\chapter{Compactness: The Next Best Thing to Being Finite}

\begin{thm}
Let $X$ be a finite topological space.  Then $X$ is compact.
\end{thm}

\oneperpage

\begin{thm}
Let $C$ be a compact subset of $\R_{\textrm{std}}$. Then $C$ has a maximum point, that is, there is a point $m \in C$ such that for every $x \in C$, $x \leq m$.
\end{thm}

\oneperpage

\begin{thm}
If $X$ is a compact space, then every infinite subset of $X$ has a limit point. 
\end{thm}

\oneperpage

\begin{cor}
If $X$ is compact and $E$ is a subset of $X$ with no limit point, then $E$ is finite.
\end{cor}

\oneperpage

\begin{thm}
A space $X$ is compact if and only if every collection of closed sets with the finite intersection property has a non-empty intersection.
\end{thm}

\oneperpage

\begin{thm}
\label{compact-closed-in-open}
A space $X$ is compact if and only if for any open set $U$ in $X$ and any collection of closed sets $\{ K_\alpha\}_{\alpha\in\lambda}$ such that $\bigcap_{\alpha\in\lambda}K_\alpha \subset U$, there exist a finite number of the $K_\alpha$'s whose intersection lies in $U$.
\end{thm}

\oneperpage

\begin{exercise}
If $A$ and $B$ are compact subsets of $X$, then $A \cup B$ is compact.  Suggest and prove a generalization.
\end{exercise}

\oneperpage

\begin{thm}     
Let $A$ be a closed subspace of a compact space.  Then $A$ is compact.
\end{thm}

\oneperpage

\begin{thm}     
Let $A$ be a compact subspace of a Hausdorff space
$X$.  Then  $A$ is closed.
\end{thm}

\oneperpage

\begin{exercise}
Construct an example of a compact subset of a topological space that is not closed.
\end{exercise}

\oneperpage

\begin{exercise}
Must the intersection of two compact sets be compact? Add hypotheses, if necessary. Extend any theorems you discover, if possible.
\end{exercise}

\oneperpage

\begin{thm}     
Every compact, Hausdorff space is normal.
\end{thm}

\oneperpage

\begin{thm} 
\label{basic-compact}
Let $\cal B$ be a basis for a space $X$. Then
$X$ is compact
if and only if every cover of $X$ by basic open sets in $\cal B$
has a finite subcover.  
   \begin{inst}
      Useful for showing tube lemma and that the product of compact spaces is compact.
   \end{inst}
\end{thm}

\oneperpage

\begin{thm}     
For any $a \leq b$ in $\R$, the subspace $[a,b]$ is compact.
\end{thm}

\oneperpage

\begin{HBthm} 
Let $A$ be a subset of $\R_{\textrm{std}}$. Then $A$ is compact if and only if
$A$ is closed and bounded.
\end{HBthm}

\oneperpage

\begin{exercise}
Consider the rationals $\Q$ with the subspace topology inherited from $\R$.  Find a set $A$ in $\Q$ that is closed and bounded but not compact.
\end{exercise}

\oneperpage

\begin{thm}
Every compact subset $C$ of $\R$ contains a maximum in the set $C$, i.e., 
there is an $m \in C$ such that for any $x \in C$, $x \leq m$.
\end{thm}

\oneperpage

\begin{thm}[The tube lemma]
Let $X\times Y$ be a product space with $Y$ compact.  If $U$ is an
open set of $X\times Y$ containing the set $x_0 \times Y$, then
there is some open set $W$ in $X$ containing $x_0$ such that
$U$ contains $W \times Y$ (called a ``tube'' around $x_0 \times Y$).
\end{thm}

\oneperpage

\begin{thm}
Let $X$ and $Y$ be compact spaces. Then $X \times Y$ is compact.
\end{thm}

\oneperpage

\begin{HBthm} 
Let $A$ be a subset
of $\R^n$ with the standard topology. Then $A$ is compact if and only if
$A$ is closed and bounded.
\end{HBthm}

\oneperpage

\begin{ASthm} 
Let $\mathcal S$ be a subbasis for a space $X$.  Then $X$ is compact if and 
only if every subbasic open cover has a finite subcover.  
%\marginpar{hint}%%%%%
   
\end{ASthm}

\oneperpage

\begin{exercise}
Use the Alexander Subbasis Theorem to prove that the space $2^X$ is compact for every $X$. 
\end{exercise}

\oneperpage

\begin{Tychonoff's Theorem}
Any product of compact spaces is compact.
\end{Tychonoff's Theorem}

\oneperpage

\begin{exercise}
Consider the set $[0,1]^\omega$ and 
show that the Tychonoff Theorem is not true if the box topology is used instead of the product topology.
%just consider $[0,1]^\omega$, and put a
%``box'' of radius 3/4 in the sup metric around every ``corner'' of
%this infinite dimensional cube.  There is no finite subcover since
%every corner is covered by exactly one set.  
\end{exercise}

\oneperpage

\begin{thm}
Every countably compact and Lindel\"of space is compact.
\end{thm}

\oneperpage

\begin{thm}     
Let $X$ be a $T_1$ space.  Then $X$ is countably
compact if and only if every infinite subset of $X$ has a limit point.
\end{thm}

\oneperpage

\begin{thm}     
If $X$ is a Lindel\"of space, then every uncountable
subset of $X$ has a limit point.
\end{thm}

\oneperpage

\begin{exercise}
Formulate and prove theorems about Lindel\"of and countably compact spaces analogous to the theorems you proved relating compactness with collections of closed sets with the finite intersection property.
\end{exercise}

\oneperpage

\begin{thm}     
If $A$ is a closed subspace of a countably compact
(respectively, Lindel\"of) space, then $A$
is countably compact (respectively, Lindel\"of).
\end{thm}

\oneperpage

\begin{thm}     
Every regular, Lindel\"of space is normal.
%  
\end{thm}

\oneperpage

\begin{thm} 
Let $\cal B$ be a basis for a space $X$. Then
$X$ is Lindel\"of 
if and only if every cover of $X$ by basic open sets in $\cal B$
has a countable subcover.  
\end{thm}

\oneperpage

\begin{cor}    
Every $2^{\textrm{nd}}$ countable space is Lindel\"of.
\end{cor}

\oneperpage

\begin{exercise}
Can you think of a topological space in which every countable open cover by basic open sets has a finite subcover and yet not every countable open cover has a finite subcover? 
\end{exercise}

\oneperpage

\begin{exercise}
Show that $\R_{LL}$ is Lindel\"of, but $\R_{LL} \times \R_{LL}$ is not Lindel\"of.
\end{exercise}

\oneperpage

\begin{thm}     
The space $\omega_1$ of countable ordinals is countably compact but not compact.
\end{thm}

\oneperpage

\begin{thm}    
The space $\omega_1+1$, which includes all countable ordinals together with the ordinal $\omega_1$,  is compact.
\end{thm}

\oneperpage

\begin{exercise}
Extend your table of spaces and properties by adding new rows for the properties compact, Lindel\"of, and countably compact and determining those properties for each of your spaces.
\end{exercise}

\oneperpage

\begin{thm}     
 Let ${\mathcal B}= \{ B_\alpha\}_{\alpha\in\lambda}$
 be a locally finite collection of subsets of a space $X$.  
 %Let $C$ be a subset of $\lambda$.  
 Then 
 $$
 \overline{ (\bigcup_{\alpha\in \lambda} B_\alpha)} = \bigcup_{\alpha\in \lambda} \overline{B}_\alpha.
 $$
 \end{thm}

\oneperpage

\begin{thm}     
Let $A$ be a closed subspace of a paracompact space. Then $A$
is paracompact.
\end{thm}

\oneperpage

\begin{thm}     
 Every paracompact space is normal.
 \end{thm}

\oneperpage

\begin{thm}     
 Every regular, $T_1$, Lindel\"of space is paracompact.
 \end{thm}

\oneperpage

