# Contents

## Idea

A locally finite cover is a cover which in a suitable sense looks locally like a finite cover.

## Definition

###### Definition

(locally finite cover)

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

A cover $\{U_i \subset X\}_{i \in I}$ of $X$ by subsets of $X$ is called locally finite if it is a locally finite set of subsets, hence 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 $i \in I$.

If $\{U_i \subset X\}_{i \in I}$ is an open cover, then it is called a locally finite open cover.

###### Remark

(alternative characterizations of local finiteness)

Let $X$ be a topological space and let $\{U_i \to X\}_{i \in I}$ be a cover by subsets. Then the following are equivalent:

1. $\{U_i \subset X\}_{i \in I}$ is locally finite (def. 1);

2. there exist an open cover $\{V_j \subset X\}_{j \in J}$ such that for each $j \in J$ there is a finite number of $i \in I$ that $V_j$ intersects $U_i$.

This is because the various $V_i$ constitute open neighbourhoods for all points $x \in X$.

Moreover, suppose that $\{V_j \subset X\}_{j \in J}$ is a cover by any subsets (not necessarily open), but that it is itself a locally finite set of subsets. Then if for all $j \in J$ there are a finite number of $i \in I$ such that $U_i$ intersects $V_j$, it follows again that also $\{U_i \subset X\}_{i \in I}$ is locally finite.

This is because by the local finiteness of $\{V_j \subset X\}_{j \in J}$ we have for every point $x \in X$ an open neighbourhood $O_x \supset \{x\}$ which intersects only a finite number of the $V_j$, and since each of these intersects only a finite number of the $U_i$, in total also $O_x$ can only intesect a finite number of the $U_i$.

## Properties

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 $\{V_j \subset X\}_{j \in J}$, be a refinement to a locally finite cover.

By definition of refinement we may choose a function

$\phi \colon J \to I$

such that

$\underset{j \in J}{\forall}\left( V_j \subset U_{\phi(j)} \right) \,.$

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 finite 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

(shrinking 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:

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

Revised on June 24, 2017 01:29:52 by Anonymous (2604:2000:e100:9200:54ec:c8c0:d6f1:6df9)