Problem 1.4

Question

Let $M$ be a topological manifold, and let $\mathcal{U}$ be an open cover of $M$.

\begin{enumerate}[label=(\alph*)]
\item Assuming that each set in $\mathcal{U}$ intersects only finitely many others, show that $\mathcal{U}$ is locally finite.
\item Give an example to show that the converse to (a) may be false.
\item Now assume that the sets in $U$ are precompact in $M$, and prove the converse: if $\mathcal{U}$ is locally finite, then each set in $\mathcal{U}$ intersects only finitely many others.
\end{enumerate}

Elu Thingol · Accepted Answer

\begin{proof}
\begin{enumerate}[label=(\alph*)]
    \item Let $p \in M$ be arbitrary. Then $p$ is contained in some set $U \in \mathcal{U}$ and $U$ is exactly the neighborhood of $p$ that intersects finitely many of the other sets in $\mathcal{U}$.
    
    \item Consider the following family of sets
    \begin{equation*}
        \mathcal{U} := \left\{ [n, n+1) \subset \mathbb{R} \mid n \in \mathbb{Z} \right\} \, \cup\, \{ \mathbb{R} \}.
    \end{equation*}
    Then $\mathcal{U}$ is obviously an open cover of $\mathbb{R}$, and it is locally finite since every point $p \in \mathbb{R}$ is only covered by at most two sets in $\mathcal{U}$. However, the neighborhood $\mathbb{R}$ intersects infinitely many other sets in $\mathcal{U}$.
    
    \item Let $U \in \mathcal{U}$. Since $\mathcal{U}$ is precompact, $\overline{U} \subseteq M$ is compact. For each $x \in \overline{U}$, pick an open set $U_x \in \mathcal{U}$ from our collection that contains $x$. Then, $\{U_x\}_{x \in \overline{U}}$ is an open cover of $\overline{U}$. Since $\overline{U}$ is compact, we have a finite collection $U_1,\ldots,U_n \in \mathcal{U}$ which covers $\overline{U}$. By assumption, each $U_1,\ldots,U_n$ intersects only finitely many sets in $\mathcal{U}$. By transitivity, $\overline{U}$, and thus $U$, intersects only finitely many sets in $\mathcal{U}$.
\end{enumerate}
\end{proof}

Problem 1.4

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Problem 1.4

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