Definition 4.16.1 (Compact). Let $X$ be a topological space, then the following are equivalent:
For every family $\seqi{U}$ of open sets with $\bigcup_{i \in I}U_{i} = X$, there exists $J \subset I$ finite such that $\bigcup_{j \in J}U_{j} = X$.
For every family $\seqi{E}$ of closed sets with $\bigcap_{j \in J}E_{j} \ne \emptyset$ for all $J \subset I$ finite, $\bigcap_{i \in I}E_{i} \ne \emptyset$.
Every filter in $X$ has a cluster point.
Every ultrafilter in $X$ converges.
If the above holds, then $X$ is compact.
Proof. (1) $\Rightarrow$ (2): For each $J \subset I$, let
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
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 Definition 4.2.12, the cluster points and the limit points of an ultrafilter coincide.
(3) $\Rightarrow$ (1): For each $J \subset I$, let
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
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.$\square$