\section{Compactness}
\label{section:compact}

\begin{definition}[Compact]
\label{definition:compact}
    Let $X$ be a topological space, then the following are equivalent:
    \begin{enumerate}
        \item For every family $\seqi{U}$ of open sets with $\bigcup_{i \in I}U_i = X$, there exists $J \subset I$ finite such that $\bigcup_{j \in J}U_j = X$.
        \item For every family $\seqi{E}$ of closed sets with $\bigcap_{j \in J}E_j \ne \emptyset$ for all $J \subset I$ finite, $\bigcap_{i \in I}E_i \ne \emptyset$.
        \item Every filter in $X$ has a cluster point.
        \item Every ultrafilter in $X$ converges.
    \end{enumerate}
    If the above holds, then $X$ is \textbf{compact}.
\end{definition}
\begin{proof}
    (1) $\Rightarrow$ (2): For each $J \subset I$, let
    \[
    U_J = \bigcup_{j \in J}E_j^c
    \]

    then $U_J \subset X$ is open. For any $J, J' \subset I$, $U_J \cup U_{J'} = U_{J \cup J'}$.

    Suppose for contradiction that $\bigcap_{i \in I}E_i = \emptyset$, then
    \[
    \mathbf{U} = \bracs{U_J|J \subset I \text{ finite}}
    \]

    is an open cover for $X$. By assumption, $U_J \subsetneq X$ for all $J \subset I$ finite. Thus $\mathbf{U}$ admits no finite subcover, contradiction.

    (2) $\Rightarrow$ (3): Let $\fF \subset 2^X$ be a filter, then $\bracsn{\overline{E}| E \in \fF}$ satisfies the hypothesis of (2).

    (3) $\Leftrightarrow$ (4): By \autoref{definition:accumulation-point}, the cluster points and the limit points of an ultrafilter coincide.

    (3) $\Rightarrow$ (1): For each $J \subset I$, let
    \[
    E_J = \bigcap_{j \in J}U_j^c
    \]

    then for each $J, J' \subset I$, $E_J \cap E_{J'} = E_{J \cup J'}$. Assume for contradiction that $\mathbf{U}$ admits no finite subcover. Let
    \[
    \fB = \bracs{E_H|J \subset I \text{ finite}}
    \]

    then $\fB$ is a filter base consisting of closed sets. By assumption, there exists $x \in \bigcap_{i \in I}U_j^c$, so $\mathbf{U}$ is not an open cover, contradiction.
\end{proof}

\begin{proposition}
\label{proposition:compact-extensions}
    Let $X$ be a topological space and $E, F \subset X$ be compact, then the following sets are compact:
    \begin{enumerate}
        \item $E \cup F$.
        \item $G \subset E$ closed.
        \item For any topological space $Y$ and $f \in C(E; Y)$, $f(E)$.
    \end{enumerate}
\end{proposition}
\begin{proof}
    (1): Let $\seqi{U}$ be an open cover of $E \cup F$, then there exists $J_1, J_2 \subset I$ finite such that $\bigcup_{j \in J_1}U_j \supset E$ and $\bigcup_{j \in J_2}U_j \supset F$. In which case, $\bigcup_{j \in J_1 \cup J_2}U_j \supset E \cup F$.

    (2): Let $\seqi{U}$ be an open cover of $G$. Since $G$ is closed, $\seqi{U} \cup \bracs{G^c}$ is an open cover of $E$. In which case, there exists $J \subset I$ finite such that $\bigcup_{j \in J}U_j \cup G^c \supset E$, so $\bigcup_{j \in J}U_j \supset G$.

    (3): Let $\seqi{U}$ be an open cover of $f(E)$, then $\bracs{f^{-1}(U_i)}_{i \in I}$ is an open cover for $E$. Thus there exists $J \subset I$ finite such that $\bigcup_{j \in J}f^{-1}(U_j) \supset E$, so $\bigcup_{j \in J}U_j \supset f(E)$.
\end{proof}

\begin{proposition}
\label{proposition:compact-closed}
    Let $X$ be a Hausdorff space and $E \subset X$ be compact, then $E$ is closed.
\end{proposition}
\begin{proof}
    Let $x \in E^c$. For each $y \in E$, there exists $U_y \in \cn(y)$ and $V_y \in \cn(x)$ such that $U_y \cap V_y = \emptyset$. By compactness, there exists $E_0 \subset E$ finite such that $\bigcup_{y \in E_0}U_y \supset E$. In which case, $E \cap \bigcap_{y \in E_0}V_y = \emptyset$ and $\bigcap_{y \in E_0}V_y \in \cn(x)$. Thus $E^c$ is open.
\end{proof}

\begin{proposition}
\label{proposition:compact-hausdorff-normal}
    Let $X$ be a Hausdorff space, $A, B \subset X$ be compact with $A \cap B = \emptyset$, then there exists $U \in \cn(A)$ and $V \in \cn(B)$ such that $U \cap V = \emptyset$.

    In particular, if $X$ is compact and Hausdorff, then $X$ is normal.
\end{proposition}
\begin{proof}
    For each $x \in A$ and $y \in B$, there exists $U_{x, y} \in \cn(x)$ and $V_{x, y} \in \cn(y)$ with $U_{x, y} \cap V_{x, y} = \emptyset$.

    Fix $x \in A$. By compactness of $B$, there exists $B_0(x) \subset B$ finite such that $\bigcup_{y \in B_0(x)}V_{x, y} \in \cn(B)$. Let $U_x = \bigcap_{y \in B_0}U_{x, y} \in \cn(x)$ and $V_x = \bigcup_{y \in B_0(x)}V_{x, y} \in \cn(B)$, then $U_x \cap V_x = \emptyset$.

    By compactness of $A$, there exists $A_0 \subset A$ finite such that $\bigcup_{x \in A_0}U_x \in \cn(A)$. Let $U = \bigcup_{x \in A_0}U_x \in \cn(A)$ and $V = \bigcap_{x \in A_0}V_x \in \cn(B)$, then $A \cap B = \emptyset$.
\end{proof}