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\begin{document}

\title{Real Analysis Lectures  -- Integration workshop 2016}
\author{Shankar Venkataramani} 
\maketitle

\begin{abstract}
Lecture notes from the Integration Workshop at University of Arizona, 
August 2016.
These notes borrow heavily from notes for previous workshops,
written and revised by many people. There is more material than I can hope to cover in the lectures. I would urge you to go through all of it, even the parts that I don't get to in the lectures, because this background will be very useful in your core classes.
\end{abstract}




\section{ Lecture 1: Sequences and series}

\subsection{Definitions, convergence tests}

%The following is a special case of definitions we made in a metric space:

\begin{definition} Let $a_n$ be a sequence of real or complex numbers. 
We say that $a_n$ {\bf converges} to $a$ if for every $\epsilon > 0$ 
there is an $N$ such that $n \ge N$ implies $|a_n-a|<\epsilon$. 
We say that $a_n$ is a Cauchy sequence if for every 
$\epsilon > 0$ there is an $N$ such that $n,m \ge N$ 
implies $|a_n-a_m|<\epsilon$. 
\end{definition}

\begin{theorem} The following statements are equivalent: \\
(1) $\reals$ is complete.\\
(2) Every bounded sequence in $\reals$ has a least upper bound.
\end{theorem}

The second statement is taken as an axiom about $\reals$. By the completeness axiom of the real numbers, a monotone sequence 
converges if and only if it is bounded. 
Given a sequence $a_n$, define $b_n = \sup \{ a_k : k \ge n \}$.
Then $b_n$ is a nonincreasing sequence and so has a limit.
Call this the limit superior or just lim sup $a_n$ and write
$\limsup_{n \ra \infty} a_n$.
Similarly, the limit inferior or lim inf is defined by 
\beann
\liminf_{n \ra \infty} a_n = \lim_{n \ra \infty}
\inf \{ a_k : k \ge n \}
\eeann

The $\liminf$ and $\limsup$ always exist although the $\liminf$ 
and the $\limsup$ can  be $\pm\infty$. 
We always have $\liminf_{n\to\infty}a_n \le \limsup_{n\to\infty} a_n$. 
They are equal if and only if $a_n$ converges. In this case they are 
equal to the limit of $a_n$ and write
$$\lim_{n\to\infty} a_n.$$

Given a sequence $a_n$ of real or complex numbers we can form the 
series $\sum_{n=1}^\infty a_n$. The partial sums are 
\beann
s_n=\sum_{k=1}^n a_k
\eeann
We can that the series converges to $s$ if the partial sums $s_n$ converges
to $s$. We say it converges absolutely if $\sum_{n=1}^\infty |a_n| < \infty$.

\begin{proposition}
({\bf Integral test}): Let $f$ be a positive decreasing 
function defined on $[1,\infty)$ such that $lim_{x \ra \infty} f(x) = 0$.
For $n = 1, 2, \cdots$, define
\beann
s_n = \sum_{k=1}^n f(k), \quad t_n = \int_1^n \, f(x) \, dx, \quad
d_n = s_n - t_n
\eeann
Then 
\begin{itemize}
\item $0 < f(n+1) \le d_{n+1} \le f(1)$, for $n=1,2,\cdots$. 
\item $d= \lim_{n \ra \infty} d_n$ exists
\item $s_n$ converges if and only if $t_n$ converges. 
\item $0 \le d_k-d \le f(k)$ for $k=1,2,\cdots$. 
\end{itemize}
\end{proposition}

\begin{proposition}
({\bf Ratio and root tests}): Given a series
$\sum_{n=1}^\infty a_n$ of nonzero complex terms, let 
\beann
r_- = \liminf_{n \ra \infty} \left| \frac{a_{n+1} }{ a_n } \right|, \quad
r_+ = \limsup_{n \ra \infty} \left| \frac{a_{n+1} }{ a_n } \right|, \quad
\rho = \limsup_{n \ra \infty} |a_n|^{1/n}
\eeann
\begin{itemize}
\item The series converges absolutely if either $r_+ < 1$ or $\rho< 1$.
\item The series diverges if either $r_- > 1$ or $\rho> 1$.
\item In all other cases the tests are inconclusive.
\end{itemize}
\end{proposition}

\subsection{Infinite products}


Let $u_n$ be a sequence of complex numbers. The infinite product
$\prod_{n=1}^\infty u_n$ is said to converge if there is an $N$ such 
that $u_n \ne 0$ for $n \ge N$ and the sequence $p_k = \prod_{n=N+1}^k$
has a nonzero limit $p$ as $k \ra \infty$.

In the case of convergence, 
$\prod_{n=1}^\infty u_n$ is defined to be $p p_1 \cdots p_N$. 

There is a connection betweeen convergence of sums and of products.
For $a_n>0$, the product $\prod_{n=1}^\infty (1+a_n)$ 
converges if and only if the series $\sum_{n=1}^\infty a_n$ 
converges. 

We say that the product $\prod_{n=1}^\infty (1+a_n)$ 
converges absolutely if $\prod_{n=1}^\infty (1+|a_n|)$ converges.
Absolute convergence of the infinite product implies convergence of 
the product.

\subsection{Sequences of functions}

\begin{definition}
A sequence of functions is said to {\bf converge pointwise} 
to a limit function $f$ on a set $S$ provided that for
every $x \in S$, and each $\epsilon > 0$, there exists $N$, 
depending possibly on both $x$ and $\epsilon$ such that
$n > N$ implies $|f_n(x) - f(x)| < \epsilon$. 
If the choice of $N$ does not depend on $x$, the sequence of functions 
is said to {\bf converge uniformly.}
Let $f_n$ be a sequence of functions defined on a set $S$. 
For each $x \in S$, set 
\beann
s_n(x) = \sum_{k=1}^n \, f_k(x) 
\eeann 
If $s_n \rightarrow s$ uniformly on $S$, then we say that 
$\sum_{n=1}^\infty f_n(x)$ converges uniformly on $S$.  
\end{definition}

\begin{proposition}
({\bf Uniform convergence and continuity})
If $f_n  \ra f$ uniformly on $S$ and each $f_n$ is continouous at a point 
$c$, then $f$ is continuous at $c$.
\end{proposition}

\begin{theorem} (Weierstrass M-test) 
Let $M_n$ be a sequence of nonnegative numbers such that
$0 \le |f_n(x)| \le  M_n $ for $n = 1, 2, \cdots$ and every $x \in S$.
If $\sum_{n=1}^\infty M_n$ converges, then 
$\sum_{n=1}^\infty f_n(x)$ converges uniformly on $S$.
\end{theorem} 

\medskip
\noindent {\bf The $L^\infty$ norm }
\smallskip

Consider the vector space $C(S)$, the real valued bounded continuous functions 
on a metric space $S$, and define the infinity norm (or sup norm) by 
\beann
||f||_\infty = \sup_{x \in S} |f(s)|
\eeann
Then $|| \quad  ||_\infty$ is a norm
This norm induces a metric $d(f, g) = ||f - g||_\infty$ in the usual way. 
The theorems above on uniform continuity show that
$C(S)$ with this metric is a complete metric space.

\medskip
\noindent {\bf  Integration and differentiation}
\smallskip

Many of the theorems on uniform convergence permit the reversal of the order 
of taking of limits.

\begin{theorem} (Integration) Let $f_n$ be Riemann integrable functions 
on $[a,b]$ for $n=1,2,\cdots$. 
Define 
\beann
g_n(x) = \int_a^x \, f_n(t) dt, \quad x \in [a, b]
\eeann
and
\beann
g(x) = \int_a^x \, f(t) dt, \quad x \in [a, b]
\eeann
Assume there exists $f$ so that $||f_n-f||_\infty \ra 0$. Then 
\begin{itemize}
\item $f$ is Riemann integrable and 
\item $||g_n - g||_\infty \ra 0$
\end{itemize}
\end{theorem}

\begin{theorem} (Differentiation) 
Assume that $f_n$ is differentiable on $(a,b)$,
$f_n^\prime$ is Riemann integrable and that there exist a function
$g$ so that $d(f_n^\prime,g) \ra 0$ and a point $c \in (a,b)$ so that 
$f_n(c)$ converges. Then 

(a) there exists f so that $d(f_n,f) \ra 0$, and

(b) $f$ is differentiable with derivative $g$.
\end{theorem}

%\subsection{Power series}
%
%An infinite series of the form
%\beann
%f(z) = \sum_{n=0}^\infty a_n (z - z_0)^n
%\eeann
%is called a power series centered at $z_0$. Define $r$ by 
%$ 1/r = \limsup_{r \ra \infty} |a_n|^{1/n}$
%(We make the conventions that  $1/0 = \infty$ and $1/\infty = 0$.) 
%Then by the root test, the series converges absolutely if $|z - z_0| < r$
%and diverges if $|z - z_0| > r$. 
%Furthermore:
%
%\noindent 1. The series converges uniformly on every compact subset 
%of $B(z_0, r)$.
%
%\noindent 2. The function $f$ can be differentiated term by term for 
%any $z \in B(z_0,r)$, 
%\beann
%f^\prime(z) = \sum_{n=1}^\infty n a_n (z - z_0)^{n-1}
%\eeann
%
%\noindent 3. The power series for $f^\prime$ has radius of convergence $r$.
%
%\noindent 4. Repeated differentiation and evaluation of this yields
%$a_k = f^{(k)}(z_0)/k!$.
%
%\begin{theorem} 
%Suppose that the power series 
%\beann
%f(z) = \sum_{n=0}^\infty a_n (z - z_0)^n
%\eeann
%has a nonzero radius $r$ of convergence. Then $f(z)$ is analytic on 
%$B(z_0,r)$. Conversely, if $f(z)$ is analytic at $z_0$, then 
%there is a power series with a nonzero radius of convergence that 
%converges to $f$ in a neighborhood of $z_0$.
%\end{theorem} 

\section{Introduction to metric spaces}

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

For motivation, we recall what open and closed sets look like in $\reals^n$. 
We use $|| \quad ||$ to denote the usual distance function in $\reals^n$.
A set $U \subset \reals^n$ is open if for any $x\in U$ 
there is an $\epsilon > 0$ such that $||x-y||<\epsilon$ implies
$y \in U$. In other words, there 
is a ball centered at $x$ that is entirely contained in $U.$ 
A sequence $x_n$ converges to $x$ if $||x_n-x||$ goes to zero. 
A set $F$ is closed if whenever a sequence $x_n$ in $F$ converges, 
the limit point must be in $F$. 

If $f$ is a function from an open subset $U$ of $\reals^n$ 
to $\reals^k$, then $f$ is continuous at $x_0$ if 
$\forall \epsilon>0$, \ 
$\exists \delta > 0$ such that $||x-x_0||<\delta$ 
implies $||f(x)-f(x_0)||<\epsilon$. 

The only structure of $\reals^n$ that we use in the above is the 
distance $||x-y||$ between two points. So we can abstract this 
by just starting with a set which has a distance function on it.
Of course, the distance functions will need to have some properties.

\subsection{Definition of a metric space and basic results}

\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}

\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}


\begin{proposition}
Let $(X,d)$ be a metric space and let $\topo$ be the collection 
of open sets in $X$. Then
\begin{enumerate}
\item $X\in\topo$ and $\varnothing\in\topo$,
\item Arbitrary unions of sets $U\in\topo$ are in $\topo$, i.e.,
for any indexing set $I,$ if $U_{i}\in\topo$ for all $i\in I$ then $%
%TCIMACRO{\dbigcup \limits_{i\in I}}%
%BeginExpansion
{\displaystyle\bigcup\limits_{i\in I}}
%EndExpansion
U_{i}\in\topo$,
\item If $U,V\in\topo$ then $U\cap V\in\topo$.
\end{enumerate}
\end{proposition}

For any set $X$, a collection of subsets $\topo$ of $X$ is said to 
be a {\bf topology} for $X$ if it satisfies the three properties above.
Note that property 3 immediately implies by induction that 
a finite intersection of open sets produces an open set. 

\begin{definition}
A sequence $x_n$ in a metric space $(X,d)$ {\bf 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}

\begin{definition}
A subset $F$ of a metric space $(X,d)$ is {\bf closed} if for every 
sequence $x_n$ in $F$ which converges to some $x$ in $X$ we have $x \in F$. 
\end{definition}

\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{definition}
The interior of a set $A,$ denoted $\overset{\circ}{A}$ 
or $int(A)$, is 
\beann
int(A) = \{ x: \exists \epsilon>0 \quad s.t. \quad 
d(x,y) < \epsilon \implies y \in A\}
\eeann
The closure of a set $A,$ denoted $\bar{A}$ or 
$cl(A)$, is 
\beann
cl(A) = \{ x: \forall \epsilon>0 \quad \exists y \in A \quad s.t. \quad
d(x,y) < \epsilon \}
\eeann
\end{definition}

\begin{proposition}
The interior of a set $A$ is the union of all open
sets contained in $A.$ The closure of a set $A$
is the intersection of all closed sets containing $A$.
\end{proposition}

Note that the proposition shows that 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 {\bf limit point} of a set $A\subset X$ if 
%every open set $U$
%containing $x$ also contains a point $y\in A\setminus\left\{  x\right\}  .$
%$x \in X$ is an {\bf accumulation point} of 
%$A$ if there exists a sequence in $A \setminus \{x\}$ that converges to $x$.
%$x \in A$ is an {\bf isolated point} (of $A$) if there is an open set 
%$O$ such that $O \cap A = \{x\}$. 
%\end{definition}
% 
%\begin{proposition}
%In a general topological space an accumulation point is always a limit point. 
%In a metric space the converse is true and so the two notions are equivalent.
%\end{proposition}
%
%We can characterize the closure in terms of limit points.
%
%\begin{proposition}
%In a metric space, $\bar{A}$ is equal to the union of $A$ and its 
%accumulation points.
%\end{proposition}
%
%In a metric space we can substitute ``limit points'' for 
%``accumulation points'' in the above proposition.
%In a general topological space the closure is the union of the set and 
%its limit points.
%
%\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 $x \in X$, $||x|| \ge 0$, and $||x|| = 0$ if and only if 
%$x = 0$. 
%\item $||a x|| = |a|||x||$ for every $a \in \reals$ 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 {\bf Euclidean metric on $\reals^n$:}
%The usual Euclidean norm gives a metric on  $\reals^n$.
%\beann
%d(x,y)=||x-y||= \left[ \sum_{j=1}^n |x_j-y_j|^2 \right]^{1/2}
%\eeann
%
%\item {\bf $\reals$ with a different topology:}
%Define a metric on $X$ by 
%\beann
%d(x,y) = \frac{|x-y| }{ 1 + |x-y|}
%\eeann
%One of the homework problems is to check that this is a metric and 
%that it gives a different topology for $\reals$ from the standard one.
%
%\item {\bf Space of functions:}  Let $D$ be any set and let $X$ be 
%the set of all bounded real-valued functions on $D$. Define 
%\beann
%d(f,g) = \sup_{x \in D} |f(x)-g(x)|
%\eeann
%Then $(X,d)$ is a metric space.
%
%\item {\bf $l^p$ norm on $\reals^n$:}
%There are other norms we can put on $\reals^n$ and hence other metrics. 
%For $1 \le p < \infty$, define 
%\beann
%||x||_p = \left[ \sum_{j=1}^n |x_j|^p \right]^{1/p}
%\eeann
%(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 $p \ra \infty$ we get
%
%\item {\bf sup norm on $\reals^n$:}
%The function 
%\beann
%d(x,y) = \max_{1 \le j \le n} |x_j-y_j|
%\eeann
%is another metric on $\reals^n$ that defines the same toplogy. 
%
%\item {\bf $l^p(\naturals)$:}
%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
%\beann
%|| (x_n)_{n=1}^\infty ||_p =  \left[ \sum_{n=1}^\infty |x_n|^p  \right]^{1/p}
%\eeann
%Consider the two sets 
%\beann
%F= \{ (x_n)_{n=1}^\infty \in l^p : x_n \ge 0 \, \forall n \} \\
%U= \{ (x_n)_{n=1}^\infty \in l^p : x_n > 0 \, \forall n \} 
%\eeann
%Is $F$ closed? Is $U$ open?
%
%\item {\bf $l^\infty(\naturals)$:} The space is now the set of bounded 
%infinite sequences. The norm is 
%\beann
%|| (x_n)_{n=1}^\infty ||_\infty =  \sup_{1 \le n < \infty} |x_n|
%\eeann
%
%\end{enumerate}

\subsection{Continuous maps}

\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 $d^\prime(f(x),f(y))< \epsilon$. 
We say the map is globally continuous (or just continuous) if 
it is continuous at every point in $X$.
\end{definition}

\begin{proposition} Let $(X,d)$ and $(Y,d^\prime)$ be metric spaces and 
$f:X \rightarrow Y$ a function. Then $f$ is continuous on $X$ if and only 
if for every open set $U$ in $Y$, $f^{-1}(U)$ is open.
\end{proposition}

Show this proposition follows from the preceding definition! Note also that for a continuous function the inverse image of a closed set 
is closed.

\begin{definition}
Let $(X,d)$ and $(Y,d^\prime)$ be metric spaces. 
A function  $f:X \rightarrow Y$ is 
{\bf 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} %In a topological space every continuous function 
%is sequentially continuous. 
In a metric space continuous = sequentially continuous. 
\end{proposition}

\subsection{Additional material: Completeness and Uniform continuity}

\begin{definition}
A sequence $x_n$ in a metric space is {\bf 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 {\bf 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}

Complete metric spaces are the natural setting for analysis, because, in these spaces Cauchy sequences (i.e sequences that ``morally" should converge) are indeed convergent. They come up frequently in applications, because they allow one to assert the existence of a solution as the limit of an appropriately constructed sequence of approximate solutions. For this reason, it is useful to identify if a given metric space is complete, and if not (as for e.g $\mathbb{Q}$ is not complete), develop an approach to make a ``natural" complete metric space (e.g. going from $\mathbb{Q}$ to $\reals$.)

Completeness is {\bf not} a topological property. Two metric spaces can be homeomorphic (topologically the same) while one if complete, and the other is not. The issue here is that continuous functions can map Cauchy sequences into sequences that are not Cauchy, and vice-versa. 

A stronger notion than continuity is {\em uniform continuity}. 

\begin{definition}
A function $f:(M,d) \to (N,\rho)$ between metric spaces  is {\em uniformly continuous} if for all $\epsilon > 0$, there is a $\delta > 0$ such that {\em for all } $ x \in M$, $d(y,x) < \delta \implies \rho(f(y),f(x)) < \epsilon$.
\end{definition}

If $f$ is uniformly continuous, then it will map Cauchy sequences into Cauchy sequence.

\begin{proposition} Every closed subset of a complete metric space is a complete metric space with the induced metric.
\end{proposition}

This gives an approach to ``completing" any metric space, that you will explore in the problem set. 

\section{Additional material: Compactness}

\subsection{Definitions of compactness}

In a topological space there are two different notions of compact. 

\begin{definition} 
A set $F$ in a topological space 
$(X,\topo)$ is {\bf 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).
\end{definition}

\begin{definition} 
A set $F$ in a topological space $(X,\topo)$ is {\bf 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}

\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.


\subsection{Properties of compact spaces}

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

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

\begin{proof}
We prove that $f$ attains its min. Let 
\beann
m=\inf_{x\in X}f\left(  x\right)
\eeann
By the definition of $\inf$, 
we can find $x_n \in X$ such that $f(x_n) <  m + 1/n$.
The sequence $x_n$ has a subsequence $x_{n_k}$ 
that converges to some $x_{min} \in X$. 
By continuity $f(x_{min})=\lim_k f(x_{n_k})$. So we have 
\beann
m \le f(x_{min}) = \lim_k f(x_{n_k}) \le \limsup_k [m + 1/n_k] = m
\eeann
Hence $f(x_{min})=m$. 
\end{proof}


\subsection{Heine-Borel theorem}

Notice that $(0,1]$, which is not closed, 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 subcover.
And notice that $[0,\infty)$, which is not bounded, is not compact because
the cover $[0,n]$, $n=1,2,\cdots$ has no finite subcover. 

\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}
%

Since compactness and sequential compactness are equivalent in $\reals^n$, 
we have 
\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 The finite-dimensional sphere $\left\{  x\in\mathbb{R}^{n}:
%||x||=1\right\}  $ is compact. 
%
%\item
%Define a metric on $X$ by 
%\beann
%d(x,y) = \frac{|x-y| }{ 1 + |x-y|}
%\eeann
%The entire space is bounded. In fact, $d(x,y) \le 1$ for all $x,y$. 
%And the entire space is closed. 
%But one of the homework problems is to show that $\reals$ is not compact with 
%this metric. 
%
%\item Recall the definiton of $l^2(\naturals)$. 
%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. 



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

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

\section{Lecture 2: Calculus}

\subsection{Multivariate differential calculus}

\begin{definition} 
Let $O \subset \reals^n$ be open and $f:O \ra \reals^m$. 
Let $c \in O$.
The function $f$ is said to be {\bf differentiable}
at $c$ if there is a linear function, called the {\bf total derivative},
$T^f_c : \reals^n \ra \reals^m$, such that 
\bea
\lim_{v \ra 0} \frac{||f(c+v) - f(c) - T^f_c(v)||}{||v||} =0
\eea
We say $f$ is differentiable on $O$ if it is differentiable at every 
point in $O$. 
\end{definition} 

If the total derivative exists, then for all directions 
the directional derivative exists 
\beann
D_v f(c) = \lim_{\epsilon \ra 0} \frac{f(c+\epsilon v)-f(c)}{\epsilon}
\eeann
and equals $T^f_c(v)$. 

Let $e_1,e_2,\cdots,e_n$ be the standard basis for $\reals^n$. Then 
the partial derivatives are the directional derivative in the directions
of the standard basis:
\beann
\frac{\partial f}{\partial x_k} = D_{e_k} f
\eeann
Note that both sides of this equation are vectors in $\reals^m$. 
The components are 
\beann
\frac{\partial f_j}{\partial x_k} = D_{e_k} f_j
\eeann
where $f=(f_1,f_2,\cdots,f_m)$.
The matrix representation of $T$ in this basis is called the 
{\bf Jacobian matrix} 
\beann
Df(c) = 
\begin{pmatrix}
\frac{\partial f_1 }{ \partial x_1}(c) & 
\frac{\partial f_1 }{ \partial x_2}(c) & \cdots & 
\frac{\partial f_1 }{ \partial x_n}(c) \\
\frac{\partial f_2 }{ \partial x_1}(c) & 
\frac{\partial f_2 }{ \partial x_2}(c) & \cdots & 
\frac{\partial f_2 }{ \partial x_n}(c) \\
. & . &  & . \\
. & . &  & . \\
. & . &  & . \\
\frac{\partial f_m }{ \partial x_1}(c) & 
\frac{\partial f_m }{ \partial x_2}(c) & \cdots & 
\frac{\partial f_m }{ \partial x_n}(c) 
\end{pmatrix}
\eeann

\begin{theorem} (Chain rule)
Suppose that $f: O \subset \reals^n \ra \reals^m$ and 
$g: U \subset \reals^k \ra O$ are both differentiable. Then 
$h=f \compose g$ is differentable and 
\beann
T^h_p = T^f_{g(p)} \compose T^g_p
\eeann
In matrix form
\beann
Dh(p) = Df(g(p)) \, Dg(p)
\eeann
\end{theorem}

Let $L(x_1, x_2) = \{\lambda x_1 + (1 - \lambda) x_2 : 0 \le \lambda \le 1\}$ 
be the line segment connecting $x_1$ and $x_2$ in $\reals^n$.

\begin{theorem} (Mean Value Theorem) 
Let $O$ be an open subset of $\reals^n$ and assume that $f : O \ra \reals^m$ 
is differentiable on $O$. Choose $x_1$ and $x_2$ so that 
$L(x_1, x_2) \subset O$.
Then for every vector $a \in \reals^m$, 
there is a point $c \in L(x_1, x_2)$ such that 
\beann
a \cdot (f(x_2) - f(x_1)) = a \cdot T^f_c (x_2 - x_1)
\eeann
\end{theorem}

We now consider higher order derivatives.

\begin{theorem} Let $f : \reals^n  \ra \reals^m$.
Then the following conditions are sufficient for the equality of the mixed
partial derivatives
\beann
\frac{\partial^2 f }{ \partial x_i \partial x_j}(c)= 
\frac{\partial^2 f }{ \partial x_j \partial x_i}(c)
\eeann 

1. Both ${\partial f / \partial x_i}$ and ${\partial f / \partial x_j}$ 
exist in an $n$-ball $B(c,\delta)$ and are differentiable at $c$.

2. Both ${\partial f /\partial x_i}$ and ${\partial f / \partial x_j}$ 
exist in an $n$-ball $B(c,\delta)$  and 
${\partial^2 f / \partial x_i \partial x_j}$ and 
${\partial^2 f / \partial x_j \partial x_i}$ 
are both continuous at c.
\end{theorem}

Call $\alpha = (\alpha_1, \cdots , \alpha_n)$ 
a multi-index if each of its entries are non-negative integers. 
Write $|\alpha| = \alpha_1 + \cdots + \alpha_n$. 
This allows for the notational abbreviations
\beann
x^\alpha = x^{\alpha_1} \cdots x^{\alpha_n}, \quad 
D^\alpha = \frac{\partial^{\alpha_1} }{ \partial x^{\alpha_1}}
\cdots
\frac{\partial^{\alpha_n} }{ \partial x^{\alpha_n}}
\eeann
and provides for a compact notation for Taylor’s formula for functions 
$f$ from $\reals^n$ to $\reals$. Write 
\beann
f^{(k)}(x; t) = \sum_{\alpha: |\alpha|=k} \,  D_\alpha f(x) t^\alpha
\eeann
and assume that $f$ 
and all of its partial derivatives of order up to $m-1$ are differentiable
at each point of an open set $S \subset \reals^n$. Choose $x$ and $a$ so that 
$L(a, x) \subset S$. Then for some $c \in L(a, x)$, 
\beann
f(x) = f(a) + \sum_{k=1}^{m-1} \, \frac{1}{k!} f^{(k)}(x; t) + 
\frac{1}{m!} f^{(m)}(c; t)
\eeann

\subsection{Implicit functions}
Let $A$ be an $n \times n$ matrix. Then, for $y \in \reals^n$, 
$Ax = y$ has a unique solution $x$ whenever $A$ has nonzero determinant. 
This suggests that in looking for a unique solution
to $f(x) = y$, we consider the Jacobian determinant, 
the determinant of the Jacobian matrix,
\beann
J_f(x) = det Df(x) = \frac{\partial{(f_1,\cdots,f_n)} }
{\partial{(x_1,\cdots,x_n)}}
\eeann

\begin{theorem}(Inverse function theorem)
Let $f : S \ra  \reals^n$ be continuously differentiable on an open set
$S \subset \reals^n$. If the Jacobian determinant
$J_f(a) \ne 0$ for some point $a \in S$, 
then there exists two open sets $X \subset  S$ and $Y \subset f(S)$ 
and a unique function $g$ defined on $Y$ such that

1. $a \in X$ and $f(a) \in Y$ 

2. $Y = f(X)$

3. $f$ is one-to-one on $X$

4. $g(Y ) = X$

5. $g(f(x)) = x$ for every $x \in X$

6. $g$ is continuously differentiable on $Y$ .

\end{theorem}
Note that for $y = f(x)$, $Dg(y) \, Df(x)$ is the identity matrix.

\begin{theorem}(Implicit function theorem)
Let $S  \subset \reals^n \times \reals^k$ and suppose that 
$f : S \ra \reals^n$ is continuously differentiable. 
Assume that
$f(x_0, y_0) = 0$ and that the $n \times n$ determinant 
$det[\partial f_j / \partial x_i (x_0,y_0)] \ne 0$. 
Then there exists a $k$-dimensional
set $Y_0$ containing $y_0$ and a unique vector valued function
$ g : Y_0 \ra \reals^n$ such that

1. $g$ is continuously differentiable

2. $g(y_0) = x_0$

3. $f(g(y), y) = 0$ for every $y \in Y_0$.
\end{theorem}


\subsection{Multivariable Riemann integrals}

Let $A=[a_1,b_1]\times\cdots\times[a_n,b_n]\subset\reals^n$ and let ${\cal P}_k$ be a partition of $[a_k,b_k]$ into $m_k$ intervals. Then

\beann
{\cal P} = {\cal P}_1\times\cdots\times{\cal P}_n
\eeann
divides $A$ into $m_1\times\cdots\times m_n$ $n$ dimensional intervals. A partition ${\cal Q}$ is called a refinement of ${\cal P}$ if ${\cal P}\subset{\cal Q}$.

For $I\in {\cal P}$ let $\hbox{vol}(I)$ be the product of the lengths of the one dimensional intervals that determine $I$. Let $f$ be a real valued function defined on $A$. For any choice of sample points $t_I \in I$, define the Riemann sum
\beann
S(f,\mathbf{t}, {\cal P})= \sum_I f(t_I)\hbox{vol}(I)
\eeann

\begin{theorem} The function $f$ is Riemann integrable on $A$ if there exist a number, $r$, having the following property:

For every $\epsilon > 0$, there exists a partition ${\cal P}_\epsilon$ such that for any refinement $\cal Q$ of ${\cal P}_\epsilon$ and any choice of sample points $t_I \in I$. 

\beann
|S(f,\mathbf{t}, {\cal P}) -  r| < \epsilon.
\eeann
\end{theorem}

We typically write the integral
\beann
\int_A f(x_1,\ldots,x_n)\ d(x_1,\ldots,x_n)= \int_A f(x)\ dx.
\eeann


\subsection{Change of variable formulas for integrals}
Let $T$ be a one-to-one continuously differentiable mapping of an open set 
$V \subset \reals^k$ into $\reals^k$ such that the
Jacobian determinant $J_T(x) \ne 0$ for all $x \in V$. Let $f$ 
be a continuous function on $\reals^k$ whose support is
compact and lies in $T(V)$. Then
\beann
\int_{\reals^k} \, f(y) \, dy = 
\int_{\reals^k} \, f(T(x)) \, |J_T(x)| \, dx 
\eeann

\subsection{Differential forms and Stokes theorem}

Let $K \subset \reals^k$ be compact and let $V \subset \reals^n$ 
be open.  A $k$-surface is a continuously differentiable mapping
$\Phi : K \ra V$. For example, each component of a 1-surface is called a curve.

A {\it differential form of order $k$}, or briefly, a $k$-form, 
is a function $\omega$, represented symbolically by
\beann
\omega= \sum a_{i_1 \cdots i_k}(x) \, dx_{i_1} \wedge \cdots \wedge dx_{i_k}
\eeann
that assigns to each $k$-surface $\Psi$ in $V$ a number 
\beann
\int_\Phi \omega = \int_K \, \sum a_{i_1 \cdots i_k}(\Phi(u)) \, 
\frac{\partial(x_{i_1},\cdots,x_{i_k}) }{ \partial(u_1,\cdots,u_k)} du
\eeann

A 0-form is defined to be a continuous function of $V$ . 
Integrals of 1-forms are called line integrals.
Let $c \in \reals$ and let $\omega, \omega_1, \omega_2$ be $k$-forms on $V$.
Then 
\beann
\int_\Phi \, c \omega = c \int_\Phi \, \omega 
\eeann

\beann
\int_\Phi \, (\omega_1 + \omega_2)  = 
\int_\Phi \, \omega_1 + \int_\Phi \, \omega_2
\eeann

For $\omega=a_{{i_1} \cdots i_k}(x) \, dx_{i_1} \wedge \cdots \wedge dx_{i_k}$ 
and for $\bar{\omega}$ obtained from $\omega$ by interchanging some pair 
of subscripts, $\bar{\omega} = - \omega$. 

Write the basic $k$-form $dx_I = dx_{i_1} \wedge \cdots \wedge dx_{i_k}$, for 
$1 \le i_1 < \cdots < i_k$, giving the standard presentation
\beann
\omega = \sum_i a_I(x) dx_I
\eeann

\subsection{Differentiation of forms}

\smallskip

The operator $d$ is a mapping from $k$-forms to $(k + 1)$-forms 
defined as follows:

1. For a class $C^1$ $0$-form $f$ , 
\beann
df = \sum_{i=1}^n \frac{\partial f }{ \partial x_i} dx_i
\eeann

2. For the class $C^1$ $k$-form $\omega$ above in the standard presentation, 
\beann
d \omega = \sum_I \, (da_I) \wedge dx_I
\eeann

For i = 1, 2, let $\omega_i$ be class $C^1$ $k_i$-forms.
Then 
\beann
d(\omega_1 \wedge \omega_2) = (d \omega_1) \wedge \omega_2 
+ (-1)^{k_1} \omega_1 \wedge (d \omega_2)
\eeann
If $\omega$ is of class $C^2$, $d(d\omega) = 0$.

\begin{definition}
A $k$-form $\omega$ is called exact if $\omega = d \eta$
for some $(k - 1)$-form $\eta$.
A class $C^1$ $k$-form is called closed if $d \omega = 0$.
\end{definition}

Every exact class $C^1$ form is closed. 
If the domain is a convex set, then the Poincare lemma states that 
the converse is true.

\begin{theorem}[General Stokes' theorem]
If $\Psi$ is a $k$-chain of class $C^2$ in an open set $V \subset \reals^m$
and if $\omega$ is a $(k - 1)$-form of class $C^1$ in $V$ , then
\beann
\int_\Psi \, d \omega = \int_{\boundary \Psi} \, \omega
\eeann
\end{theorem}

Various theorems from calculus and vector calculus are special cases 
of this general theorem as indicated in the following table. 

\begin{table}
\begin{center}
\begin{tabular}{|r|r|l|}
\hline
$k$ & $m$ & theorem \\
\hline
1 &  1  & fundamental theorem \\
2 &  2  & Green's theorem \\
3 &  3  & divergence theorem \\
2 &  3  & classical Stokes theorem \\
\hline
\end{tabular}
\end{center}
\end{table}

\begin{theorem} 
[Green's Theorem] Let $C$ be a simple closed curve in the $xy$-plane 
Let $M(x,y)$ and $N(x,y)$ be continuously differentiable on an 
open set containing $C$ and $R$, the region it encloses. Then 
\beann
\int_C (M dx + N dy) = \int_R (
\frac{\partial N }{ \partial x} - 
\frac{\partial M }{ \partial y}) dx dy
\eeann
\end{theorem}

\begin{theorem} 
[Divergence Theorem] Let $F$ be a continuously 
differentiable vector field on an open set $V \subset \reals^3$,
and let $C \subset V$ be closed with positively oriented boundary
$\boundary C$. Then 
\beann
\int_C (\nabla \cdot F) dV =  \int_{\boundary C} (F \cdot \mathbf{n} ) \, dA
\eeann
where $\mathbf{n}$ is a unit normal vector, pointing outwards. 
\end{theorem}


\begin{theorem} 
[classical Stokes' Theorem] Let $F$ be a continuously 
differentiable vector field on an open set $V \subset \reals^3$,
and let $S \subset V$ be a 2-surface of class $C^2$. Then
\beann
\int_S (\nabla \times F) \cdot \mathbf{n} \, dV =  \int_{\boundary S}
(F \cdot \mathbf{t}) \, ds
\eeann
where $\mathbf{t}$ is a oriented unit tangent vector.
\end{theorem}



\section{Lecture 3: Complex Analysis} 

\subsection{Analytic functions}

Let $O$ be an open subset of $\complex$. Let $f:O \ra \complex$. 
We say $f$ is analytic at $z_0$ if the following complex limit exists:
\beann
f^\prime(w) = \lim_{z \ra w} {f(z)-f(w) \over z - w}
\eeann
for all $w$ in a neighborhood of $z_0$.
One can think of a function from $\complex$ to $\complex$ as 
a function from $\reals^2$ to $\reals^2$. We write $z=x+iy$ and 
$f(z) = u(x,y) + i v(x,y)$. Then the function is $F(x,y)=(u(x,y),v(x,y))$. 
It is important to understand that analyticity is a much stronger
property than requiring that $F$ have a total derivative. 
The above limit involves complex 
numbers and so it includes as special cases $z$ approaching
$z_0$ along any direction. These  ``directional limits'' must 
all give the same complex number as the limit. 
In particular, by considering taking the limit in the coordinate 
directions, one obtains the Cauchy Riemann equations.

\begin{theorem} $f$ is analytic at $z_0=(x_0,y_0)$ if and only if for all $(x,y)$ in a 
neighborhood of $(x_0,y_0)$ the total derivative of $F$ exists  and 
\beann
{\partial u \over \partial x }(x,y) = 
{\partial v \over \partial y }(x,y), \quad 
{\partial u \over \partial y }(x,y) = 
- {\partial v \over \partial x }(x,y).
\eeann
\end{theorem}

\subsection{Power series}

An infinite series of the form
\beann
f(z) = \sum_{n=0}^\infty a_n (z - z_0)^n
\eeann
is called a power series centered at $z_0$. Define $r$ by 
$ 1/r = \limsup_{r \ra \infty} |a_n|^{1/n}$
(We make the conventions that  $1/0 = \infty$ and $1/\infty = 0$.) 
Then by the root test, the series converges absolutely if $|z - z_0| < r$
and diverges if $|z - z_0| > r$. 
Furthermore:

\noindent 1. The series converges uniformly on every compact subset 
of $B(z_0, r)$.

\noindent 2. The function $f$ can be differentiated term by term for 
any $z \in B(z_0,r)$, 
\beann
f^\prime(z) = \sum_{n=1}^\infty n a_n (z - z_0)^{n-1}
\eeann

\noindent 3. The power series for $f^\prime$ has radius of convergence $r$.

\noindent 4. Repeated differentiation and evaluation of this yields
$a_k = f^{(k)}(z_0)/k!$.

\begin{theorem} 
Suppose that the power series 
\beann
f(z) = \sum_{n=0}^\infty a_n (z - z_0)^n
\eeann
has a nonzero radius $r$ of convergence. Then $f$ is analytic on 
$B(z_0,r)$. Conversely, if $f$ is analytic at $z_0$, then 
there is a power series with a nonzero radius of convergence that 
converges to $f$ in a neighborhood of $z_0$.
\end{theorem} 

\subsection{Integration}

\begin{definition} A domain $D$ is simply connected if the region bounded by 
every simple closed curve in $D$ is contained in $D$, i.e., every 
simple closed curve in $D$ may be continuously contracted to a 
point without leaving $D$. 
\end{definition}

\begin{theorem} [one of many ``Cauchy's theorems''] 
If  $D$  is a simply connected open set and $f$ is analytic on $D$ and 
$\gamma$ is a differentiable closed curve in $D$, then 
\be
\oint_\gamma f(z) dz =0
\ee
\end{theorem}

\subsection{Zeroes, poles and residues}

We say $f$ has a zero at $z_0$ if $f(z_0)=0$. 
In this case it is possible to write it in the form 
$f(z) = (z-z_0)^n g(z)$ in a neighborhood of $z_0$ where $g$ does not 
vanish on this neighborhood. The integer $n$ is unique and called the 
{\it order} of the zero.

A neigborhood of a point $z_0$ means an open set containing $z_0$.
By a deleted neighborhood of $z_0$ we will mean a neighborhood of 
$z_0$ with $z_0$ removed. A function
$f$ has an isolated singularity at $z_0$ if it is analytic on
a deleted neighborhood of $z_0$. 

If $f$ has an isolated singularity at $z_0$, and we can redefine it 
at $z_0$ so that the function is analytic at $z_0$, then we say 
$f$ has a removable singularity at $z_0$. 
Otherwise we consider $1/f$ where $1/f$ is defined to be $0$ at 
$z_0$. If this is anayltic at $z_0$  we say $f$ has a pole at $z_0$.
The order of the pole is defined as the order of the zero of 
$1/f$ at $z_0$. If it does not have a pole we say it has an essential 
singularity. 

\begin{theorem} If $f$ has a pole of order $n$ at $z_0$ then 
\beann
f(z) = {a_{-n}\over (z-z_0)^n} + {a_{-(n-1)}\over (z-z_0)^{n-1}}
+ \cdots + {a_{-1}\over z-z_0} + g(z) 
\eeann
where $g$ is analytic at $z_0$. 
\end{theorem}
The {\it principal part} of $f(z)$ (at $z_0$) is 
\beann
{a_{-n}\over (z-z_0)^n} + {a_{-(n-1)}\over (z-z_0)^{n-1}}
+ \cdots + {a_{-1}\over z-z_0} 
\eeann
The {\it residue} of $f$ at $z_0$ is $a_{-1}$ 

%\section{Differential Equations}
%
%\subsection{First order ordinary differential equations}
%
%Let $V \subset \reals^n$ and $I = [t_0, t_f ]$  
%and $\phi: I \times V \ra \reals^n$. A solution to the initial value problem
%\beann
%y^\prime = \phi(t, y),\ y(t_0) = y_0,
%\eeann
%is a differentiable function $f$ on $I$ such that 
%$f(t_0) = y_0$, $f(t) \in V $ and $f^\prime(t) = \phi(t, f(t))$. 
%The ODE is called autonomous if $\phi$ is independent of $t$.
%The basic estimate used to study the dependence of solutions on 
%initial conditions is Gronwall's inequality.
%\begin{theorem} [Gronwall's inequality]
%Let $f, g : [a, b) \ra \reals$ be continuous and nonnegative. Suppose
%\beann
%f(t) \le K + \int_a^t f(s) g(s) \, ds, \quad K \ge 0
%\eeann
%Then
%\beann
%f(t) \le K \exp\left( \int_a^t g(s) \, ds \right)
%\eeann
%for $t \in [a, b)$.
%\end{theorem}
%
%\subsection{Linear ODE's}
%
%A {\it linear system} is one in which $\phi(t, y) = A(t)y + g(t)$. 
%It is called {\it homogeneous}  if $g(t) = 0$ for all $t$. For
%a homogeneous autonomous linear system $y^\prime = Ay$, 
%a solution is $y(t) = \exp(t - t_0) A y_0$.
%By Gronwall’s inequality, this solution is unique.
%
%For nonautomonous systems having continuous $A$, 
%the solutions of $y^\prime = A(t)y$ form a vector space of
%dimension $n$ over the complex numbers. 
%An $n \times n$ matrix whose columns are linearly independent solutions
%is called a fundamental matrix. Once this matrix valued function 
%has been found, we can find a solution to
%the non-homogeneous system by variation of constants.
%
%\begin{theorem} 
%If $\Phi$ is a fundamental matrix of $y^\prime = A(t)y$ on $I$, 
%then the function
%\beann
%\psi(t) = \Phi(t) \int_{t_0}^t \Phi^{-1}(s)g(s) \, ds
%\eeann
%is the unique solution of the nonhomogenous linear system above with 
%initial condition  $\psi(t_0) = 0$.
%\end{theorem} 
%
%\subsection{Existence of solutions and iteration techniques}
%
%\begin{lemma} Let $\phi$ be Lipschitz in $y$ uniformly in $t$ and 
%assume that $B(y_0,r) \subset V$ . Choose $M$ so that
%$|\phi(t, y)| \le M$ for $(t,y) \in I \times B(y_0,r)$. 
%Set $t_0 \in I$ and $\delta = r/M$ . Then there is a unique $C^1$ function
%$y(t)$, $t \in (t_0 - \delta, t_0 + \delta)$ satisfying $y(t) \in B(y_0,r)$ 
%that is a solution to the ordinary differential equation.
%\end{lemma}              
%
%The differential equation is equivalent to 
%\beann
%y(t) = y_0 +  \int_{t_0}^t \phi(t,y(s)) \, ds
%\eeann
%The {\it Picard iteration technique} 
%begins by seting $y_0(t) = y_0$ and defining inductively
%\beann
%y_{n+1}(t)=y_0 + \int_{t_0}^t \, \phi(t,y_n(s)) \, ds
%\eeann
%Let $K$ be the Lipschitz constant. 
%By induction, we find that
%\beann
% |y_{n+1}(t) - y_n(t)| \le {M K^n |t - t_0|^{n+1}  \over (n + 1)!}
%\eeann
%Thus, $y_n$ is a Cauchy sequence in the $L^\infty$ norm. 
%Consequently, the limit $y$ is a continuous curve
%that is a solution. To check uniqueness, let $\hat{y}$ 
%be another solution. Then check that 
%\beann
%|y_n(t) - \hat{y}(t)| \le {M K^n |t - t_0|^{n+1}  \over (n + 1)!}
%\eeann
%to see that $\hat{y}=y$.  
%

\end{document}
\end

{\bf MAKE THIS A PROBLEM ??????????}
In a metric space we defined continuity at a point. We can do this
in a general topological space as well. 
We define a {\bf 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.) 

