Definition 3.10.1 (Compact).label Let $X$ be a topological space, then the following are equivalent:
- (1)
For every open covering $\curl{U_i}_{i\in I}$ of $X$ there exists a finite subcover $\curl{U_j}_{j\in J}$.
- (2)
For every family of closed sets $\curl{C_i}_{i\in I}$ satisfying the finite intersection property
For each finite subset $J\suf I$, $\caps{}{j\in J}C_{j}\neq\emp$
we have $\caps{}{i\in I }C_{i}\neq\emp$.
- (3)
Every filter in $X$ has a cluster point.
- (4)
Every ultrafilter in $X$ converges.
If the above holds, then $X$ is compact
Proof. (1)$\implies$(2): For each $J\suf I$, let
then $U_{J}\suf X$ is open. Suppose for contradiction that $\caps{}{i\in I}E_{i}=\emp$ then
isan open cover for $X$. By assumption, $U_{J}\sub X$ for all $J\suf I$ finite hence $\textbf{U}$ admits no finite subcover, a contradiciton.
(2)$\implies$(3): Let $\scr{F}\suf \cali{P}(X)$ be a filter, then $\curl{\cl{E}:E\in\scr{F}}$ satisfies the hypothesis of (2).
(3)$\iff$(4): The cluster points and limit points of an ultrafilter coincides. (3)$\implies$(1): For each $J\suf I$, let
then for each $J,J'\suf I,E_{J}\cap E_{J'}=E_{J\cup J'}$. Assume for contradiction that $\curl{U_i}_{i\in I}$ admits no finite subcover. Let
then $\scr{B}$ is a filter base consisting of closed sets. By assumption, there exists $x\in \caps{}{i\in I}U_{i}^{c}$, so $\curl{U_i}_{i\in I}$ is not an open cover, contradiction.$\square$
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