. By local compactness of $X$, for each $x$ there is a smaller open neighbourhood $V_x$ with

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

So $\{V_x \subset X\}_{x \in Q_n}$ is still an open cover of $Q_n$. By compactness of $Q_n$, there exists a finite set $J_n \subset Q_n$ such that $\{V_x \subset X\}_{x \in J_n}$ is a finite open cover. The union

$U_{n + 1} \coloneqq \underset{x \in J}{\cup} V_x$

is an open neighbourhood of $Q_n$, hence in particular of $Cl(U_n)$. Moreover, since finite unions of compact spaces are compact (this prop.) and since the closure of a finite union is the union of the closures (this prop.), the closure of $U_{n+1}$ is compact:

\begin{aligned} Cl(U_{n+1}) &= Cl\left( \underset{x\in J_n}{\cup} V_x \right) \\ & = \underset{x \in J_n}{\cup} \underset{\text{compact}}{Cl( V_x )} \end{aligned} \,.

This produces by induction a set $\{U_i \subset X\}_{i \in \mathbb{N}}$ with $Cl(U_i)$ compact and $Cl(U_i) \subset U_{i+1}$ for all $i \in \mathbb{N}$. It remains to see that this is a cover. This follows since by construction each $U_{i+1}$ is an open neighbourhood not just of $Cl(U_{i})$ but in fact of $Q_i$, hence in particular of $K_i$, and since the $K_i$ form a cover:

$\underset{i \in \mathbb{N}}{\cup} U_i \supset \underset{i \in \mathbb{N}}{\cup} K_i = X \,.$
###### Proposition

(locally compact and sigma-compact spaces are paracompact)

Let $X$ be a topological space which is

Then $X$ is also paracompact.

###### Proof

Let $\{U_i \subset X\}_{i \in I}$ be an open cover of $X$. We need to show that this has a refinement by a locally finite cover.

By lemma there exists a countable open cover $\{V_n \subset X\}_{n \in \mathbb{N}}$ of $X$ such that for all $n \in \mathbb{N}$

1. $Cl(V_n)$ is compact;

2. $Cl(V_n) \subset V_{n+1}$.

Notice that the complement $Cl(V_{n+1}) \setminus V_n$ is compact, since $Cl(V_{n+1})$ is compact and $V_n$ is open (by this lemma).

By this compactness, the cover $\{U_i \subset X\}_{i \in I}$ regarded as a cover of the subspace $Cl(V_{n+1})\setminus V_n$ has a finite subcover $\{U_i \subset X\}_{i \in J_n}$ indexed by a finite set $J_n \subset I$, for each $n \in \mathbb{N}$.

We consider the sets of intersections

$\mathcal{U}_n \coloneqq \{ U_i \cap ( V_{n+2} \setminus Cl(V_{n-1}) ) \} \,.$

Since $V_{n+2} \setminus Cl(V_{n-1})$ is open, and since $Cl(V_{n+1}) \subset V_{n+2}$ by construction, this is still an open cover of $Cl(V_{n+1})\setminus V_n$. We claim now that

$\mathcal{U} \coloneqq \underset{n\in \mathbb{N}}{\cup} \mathcal{U}_n$

is a locally finite refinement of the original cover, as required:

1. $\mathcal{U}$ is a refinement, since by construction each element in $\mathcal{U}_n$ is contained in one of the $U_i$;

2. $\mathcal{U}$ is still a covering because by construction it covers $Cl(V_{n+1}) \setminus V_n$ for all $n \in \mathbb{N}$, and since by the nested nature of the cover $\{V_n \subset X\}_{n \in \mathbb{N}}$ also $\{Cl(V_{n+1}) \setminus V_n\}_{n \in \mathbb{N}}$ is a cover of $X$.

3. $\mathcal{U}$ is locally finite because each point $x \in X$ has an open neighbourhood of the form $V_{n+2} \setminus Cl(V_{n-1})$ (since these also form an open cover, by the nestedness) and since by construction this has trivial intersection with $\mathcal{U}_{\geq n+3}$ and since all $\mathcal{U}_n$ are finite, so that also $\underset{k \lt n+3}{\cup} \mathcal{U}_k$ is finite.