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\begin{document}

{\bf Page 82, Exercises 3.2} \hfill Solutions by Dave Lantz

\begin{enumerate}

\item[3.2.5.] Note first that there is a typo in this question:
Rather than $V_\epsilon(x)\cap A = \{a\}$, it should say
$V_\epsilon(a)\cap A = \{a\}$.  Also, the relevant definitions
here are Definitions~3.2.4 and 3.2.6, but as usual, I won't cite
them (and I won't demand that you cite them) explicitly --- I 
expect you to know what they say, but not what number our text
gives them.

Now suppose first that $a$ is an isolated point of $A$, i.e.,
it is in $A$ but not a limit point of $A$.  Because it is not a 
limit point, there is some $\epsilon$-neighborhood $V_\epsilon(a)$ 
that does not intersect $A$ in any other point than $a$ --- but 
that neighborhood \underbar{does} have $a$ in it (at its center),
so $V_\epsilon(a)\cap A = \{a\}$.

Conversely, suppose that there is an $\epsilon$-neighborhood 
$V_\epsilon(a)$ such that $V_\epsilon(a)\cap A = \{a\}$.  Then
$a$ is not a limit point of $A$, but we were told it was an 
element of $A$, so it is an isolated point of $A$.

\item[3.2.10.]\begin{itemize}
\item[(a)] For each $x$ in the universal set of which the 
$E_\lambda$'s are all subsets:
\begin{eqnarray*}
x\in \left(\bigcup_{\lambda\in\Lambda} E_\lambda\right)^c 
   &\iff& x\notin \bigcup_{\lambda\in\Lambda} E_\lambda \\
   &\iff& x\notin E_\lambda {\rm\ for\ all\ }\lambda\in\Lambda \\
   &\iff& x\in E_\lambda^c {\rm\ for\ all\ }\lambda\in\Lambda \\
   &\iff& x\in \bigcap_{\lambda\in\Lambda} E_\lambda^c\ , 
\end{eqnarray*}
so $\left(\bigcup_{\lambda\in\Lambda} E_\lambda\right)^c = 
\bigcap_{\lambda\in\Lambda} E_\lambda^c$.  Similarly, 
for each $x$ in the universal set, 
\begin{eqnarray*}
x\in \left(\bigcap_{\lambda\in\Lambda} E_\lambda\right)^c 
   &\iff& x\notin \bigcap_{\lambda\in\Lambda} E_\lambda \\
   &\iff& x\notin E_\lambda {\rm\ for\ at\ least\ one\ }
        \lambda\in\Lambda \\
   &\iff& x\in E_\lambda^c {\rm\ for\ at\ least\ one\ }
        \lambda\in\Lambda \\
   &\iff& x\in \bigcup_{\lambda\in\Lambda} E_\lambda^c\ , 
\end{eqnarray*}
so $\left(\bigcap_{\lambda\in\Lambda} E_\lambda\right)^c = 
\bigcup_{\lambda\in\Lambda} E_\lambda^c$.
\item[(b)] (i) Take a finite collection 
$\{E_\lambda\}_{\lambda\in\Lambda}$ of closed sets.  Then 
each $E_\lambda^c$ is an open set, so by Theorem~3.2.3(ii) 
their intersection is open.  But by De~Morgan's Law, that 
intersection, $\bigcap_{\lambda\in\Lambda}E_\lambda^c$, is
equal to $\left(\bigcup_{\lambda\in\Lambda}E_\lambda\right)^c$,
so by Theorem~3.2.13, the union 
$\bigcup_{\lambda\in\Lambda}E_\lambda$ is closed.

(ii) Take an arbitrary collection 
$\{E_\lambda\}_{\lambda\in\Lambda}$ of closed sets.  Then 
each $E_\lambda^c$ is an open set, so by Theorem~3.2.3(i) 
their union is open.  But by De~Morgan's Law, that 
union, $\bigcup_{\lambda\in\Lambda}E_\lambda^c$, is
equal to $\left(\bigcap_{\lambda\in\Lambda}E_\lambda\right)^c$,
so by Theorem~3.2.13, the intersection 
$\bigcap_{\lambda\in\Lambda}E_\lambda$ is closed.
\end{itemize}

\end{enumerate}

\vspace{.2truein}

{\bf Page 87, Exercises 3.3}

\begin{enumerate}

\item[3.3.2.] Suppose $K$ is a a closed and bounded subset of
$\R$, and let $(x_n)_{x\in\N}$ be a sequence in $K$; we must 
show that $(x_n)$ has a subsequence that converges to an element
of $K$:  Because
$K$ is bounded, so is the sequence $(x_n)$.  By the 
Bolzano-Weierstrass Theorem (Theorem~2..5), $(x_n)$ has a 
convergent subsequence $(y_n)$, still in $K$.  Because $K$ is
closed, the limit of $(y_n)$ is also in $K$.  Therefore, $K$
is compact.  

\item[3.3.10.] Suppose $E$ is a ``clompact'' subset of $\R$. 
All one-element sets are closed, so one cover of $E$ by closed
sets is $\{\{x\} : x\in E\}$.  Because $E$ is clompact, there is
a finite subset $F$ of $E$ for which $\{\{x\} : x\in F\}$ is a 
cover of $E$, i.e., the union of these one-element sets is $E$.
But it is also the finite set $F$, so $E = F$.  Thus, the only 
clompact subsets of $\R$ are the finite subsets.  

\end{enumerate}

\end{document}

FROM THIS POINT ON, NOTHING SHOWS UP IN THE .pdf FILE, SO I KEEP
SOME FREQUENTLY USED LaTeX-ISMS HERE TO COPY AND PASTE IN WHERE 
I NEED THEM:

\item[...] \begin{itemize}
\item[(a)]
\item[(b)]
\end{itemize}

\begin{eqnarray*}

   &=&  \\
   &\leq& \ .
\end{eqnarray*}

\newline\centerline{\includegraphics{.png}}

AND IF THERE ARE PROBLEMS FROM MORE THAN ONE SECTION IN THE BOOK:

\vspace{.2truein}

{\bf Page , Exercises }

\begin{enumerate}

\end{enumerate}