Introduction

Theorems

# Contents

## Idea

A cover which is locally a finite cover.

## Definition

###### Definition

(locally finite cover)

Let $(X,\tau)$ be a topological space.

An open cover $\{U_i \subset X\}_{i \in I}$ of $X$ is called locally finite if for all points $x \in X$, there exists a neighbourhood $U_x \supset \{x\}$ such that it intersects only finitely many elements of the cover, hence such that $U_x \cap U_i \neq \emptyset$ for only a finite number of $i \in I$.

## Properties

Any open cover defined by a (generalized) partition of unity has a locally finite shrinking. (…details…)

The following says that if there exists a locally finite refinement of a cover, then in fact there exists one with the same index set as the original cover.

###### Lemma

(locally finite refinement induces locally finite cover with original index set)

Let $(X,\tau)$ be a topological space, let $\{U_i \subset X\}_{i \in I}$ be an open cover, and let $(\phi \colon J \to I, \{V_j \subset X\}_{j \in J})$, be a refinement to a locally finite cover.

Then $\left\{ W_i \subset X \right\}_{i \in I}$ with

$W_i \;\coloneqq\; \left\{ \underset{j \in \phi^{-1}(\{i\})}{\cup} V_j \right\}$

is still a refinement of $\{U_i \subset X\}_{i \in I}$ to a locally finite cover.

###### Proof

It is clear by construction that $W_i \subset U_i$, hence that we have a refinement. We need to show local finiteness.

Hence consider $x \in X$. By the assumption that $\{V_j \subset X\}_{j \in J}$ is locally finite, it follows that there exists an open neighbourhood $U_x \supset \{x\}$ and a finitee subset $K \subset J$ such that

$\underset{j \in J\backslash K}{\forall} \left( U_x \cap V_j = \emptyset \right) \,.$

Hence by construction

$\underset{I \in I\backslash \phi(K)}{\forall} \left( U_x \cap W_i = \emptyset \right) \,.$

Since the image $\phi(K) \subset I$ is still a finite set, this shows that $\{W_i \subset X\}_{i \in I}$ is locally finite.

###### Lemma

Let $(X,\tau)$ be a normal topological space, and let $\{U_i \subset X\}_{i \in I}$ be a locally finite open cover. Then there exists a shrinking to a locally finite open cover $\{V_i \subset X\}_{i \in I}$ whose closures $Cl(-)$ are still contained in the original cover:

$V_i \suubset Cl(V_i) \subset U_i \,.$

Revised on April 28, 2017 08:19:43 by Urs Schreiber (92.218.150.85)