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\begin{document}

\title{Topology Lectures  -- Integration workshop 2011}
\author{Tom Kennedy} 
\maketitle

\begin{abstract}
Lecture notes from the Integration Workshop at University of Arizona, 
August 2010.
These notes are based heavily on notes from the 2009 workshop
written originally by David Glickenstein with minor
additions by Shankar Venkataramani and on notes from an earlier workshop
written by Philip Foth.
\end{abstract}

\section{Introduction to topology}

\subsection{Topology of $\reals^n$}

For motivation, we recall what open and closed sets look like in $\reals^n$. 
A set $U \subset \reals^n$ is open if for any $x\in U$ 
there is a ball centered at $x$ contained in $U.$ 
A set $F$ is closed if it contains all of its limit points, i.e., for every 
convergent sequence that is in $F$, the limit is in $F$.
It can be shown that a set in $\reals^n$
is closed if and only if its complement is open. 

Let $f$ be a function from $\reals^n$ to $\reals^k$ or from $U$ to 
 $\reals^k$ where $U$ is an open set in $\reals^n$. 
Then the $\epsilon-\delta$ definition of continuity for $f$ at $x_0$ is that 
$\forall \epsilon>0$,$\exists \delta > 0$ such that $||x-x_0||<\delta$ 
implies $||f(x)-f(x_0)||<\epsilon$. 
(Here $|| \quad ||$ denotes the usual distance function in $\reals^n$ or 
$\reals^k$.) If $f$ is continuous at every 
point in its domain we say it is continuous. It then follows that
$f$ is continuous under this $\epsilon-\delta$ definition if and only 
if, for all open subsets $U$ in $\reals^k$, $f^{-1}(U)$ is open in $\reals^n$. 

The aboves suggest that we can do a lot if we just know what the open 
sets are, not the metric they came from. So we can abstract 
things by just looking at the collection of open sets. 
Note that the open sets in $\reals^n$ have some obvious properties.
The empty set and the whole space are open. Any union of open 
sets is an open set. Any finite intersection of open sets is open. 
These observations will be the basis for the definition of a 
topology. 


\subsection{Definition of a topology and basic results}

We define a topological space by specifying which sets are open. 
In order for this to be useful, we must put a few conditions on the 
collection of open sets. 

\begin{definition}
A topological space $\left(  X,\mathcal{T}\right)  $ is a set $X$ together
with a collection $\mathcal{T}$ of subsets of $X$ which satisfy:

\begin{enumerate}
\item $X\in\mathcal{T}$ and $\varnothing\in\mathcal{T}$,

\item Arbitrary unions of sets $U\in\mathcal{T}$ are in $\mathcal{T}$, i.e.,
for any indexing set $I,$ if $U_{i}\in\mathcal{T}$ for all $i\in I$ then $%
%TCIMACRO{\dbigcup \limits_{i\in I}}%
%BeginExpansion
{\displaystyle\bigcup\limits_{i\in I}}
%EndExpansion
U_{i}\in\mathcal{T}$,

\item If $U,V\in\mathcal{T}$ then $U\cap V\in\mathcal{T}$.
\end{enumerate}
\end{definition}

Instead of explicitly writing $U\in\mathcal{T}$, we usually say that $U$ is
open. Note that property 3 immediately implies by induction that 
a finite intersection of open sets produces an open set. 

\begin{definition}
A set $F$ is closed if $F^{C}=X\setminus F\in\mathcal{T}$, i.e. if $F^{C}$ is open.
\end{definition}

\begin{proposition}
Arbitrary intersections and finite unions of closed sets are closed.
\end{proposition}

\begin{definition}
The interior of a set $A,$ denoted $\overset{\circ}{A}$ 
or $int(A)$, is the union of all open
sets contained in $A.$ The closure of a set $A,$ denoted $\bar{A}$ or 
$cl(A)$, is the intersection of all closed sets containing $A$.
\end{definition}

The interior is open since it is a union of open sets, and 
the closure is closed since it is an intersection of closed sets.

\begin{proposition} The interior of $A$ is the largest open set contained in 
$A$. This means that $int(A)$ is open and if $B$ is another open set 
contained in $A$, then $B \subset int(A)$. 
The closure of $A$ is the smallest closed set containing 
$A$. This means that $\bar{A}$ is closed and if $B$ is another closed set 
containing $A$, then $\bar{A} \subset B$.
\end{proposition}

\begin{definition}
A point $x\in X$ is a limit point of a set $A\subset X$ if every open set $U$
containing $x$ also contains a point $y\in A\setminus\left\{  x\right\}  .$
\end{definition}

We can characterize the closure in terms of limit points.

\begin{proposition}
$\bar{A}$ is equal to the union of $A$ and its limit points.
\end{proposition}

\begin{proof}
Let $F$ be a closed set containing $A.$ Then $X\setminus F$ is an open set
disjoint from $A,$ so if $x$ is a limit point of $A$ it cannot be in
$X\setminus F,$ thus all limit points are contained in $\bar{A}$ 
(which is the intersection of all closed sets containing $A$). 
Conversely, if $x\in\bar
{A}\setminus A$ then if there were an open set $U$ containing $x$ but disjoint
from $A,$ then $\bar{A}\cap U^{C}$ is closed set strictly contained in
$\bar{A}$ containing $A,$ a contradiction since $\bar{A}$ is the smallest such set.
\end{proof}

\begin{definition}
A sequence $x_n \in X$ converges to a point $y \in X$ if for all open 
sets $O \ni y$, there exists an index $N < \infty$ such that for all $n > N$,
$x_n \in O$. 
\end{definition}

The above definition is a direct generalization of the notion 
of convergence in metric spaces.
However, there is no generalization of the notion of a Cauchy sequence to 
a topological space, since this requires the ability to compare the 
``size'' of neighbourhoods at distinct points, 
and a topological structure does not allow for this comparison.

\begin{definition}
$x \in A$ is an {\em isolated point} (of $A$) if there is an open set 
$O$ such that $O \cap A = \{x\}$. $x \in X$ is an {\em accumulation point} of 
$A$ if there exists a sequence in $A \setminus \{x\}$ that converges to $x$.
\end{definition}
 
In a metric space limit points and accumulation points are the same. 
But in topological spaces they need not be. In general one should 
be cautious using sequences when working in a topological space. 
There are many characterizations of topogical properties in a metric
space using sequences that do not carry over to topological spaces. 

\smallskip

We end this section with some examples of topological spaces. 

\smallskip

\noindent {\bf Example (metric spaces):}
Recall that a function $d:X\times X\rightarrow\mathbb{R}$ is a metric if for all $x,y,z\in
X,$ it satisfies:

\begin{enumerate}
\item (positive definite) $d\left(  x,y\right)  \geq0$ with $d\left(
x,y\right)  =0$ if and only if $x=y$

\item (symmetric) $d\left(  x,y\right)  =d\left(  y,x\right)  $

\item (triangle inequality) $d\left(  x,y\right)  +d\left(  y,z\right)  \geq
d\left(  x,z\right)  $
\end{enumerate}

\noindent We now define a set $U$ to be open if $\forall x \in U$,
$\exists \epsilon >0$ such that $d(y,x)< \epsilon$ implies $y \in U$. 
It is easy to check this defines a topology which we will refer to 
as the {\it metric topology}.
\smallskip

\noindent {\bf Example (discrete topology):}
We define all sets to be open.
\smallskip

\noindent {\bf Example (coarse or indiscrete topology):} 
The only open sets are $X$ and $\varnothing.$
\smallskip

\noindent {\bf Example (finite complement topology):} 
A set is defined to be open if its complement is finite or the set is
the empty set. 

\subsection{Continuous maps}

We start with metric spaces.
\begin{definition} Let $(X,d)$ and $(Y,d^\prime)$ be metric spaces
and $f:X \rightarrow Y$ a function. For $x \in X$, we say $f$ is
continuous at $x$ if $\forall \epsilon >0$, $\exists \delta>0$ such that
$d(x,y) < \delta$ implies $d(f(x),f(y))< \epsilon$. 
We say the map is globally continuous (or just continuous) if 
it is continuous at every point in $X$.
\end{definition}

We now give the definition for topological spaces. 

\begin{definition} Let $X$ and $Y$ be topological spaces and 
$f:X \rightarrow Y$ a function. We say $f$ is continuous if for any
open set $U$ in $Y$, $f^{-1}(U)$ is open.
\end{definition}

Of course metric spaces are topological spaces, so when $X$ and $Y$
are metric spaces we have two definitions of continuity.

\begin{proposition}
If $(X,d)$ and $(Y,d^\prime)$ are metric spaces and $f:X \rightarrow Y$,
then the above two definitions of continuity for $f$ are equivalent.
\end{proposition}

Note that for a continuous function the inverse image of a closed set 
is closed.

In a metric space we defined continuity at a point. We can do this
in a general topological space as well. 
We define a {\it neighborhood} of a point $x$ to be a set $N$ containing $x$ 
such that there is an open set $U$ with $x \in U \subset N$. 
Note that an open set is a neighborhood of each of its members. 
Also note that a neighborhood does not have to be open.
Now define a function $f:X \rightarrow Y$ from one topological 
space to another to be continuous at $x \in X$ if for every 
neighborhood $V$ of $f(x)$, $f^{-1}(V)$ is a neighborhood of $x$. 
(Note that if $V$ happens to be open, we are not asserting that 
$f^{-1}(V)$ is open.) 

Continuous maps allow us to define equivalence of topological spaces. 
\begin{definition} 
We say that two topological spaces are homeomorphic if 
there exists a continuous bijection between them with a 
continuous inverse. Such a map is called a homeomorphism.
\end{definition}

\noindent {\bf Example:} One of the problems is to show that 
$\reals$ and $(0,1)$ are homeomophic. One of the problems in section
three is to show that $\reals$ and $\reals^2$ are not homeomorphic. 


\subsection{Construction of topologies}

Let $X$ be a topological space and $Y\subset X$ be a subset. We can give $Y$
the \emph{subspace topology} by saying a set $U\subset Y$ is open if $U=V\cap
Y$ for some open set $V\subset X.$ It is easy to show that this gives a
topology. Think about how this gives a topology on the sphere $S^{n}%
\subset\mathbb{R}^{n+1}.$

If $\topo_1$ and $\topo_2$ are topologies on $X$, we say 
$\topo_1$ is {\it finer}
(or {\it stronger}) than $\topo_2$ if $\topo_2 \subset \topo_1$. 
It is coarser or weaker if the inclusion goes the other way.

\begin{proposition} Let $\mathcal{S}$ be a collection of subsets of $X$. 
Then there is a unique topology $\topo$ which is the weakest topology 
containing $\mathcal{S}$. This means that if $\topo^\prime$ is another 
topology containing $\mathcal{S}$, then $\topo^\prime$ is stronger than 
$\topo$.
\end{proposition} 

\begin{proof}
Consider all the topologies that contain $\mathcal{S}$. (There
is at least one - the discrete topology.) Define $\topo$ to be their
intersection. (Think carefully about what this means. The elements of 
a topology are subsets of $X$.) In other words, $\topo$ is the 
collection of subsets $U$ of $X$ such that for every topology $\topo^\prime$
that contains $\mathcal{S}$ we have $U \in \topo^\prime$. 
It is now a matter of definition chasing to check that this works. 
\end{proof}

We can think of the topology in the proposition as being formed by 
starting with $\mathcal{S}$ and adding ``just enough'' sets to get a 
topology.

Let $X$ and $Y$ be topological spaces. We can give $X\times Y$ a topology by
taking the weakest topology that contains all sets of the form 
$U\times V$ where $U\subset X$ and $V\subset Y$ are
open sets. (Note that not all open sets can be written as $U\times V$ for some
$U\subset X$ and $V\subset Y.$) 
This construction is the {\it product topology}.
The above immediately generalizes to a finite 
cartesian product. 

We now have two ways to put a topology on $\reals^n$. The metric 
topology we have already seen and the product topology that you get 
by thinking of $\reals^n$ as the product of $n$ copies of $\reals$.
Check that they are the same.

\begin{proposition} If $Y$ is a set, $(X,\topo)$ is a topological space, 
and $f:X \rightarrow Y$ is a function, then we can define a topology 
$\topo^\prime$ on $Y$ by taking $\topo^\prime$ to be all subsets $U$ of $Y$ 
such that $f^{-1}(U) \in \topo$.
\end{proposition}
With this construction $f$ is a continuous function.
It is important to note that this construction works because of the 
set identities
\bea
f^{-1}(\cup_\alpha U_\alpha) = \cup_\alpha f^{-1} (U_\alpha) \nonumber \\
f^{-1}(\cap_\alpha U_\alpha) = \cap_\alpha f^{-1} (U_\alpha)
\label{setid}
\eea

Let $X$ be a topological space and let $\sim$ be an equivalence relation.
Recall that an equivalence relation $\sim$ is a relation satisfying the
following properties:

\begin{enumerate}
\item (reflexivity) $x\sim x.$

\item (symmetry) $x\sim y$ implies $y\sim x$

\item (transitivity) $x\sim y$ and $y\sim z$ implies $x\sim z.$
\end{enumerate}

\noindent Then $Q=X/\sim$ denotes the set of equivalence classes of the
relation. For $x \in X$, we denote the equivalence class containing
$x$ by $[x]$.
There is a natural quotient map $q:X\rightarrow Q$ given by
$q\left(  x\right)  =\left[  x\right]$. 
We now use the previous proposition to define a topology 
on the quotient space $Q$: the open sets in $Q$ are the sets $U$ such 
that $p^{-1}(U)$ is open in $X$.
We call this the \emph{quotient topology} 

\noindent {\bf Example :}
The circle is a subset of the plane and so inherits a natural topology 
from the usual topology on the plane. Equivalently, the usual 
distance function on the circle is a metric which defines this topology. 
We can also think of the circle as the interval $[0,2 \pi]$ 
with the two endpoints identified. 
To be more precise, we define an equivalence relation by defining
$0 \sim 2 \pi$, and no other distinct points are equivalent. 
Since $[0,2 \pi]$ has a topology, we can consider the quotient 
topology on $[0,2 \pi]/\sim$. Show that $[0,2 \pi]/\sim$ is homeomorphic
to the circle with the usual topology. 

\medskip

Given a map $f: X \rightarrow Y$ and a topology on $X$ we have defined 
a topology on $Y$ by taking advantage of the set identities \reff{setid}. 
If instead we have a topology on $Y$, we might try to use it to define
a topology by $X$ by taking the collection of all sets of the form 
$f^{-1}(U)$ where $U$ is open. We leave it to the reader to check that 
this does not work in general - the resulting collection of sets 
need not have the properties of a topology. 
We can define a toplogy by taking the weakest topology that 
includes all sets of the form $f^{-1}(U)$ where $U$ is open in $Y$.
With this definition $f$ is a continuous function. In fact, this topology
on $X$ is weakest topology with this property. 

We can generalize this construction. Suppose that the index $\alpha$ 
ranges over some index set $\mathcal{A}$ and for each $\alpha$ 
we have a topological space $(Y_{\alpha},\mathcal{S}_{\alpha})$ and 
a function $f_\alpha : X \rightarrow Y_\alpha$. 
Then we can define the induced (or {\em weak}) topology on $X$ 
to be the weakest topology containing all sets of the form 
$f_{\alpha}^{-1} (U)$ where $U$ is open in $Y_\alpha$.  

\begin{proposition}
The {\em weak} topology constructed above is the weakest topology on 
$X$ that makes all the functions $f_{\alpha}$ continuous.
\end{proposition}

\subsection{More exotic examples}

\begin{itemize}
\item Line with two origins. We consider two copies of the real line.
We denote elements of one of them by $(x,1)$ where $x \in \reals$ and 
the elements of the other by $(x,2)$ where $x \in \reals$.
We define an equivalence relation by $(x,1) \sim (x,2)$ if $x \neq 0$. 
(Of course, all points are defined to be equivalent to themselves.)
Note that $(0,1)$ and $(0,2)$ are not equivalent (hence the name). 
Their equivalence classes just contain one element. All other equivalence
classes contain two elements. 
The line with two origins is the quotient
$\mathbb{R\cup\mathbb{R}}^{\prime}/\sim$ where $x\sim x^{\prime}$ if $x\neq0.$

\item Order topology. A total ordering on a set $X$ is a relation $\leq$ such
that for any $x,x^{\prime}\in X$ we have either that $x\leq x^{\prime}$ or
$x^{\prime}\leq x$ and both are true if and only if $x=x^{\prime},$ and the
relation is transitive. The order topology is the weakest topology 
that contains the ``intervals''
$$
\left(  a,b\right)  =\left\{  x\in X:a <  x\text{ and
}x < b\right\}
$$
Products of ordered sets can be given the dictionary
order. What do you think the definition of the dictionary order is?

\item Zariski topology. Consider the following topology on $\mathbb{R}^{n}.$
We take as the closed sets the sets
\[
F\left(  S\right)  =\left\{  x\in\mathbb{R}^{n}:f\left(  x\right)  =0 \quad
\forall
f\in S\right\}
\]
where $S$ is a set of polynomials in $n$ variables. Show that this is a
topology on $\mathbb{R}^{n}.$ Show that any two open sets must intersect, and
hence the topology cannot be Hausdorff. (The definition of Hausdorff 
appears later.)
\end{itemize}


\subsection{Local bases, basis, subbasis}

Another way to specify a topology is with a local base 
(system of neighborhoods).

\begin{definition} Let $X$ be a set, and for every $x \in X$, 
let there be given a collection $\mathcal{N}(x)$ of subsets of $X$ satisfying
\begin{enumerate}
\item $V \in \mathcal{N}(x) \implies x \in V$.
\item If $V_1,V_2 \in \mathcal{N}(x)$, then $\exists V_3 \in \mathcal{N}(x)$ 
such that $V_3 \subseteq V_1 \cap V_2$.
\item If $V \in \mathcal{N}(x)$, then there exists a $W \in \mathcal{N}(x)$ 
such that $W \subset V$ and the following holds. If $y \in W$, 
then there exists  $U \in \mathcal{N}(y)$ such that $U \subset V$.
\end{enumerate}
The collection $\{\mathcal{N}(x) | x \in X\}$ is a {\em local base}.

Given a local base, we can define a topology $\mathcal{T}$ by 
$O \in \mathcal{T}$ iff for all $x \in O$, there 
exists $V \in \mathcal{N}(x)$ such that $x \in V \subseteq O$.
\end{definition}

Note that the neighborhoods of $x$ in $\mathcal{N}(x)$ 
{\em do not have to be open}! However, given any local base, by 
``shrinking'' the neighbourhoods a little if necessary, we can obtain a 
local base which generates the same topology, all of whose elements are open 
sets.  In this case, condition 3 above simplifies to 

{\em $3'$.  If $V \in \mathcal{N}(x)$ and $y \in V$, then there 
exists $U \in \mathcal{N}(y)$ such that $U \subset V$.}

\medskip

We can also specify a topology with a basis or a subbasis.

\begin{definition}
A basis $\basis$ is a collection of subsets of $X$ such that \\
\noindent (1) for all $x\in X,$ there exists $U\in \basis$ such that 
$x\in U$  \\
\noindent (2) if $U,U^{\prime}\in \basis$ and $x\in U\cap U^{\prime},$ 
then there is a set $U^{\prime\prime}\in \basis$
such that $x\in U^{\prime\prime}$ and $U^{\prime\prime}\subset U\cap
U^{\prime}.$  \\
\noindent A basis generates a topology by taking the open sets to be 
all sets we can form by taking a union of a collection of sets in $\basis$. 
Equivalently we can define a set $V$ to be open if every
point $x\in V$ has a set $U\in \basis$ such that $x\in U\subset V.$ 
\end{definition}

An example of a basis is the open intervals for $\mathbb{R}$. Note that
the basis determines the topology. The sets in the basis have to be open, 
but the basis itself need not be a topology since unions of elements 
of the basis are not necessarily in the basis.

We can also specify a topology through a subbasis.

\begin{definition}
A subbasis $\basis^{\prime}$ (for a topology on $X$) is a collection of 
sets whose union is $X$. We define a topology by taking the open sets 
to be all sets which are the union of
finite intersections of elements of $B^{\prime}.$
\end{definition}

Given a subbasis $\basis^{\prime}$, define $\basis$ to be all finite
intersections of sets from $\basis^{\prime}$. Then $\basis$ is a basis
that generates the same topology as the subbasis. 

\begin{proposition}
Let $X$ and $Y$ be topological space. Then a basis for the product 
topology on $X \times Y$ is the collection of sets of the form 
$U \times V$ where $U$ is an open set in $X$ and $V$ is an open set in $Y$. 
\end{proposition}

\begin{proof}
First we check that this collection of sets is a basis. 
Property 1 is immediate. For property 2, if 
$(x,y) \in (U \times V) \cap (U^\prime \times V^\prime)$, then 
$(x,y) \in (U \cap U^\prime) \times (V \cap V^\prime)$ and 
$(U \cap U^\prime) \times (V \cap V^\prime)$ is in the basis. 
The product topology is the weakest topology containing the sets in 
the basis, and so coincides with the topology defined using the basis. 
\end{proof}

\subsection{Separation and countability}

Here we simply list some of the separation and countability properties.

Separation:

\begin{itemize}
\item Hausdorff. A space is Hausdorff if for every two points $x,y\in X,$
there are disjoint open sets $U$ and $V$ such that $x\in U,$ $y\in V$.
Note that a subspace of a Hausdorff space is
Hausdorff but the quotient of a Hausdorff space may not be Hausdorff 
One of the problems is to show that the line with two origins is an 
example of this.

\item Regular. A space is regular if one point sets are closed and for each
pair of a point $x$ and a closed set $B$ disjoint from $x$ there are disjoint
open sets containing $x$ and $B.$

\item Normal. A space is normal if one point sets are closed and for each pair
of disjoint closed sets $A,B$ there are disjoint open sets containing $A$ and
$B.$
\end{itemize}

Hausdorff is the most important. One reason is the following.

\begin{proposition}
Finite point sets in Hausdorff spaces are closed.
\end{proposition}
%
%\begin{proof}
%It is sufficient to show that one point sets are closed since finite unions of
%closed sets are closed. We need only show that there are no limit points of a
%one point set $x$. Recall that a limit point of $x$ is a point $y$ such that
%every open set containing $y$ must also contain $x.$ But since the space is
%Hausdorff, there is always an open set containing $y$ not containing $x,$ so
%if $y\neq x$ then $y$ is not a limit point. Hence $\left\{  x\right\}  $ is
%equal to its closure.
%\end{proof}

Countability. A set is countable if there is a bijection between it and the
natural numbers. It is easy to see that the integers, the even integers, and
the rational numbers are all countable sets. It is also possible to see that
the real numbers between $0$ and $1$ form an uncountable set using Cantor's
diagonal argument. Topological spaces have the following countability axioms:

\begin{itemize}
\item First countable. A space is first countable if every point has a
countable basis, i.e. given $x\in X$ there is a countable collection of open
sets $U_{1},U_{2},U_{3},\ldots$ such that for any neighborhood $V$ of $x,$
there is $k\in\mathbb{N}$ such that $U_{k}\in V.$

\item Second countable. A space is second countable if it has a countable
basis for the topology. (Long line is an example which is not second countable.)
\end{itemize}

\section{Compactness}

\subsection{Definition of compactness in general topological space}

For a subset $X$ of $\reals^n$, the following three properties are 
equivalent :

\begin{itemize}
\item $X$ is closed and bounded. (A subset of $\reals^n$ is bounded if 
there is an $R>0$ such that $||x|| \le R$ for $x \in X$.) 

\item Every sequence contained in $X$ has a limit point in $X$. That is, every
sequence has a subsequence which converges to a point in $X.$

\item Given any collection of open sets whose union contains $X$ 
(an open cover of $X$), there is a finite subcollection whose union 
still contains $X$ (finite subcover).

\end{itemize}

We shall use the last property to define compactness in a general topology.

\begin{definition}
Let $X$ be a topological space and $F \subset X$. 
We say $F$ is \emph{compact} if any open cover of $F$ has a
finite subcover, i.e. for any collection of open sets 
$\left\{  U_{i}\right\}_{i\in I}$ with $F \subset%
%TCIMACRO{\dbigcup \limits_{i\in I}}%
%BeginExpansion
{\displaystyle\bigcup\limits_{i\in I}}
%EndExpansion
U_{i},$ there exists a finite subcollection $\left\{  U_{i_{1}},U_{i_{2}%
},\ldots,U_{i_{k}}\right\}  \subset\left\{  U_{i}\right\}  _{i\in I}$ such
that $F\subset%
%TCIMACRO{\dbigcup \limits_{j=1}^{k}}%
%BeginExpansion
{\displaystyle\bigcup\limits_{j=1}^{k}}
%EndExpansion
U_{i_{j}}.$
\end{definition}

This may seem a rather abstract definition but it has many important
properties which follow immediately. 

\smallskip
\noindent{\bf Remark:} 
It is easily checked that a set $F \subset X$ is 
compact according to the above definition if and only if 
the $F$ with the relative topology is a compact topological space.


\subsection{Properties of compact spaces}

In this section we see some properties of a compact set.

\begin{proposition}
If $F$ is compact and $A \subset F$ is closed, then $A$ is compact.

\begin{proof}
Let $U_\alpha$ be an open cover of $A$. Since $A$ is closed, $A^c$ is open.
Since $\cup_\alpha U_\alpha$ contains $A$, 
$A^c \cup (\cup_\alpha U_\alpha)$ contains $F$. (In fact it equals the 
entire space.) So this open cover of $F$ admits a finite subcover. 
The finite subcover may or may not contain $A^c$, but if it doesn't
we can add it to the finite subcover. So 
\bea 
A \subset A^c \cup (\cup_{i=1}^n U_{\alpha_i})
\eea
Since $A^c$ does not cover any of $A$, this implies 
\bea 
A \subset \cup_{i=1}^n U_{\alpha_i}
\eea
so we have a finite subcover of $A$. 
\end{proof}
\end{proposition}

\begin{proposition}
If $f:X\rightarrow Y$ is continuous and $X$ is compact, then $f\left(
X\right)$ is compact.

\begin{proof}
If $\left\{  U_{i}\right\}  _{i\in I}$ is an open cover of $f\left(  X\right),$
then $\left\{  f^{-1}\left(  U_i\right)\right\}_{i \in I}$
is an open cover of $X$.  So it must have a finite
subcover $\left\{  f^{-1}\left(U_{i_{j}}\right)  \right\}_{j=1}^{k}.$ 
But then $\left\{  U_{i_{j}}\right\}_{j=1}^{k}$ must
cover $f\left(  X\right)$.
\end{proof}
\end{proposition}

\begin{proposition}
If $f:X\rightarrow\mathbb{R}$ is continuous and $X$ is compact, then there
exist $x_{m}$ and $x_{M}$ in $X$ such that
\begin{align*}
f\left(  x_{m}\right)   &  =\inf_{x\in X}f\left(  x\right) \\
f\left(  x_{M}\right)   &  =\sup_{x\in X}f\left(  x\right)  .
\end{align*}
In words, $f$ attains its minimum and maximum.

\begin{proof}
Since $f\left(  X\right)  $ is compact, it must be a closed and bounded 
subset of $\reals$ and thus
it contains its lower and upper bounds.
\end{proof}
\end{proposition}

\begin{proposition}
If $X$ and $Y$ are compact topological spaces, then $X\times Y$ with the 
product topology is compact.
\end{proposition}

\begin{proof}
Let $\mathcal{U}$ be an open cover of $X \times Y$. Define a subset $A$ of
$X$ to be ``good'' if there is a finite subcover (from $\mathcal{U}$ ) 
for $A \times Y$. Our goal is to show $X$ is good. 

Consider an arbitary $x \in X$.  We claim 
there is an open neighborhood $V_x$ 
of $x$ that is good. $\forall y \in Y$, $(x,y)$ is a point in $X \times Y$
and so there is a $U_y \in \mathcal{U}$ with $(x,y) \in U_y$. 
Recall that a basis for the product is given by the products of 
an open set in $X$ with an open set in $Y$. 
So there are open sets $V_y$ in $X$ and 
$W_y$ in $Y$ such that $(x,y) \in V_y \times W_y \subset U_y$. 
Now $\{W_y\}_{y \in Y}$ is a cover of $Y$ and so has a finite subcover
$\{ W_{y_i} : i=1,\cdots,n\}$. 
Define $V_x= V_{y_1} \cap \cdots \cap V_{y_n}$. 
Then $V_x$ is an open neighborhood
of $x$.  $V_x \times Y$ is covered by $\cup_{i=1}^n (V_{y_i} \times W_{y_i})$
which is itself covered by $\cup_{i=1}^n U_{y_i}$, i.e.,  finite
subcover of $\mathcal{U}$. So $V_x$ is good. 

Now $\forall x \in X$, let $V_x$ be a good open neighborhood of $x$. 
Then $V_x$ is a cover of $X$ and so has a finite subcover, i.e., 
\beann
X \subset \cup_{i=1}^n V_{x_i}
\eeann
Each $V_{x_i} \times Y$ has a finite subcover from $\mathcal{U}$, and the 
union of these finite subcovers will be a finite subcover of $X \times Y$. 
\end{proof}

\begin{proposition}
Compact subsets of a Hausdorff space are closed.
\end{proposition}

%\begin{proof}
%Let $K$ be a compact subset of a Hausdorff space $X.$ We wish to show that $X$
%contains its limit points. Let $x\in\bar{K}\setminus K.$ Since $X$ is
%Hausdorff, for every $y\in K$ there are disjoint open sets $U_{y},$ $V_{y}$
%containing $y$ and $x$ respectively. The sets $U_{y}$ cover $K$ and since $K$
%is compact, there is a finite subcover $\left\{  U_{1},\ldots,U_{n}\right\}
%.$ This implies that $%
%%TCIMACRO{\dbigcap \limits_{j=1}^{n}}%
%%BeginExpansion
%{\displaystyle\bigcap\limits_{j=1}^{n}}
%%EndExpansion
%V_{y}$ is an open set (since it is a \emph{finite} intersection) containing
%$x$ disjoint from $K,$ contradicting the fact that $x$ is a limit point. Thus
%$K=\bar{K}.$
%\end{proof}

\subsection{Heine-Borel theorem}

Notice that $(0,1]$ is not compact, because the cover $%
%TCIMACRO{\dbigcup \limits_{k\in\mathbb{N}}}%
%BeginExpansion
{\displaystyle\bigcup\limits_{k\in\mathbb{N}}}
%EndExpansion
\left\{  (1/k,1]\right\}  $ has no finite subcovers.

\begin{definition}
A subset $X$ of $\mathbb{R}^{n}$ is said to be \emph{bounded }if there exists
$r>0$ such that $X\subset B\left(  0,r\right)  =\left\{  x\in\mathbb{R}%
^{n}:\left\vert x\right\vert <r\right\}  $.
\end{definition}

\begin{theorem}
Subsets of $\reals^n$ are compact if and only if they are closed and bounded.
\end{theorem}

\begin{proof}
If a subset of $\reals^n$ is compact, it must be closed since 
$\reals^n$ is Hausdorff. The subset must be bounded 
because we can take the cover of $\left(
-k,k\right)  ^{n}$ for $k=1,2,\ldots$ and it must have a finite subcover.

To prove the other direction, 
we start by showing that $[a,b]$ is compact in $\reals$. 
For convenience we take $[a,b]=[0,1]$. 
Let $\mathcal{U}$ be a cover of $\left[  0,1\right]  .$ We let
\[
S=\left\{  x\in\left[  0,1\right]  :\left[  0,x\right]  \text{ has a finite
subcover in }\mathcal{U}\right\}  .
\]
Now we show that $y=\sup S$ must be $1.$ Observe since $\mathcal{U}$ is a
cover, $y$ is contained in some open set $U \in \mathcal{U}$, 
and hence the interval 
$\left(y-\varepsilon,y+\varepsilon\right)  \subset U$ for some small $\varepsilon>0.$
This implies both that $y\in S$ since there must be some $y^{\prime}\in S,$
$y^{\prime}>y-\varepsilon$ since $y$ is the sup, so take the finite cover of
$\left[  0,y^{\prime}\right]  $ and add in $U.$ But this also implies that
$y+\varepsilon/2\in S$ if $y+\varepsilon/2\in\left[  0,1\right]  ,$ so $y=1.$

Since the topology on $\reals^n$ is the same as the product topology 
it gets by thinking of it as the product of $n$ copies of $\reals$, 
$[a,b]^n$ is compact in $\reals^n$.  
A closed and bounded set is a closed subset of some compact set $\left[
-k,k\right]  ^{n},$ and thus is compact. 

\end{proof}

\subsection{Sequences, continuity  and compactness}

\begin{definition}
A function  $f:(X,\mathcal{T}) \rightarrow (Y,\mathcal{S})$ is {\em sequentially continuous} if for every convergent sequence $x_n \rightarrow x$ in $X$, we have $f(x_n) \rightarrow f(x)$.
\end{definition}

\begin{proposition} Every continuous function is sequentially continuous. In a first countable space (for example, in a metric space), the converse is also true.
\end{proposition}

\begin{definition}
A space $X$ is {\em sequentially compact} if every sequence in $X$ has a convergent subsequence.
\end{definition}

\begin{proposition} Every compact space is necessarily sequentially compact. In a metric space, the converse is also true.
\end{proposition}

Taken in conjunction with the Heine-Borel theorem, this implies that
\begin{theorem} (Bolzano-Weierstrass)  Every bounded sequence in $\mathbb{R}^n$ has a convergent subsequence.
\end{theorem}

\subsection{Examples and non-examples of compact spaces}

\begin{itemize}
\item Any finite topological space is compact.

\item The finite-dimensional sphere $\left\{  x\in\mathbb{R}^{n}:\left\vert
x\right\vert ^{2}=1\right\}  $ is compact. 

\item The Cantor set is compact.

\item For any space with the finite-complement topology, every subset 
is compact.
The is a reflection of the fact that there are not very many open 
sets in this topology. (The stronger the topology, the harder it is 
for a set to be compact.)

\item Let $l^2$ be the set of sequences $(x_n)_{n=1}^\infty$ 
with $\sum_{n=1}^\infty x_n^2 < \infty$. For such a sequence we define
\bea
|| (x_n)_{n=1}^\infty || =  \left[ \sum_{n=1}^\infty x_n^2  \right]^{1/2}
\eea
Then we define a metric by 
\bea
d((x_n)_{n=1}^\infty,(y_n)_{n=1}^\infty) = ||(x_n-y_n)_{n=1}^\infty|| 
\eea
The unit ball is this space is a closed and bounded set, but it is 
not compact. In fact, it is not sequentially compact. For example, 
let $e_n$ be the sequence which is $1$ in the $n$th place and $0$ 
elsewhere. Then $d(e_n,e_m)=\sqrt{2}$ for $n \neq m$, so $e_n$ cannot
have a convergent subsequence. One of the exercises is to prove
that in this space every compact set has empty interior. 
Another exercise is to prove that the following set is compact:
\bea
\{ (x_n)_{n=1}^\infty: \sum_{n=1}^\infty n x_n^2 \le 1\}
\eea

\end{itemize}

\subsection{Compactness in metric spaces}

In $\reals^n$, compactness is equivalent to being closed and bounded. 
We saw in one of the examples that this is not true in all metric
spaces. There is a characterization of compactness in 
metric spaces. We have to replace boundedness by a stronger property 
and we also have to replace closed. 

To see why we have to replace closed, consider the following example. 
Let $Q$ be the rationals. Let $I= [0,1] \cap Q$. Then $I$ is a closed 
set in $Q$. It is not sequentially compact and so is not compact. 
The problem in this example is that the space has ``missing points.''

\begin{definition}
A sequence $x_n$ in a metric space is {\it Cauchy} if for any $\epsilon>0$ 
there exist an integer $N$ such that for $n,m \ge N$ we have 
$d(x_n,x_m)<\epsilon$. A metric space $X$ is {\it complete} if 
every Cauchy sequence converges, i.e., for every Cauchy sequence
$x_n$ there is $ x \in X$ such that $x_n$ converges to $x$.
\end{definition}

\begin{definition}
In a metric space a set is totally bounded if for any $\epsilon >0$, 
it can be convered by a finite number of balls of radius 
$\epsilon$.
\end{definition}

\begin{theorem}
A metric space is compact if and only if it is complete and totally 
bounded.
\end{theorem}

\section{Connectedness}

\subsection{Connected and disconnected sets in $\mathbb{R}^{n}$}

The key property of a connected set is the intermediate value theorem, which
states that if $f:\left[  a,b\right]  \mathbb{\rightarrow\mathbb{R}}$ is a
continuous function and $f\left(  a\right)  \leq r\leq f\left(  b\right)  $
then there exists $c\in\left[  a,b\right]  $ such that $f\left(  c\right)
=r.$ Notice this is not true for functions on disconnected sets such as
$\left(  0,1\right)  \cup\left(  1,2\right)  .$

\subsection{Definition of connected}

\begin{definition}
A separation of a space $X$ is a pair $U,V$ of disjoint open subsets of $X$
such that $X=U\cup V.$ Note that the two sets $U$ and $V$ are both open and
closed since $U=X\setminus V$ and $V=X\setminus U.$ The trivial separation
consists of $X$ and $\varnothing.$
\end{definition}

\begin{definition}
A space $X$ is \emph{connected} if there exist no nontrivial separations of
$X.$ Equivalently, $X$ is connected if the only open and closed subsets of $X$
are $X$ and $\varnothing$ (since if $A\subset X$ is open and closed, then
$X=A\cup A^{C}$ is a separation if neither is empty). A space which is not
connected is said to be \emph{disconnected}.
A subset of a topological space is connected if it is connected as 
a toplogical space itself when we endow it with the subspace topology.
\end{definition}

\begin{example}
$\left(  0,1\right)  $ is connected.
\end{example}

\begin{example}
$\left(  0,2\right)  \setminus\left\{  1\right\}  $ is disconnected since
$\left(  0,1\right)  $ and $\left(  1,2\right)  $ form a nontrivial separation.
\end{example}

Using the definition of the subspace topology, a subset $A$ of $X$ is 
not connected if we can find open sets $U$ and $V$ in $X$ such that 
$A \subset U \cup V$, $A \cap U \ne \emptyset$ $A \cap V \ne \emptyset$ 
and $U \cap V \cap A = \emptyset$. One might ask if it is always possible
to choose these sets so that $U \cap V = \emptyset$. It is not.

\begin{example}
Let $X=\{a,b,c\}$. Let 
\bea
\topo=\{ \emptyset, \{b\},\{a,b\},\{b,c\},X\}
\eea
and let $S=\{a,c\}$. The subpace topology on $S$ is 
\bea
\topo'=\{ \emptyset, \{a\},\{c\},\{a,c\}\}
\eea
i.e., the discrete topology. So $S=\{a\} \cup \{c\}$ is a separation
that shows $S$ is not connected. But there are no disjoint open 
sets $U,V$ in $X$  with $U \cap S=\{a\}$ and  $V \cap S=\{c\}$.
\end{example}



A note on the proof of Heine-Borel: we essentially used that $\left[
0,1\right]  $ is connected and showed that the set of points $y\in\left[
0,1\right]  $ such that $\left[  0,y\right]  $ can be covered by a finite
subcover is both open and closed, and hence must be everything.




\subsection{Properties of connected sets}

\begin{proposition}
The union of a collection of connected sets whose intersection is 
not empty is a connected set.
\end{proposition}

\begin{proof}
Let $Y$ be the topological space. Let $X_i \subset Y$ be connected.  
Let $X = \cup_i X_i$. We give $X$ the subspace topogy. 
Note that there are two ways to put a topology on $X_i$, the 
subspace toplogy we get by thinking of it as a subset of $Y$ and the 
subspace toplogy we get by thinking of it as a subset of $X$.
We leave it to the reader to check they are the same. So the original 
assumption that $X_i$ is connected as a subset of $Y$ means it is 
connected as a subset of $X$. 

Let $x \in \cap_i X_i$. 
Now suppose $X = U \cup V$ where 
$U$ and $V$ are disjoint open subsets of $X$. $x$ must belong to one of 
$U$ and $V$. Assume it belongs to $U$. Now $X_i \cap U$ and 
$X_i \cap V$ are open sets in $X_i$ in the subspace topology 
and $X_i \cap U$ is not empty since it contains $x$. So $X_i \cap V$ 
must be empty, i.e., $X_i \subset U$. This is true for all $i$, 
so $X=U$, and so $V$ is empty. 
\end{proof}

\begin{proposition}
Let $A$ be a connected subset of $X.$ If $A\subset B\subset\bar{A}$ then $B$
is connected.
\end{proposition}

\begin{proof}
Suppose $B=U\cup V,$ where $U$ and $V$ are disjoint and open in 
the subspace topology for $B$. We leave it to the reader to check that 
$U \cap A$ and $V \cap A$ are open in $A$ with the subspace topology.
They are clearly disjoint and cover $A$. Since $A$ is connected, one of 
them must be empty. Assume that $V \cap A$ is empty. 
So $A$ is entirely contained in $U$. 

Since $U,V$ are open in $B$, there are open sets $U',V'$ in $X$ 
with $U=B \cap U'$ and $V=B \cap V'$ 
Since $A \subset U$, $A \subset (V')^c$. Since 
$(V')^c$ is closed in $X$, $\bar{A} \subset (V')^c$. This implies
$B \subset (V')^c$. If $x \in V$, then $x \in B$, and so $x \notin V'$.
But $V \subset V'$, a contradiction. So $V$ is empty. This shows $B$ is
connected.
\end{proof}

\begin{proposition}
The product of connected sets is connected.
\end{proposition}

\begin{proof}
We see that $\left\{  x\right\}  \times Y$ and $X\times\left\{  y\right\}  $
are connected. Since both contain $\left(  x,y\right)  ,$ $V_{x,y}=\left(
\left\{  x\right\}  \times Y\right)  \cup\left(  X\times\left\{  y\right\}
\right)  $ is connected. Now pick some $y_0 \in Y$. We see that
\[%
%TCIMACRO{\dbigcup \limits_{x\in X}}%
%BeginExpansion
{\displaystyle\bigcup\limits_{x\in X}}
%EndExpansion
V_{x,y_0}=X\times Y
\]
and
\[%
%TCIMACRO{\dbigcap \limits_{x\in X}}%
%BeginExpansion
{\displaystyle\bigcap\limits_{x\in X}}
%EndExpansion
V_{x,y_0}=X\times\left\{  y_0 \right\}  \neq\varnothing.
\]
Thus $X\times Y$ is connected.
\end{proof}

\begin{proposition}
If $f:X\rightarrow Y$ is continuous and $X$ is connected then $f\left(
X\right)  $ is connected.
\end{proposition}

\begin{proof}
Exercise
\end{proof}

\begin{proposition}
(Intermediate value theorem) If $X$ is connected, $f:X\rightarrow\mathbb{R}$
is continuous, and $f\left(  a\right)  \leq r\leq f\left(  b\right)  $ then
there exists $c\in X$ such that $f\left(  c\right)  =r.$
\end{proposition}

\begin{proof}
We know that $f\left(  X\right)  $ is a connected subset of $\mathbb{R}$. Now
if there is no $c$ such that $f\left(  c\right)  =r,$ then we can cover
$f\left(  X\right)  $ by the sets $\left(  -\infty,r\right)  \cap f\left(
X\right)  $ and $\left(  r,\infty\right)  \cap f\left(  X\right)  ,$ which are
disjoint open sets. They are nonempty since one contains $f\left(  a\right)  $
and the other $f\left(  b\right)  .$ This is a separation, contradicting that
$f\left(  X\right)  $ is connected.
\end{proof}

\subsection{Path connected}

\begin{definition}
A path in $X$ is a continuous map $\gamma:\left[  a,b\right]  \rightarrow X.$
\end{definition}

\begin{definition}
A space $X$ is \emph{path connected} if any two points can be joined by a path.
\end{definition}

One of the problems gives an example of a set that is 
connected but not path connected. So these two notions are not 
equivalent. However, one is stronger than the other. 

\begin{proposition}
If $X$ is path connected, then it is connected.
\end{proposition}

\begin{proof}
We shall show that if $X$ is not connected, then it is not path connected. If
$X$ is not connected, then there is a separation $\left\{  U,V\right\}  .$
Given $x,y\in X$ if there were a path $\gamma:\left[  a,b\right]  \rightarrow
X$ between them, then $\gamma\left(  \left[  a,b\right]  \right)  $ would be
connected, which implies the path must lie entirely in $U$ or $V$ (otherwise
$U$ and $V$ would form a separation for $\gamma\left(  \left[  a,b\right]
\right)  $), which says that there are no paths between points in $U$ and
points in $V.$ Hence $X$ is not path connected.
\end{proof}

\subsection{Components}

\begin{definition}
Given $X$, we can define an equivalence relation on $X$ by setting $x\sim y$
if there is a connected subset containing both $x$ and $y.$ The equivalence
classes are called \emph{components} or \emph{connected components} of $X.$
\end{definition}

Show that this is an equivalence relation.

\begin{proposition}
The components of $X$ are connected disjoint subsets of $X$ whose union is
$X,$ such that each connected subset of $X$ intersects only one component.
\end{proposition}

\begin{proof}
Let $\left\{  C_{i}\right\}  _{i\in I}$ be the components. Since the
components are equivalence classes, they must be disjoint and must cover. If
$U$ is connected and $x_{i}\in U\cap C_{i}$ and $x_{j}\in U\cap C_{j}$ then
$x_{i}\sim x_{j},$ which implies that $C_{i}=C_{j}$ by the definition of
components. Now we must show that components are connected. Fix $x_{0}\in
C_{i}.$ For any $x\in C_{i},$ there is a connected set $A_{x}$ containing both
$x_{0}$ and $x$ since $x\sim x_{0}.$ Thus $C_{i}=%
%TCIMACRO{\dbigcup \limits_{x\sim x_{0}}}%
%BeginExpansion
{\displaystyle\bigcup\limits_{x\sim x_{0}}}
%EndExpansion
A_{x},$ which implies that $C_{i}$ is connected since it is the union of
connected sets with a common intersection point $x_{0}.$
\end{proof}

We can also look at path components.

\begin{definition}
Define an equivalence relation on $X$ by $x\sim y$ if there is a path from $x$
to $y.$ The equivalence classes are called \emph{path components} of $X.$
\end{definition}


\end{document}