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\pageno=31
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\hsize 7.1truein
\vsize 9.7truein
\def\u{\vskip -10pt}
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26.  Compact Spaces

A family of sets, ${\cal F} = \{F_\alpha ~|~ \alpha \in A\}$,
{\bf covers}
the
set $S$ if 
\s
\centerline{$S \subset \cup_{\alpha \in A}F_\alpha$}
\s
${\cal F}$  is said to be a cover of $S$.  



\vskip 5pt
\hrule

${\cal F}_1 = \{~ (x-1, x+1) ~|~ x \in S\}$ is a cover of $S$ since
$$S
\subset
\cup_{x \in S} (x-1, x+1)$$

\hrule

If $S \not= \emptyset$, take $x_0 \in S$ \hfil \break
${\cal F}_2 = \{~ (x_0 - r, x_0 + r) ~|~ r > 0 \}$ is a cover of $S$
since $$S \subset
\cup_{r > 0} (x_0 - r, x_0 + r)$$


\hrule
\vskip -5pt


${\cal F}'$ is a {\bf subcover} of ${\cal F}$ if ${\cal F}' \subset
{\cal
F}$ and ${\cal F}'$ covers $S$.


${\cal F}'$ is a {\bf finite subcover} of ${\cal F}$ (or a finite
subfamily of ${\cal F}$) if ${\cal
F}'$ is a
subcover of ${\cal F}$ and ${\cal F}'$ is finite. 


${\cal F} = \{~ ({1 \over n}, 1)~|~ n = 1, 2, 3, ... \}$
is a cover of $(0, 1)$ since $(0, 1) \subset \cup_{n=1}^\infty({1
\over n},
1)$.  

${\cal F}' = \{~ ({1 \over n}, 1)~|~ n = 5, 6, 7, ... \}$
is a subcover of ${\cal F}$ since ${\cal F}' \subset {\cal F}$ and
$(0, 1) \subset \cup_{n=5}^\infty({1 \over n},   
1)$.


Does there exist a finite subcover?

Defn: A family of sets, ${\cal F} = \{F_\alpha ~|~ \alpha \in A\}$,
is an {\bf open cover} of $S$ if ${\cal F}$ covers $S$ and if
$F_\alpha$ is
open for all $\alpha \in A$.

Defn:  A space $S$ is {\bf compact} if every open cover of $S$ has a
finite
subcover.

Lemma:  If $X$ is finite, then $X$ is compact.

Lemma 26.1:  Let $Y$ be a subspace of $X$.  Every cover of $Y$
consisting of open sets in $X$ has a finite subcover if and only if
every cover of $Y$ consisting of open sets in $Y$ has a finite
subcover


Thm 26.2:  Every closed subspace of a compact space is compact.  

Lemma 26.4:  If $Y$ is a compact subspace of the Hausdorff space $X$,
and $x_0 \not\in Y$, then there exist disjoint open sets $U$ and $V$
of $X$ such that $x_0 \in U$ and $Y \in V$.

Thm 26.3:  Every compact subspace of a Hausdorff space is closed.

Thm 26.5:  The image of a compact space under a continuous map is
compact.

\eject
Thm 26.6:  If $f:X \rightarrow Y$ is continuous and a bijection and
if
$X$
is compact and $Y$ is Hausdorff, then $f$ is a homeomorphism.




Note: $f: [0, 1) \rightarrow \{ (x,y) ~|~ x^2 + y^2 = 1\}, \break 
f(x)
= e^{2\pi i x}$ is continuous and a bijection, but $f^{-1}$ is NOT
continuous.


Note:  $f: \{1, 2\} \rightarrow \{1, 2\}$, $f(n) = n$ is a bijection.
$X =  \{1, 2\}$ is compact since $X$ is finite.
\s
If $X$ has the $\underline{\hskip 1.6in}$ topology and
\s
 $Y$ has the
$\underline{\hskip 1.6in}$ topology, then 
\s
 $f$ is continuous, but not
a homeomorphism.  
\s
$Y = \{1, 2\}$ is not $\underline{\hskip 1.6in}$.

Lemma 26.8 (The tube lemma).  Suppose $N$ is an open set in $X
\times Y$ where $Y$ is compact.  If there exists an $x_0 \in X$ such
that $x_0 \times Y \subset N$, then there exists an open set $W$ such
that $x_0 \in W$ and $W \times Y \subset N$.

Thm 26.7:  The product of finitely many compact spaces is compact.

Thm 37.3 (Tychonoff theorem).  An arbitrary product of compact spaces
is compact in the product topology.  


\eject

Defn:  A collection {\cal C} is said to have the {\bf finite intersection
property} if for every finite subcollection \break
$\{C_1, ..., C_n\} \subset
{\cal C}$, $\cap_{i = 1}^n C_i \not= \emptyset$.

\vfill

Example 1: $\{(-n, n) ~|~ n = 1, 2, 3, ... \}$ has/does not have finite 
intersection
property.

\vfill

Example 2:  $\{(n, n + 2) ~|~ n \in {\cal Z} \}$ has/does not have finite 
intersection
property.

\vfill

Example 3:  $\{(0, {1 \over n}) ~|~ n = 1, 2, 3, ... \}$ has/does not have
finite intersection
property.

\vfill

Thm 26.9:  $X$ is compact if and only if for every collection ${\cal C}$ of closed sets in 
$X$ having the finite intersection property, $\cap_{C \in {\cal C}} C \not= \emptyset$.

\end
\eject
27:  Compact subspaces of the real line.

Thm 27.1:  Let $X$ be a simply ordered set with the lub property.  If $X$ has the order 
topology, then $[a, b]$ is compact.


Cor:  $[a, b] \subset {\bf R}$ is compact.  $\Pi_{i = 1}^n [a_i, b_i] \subset {\bf R}^n$ is 
compact.  

Thm 27.3:  A subspace $A$ of $R^n$ (with standard topology) is compact if and only if it is 
closed and bounded in the euclidean or square metric.

\end


\end




Defn:  A collection ${\cal C}$ of subsets of $X$ is said to have the
{\bf finite intersection property} if for every finite subcollection
$\{C_1, ..., C_n\}$ of ${\cal C}$, $C_1 \cup ...\cup C_n \not=
\emptyset$

\end
Thm 4:  If $f:D \rightarrow R$ is continuous and if $D$ is compact,
then $M = sup_{\{p \in X\}} f(p)$ and $m = inf_{\{p \in X\}} f(p)$
exist and there exist points in $p, q \in D$ such that $f(p) = M$ and
$f(q) = m$.


Defn:  $f: D
\rightarrow Y$.
is {\bf uniformly continuous} if
all $\epsilon > 0$, there exists a $\delta > 0$ such that

\centerline{
$d_X(x, p) < \delta$ implies $d_Y(f(x), f(p)) < \epsilon$}
\vskip -10pt
 for all $x \in
D$.

Defn:  If $f:D \rightarrow Y$ is continuous and if $D$ is compact,
then $f$ is uniformly continuous.


\end

Thm 3.12 (Heine-Borel Thm)  Suppose that a family ${\cal F}$ of open
intervals covers the closed interval $I = [a, b]$.  Then a finite
subfamily of ${\cal F}$ covers $I$.


A family of sets, ${\cal F} = \{F_\alpha ~|~ \alpha \in A\}$, covers
the
set $S$ if 

\centerline{$S \subset \cup_{\alpha \in A}F_\alpha$}

${\cal F}$  is said to be a cover of $S$.  


Defn:  In a metric space $X$, a set $A$ is compact if
every sequence contains a
subsequence which
converges to a point in $A$.

A set $A$ is bounded if there exists an open ball, $B(p, r)$ such
that $A \subset B(p, r)$.

Thm 6.13:  Compact metric implies closed \& bounded.

Thm 6.14:  In ${\cal R}^n$ with standard metric, 
{closed \& bounded implies
compact.}

Thm: Convergence implies Cauchy.

Thm:  In a compact metric space, Cauchy implies convergence.

Thm:  In $R^n$, Cauchy implies convergence (even though $R^n$ is not
compact).



Thm 6.15 (Heine-Borel Theorem)
\v
Suppose $X$ is a metric space and $A \subset X$, then the following
are
equivalent:
\v
(1.)  Every sequence contains a
subsequence which
converges to a point in $A$.
\v
(2.)  Every open cover of $A$ contains a finite subcover.

\end