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\begin{document}
 \parindent=0cm

\section*{II.2 Compactness}

\begin{defn}
Let $X$ be a topological space and $S$ be a subset of $X$. A
collection  $\mathcal{U}$ is a covering of $S$ if $S \subset \bigcup_{U
\in \mathcal{U}} U$. If each $ U \in \mathcal{U} $ is an open set,
$\U$ is called an open covering of $S$.
\end{defn}

\begin{defn}
A topological space $X$ is compact if every open covering of X has a
finite subcovering, i.e., for each open covering $\mathcal{U}$ there
exists a finite number of sets $\{U_{1}, U_{2}, \ldots, U_{n}\}
\subset \mathcal{U}$ such that $ X = \bigcup U_{i}$.
\end{defn}

Note that a subset $A$ of $X$ is compact iff an open covering of $A$ in $X$ has a finite subcovering.


%\begin{defn}
%$x_{0}$ is a cluster point of $(x_{n})$ if for any open
%neighborhood $U$ of $x_{0}$ and for any $N \in \mathbb{N}$, there exists $n > N$
%such that $x_{n}$ is in $U$.
%\end{defn}

%\begin{defn}
%If $\mu : \N \ra \N$ is a monotone increasing function($n>m
%\Ra \mu(n) > \mu(m)$), then $x \circ \mu: \N \ra X$ is called a
%subsequence of $(x_{n})$
%\end{defn}

\begin{ex}

1. $X$ has a indiscrete topology. $\Ra$ $X$ is compact. \\
2. $X$ is a finite set. $\Ra$ $X$ is compact. \\
3. A finite union of compact sets is compact. \\
4. $\R$ is not compact. \\
5. $(a,b]$ is not compact. Consider $(0,1]$ and its open covering
    \[
    \{(1/n,1] | n=1,2,3, \ldots\}.
    \]
6. (Heine-Borel property) Every closed and bounded interval $[a,b]$ in $\R$ is
compact.\\
  \begin{proof}
  Let $\mathcal{U}$ be an open covering of $I=[a,b]$.\\
  Let $S:=\{x \in I|[a,x]\,\, can \,\,be\,\, covered\,\, by\,\, finitely\,\, many\,\, open\,\,
  sets\,\,
  in\,\, \U\}$. Then $S\neq\emptyset$ since $a \in S$. Let $c:=\sup S$.
  Since $\U$ is a covering, $\exists U\in \U$ such that $c\in U$. If
  $c\neq b$, choose $\eps>0$ so that $(c-2\eps,c+2\eps)\subset U$.
  Since $[a,c-\eps]$ is covered by finitely many open sets in $\U$,
  say $U_1,\cdots U_n$, we have $[a,c+\eps]\subset
  (\bigcup^{n}_{i=1}U_i)\cup U$, i.e., $[a,c+\eps]$ is covered by
  finitely open sets, and hence $c+\eps\in S$. This is a contradiction
  to the fact $c=\sup S$, and we conclude that $c=b$. But in this
  case, by the same argument as above, $c=b\in S$.
  \end{proof}
\end{ex}


\begin{prop} %prop1
X is compact. $\LRa$ If
$\mathcal{C}$ is a family of closed subsets of $X$ with finite
intersection property, i.e., $C_{1} \bigcap C_{2} \bigcap \ldots
\bigcap C_{n} \neq \emptyset$ for any finite subcollection $\{C_{1},
\ldots, C_{n} \} \subset \mathcal{C}$, then $\bigcap_{C \in
\mathcal{C}} C \neq \emptyset$.
\end{prop}

\begin{proof}
Every collection of open subsets whose union is $X$ contains a
finite subcollection whose union is X. $\LRa$ Every collection of
closed subsets whose intersection is empty contains a finite
subcollection whose intersection is empty.
\end{proof}

\begin{ex}
$\{(-\infty,n] ~|~ n \in \Z\}$ satisfies FIP but $\bigcap \{ (-\infty,n] ~|~ n \in \Z \} = \emptyset$.
$\Ra \R$ is not compact.
\end{ex}

\begin{prop} %prop2
A closed subset of a compact space is compact.
\end{prop}

\begin{proof}
Suppose $\mathcal{U}$ is an open covering of closed subset A.
Then $\mathcal{U} \cup \{A^{c} \} $ is an open covering of X and hence $\exists$
a finite open subcovering  $\mathcal{V} \subset \mathcal{U}$ of $X$.
Now $\mathcal{V}- \{ A^{c} \}$. is the desired finite subcovering for $A$.
\end{proof}

\begin{lem} %lemma3
Suppose $X$ is Hausdorff. If $C$ is a compact subset of
$X$ and $x$ is a point disjoint from $C$, then
there exist two disjoint open neighborhoods $U$ and $V$ of $C$ and $x$ respectively. In other words, a compact set and a point can be separated by open sets in a Hausdorff space.
\end{lem}

\begin{proof}
For each $y \in C$, the Hausdorff property implies that
there are disjoint open neighborhoods $U_{y} \ni y$ and $V_{y} \ni
x$. Since C is compact, $C \subset U_{y_1} \bigcap U_{y_2} \bigcap
\ldots \bigcap U_{y_n} = U.$ Set $V = V_{y_1} \bigcap V_{y_2}
\bigcap \ldots \bigcap V_{y_{n}}.$
\end{proof}

\begin{prop}
A compact subset of a Hausdorff space is closed.
\end{prop}

\begin{proof}
$C^{c}$ is open since, for any point $x \in C^c$, there exists an open neighborhood $O_x$ disjoint with $C$.
\end{proof}


\begin{prop}
Two disjoint compact subsets of a Hausdorff space have disjoint open
neighborhoods.
\end{prop}

\begin{proof}
Suppose C and D are compact subsets. For each $y \in D$, there exist
disjoint open neighborhoods $U_{y}, V_{y}$ of $C$ and $y$ respectively. By
compactness, $D$ is covered by finitely many $V_{y}$'s. Also
$C$ is contained in the intersection of the corresponding finitely
many open sets $U_{y}$'s.
\end{proof}

\begin{thm}  %thm6
The product of finitely many compact spaces is compact.
\end{thm}

\begin{proof}
It is sufficient to show that the product of two compact spaces is compact.
Let $X$ and $Y$ be two compact spaces.
Suppose $\mathcal{U} = \{U_{\alp} ~|~ \alp \in J\}$ is an open covering of $X \times Y$.
Fix $x \in X$. Then for each $y \in Y$, there is an open set $ U \in \mathcal{U}$ which
contains $(x,y)$, and a basic open neighborhood of $(x,y)$,  $V \times W \subset \mathcal{U}$.
The collection of all such basic open sets covers the compact set
 $ \{x \} \times Y $ and hence has
a finite subcovering $ \{ V_1 \times W_1, \cdots, V_n \times W_n \}$
where each $V_i \times W_i \subset
U_i$ for some $U_i \in \mathcal{U} $.

Let $V_{x} = \bigcap_{i=1}^n V_{i}$, then
    $
    V_{x} \times Y
    \subset \bigcup_{i=1}^{n}(V_{i} \times W_{i}) \subset \bigcup^n_{i=1} U_i
    $ is an open set containing $\{x\} \times Y$.
Now $\mathcal{V} = \{V_{x} | x \in X \}$ is an open covering of $X$.
Since $X$ is compact, there is a finite subcovering
$\{V_{x_1}, \ldots V_{x_{n}}\}$ of $\mathcal{V}$.
Also each $V_{x_i} \times Y$ is covered by a
finite number of elements in $\mathcal{U}$.
Thus $X \times Y$ is covered by
finitely many open sets in $\mathcal{U}$.
\end{proof}


\begin{thm} %thm7
(Generalized Heine-Borel) Suppose $A$ is a subset of $\R^n$, then $A$ is compact if and only if $A$ is closed and bounded.
\end{thm}

\begin{proof}
Suppose that $A$ is compact, then $A$ is closed since $\R^n$ is Hausdorff. Consturct an open covering $\mathcal{B}$ of $A$ as
    \[
    \mathcal{B} = \{B_n(0) ~|~ n = 1,2, \ldots \}.
    \]
Since $A$ is compact, $\mathcal{B}$ has a finite subcovering and hence $A$ is contained in $B_n(0)$ for some $n \in \N$.

Conversely suppose that $A$ is closed and bounded, then there exists $r \in \R$ such that $A $ is contained in $B_r(0)$. Note that $A \subset B_r(0) \subset [-r, r]^n$. Thus $A$ is a closed subset of the compact space $[-r,r]^n$. Hence $A$ is compact.
\end{proof}

\begin{ex}

1. Closed balls and spheres are compact.

2. Cantor set is compact.

3. $I \bigcap \Q$ is not compact.

4. $\R \subset \R^2$ is not compact.
\end{ex}

\begin{rem}
Let $X$ be a metric space. If $A \subset X$ is compact, then $A$ is closed and bounded.
But not conversely.
\end{rem}

\begin{ex}
Consider a metric space $X$ with a metric
    \[
    d(x,y) =
    \left\{
        \begin{array}{ll}
        1 & \textrm{if~} x = y\\
        0 & \textrm{otherwise}.\\
        \end{array}
    \right.
    \]
It is clear that the topology induced by $d$ is discrete.
Thus, if $X$ is an infinite set, $X$ cannot be compact. However $X$ is bounded.
\end{ex}

\begin{thm}
Let $f : X \ra Y$ be a continuous function from a space $X$ to a space $Y$. Then
the followings hold.

1. If $X$ is compact, $f(X)$ is compact.

2. If $X$ is compact and $Y$ is Hausdorff, $f$ is a closed map.

3. If $X$ is compact, $Y$ is Hausdorff and $f$ is bijective, then $f$ is a homeomorphism.
\end{thm}

\begin{proof}

1. Let $\U$ be an open covering of $f(X)$, then $\V = \{f^{-1}(U) ~|~ U \in \U \}$
is an open covering of $X$ and has a finite subcovering
$\{f^{-1}(U_1), f^{-1}(U_2), \ldots, f^{-1}(U_n) \}$.
Since $X \subset f^{-1}(U_1) \bigcup f^{-1}(U_2) \bigcup \ldots \bigcup f^{-1}(U_n) = f^{-1}(U_1 \bigcup \ldots \bigcup U_n)$, $f(X)$ is covered by a finite subcovering $\{U_1, \ldots, U_n \}$.

2. Suppose $A$ is a closed subset of $X$. Since $X$ is compact, $A$ is compact and $f(A)$ is compact. Now $f(A)$ is a compact subset of a Hausdorff space $Y$. Hence $f(A)$ is closed.

3. It is clear from the above result.
\end{proof}

\textbf{(exercise)} Let $f:X \ra \R$ be a continuous map from a compact space $X$ to the real line $\R$. Show that $f$ attains its maximum and minimum.

\begin{cor}
Let $f : X \ra Y$ be a continuous function from a compact space
$X$ to a Hausdorff space $Y$. If $f$ is continuous and injective, $f$ is an imbedding.
\end{cor}

\begin{ex}

1. The function $f : \prod A_i \ra [0,1]$, given by $ f((a_i)) = \sum \frac{a_i}{3^i}$, is an imbedding.

2. A wild arc in $\R^3$ is an imbedding of an unit interval $[0,1]$ into $\R^3$.

\end{ex}

\begin{defn}
$X$ is countably compact if every countable covering has a finite subcovering.
\end{defn}

\begin{defn}
$X$ has the Bolzano Weierstrass Property(BWP) if every infinite subset of $X$ has an accumulation point.
\end{defn}

\begin{thm}
If $X$ is countably compact, the $X$ has BWP.
\end{thm}

\begin{proof}
It suffices to show that every countably infinite subset has an accumulation point.
Suppose that a countably infinite set $A = \{a_1, a_2, \ldots \}$ does not have an accumulation point. Then
    \begin{itemize}
    \item[(i)] $A$ is closed since $\bar{A} = A \bigcup A' = A$,
    \item[(ii)] each $a_i$ is an isolated point of $A$. Thus, for each $a_i$, there is an open neighborhood $O_i$ such that $O_i \bigcap A = \{a_i\}$.
    \end{itemize}
Thus $\{O_i\} \bigcup \{X-A\}$ is an countable open covering of $X$. By the countable compactness there is a finite subcovering $\{O_{i_1}, \ldots, O_{i_n}\} \bigcup \{X-A\}$. Therefore some $O_{i_k}$ must contain infinitely many $a_i$'s in $A$. This is a contradiction.
\end{proof}

\begin{thm}
If a space $X$ is Hausdorff and has BWP, then $X$ is countably compact.
\end{thm}

\begin{proof}
Let $\U = \{U_1, \ldots\}$ be a countable open covering of $X$. We may assume that $\U$ is not redundant, i.e., $U_{n+1}$ is not contained in $U_1 \bigcup \ldots \bigcup U_n$ for each $n$.
Suppose that $\U$ does not have a finite subcovering. Then there is a set $A=\{x_n \in X ~|~ n=1,2,\ldots\}$ such that each $x_n$ is in $U_n - (U_1 \bigcup \ldots \bigcup U_{n-1})$. $A$ is an infinite set since $x_i \neq x_j$ if $i \neq j$.
By BWP, $A$ has an accumulation point $x \in A$.
There is $U_n \in \U$ which contains $x$. From the Hausdorff condition,
$U_n$ should contain infinitely many $x_i$'s. This is a contradiction.
\end{proof}

\begin{rem}
Compact $\Ra$ Countably compact $\Ra$ BWP
\end{rem}


\end{document}
