Exercise 1.14 (Closure of locally finite collection of subsets)

Question

extbf{Lemma 1.13.} Suppose $\mathcal{X}$ is a locally finite collection of subsets of a topological space $M$.
\begin{enumerate}[label=(\alph*)]
    \item The collection $\{\overline{X}: X \in \mathcal{X}\}$ is also locally finite.
    \item $\overline{\bigcup_{X \in \mathcal{X}} X} = \bigcup_{X \in \mathcal{X}} \overline{X}$.
\end{enumerate}

	extbf{Exercise 1.14.} Prove the preceding lemma.

Elu Thingol · Accepted Answer

\begin{proof} We first demonstrate (b) and then use it to demonstrate (a).
\paragraph{Part (b)} We show that each set is a subset of the other one.
\begin{itemize}
    \item[($\supseteq$)] Note that  
        \begin{equation*}
            \forall X \in \mathcal{X}: \quad X \subseteq \bigcup_{X \in \mathcal{X}} X,
        \end{equation*}
        thus,
        \begin{equation*}
            \forall X \in \mathcal{X}: \quad \overline{X}\subseteq\overline{\bigcup_{X \in \mathcal{X}} X}.
        \end{equation*}
        Taking the union on the left side, we have 
        \begin{equation*}
            \bigcup_{X \in \mathcal{X}} \overline{X} \subseteq \overline{\bigcup_{X \in \mathcal{X}} X}.
        \end{equation*}
    \item[($\subseteq$)] For the converse, notice that
    \begin{equation*}
        \overline{\bigcup_{X \in \mathcal{X}} X} \supseteq \bigcup_{X \in \mathcal{X}} \overline{X} \supseteq \bigcup_{X \in \mathcal{X}} X.
    \end{equation*}
    This, it suffices to demonstrate that $\bigcup_{X \in \mathcal{X}} \overline{X}$ is closed. 
    For that purpose, pick $x 
ot\in \bigcup_{X \in \mathcal{X}} \overline{X}$. Due to local finiteness, there is an open neighborhood $U$ of $x$ which only intersects finitely many $X \in \mathcal{X}$. We denote them by $X_{1},\ldots,X_{n}$. Now remove these sets from $U$ by defining a new neighborhood
    \begin{equation*}
        V := U\setminus(\overline{X_1}\cup\ldots\cup\overline{X_n}).
    \end{equation*}
    The new neighborhood $V$ is an open set that contains $x$ and has an empty intersection with $X_1,\ldots,X_n$ and therefore with all $X \in \mathcal{X}$. In other words, the complement of $\bigcup_{X \in \mathcal{X}} \overline{X}$ is open, or equivalently, $\bigcup_{X \in \mathcal{X}} \overline{X}$ is closed.
\end{itemize}

\paragraph{Part (a)} Consider the collection of subsets $\mathcal{Y}:=\{\overline{X}: X \in \mathcal{X}\}$ and fix an open neighborhood $U$ of some point $x \in X$. By assumption, $x$ only intersections finitely many $X_1,\ldots,X_n \in \mathcal{X}$. Since $U$
\begin{equation*}
    V := U \setminus \left(\overline{\bigcup_{X \in \mathcal{X}\setminus\{X_1,\ldots,X_n\}} X} ight)
    = U \setminus \bigcup_{X \in \mathcal{X}\setminus\{X_1,\ldots,X_n\}} \overline{X}
    = U \setminus \bigcup_{Y \in \mathcal{Y}\setminus\{\overline{X_1},\ldots,\overline{X_n}\}} Y
\end{equation*}
We see that $V$ is an open set that contains $x$ and intersects at most $\overline{X_1},\ldots,\overline{X_n}$ but none others in $\mathcal{Y}$. Since our choice of $x$ and $U$ was arbitrary, $\mathcal{Y}$ is a locally finite collection of subsets.
\end{proof}

Exercise 1.14 (Closure of locally finite collection of subsets)

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Exercise 1.14 (Closure of locally finite collection of subsets)

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