%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\documentclass[12pt]{article}
\usepackage{amssymb}
\topmargin-2cm
\textheight 25cm
\textwidth17cm
\oddsidemargin0cm
\evensidemargin0cm
\pagestyle{empty}
\parindent0cm
\parskip 2.5ex
\newcommand{\re}{\rm Re}
\newcommand{\im}{\rm Im}
\newcommand{\C}{\Bbb C}
\begin{document}
{\bf\large Math 425}\hfill{\bf \large Fall 1997}

{\bf Example for Section 1.3:}

Let
\[
   D=\{z\in \C\,|\, \re(z)\leq 6, \im(z)>0\}.
\]
What are the interior points and the boundary points of $D$?

a) Let $w_0=x+iy\in D$ with $x<6$ and $y>0$.
\\[1ex]
To prove that $w_0$ is an interior point of $D$, we have to find a disk
\\
 with center $w_0$ lying entirely in $D$.
\\
To do so, let $r>0$ so that $6-x>r$ and $y>r$.
\\
\underline{Claim:} The disk $\hat{D}\{ z\in\C\,|\, |z-w_0|<r\}$ is contained in
$D$.
\\
\underline{Proof:}
Choose $z\in\hat{D}$, then we calculate
\[
  \re(z)=\re(z-w_0+w_0)=\re(z-w_0)+\re(w_0)<|z-w_0| + 6-r < r +6-r=6
\]
and
\[
   \im(z)=\im(z-w_0)+\im(w_0)>-|z-w_0| +y > -r+r =0.
\]
Therefore, $z\in D$ and thus $\hat{D}\subseteq D$.
Thus we have proven, that $w_0$ is an interior point of $D$.

b) Let $z_0=6+iy\in D$ with $y>0$.
\\
To prove that $z_0$ is a boundary point of $D$, we have to show, that each disk
centered at $z_0$ contains points of $D$ and of $\C\backslash D$.
\\
So, let $\hat{D}=\{z\in\C\,|\, |z-z_0|<\epsilon\}$ be a disk centered at $z_0$.
But then we simply get
\[
\begin{array}{ll}
       6+\frac{\epsilon}{2} +iy &\in \hat{D}\backslash D \\[.7ex]
       6-\frac{\epsilon}{2} +iy &\in \hat{D}\cap D,
\end{array}
\]
which means, that $\hat{D}$ contains points of $D$ as well as of the
complement of $D$.
Hence $z_0$ is a boundary point of $D$.

c) In the same way as in b) one can see that the points $z_1=x$ with $x\leq 6$
are boundary points of $D$, too.

As a result:
\\[1ex]
\mbox{}\hspace*{1.5cm}\begin{tabular}{rcl}
   interior of $D$ &=&$\{z\in\C\,|\, \re(z)<6,\,\im(z)>0\}$\\[1ex]
   boundary of $D$ &=&$\{z\in\C\,|\, \re(z)=6\;\mbox{\rm and }
\im(z)\geq0\;\mbox{\rm or }
                                       \re(z)\leq6\;\mbox{\rm and } \im(z)=0\}$
\end{tabular}

By b) not every point of $D$ is an interior point, thus $D$ is not open.
By c) there are boundary points not contained in $D$, therefore $D$ is not
closed.

Note that the interior of $D$ is an open set, and
$\{z\in\C\,|\, \re(z)\leq6, \im(z)\geq0\}$ is a cloed set.

\end{document}
