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\begin{document}
\title{Topology Lectures -- Integration workshop 2019}
\author{David Glickenstein}
\maketitle
\begin{abstract}
Lecture notes from the Integration Workshop at University of Arizona, August
2019. These notes are based heavily on notes from previous integration
workshops written by Philip Foth, Tom Kennedy, Shankar Venkataramani and others.
\end{abstract}
\section{Introduction to topology}
\subsection{Topology theorems}
\subsubsection{Theorems on $\mathbb{R}$}
The basics of point set topology arise from trying to understand the following
theorems from basic calculus: (in the following, we assume intervals written
$\left[ a,b\right] $ have the property $a**y,$ then by
continuity there exists $\delta>0$ such that $\left\vert f\left( s\right)
-f\left( x\right) \right\vert y.$
Similarly, we must have $f\left( s\right) \geq y$ since if $f\left(
s\right) 0$ such that if $\left\vert
x-s\right\vert <\delta$ then $\left\vert f\left( s\right) -f\left(
x\right) \right\vert 0$,$\exists\delta> 0$ such that $||x-x_{0}||<\delta$ implies
$||f(x)-f(x_{0})||<\epsilon$. (Here $|| \quad||$ denotes the usual distance
function in $\mathbb{R}^{n}$ or $\mathbb{R}^{k}$.) 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 $\mathbb{R}^{k}$, $f^{-1}(U)$ is open in $\mathbb{R}^{n}$.
The only structure of $\mathbb{R}^{n}$ that we need in the above is the
ability to measure the distance between two points in the space. So we can
immediately generalize the above to a metric space. The above shows more,
namely 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 $\mathbb{R}^{n}$, and more
generally the open sets in a metric space, 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{Metric spaces}
A metric space is one generalization of $\mathbb{R}^{n}.$
\begin{definition}
A metric space $(X,d)$ is a set $X$ and a function (called the metric)
$d:X\times X\rightarrow\mathbb{R}$ such that for all $x,y,z\in X,$ the metric 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,z\right) \leq d\left( x,y\right)
+d\left( y,z\right) $
\end{enumerate}
\end{definition}
We have a natural notion of convergence in a metric space.
\begin{definition}
A sequence $x_{n}$ in a metric space $(X,d)$ \textbf{converges} to a point
$x\in X$ if $\forall\epsilon>0$, there exists an index $N<\infty$ such that
$n>N\,\implies\,d(x_{n},x)<\epsilon$.
\end{definition}
This leads to a definition of closed sets.
\begin{definition}
A subset $F$ of a metric space $(X,d)$ is \textbf{closed} if for every
sequence $x_{n}$ in $F$ which converges to some $x$ in $X$ we have $x\in F$.
\end{definition}
Note that we were also able to describe convergence in terms of open balls and
open sets. We have a similar notion in a metric space.
\begin{definition}
In a metric space $(X,d)$, a set $U$ is said to be open if $\forall x\in U$,
$\exists\epsilon>0$ such that $d(y,x)<\epsilon$ implies $y\in U$.
\end{definition}
Open sets satisfy the following properties.
\begin{proposition}
Let $(X,d)$ be a metric space and let $\mathcal{T}$ be the collection of open
sets in $X$. Then
\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{proposition}
Note that property 3 immediately implies by induction that a finite
intersection of open sets produces an open set. Note the relationship to
closed sets:
\begin{proposition}
A set $F$ is closed if and only if $F^{C}=X\setminus F$ is open.
\end{proposition}
\begin{corollary}
Arbitrary intersections and finite unions of closed sets are closed.
\end{corollary}
%\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}
\smallskip
If the space $X$ is a vector space, then one way to get a metric on $X$ is to
start with a norm.
\begin{definition}
A function $|| \quad||$ on $X$ is a norm if
\begin{itemize}
\item For all $x \in X$, $||x|| \ge0$, and $||x|| = 0$ if and only if $x = 0$.
\item $||a x|| = |a|||x||$ for every $a \in\mathbb{R}$ and $x \in X$.
\item $|| x + y || \le|| x|| + || y||$ for every $x,y \in X$.
\end{itemize}
A normed space is a vector space with a norm defined on it.
\end{definition}
\begin{proposition}
Let $(X,|| \quad||)$ be a normed space. Define $d(x,y)=||x-y||$. Then $(X,d)$
is a metric space.
\end{proposition}
We end this section with some examples of metric spaces.
\begin{enumerate}
\item \textbf{Euclidean metric on $\mathbb{R}^{n}$:} The usual Euclidean norm
gives a metric on $\mathbb{R}^{n}$.
\begin{align*}
d(x,y)=||x-y||= \left[ \sum_{j=1}^{n} |x_{j}-y_{j}|^{2} \right] ^{1/2}%
\end{align*}
\item \textbf{$\mathbb{R}$ with a different metric:} Define a metric on $X$
by
\[
d(x,y)=\frac{|x-y|}{1+|x-y|}%
\]
One of the homework problems is to check this is a metric and to determine if
it gives a different topology for $\mathbb{R}$ from the standard one.
\item \textbf{Space of functions:} Let $D$ be any set and let $X$ be the set
of all bounded real-valued functions on $D$. Define
\begin{align*}
d(f,g) = \sup_{x \in D} |f(x)-g(x)|
\end{align*}
Then $(X,d)$ is a metric space.
\item \textbf{$l^{p}$ norm on $\mathbb{R}^{n}$:} There are other norms we can
put on $\mathbb{R}^{n}$ and hence other metrics. For $1\leq p<\infty$, define
\[
||x||_{p}=\left[ \sum_{j=1}^{n}|x_{j}|^{p}\right] ^{1/p}%
\]
(The case $p=2$ is the usual Euclidean metric.) It is not hard to show that we
get the same collection of open sets, i.e., the same topology, for all the
value of $p$. As $p\rightarrow\infty$ we get:
\item \textbf{sup norm on $\mathbb{R}^{n}$:} The function
\[
d(x,y)=\max_{1\leq j\leq n}|x_{j}-y_{j}|
\]
is another metric on $\mathbb{R}^{n}$ that defines the same topology.
\item \textbf{$l^{p}(\mathbb{N})$:} If we look at infinite sequences instead
of just vectors, things are more interesting. Let $l^{p}$ be the set of
sequences $(x_{n})_{n=1}^{\infty}$ with $\sum_{n=1}^{\infty}|x_{n}|^{p} <
\infty$. For such a sequence we define
\begin{align*}
|| (x_{n})_{n=1}^{\infty}||_{p} = \left[ \sum_{n=1}^{\infty}|x_{n}|^{p}
\right] ^{1/p}%
\end{align*}
Consider the two sets
\begin{align*}
F= \{ (x_{n})_{n=1}^{\infty}\in l^{p} : x_{n} \ge0 \, \forall n \}\\
U= \{ (x_{n})_{n=1}^{\infty}\in l^{p} : x_{n} > 0 \, \forall n \}
\end{align*}
Is $F$ closed? Is $U$ open?
\item \textbf{$l^{\infty}(\mathbb{N})$:} The space is now the set of bounded
infinite sequences. The norm is
\[
||(x_{n})_{n=1}^{\infty}||_{\infty}=\sup_{1\leq n<\infty}|x_{n}|
\]
\end{enumerate}
Notice that some of these have the same convergence properties. Suppose
$\left\{ x_{k}\right\} _{k=1}^{\infty}\subseteq\mathbb{R}^{n}$ converges to
$x_{\infty}$ in the $p$-norm, i.e., $\left\Vert x_{k}-x_{\infty}\right\Vert
_{p}\rightarrow0$ as $k\rightarrow\infty.$ Note that%
\[
\left\Vert x\right\Vert _{\infty}\leq\left\Vert x\right\Vert _{p}\leq
n^{1/p}\left\Vert x\right\Vert _{\infty}.
\]
It follows that for any $p$ and $q,$
\[
n^{-1/q}\left\Vert x\right\Vert _{q}\leq\left\Vert x\right\Vert _{\infty}%
\leq\left\Vert x\right\Vert _{p}\leq n^{1/p}\left\Vert x\right\Vert _{\infty
}\leq n^{1/p}\left\Vert x\right\Vert _{q}%
\]
so we have that the $p$ and $q$ norms are equivalent. It follows that a
sequence converges in $p$-norm if and only if it converges in $q$-norm. So
\textquotedblleft convergence\textquotedblright\ is not just a norm property,
but something more general. The same can be said for equivalent metric spaces.
So what is the most general object for which convergence makes sense?
\subsection{Topological spaces}
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}
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 also have the notion of an accumulation point, which is related to sequences:
\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 $n>N$, $x_{n}\in
O$. A point $x\in A$ is an \emph{isolated point} (of $A$) if there is an open
set $O$ such that $O\cap A=\{x\}$. $x\in X$ is an \emph{accumulation point} of
$A$ if there exists a sequence in $A\setminus\{x\}$ that converges to $x$.
\end{definition}
\begin{proposition}
In a metric space limit points and accumulation points are the same.
\end{proposition}
However, in topological spaces limit points and accumulation points need not
be the same. In general one should be cautious using sequences when working in
a topological space. There are many characterizations of topological
properties in a metric space using sequences that do not carry over to
topological spaces.
\smallskip
Of course all our previous examples of metric spaces are topological spaces.
We end this section with a few trivial examples of topological spaces.
\noindent\textbf{Example (discrete topology):} We define all sets to be open.
Does this topology come from a metric? \smallskip
\noindent\textbf{Example (coarse or indiscrete topology):} The only open sets
are $X$ and $\varnothing.$ Does this topology come from a metric? \smallskip
\noindent\textbf{Example (finite complement topology):} A set is defined to be
open if its complement is finite or the set is the empty set.
Can you think of more topological spaces? More help is in the problems.
\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}
\bigskip
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 \textit{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\textbf{Example:} One of the problems is to show that $\mathbb{R}$
and $(0,1)$ are homeomophic. One of the problems in section three is to show
that $\mathbb{R}$ and $\mathbb{R}^{2}$ are not homeomorphic.
\begin{definition}
A function $f:(X,\mathcal{T}) \rightarrow(Y,\mathcal{S})$ is
\emph{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}
\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 $\mathcal{T}_{1}$ and $\mathcal{T}_{2}$ are topologies on $X$, we say
$\mathcal{T}_{1}$ is \textit{finer} (or \textit{stronger}) than $\mathcal{T}%
_{2}$ if $\mathcal{T}_{2} \subset\mathcal{T}_{1}$. 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 $\mathcal{T}$ which is the weakest topology containing $\mathcal{S}$.
This means that if $\mathcal{T}^{\prime}$ is another topology containing
$\mathcal{S}$, then $\mathcal{T}^{\prime}$ is stronger than $\mathcal{T}$.
\end{proposition}
\begin{proof}
Consider all the topologies that contain $\mathcal{S}$. (There is at least one
- the discrete topology.) Define $\mathcal{T}$ to be their intersection.
(Think carefully about what this means. The elements of a topology are subsets
of $X$.) In other words, $\mathcal{T}$ is the collection of subsets $U$ of $X$
such that for every topology $\mathcal{T}^{\prime}$ that contains
$\mathcal{S}$ we have $U \in\mathcal{T}^{\prime}$. 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 \textit{product topology}. The above immediately
generalizes to a finite Cartesian product.
We now have two ways to put a topology on $\mathbb{R}^{n}$. The metric
topology we have already seen and the product topology that you get by
thinking of $\mathbb{R}^{n}$ as the product of $n$ copies of $\mathbb{R}$.
Check that they are the same.
\begin{proposition}
If $Y$ is a set, $(X,\mathcal{T})$ is a topological space, and $f:X\rightarrow
Y$ is a function, then we can define a topology $\mathcal{T}^{\prime}$ on $Y$
by taking $\mathcal{T}^{\prime}$ to be all subsets $U$ of $Y$ such that
$f^{-1}(U)\in\mathcal{T}$. This is the strongest topology on $Y$ that makes
$f$ continuous.
\end{proposition}
With this construction $f$ is a continuous function. It is important to note
that this construction works because of the set identities
\begin{align}
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}%
\end{align}
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\textbf{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 $0 \sim2 \pi$, 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 (\ref{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 topology 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 \emph{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 \emph{weak} topology constructed above is the weakest topology on $X$ that
makes all the functions $f_{\alpha}$ continuous.
\end{proposition}
\smallskip\noindent\textbf{Remark:} If $Y \subset X$ and $X$ is a topological
space, then $Y$ inherits a natural topology (the subspace topology) from $X$.
Another way to define this topology is that it is the weakest topology that
make the inclusion map from $Y$ to $X$ continuous.
\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 $x \in\mathbb{R}$ and the
elements of the other by $(x,2)$ where $x \in\mathbb{R}$. We define an
equivalence relation by $(x,1) \sim(x,2)$ if $x \neq0$. (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 Long line : This is a particular example of the previous example. Let
$X=[0,1) \times\mathbb{R}$. The ``dictionary order'' is a total order defined
as follows. Given $(x_{1},y_{1})$ and $(x_{2},y_{2})$, to determine which is
larger, we first look at the first component. If $x_{1}x_{2}$ we define $(x_{1},y_{1})
> (x_{2},y_{2})$. If $x_{1}=x_{2}$ we look at the second coordinate. In this
case if $y_{1}y_{2}$ we define $(x_{1},y_{1}) > (x_{2},y_{2})$. Then we use this
total order to put the order topology on $X$. We can think of $X$ as an
uncountable number of copies of $[0,1)$ glued together end to end.
\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), a generalization of metric topology.
\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 \emph{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 $\mathcal{N}(x)$ \emph{do not have to be
open}! However, given any local base, by \textquotedblleft
shrinking\textquotedblright\ the neighborhoods 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
\emph{$3^{\prime}$. 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 $\mathcal{B}$ is a collection of subsets of $X$ such that
\newline\noindent(1) for all $x\in X,$ there exists $U\in\mathcal{B}$ such
that $x\in U$ \newline\noindent(2) if $U,U^{\prime}\in\mathcal{B}$ and $x\in
U\cap U^{\prime},$ then there is a set $U^{\prime\prime}\in\mathcal{B}$ such
that $x\in U^{\prime\prime}$ and $U^{\prime\prime}\subset U\cap U^{\prime}.$
\newline\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
$\mathcal{B}$. Equivalently we can define a set $V$ to be open if every point
$x\in V$ has a set $U\in\mathcal{B}$ 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 $\mathcal{B}^{\prime}$ (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 $\mathcal{B}^{\prime}$, define $\mathcal{B}$ to be all finite
intersections of sets from $\mathcal{B}^{\prime}$. Then $\mathcal{B}$ 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}
\section{Separation and countability}
Here we simply list some of the separation and countability properties.
Separation:
\begin{itemize}
\item Hausdorff. A space is \emph{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 \emph{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 \emph{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 \emph{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 $U_{k}\subset V.$
\item Second countable. A space is \emph{second countable} if it has a
countable basis for the topology. (Long line is an example which is not second countable.)
\end{itemize}
\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}
We first define a separation.
\begin{definition}
A \emph{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}
We can now define connected.
\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 topological 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
\[
\mathcal{T}=\{\emptyset,\{b\},\{a,b\},\{b,c\},X\}
\]
and let $S=\{a,c\}$. The subspace topology on $S$ is
\[
\mathcal{T}^{\prime}=\{\emptyset,\{a\},\{c\},\{a,c\}\}
\]
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}
We collect some 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 $U^{\prime},V^{\prime}$ in
$X$ with $U=B \cap U^{\prime}$ and $V=B \cap V^{\prime}$ Since $A \subset U$,
$A \subset(V^{\prime})^{c}$. Since $(V^{\prime})^{c}$ is closed in $X$,
$\bar{A} \subset(V^{\prime})^{c}$. This implies $B \subset(V^{\prime})^{c}$.
If $x \in V$, then $x \in B$, and so $x \notin V^{\prime}$. But $V \subset
V^{\prime}$, 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}
Path connected is a more clear form of connectivity that is also very useful.
\begin{definition}
A \emph{path} in $X$ is a continuous map $\gamma:\left[ a,b\right]
\rightarrow X.$ 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}
\section{Compactness}
Compactness is an important and perhaps surprisingly useful property.
\subsection{Compactness in $\mathbb{R}^{n}$ and metric spaces}
For a subset $X$ of $\mathbb{R}^{n}$, the following three properties are
equivalent. So we can take any one of them to be the definition of compact in
$\mathbb{R}^{n}$.
\begin{itemize}
\item $X$ is closed and bounded. (A subset of $\mathbb{R}^{n}$ 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}
In a general topological space they are not equivalent. In a metric space the
second and third are equivalent, but the first is not. We will first consider
compactness in metric spaces and give a characterization of compactness in a
metric space that is analogous to the first characterization of compact sets
in $\mathbb{R}^{n}$.
We start by making the second and third properties above into definitions.
\begin{definition}
A set $F$ in a topological space $(X,\mathcal{T})$ is \emph{compact} if for
any collection of open sets whose union contains $F$ (an open cover of $F$),
there is a finite subcollection whose union still contains $F$ (finite
subcover). The space $X$ is compact if $X$ is a compact subset of $X.$
\end{definition}
Note, that we can consider $F\subset X$ compact a subset of $X,$ or we can
take the subspace topology and consider $F$ as a compact subset of $F.$ Are
these the same? It turns out that these are equivalent, so referring to a
compact subset as a compact space is perfectly fine. (Convince yourself this
is true and see the exercises!)
There is another common, useful definition of compactness.
\begin{definition}
A set $F$ in a topological space $(X,\mathcal{T})$ is \emph{sequentially
compact} if every sequence contained in $F$ has a limit point in $F$. That is,
every sequence has a subsequence which converges to a point in $F.$
\end{definition}
These are generally not the same thing, but...
\begin{proposition}
In a metric space a set is compact if and only if it is sequentially compact.
\end{proposition}
In a general topological space you can have sequentially compact sets which
are not compact and compact sets which are not sequentially compact.
\smallskip\noindent\textbf{Remark:} It is easily checked that a set $F \subset
X$ is compact according to the above definition if and only if the space $F$
with the relative topology is a compact topological space.
\begin{definition}
A subset $X$ of $\mathbb{R}^{n}$ is said to be \emph{bounded }if there exists
$r>0$ such that $X\subset B\left( 0,r\right) =\left\{ x\in\mathbb{R}%
^{n}:\left\vert x\right\vert 0.$ 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 $\mathbb{R}^{n}$ is the same as the product topology it
gets by thinking of it as the product of $n$ copies of $\mathbb{R}$,
$[a,b]^{n}$ is compact in $\mathbb{R}^{n}$ (see next section). A closed and bounded set is a
closed subset of some compact set $\left[ -k,k\right] ^{n},$ and thus is compact (see next section).
\end{proof}
In $\mathbb{R}^{n}$, compactness is equivalent to being closed and bounded,
but this is not true in a general metric space. One of our examples is a
metric space in which there are bounded closed sets that are not compact.
(Which example?) 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 \textit{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 \textit{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
covered 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}
\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
\begin{align}
A \subset A^{c} \cup(\cup_{i=1}^{n} U_{\alpha_{i}})
\end{align}
Since $A^{c}$ does not cover any of $A$, this implies
\begin{align}
A \subset\cup_{i=1}^{n} U_{\alpha_{i}}%
\end{align}
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 $\mathbb{R}$ 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.,
\begin{align*}
X \subset\cup_{i=1}^{n} V_{x_{i}}%
\end{align*}
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{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
$\sum_{n=1}^{\infty}x_{n}^{2} < \infty$ and
\begin{align}
|| (x_{n})_{n=1}^{\infty}|| = \left[ \sum_{n=1}^{\infty}x_{n}^{2} \right]
^{1/2}%
\end{align}
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:
\begin{align}
\{ (x_{n})_{n=1}^{\infty}: \sum_{n=1}^{\infty}n x_{n}^{2} \le1\}
\end{align}
\item Let $L^{2}(\mathbb{R})$ be the space of functions $f$ on $\mathbb{R}%
^{2}$ with $\int f^{2}(x)dx<\infty$. We define
\[
||f||_{2}=\left[ \int f^{2}(x)\,dx\right] ^{1/2}%
\]
(There are some major issues here which we ignore. ) The closed unit ball is
not compact in this space. (Can you prove this?) For $g\in L^{2}$, define
\[
T_{g}(f)=\int f(x)g(x)\,dx
\]
Look at the weakest topology that makes all these functions $T_{g}$ (where $g$
ranges over $L^{2}$) continuous. A big theorem from functional analysis says
the closed unit ball is compact in this topology.
\end{itemize}
\end{document}**