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\begin{document}

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\section*{locally finite cover}

\hypertarget{context}{}\subsubsection*{{Context}}\label{context}

\hypertarget{topology}{}\paragraph*{{Topology}}\label{topology}

[[!include topology - contents]]

\hypertarget{contents}{}\section*{{Contents}}\label{contents}

\noindent\hyperlink{idea}{Idea}\dotfill \pageref*{idea} \linebreak
\noindent\hyperlink{definition}{Definition}\dotfill \pageref*{definition} \linebreak
\noindent\hyperlink{properties}{Properties}\dotfill \pageref*{properties} \linebreak
\noindent\hyperlink{related_concepts}{Related concepts}\dotfill \pageref*{related_concepts} \linebreak


\hypertarget{idea}{}\subsection*{{Idea}}\label{idea}

A \emph{locally finite cover} is a [[cover]] which in a suitable sense looks \emph{locally} like a [[finite cover]].

\hypertarget{definition}{}\subsection*{{Definition}}\label{definition}

\begin{defn}
\label{LocallyFiniteCover}\hypertarget{LocallyFiniteCover}{}
\textbf{(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 \emph{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 [[intersection|intersects]] only [[finite number|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 \emph{locally finite open cover}.

\end{defn}
\begin{remark}
\label{AlternativeCharacterizations}\hypertarget{AlternativeCharacterizations}{}
\textbf{(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:

\begin{enumerate}%
\item $\{U_i \subset X\}_{i \in I}$ is locally finite (def. \ref{LocallyFiniteCover});


\item there exist an \emph{[[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 $U_i$.



\end{enumerate}
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 $U_i$.

\end{remark}
\hypertarget{properties}{}\subsection*{{Properties}}\label{properties}

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

\begin{lemma}
\label{LocallyFiniteRefinementImpliesLocallyFiniteCoverWithOriginalIndexSet}\hypertarget{LocallyFiniteRefinementImpliesLocallyFiniteCoverWithOriginalIndexSet}{}
\textbf{(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 [[axiom of choice|choose]] a [[function]]

\begin{displaymath}
\phi \colon J \to I
\end{displaymath}
such that

\begin{displaymath}
\underset{j \in J}{\forall}\left( V_j \subset U_{\phi(j)} \right)
  \,.
\end{displaymath}
Then $\left\{ W_i \subset X \right\}_{i \in I}$ with

\begin{displaymath}
W_i 
    \;\coloneqq\;
  \left\{
    \underset{j \in \phi^{-1}(\{i\})}{\cup} V_j
  \right\}
\end{displaymath}
is still a [[refinement]] of $\{U_i \subset X\}_{i \in I}$ to a locally finite cover.

\end{lemma}
\begin{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 set|finite]] [[subset]] $K \subset J$ such that

\begin{displaymath}
\underset{j \in J\backslash K}{\forall} \left( U_x \cap V_j = \emptyset \right)
  \,.
\end{displaymath}
Hence by construction

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

\end{proof}
\begin{lemma}
\label{}\hypertarget{}{}
\textbf{([[shrinking lemma]])}

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

\begin{displaymath}
V_i \subset Cl(V_i) \subset U_i
  \,.
\end{displaymath}
\end{lemma}
\hypertarget{related_concepts}{}\subsection*{{Related concepts}}\label{related_concepts}

\begin{itemize}%
\item [[cover]], [[open cover]],


\item [[paracompact topological space]]



\end{itemize}
[[!redirects locally finite covers]]

[[!redirects locally finite open cover]] [[!redirects locally finite open covers]]

[[!redirects locally finite closed cover]] [[!redirects locally finite closed covers]]



\end{document}