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\section{Basic Concepts}

\subsection{Limit, Supremum, Infimum}


\begin{definition}
$x$ is the \emph{limit} if $x_n$ as $n\to\infty$, $\lim_{n\to\infty} x_n
= x$, if $\forall\, 1/m$, $\exists\, N$ such that $n \ge N \Rightarrow
|x - x_n| < 1/m$.
\end{definition}

\begin{definition}
If $E\subset \R$ is bounded above, then there exists a unique real number,
$\sup E$, called the \emph{supremum} of $E$, such that
\begin{enumerate}
\item $\sup E$ is an upper bound of $E$.
\item If $y$ is any upper bound of $E$, then $ y\ge \sup E$.
\end{enumerate}
\end{definition}

\begin{definition}
If $E\subset \R$ is bounded below, then there exists a unique real number,
$\inf E$, called the \emph{infimum} of $E$, such that
\begin{enumerate}
\item $\inf E$ is a lower bound of $E$.
\item If $y$ is any lower bound of $E$, then $ y\le \sup E$.
\end{enumerate}
\end{definition}

\begin{definition}
$\{y_n\}$ is a \emph{subsequence} of $\{x_n\}$ if
$\exists\, m:\N\to\N$ such that $y_n = x_{m(n)}$, $\forall\, n$.
\end{definition}

We sometimes denote a subsequence by $\{x_{n'}\}$ where $n'$ stands
for~$m(n)$.

\begin{definition}
$x$ is a \emph{limit point} of $\{x_n\}$ if there exists a subsequence
$\{x_{n'}\}$ of $\{x_n\}$ such that $\lim_{n'\to\infty}x_{n'} = x$.
\end{definition}

\emph{Example:} $x_n =
\left\{\begin{array}{cl}
 1  & n\ \mbox{even}\\
1/n & n\ \mbox{odd}
\end{array}\right.$

0 and 1 are limit points, but neither is the limit of $x_n$.

\emph{Note:} A convergent sequence has only one limit point.

\begin{definition}
The \emph{limsup} of a sequence $\{x_n\}$ is the limit of the sequence
$y_k = \sup_{n\ge k}\{x_n\}$,
\[ \limsup\{x_n\} = \lim_{k\to \infty} y_k = \lim_{k\to\infty}\sup_{n\ge
k}\{x_n\} \]
\end{definition}

\begin{definition}
The \emph{liminf} of a sequence $\{x_n\}$ is the limit of the sequence
$z_k = \inf_{n\ge k}\{x_n\}$,
\[ \liminf\{x_n\} = \lim_{k\to \infty} z_k = \lim_{k\to\infty}\inf_{n\ge
k}\{x_n\} \]
\end{definition}

The sequence $y_k$ is decreasing, $y_{k+1} \le y_k$, and the sequence
$z_k$ is increasing, $z_{k+1}\ge z_k$. Therefore, if we allow limits to
be $\pm\infty$, $\limsup\{x_n\}$ and $\liminf\{x_n\}$ always exist. Note
also that $\limsup\{x_n\}$ is the \emph{maximum} of all limit points,
while $\liminf\{x_n\}$ is the \emph{minimum} of all limit points.

\emph{Example:} As above, $\limsup \{x_n\} = 1$,
$\liminf \{x_n\} = 0$.

\subsection{Open Sets}

\begin{definition}
$A\subset\R$ is \emph{open} if $\forall\, x \in A$, there is an interval
$I_x = (a,b)$ such that $x \in I_x \subset A$.
\end{definition}

\emph{Example:} $A = \bigcup_{n=1}^\infty
\big(\frac{1}{n+1},\frac{1}{n}\big)$

\begin{theorem}\label{01:OPENSETS}
$A \subset \R$ is open $\iff$ $A$ is a countable disjoint union of intervals.
\end{theorem}

\proof ($\Rightarrow$) If $A$ is open there is an interval $I_x\subset A$
around each of its points $x\in A$. Thus $A = \bigcup_{x\in A} I_x$.

If two intervals $I_x = (a,d)$, $I_y = (c,d)$ are not disjoint, then
either $I_x\cup I_y  = (a,d)$ or $I_x \cup I_y = (a,b)$ so $I_x\cup
I_y$ can be combined into one interval. Therefore, $A$ is the union of
\emph{disjoint} open intervals.

How many? Each disjoint interval contains a (different) rational number,
and $\Q$ is countable, so at most a \emph{countable number}.

The converse ($\Leftarrow$) is obvious.
\qed

\begin{theorem}\label{01:OPEN}
\begin{enumerate}
\item The union of any number of open sets is open.
\item The intersection of a finite number of open sets is open.
\end{enumerate}
\end{theorem}

\proof  1. Let $x\in\bigcup U_\alpha$ where each $U_\alpha$ is
open.  Then $x \in U_\alpha$ for some $\alpha$ so there is an interval
$(a,b)$ in $U_\alpha$ containing $x$. Then $x\in(a,b)\subset\bigcup
U_\alpha$, proving that the union is open.

2. Let $x\in U_1\cap\dots\cap U_n$ where each $U_i$ is open. For each
$i$ there is an interval $(a_i,b_i)$ containing $x$ so $x \in
(a_1,b_1)\cap\dots\cap(a_n,b_n)\subset U_1\cap\dots\cap U_n$.
This proves the intersection $U_1\cap\dots\cap U_n$ is open since
$(a_1,b_1)\cap\dots\cap(a_n,b_n) = (a,b)$ where $a = \max\{a_i\}$ and
$b = \min\{b_i\}$.
\qed

\emph{Note:} $\bigcap_{n=1}^\infty \big(-\frac{1}{n},\frac{1}{n}\big) =
\{0\}$, so the assumption of finite is necessary in~2.

\subsection{Closed Sets}

\begin{definition} We say that $x$ is a \emph{limit point} of a set
$A\subset\R$ if $\exists\, \{x_n\}\subset A$ such that $x = \lim x_n$.
\end{definition}

Note that an equivalent condition for $x$ to be a limit point of $A$ is
that every open set containing $x$ contains a point of $A$.

\begin{definition}
A set $A\subset\R$ is \emph{closed} if it contains all its limit points.
\end{definition}

\emph{Examples:} $[a,b]$, $(-\infty,\infty)$ (also open!), $\emptyset$, any
finite set.

\begin{theorem}\label{01:COMPLEMENTS}
$A$ is closed $\iff$ $A^c$ is open.
\end{theorem}

\proof ($\Rightarrow$) Let $y\in A^c$. Then $y$ is not a limit point
of $A$ (closed). So there is an open interval $I_y$, $y\in I_y \subset
A^c$ (otherwise every open interval containing $y$ intersects $A$ which
implies $y$ is a limit point of $A$). $\Rightarrow$ $A^c$ is open.

($\Leftarrow$) Let $x$ be a limit point of $A$. If $x\in
A^c$ then $\exists\, I_x$ such that $x\in I_x\subset A^c$
(open). Contradiction! Therefore, $x\in A$.  \qed

Using the above characterization of open sets, we see that a closed set is
the complement of a countable union of disjoint open sets.

\begin{theorem}
\begin{enumerate}\label{01:CLOSED}
\item The union of a finite number of closed sets is closed.
\item The intersection of any number of closed sets is closed.
\end{enumerate}
\end{theorem}

\proof These statements follow from
Theorems~\ref{01:OPEN},~\ref{01:COMPLEMENTS}, and the following
identities:
\[
\big(U_1\cup\dots\cup U_n)^c = U_1^c\cap\dots\cap U_n^c \mbox{\ and\ }
\big(\bigcap U_\alpha\big)^c = \bigcup U_\alpha^c
\]
\qed

\emph{Note:} $\bigcup_{n=1}^\infty \big[-1+\frac{1}{n},1-\frac{1}{n}\big] =
(-1,1)$, so the assumption of finite is necessary in 1.

\emph{Example (Cantor Set):} Remove the middle third, $(1/3,2/3)$,
from $[0,1]$ to get $[0,1/3]\cup[2/3,1]$. Next remove the
middle third from the intervals $[0,1/3]$ and $[2/3,1]$ to get
$[0,1/9]\cup[2/9,1/3]\cup[2/3,2/9]\cup[4/9,1]$. Repeat this process
recursively with each closed subinterval. The resulting set $C$ is called
the \emph{Cantor Set}. It contains all of its limit points (in fact,
every point is a limit point), but $C$ contains no intervals! It is also
uncountable.

An equivalent way to define $C$ is the set of numbers in $[0,1]$ whose
ternary expansion contains only 0's and 2's.


\begin{definition}
Let $A\subset \R$. The \emph{closure} of $A$, $\overline{A}$, is the
union of $A$ and all the limit points of~$A$.
\end{definition}

\emph{Note:} It is not hard to prove that $\overline{A}$ is the smallest
closed set containing~$A$ (or, equivalently, that $\overline{A}$ is the
intersection of all closed sets containing $A$). In fact, this statement
could be used as the definition of $\overline{A}$.

\begin{definition}
A subset $B\subset A$ is \emph{dense} in $A$, if $A\subset \overline{B}$.
\end{definition}

\emph{Note:} $B$ is dense in $A$ if every point of $A$ is the limit of a
sequence of points in $B$.

\emph{Examples:} 1) $\Q$ is dense in $\R$.

2) $(a,b)$ is dense in $[a,b]$.

3) $(0,1)\setminus\{1/n\ |\ n\in\N\}$ is dense in $(0,1)$.

\subsection{Compact Sets}

\begin{definition}
A set $A\subset\R$ is \emph{compact} if every open cover of A has a finite
subcover:
\[
A\subset \bigcup_{x\in C} U_\alpha \Rightarrow
A\subset U_{\alpha_1}\cup\cdots\cup U_{\alpha_N}
\mbox{\ for some\ } \alpha_1,\dots,\alpha_N\in C
\]
Here $C$ is an indexing set and the $U_\alpha$ are open sets.
\end{definition}

\begin{theorem}\label{01:COMPACTNESS}
The following are equivalent:
\begin{enumerate}
\item $A$ is compact
\item $A$ is closed and bounded
\item Every sequence in $A$ has a limit point in $A$.
\end{enumerate}
\end{theorem}

\proof $1\Rightarrow2$: Let $y$ be a limit point of $A$ and suppose
$y\notin A$. The complements of $[y-1/n,y-1/n]$ are open and cover
$\R\setminus\{y\} \supset A$. Since $A$ is compact, a finite subset
of them cover $A$. But the sets are nested, so this means $\exists\, N$
such that $A$ is contained in the complement of $[y-1/N,y+1/N]$. But
this implies that $y$ is not a limit point of $A$, a contradiction.
Therefore, $y\in A$, proving that $A$ is closed.  To prove that $A$ is
bounded, consider the open cover $\{(x-1,x+1)\ |\ x\in A\}$. Since $A$
is compact, $A\subset(x_1-1,x_1+1)\cup\cdots\cup(x_n-1,x_n+1)$ for some
$x_1,\dots,x_n\in A$. So $A$ is bounded above by $\max\{x_i+1\}$ and below by
$\min\{x_i-1\}$.

$2\Rightarrow3$: Let $\{x_n\}\subset A$. Since $A$ is bounded, $y =
\limsup \{x_n\}$ exists and is a limit point of $\{x_n\}$ and $A$. Since
$A$ is closed $y\in A$.

$3\Rightarrow1$: Let $\U = \{U_\alpha\}$ be an open cover of
$A\subset\bigcup U_\alpha$. First we find a countable subset of $\U$
that still covers $A$ as follows.  Let $I_j$, $j = 1,2,3,\dots$, be the
countable collection of intervals that have rational endpoints. For
$j=1,2,3,\dots$ choose one $U_\beta\in\U$ that contains $I_j$, if
any. Let $\U' = \{U_\beta\} \subset \U$ be the resulting countable
subcollection. To see that $\U'$ still covers $A$, note that $x\in A
\Rightarrow x\in U_\alpha\in\U$ for some $\alpha$. Since $U_\alpha$
is open, it contains an interval around $x$, say $x\in(a,b)\subset
U_\alpha$. By shrinking this interval, if necessary, we may assume
$a,b\in\Q$ and hence $(a,b) = I_j$ for some $j$. Therefore $\exists\,
U_\beta\in\U'$ such that $x\in I_j\subset U_\beta$ and hence $\U'$
covers~$A$.

We now have a countable subcover, say, $U_1, U_2,\dots$. If we take $n$
large enough, then $U_1, U_2, \dots, U_n$ must already cover~$A$. Suppose
not. Then for each $n$, $\exists\, x_n\in A$ that is not contained in
$U_1,\dots,U_n$. By assumption, the sequence $\{x_n\}$ has a limit
point $x\in A$. Thus $x\in U_k$ for some~$k$. But by construction, $U_k$
does not contain $x_k, x_{k+1},\dots$ contradicting the fact that any
neighborhood of a limit point must contain an infinite number of points
in the sequence.
\qed

\emph{Example:} Let $a,b\in\R$. Then $[a,b]$ is closed and bounded, and
so is compact.  On the other hand, $(a,b)$ is bounded but not closed, so it
is not compact. The intervals, $(-\infty, b]$ and $[a,\infty)$ are closed,
but not bounded, so they are not compact.

\subsection{Cauchy Sequences}

It is useful to have a criterion for convergence that does not explicitly
involve the limit of the sequence.

\begin{definition}
A sequence $\{x_n\}$ is a \emph{Cauchy sequence} if $\forall\, 1/m$,
$\exists\, N$ such that $|x_i - x_j| < 1/m$ whenever $i, j \ge N$.
\end{definition}

\begin{theorem}
A sequence converges in the usual sense $\iff$ it is Cauchy.
\end{theorem}

\proof ($\Rightarrow$) Assume $x_n\to x$: Given $1/m$, $\exists\, N$
such that if $n \ge N$, then $|x_n - x| < 1/(2m)$. Then
$|x_i - x_j| \le |x_i - x| + |x - x_j| < 1/(2m) + 1/(2m) = 1/m$ whenever
$i, j\ge N$, so $\{x_n\}$ is Cauchy.

($\Leftarrow$) Assume $\{x_n\}$ is Cauchy. Given $1/m$, $\exists\, N$
such that $|x_i - x_j|<1/m$ whenever $i,j\ge N$. Since
\[ |x_i| = |x_i- x_N + x_N| \le |x_N| + |x_i - x_N| < |x_N| + \frac{1}{m}
\quad \forall\, i\ge N\]
we see that $\{x_n\}$ is bounded. Therefore, $s = \limsup\{x_n\}$
and $t = \liminf\{x_n\}$ are bounded. By definition of $\limsup$ and
$\liminf$, $\exists\, i, j \ge N$ such that $s - x_i < 1/m$ and $x_j -
t < 1/m$. But then
\[ 0 \le s - t = (s - x_i) + (x_j - t) + (x_i - x_j) \le
\frac{1}{m} + \frac{1}{m} + \frac{1}{m} = \frac{3}{m} \]
since $m$ is arbitrary, we must have $s = t = \lim x_n$.
\qed

\emph{Example:}  Let $x_n = \log(n)$. Even though $|x_{n+1} - x_n| =
\log(1 + 1/n) \to 0$, the sequence is \emph{not} Cauchy, because $|x_i -
x_j|$ must be small \emph{for all} $i$, $j$ large enough.  In this case,
$|x_i - x_j| = |\log(i/j)|$ can easily approach $\infty$ for large values
of $i$.

\subsection{Continuity}

\begin{definition}
A function $f$ is \emph{continuous} at $x$ if it is defined in an interval
around $x$ and if $\forall\, 1/m$, $\exists\, 1/n$ (that may depend on $1/m$
and $x$) such that $|f(x) - f(y)| < 1/m$ whenever $|x - y| < 1/n$. We
say $f$ is continuous if is continuous at each point in its domain.
\end{definition}

By changing the condition $|x - y|< 1/n$ to $0 \le x - y < 1/n$ in the
above definition we get the definition for \emph{continuous from the left}
at $x$. Similarly, changing $|x-y|<1/n$ to $0 \le y - x < 1/n$ we get the
definition for \emph{continuous from the right} at $x$.  We say that $f$
is continuous on a closed interval $[a,b]$ if $f$ is continuous on $(a,b)$,
continuous from the right at $a$, and continuous from the left at $b$.

\begin{proposition}\label{01:SEQUENCES}
$f$ is continuous at $x$ if and only if for all sequences $x_j\to x$
we have $f(x_j) \to f(x)$.
\end{proposition}

\proof ($\Rightarrow$) Given $1/m$, $\exists\, 1/n$ such that if $|x-y|<1/n$
then $|f(x)-f(y)|<1/m$. Also $\exists\, N$ such that if $j\ge N$ then
$|x-x_j|<1/n$. Putting these together gives $|f(x)-f(x_j)|<1/m$ if
$j\ge N$.

($\Leftarrow$) Suppose $f$ is not continuous at~$x$. Then, $\exists\, 1/m$
such that $\forall\, 1/n$ $\exists\, x_n$ satisfying $|x_n - x|<1/n$ and
$|f(x_n) - f(x)| \ge 1/m$. But $x_n \to x$, so by assumption,
$|f(x_n)-f(x)|\to 0$, a contradiction. So, $f$ must be continuous at~$x$.
\qed

\begin{definition}
A function $f$ is \emph{uniformly continuous} if $\forall\, 1/m$,
$\exists\, 1/n$ (that depends only on $1/m$) such that $|f(x) - f(y)|
< 1/m$ whenever $|x - y| < 1/n$.
\end{definition}

\begin{theorem}\label{01:UNIFORMCONTINUITY}
If $f$ is continuous on $[a,b]$, then $f$ is uniformly continuous.
\end{theorem}

\proof Suppose not. Then $\exists\, 1/m$ such that $\forall\, 1/n$,
$\exists\, x_n, y_n$ satisfying $|x_n - y_n| < 1/n$ but $|f(x_n) - f(y_n)|
\ge 1/m$. Since $[a,b]$ is compact, there exist convergent subsequences
$x_{n'} \to x_0$ and $y_{n'} \to y_0$. But $|x_{n'} - y_{n'}| < 1/n'$,
for an infinite number of $n'\to\infty$, so $x_0 = y_0$.  Since $f$
is continuous, $f(x_{n'}) \to f(x_0)$ and $f(y_{n'}) \to f(y_0)$, and
hence $|f(x_{n'}) - f(y_{n'})|\to 0$, a contradiction!  \qed

\begin{theorem}\label{01:CONTINUOUS}
$f$ is continuous $\iff$ $f^{-1}(A)$ is open for all open sets $A$.
\end{theorem}

\proof  ($\Rightarrow$) Let $x\in f^{-1}(A)$.  Since $A$ is open, there
is an interval contained in $A$ around the point $f(x)\in A$. Thus, by
choosing $1/m$ small enough, we may assume $(f(x)-1/m,f(x)+1/m)\subset
A$. By the definition of continuity, $\exists\, 1/n$ such that
$|f(x) - f(y)|<1/m$ whenever $|x-y|<1/n$. In other words, if
$y\in(x-1/n,x+1/n)$ then $f(y)\in (f(x)-1/m,f(x)+1/m)\subset A$. This
shows $(x-1/n,x+1/n)\subset f^{-1}(A)$ and hence that $f^{-1}(A)$ is open.

($\Leftarrow$) Fix $x$ in the domain of $f$. For any $1/m$, the set $A =
(f(x)-1/m,f(x)+1/m)$ is open, so by assumption, $f^{-1}(A)$ is open.
Therefore, $f^{-1}(A)$ contains an interval around $x\in f^{-1}(A)$. Hence
for $1/n$ small enough, $(x-1/n,x+1/n)\subset f^{-1}(A)$. This shows
that if $|x-y|<1/n$ (i.e., $y\in(x-1/n,x+1/n)$), then $|f(x)-f(y)|<1/m$
(i.e., $f(y)\in A$), proving that $f$ is continuous at~$x$.  \qed

\begin{theorem}\label{01:COMPACT}
If $f$ is continuous and $A$ is compact, then $f(A)$ is compact.
\end{theorem}

\proof
Let $f(A) \subset \bigcup U_\alpha$ be an open cover of
$f(A)$. Then $\bigcup f^{-1}(U_\alpha)$ is an open cover of $A$
by Theorem~\ref{01:CONTINUOUS}. Since $A$ is compact, there exists
a finite subcover, $A \subset f^{-1}(U_{\alpha_1})\cup\dots\cup
f^{-1}(U_{\alpha_n})$. But then $f(A) \subset U_{\alpha_1}\cup\dots\cup
U_{\alpha_n}$ is a finite subcover of~$f(A)$ proving that $f(A)$
is compact. \qed

\begin{theorem}[Extreme Value Theorem]
If $f$ is continuous on $[a,b]$, then it has a maximum and a minimum on
that interval.
\end{theorem}

\proof By Theorem~\ref{01:COMPACT}, $f([a,b])$ is compact, and hence closed
and bounded by Theorem~\ref{01:COMPACTNESS}. Therefore, $s = \sup\{f(x)\
|\ x\in[a,b]\}$ and $t = \inf\{f(x)\ |\ x\in[a,b]\}$ exist, and are finite
limit points of $f([a,b])$. Consequently, there are sequences $\{x_n\},
\{y_n\}\subset [a,b]$ such that $f(x_n)\to s$ and $f(y_n)\to t$. Since
$[a,b]$ is compact, there exists convergent subsequences $x_{n'}\to x_0$
and $y_{n'}\to y_0$. By Proposition~\ref{01:SEQUENCES}, $s = \lim f(x_n)
= f(x_0)$ and $ t = \lim f(y_n) = f(y_0)$. Therefore, $f(x_0)$ is a
maximum of $f$ and $f(y_0)$ is a minimum of $f$ on~$[a,b]$.  \qed

\begin{theorem}[Intermediate Value Theorem]
Let $f$ be continuous on $[a,b]$ and suppose $y$ is a number between $f(a)$
and $f(b)$. Then $\exists, x\in[a,b]$ such that $f(x) = y$.
\end{theorem}

\proof We may assume without loss of generality that $f(a) < f(b)$.
We repeatedly bisect the interval $[a,b]$ as follows.  Let $c = (a+b)/2$. If
$f(c) \le y$, let $[a_1,b_1] = [c,b]$ and if $f(c) > y$, let $[a_1,b_1]
= [a,c]$, so that $f(a_1) \le y < f(b_1)$. Iterating this process
we get a sequence of intervals $[a_n,b_n]$ satisfying $f(a_n)\le
y < f(b_n)$. Note that the length of these interval is shrinking
geometrically:
\[ b_n - a_n = (b_{n-1} - a_{n-1})/2 = \dots = (b - a)/2^n \]
It follows that both sequences $\{a_n\}$ and $\{b_n\}$ are Cauchy and
converge to a common limit $x\in[a,b]$. Therefore,
\[ f(x) = \lim f(a_n) \le y \le \lim f(b_n) = f(x) \]
and $f(x) = y$ as claimed. \qed

\emph{Note:} The previous two theorems immediately imply that if $f$ is
continuous on $[a,b]$ then $f([a,b]) = [c,d]$ where $c$ is the maximum of
$f$ and $d$ is the minimum of $f$.


 \end{document}