4.16 Compactness
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.$\square$
Proposition 4.16.2. Let $X$ be a topological space and $E, F \subset X$ be compact, then the following sets are compact:
$E \cup F$.
$G \subset E$ closed.
For any topological space $Y$ and $f \in C(E; Y)$, $f(E)$.
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 $\bigcup_{j \in J}U_{j} \supset f(E)$.$\square$
Proposition 4.16.3. Let $X$ be a Hausdorff space and $E \subset X$ be compact, then $E$ is closed.
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.$\square$
Proposition 4.16.4. 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.
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 $A \cap B = \emptyset$.$\square$