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\begin{document}
\begin{problems}

\item Do each of the following (4 points each).  
\begin{subproblem}
\item Define \emph{cauchy sequence}.
\vspace*{.1in}

A sequence $\{x_n\}_{n\in\N}$ in a metric space $X$ is said to be
Cauchy if for every $\epsilon>0$, there exists $N\in\N$ such that
$d(x_n,x_m)<\epsilon$ whenever $n,m\geq N$.
\vspace*{.25in}

\item Define \emph{convergent series}.
\vspace*{.1in}

A series $\sum_{n=0}^\infty a_n$ is said to be convergent if the 
corresponding sequence $\{s_n\}_{n\in\N}$ of partial sums
$$
s_n = a_1 + \dots + a_n
$$
converges.
\vspace*{.25in}

\item State the root test.
\vspace*{.1in}

Let $\sum_{n=0}^\infty a_n$ be a series of non-negative real numbers.
Let $L = \limsup |a_n|^{1/n}$.  Then
\begin{itemize}
\item if $L>1$, the series diverges;
\item if $L<1$, the series converges.
\end{itemize}
\vspace*{.25in}


\item Define \emph{compact set}.
\vspace*{.1in}

A set $E\subset X$ in a metric space $X$ is said to be compact if
for every open cover $\mathcal{U}$ of $E$, there exist finitely
many elements $U_1,\dots, U_n\in \mathcal{U}$ such that 
$E\subset U_1\cup\dots \cup U_n$.
\vspace*{.25in}

\item Define \emph{upper limit of a sequence}.
\vspace*{.1in}

Let $\{a_n\}$ be a sequence and $E$ be the set of all limits of
convergent subsequences of $\{a_n\}$.  Then $\limsup a_n = \sup E$.
\vspace*{.25in}

\end{subproblem}
\newpage

\item Five of the following ten assertions are false.  Identify them
and give counterexamples on the following page.  Note that you do not have to
justify your counterexample.  (5 points each)
\begin{subproblem}
\item A convergent sequence is bounded.  \emph{True}

\item A Cauchy sequence converges. \emph{False}

Let $X = \Q$ and $\{a_n\}\subset X$ be a sequence of positive
numbers with $a_n^2\to 2$.  Then $\{a_n\}$ is Cauchy but does not
converge.

\item If $\sum a_n$ converges then $\lim_{n\to\infty} a_n = 0$. \emph{True}

\item Every sequence in a compact set has a convergent subsequence.
	\emph{True}

\item If $A$ is a set and $B\subset A$ is a proper subset (i.e. $B\neq A$), 
then the cardinalities of $A$ and $B$ are not equal.
	\emph{False}

Let $A = \Z$ and $B = \N$, for instance.

\item If $U$ is an open set, then the interior of the closure of $U$
      is equal to $U$. \emph{False}

Let $U = (0,1)\cup (1,2)\subset\R$.


\item If $\sum_{n=0}^\infty a_n$ converges, and $\{a_{n_j}\}_{j\in\N}$ is a 
	subsequence, then $\sum_{j=0}^\infty a_{n_j}$ converges.
	\emph{True}

\item Let $\{a_n\}$ be a sequence in a metric space $X$.  Let 
      $E$ be the set of all limits of subsequences of $\{a_n\}_{n\in\N}$.
	Then $E$ is closed.
	\emph{True}

\item Let $\{a_n\}_{n\in\N}\subset\R$ be a sequence
      such that $\lim_{n\to\infty} |a_{n+1} - a_n|$ converges.
	Then $\{a_n\}$ converges. \emph{False}

Let $a_n = 1/1 + \dots 1/n$.
	
\item The product of two irrational numbers is irrational.
	\emph{False}

For example, $\sqrt{2}\sqrt{2} = 2$.

\end{subproblem}
\newpage

\item Do three of the following four problems. If you turn in solutions to
all four, I will simply grade the first three.  (10 points each)
\begin{subproblem}
\item Let $\{a_n\}_{n\in\N}\subset\R$ be a sequence such that
$\sum |a_n|$ converges.  Show that $\sum a_n$ converges.
\vspace*{.2in}

\proof
Let $s_n = a_1 + \dots + a_n$.  Then it is enough to show that
$\{s_n\}$ is Cauchy.  On the other hand, if 
$$
t_n = |a_1| + \dots + |a_n|,
$$
then $\{t_n\}$ is Cauchy by hypothesis.  So let $\epsilon > 0$ be
given, and choose $N\in\N$ such that $|t_n - t_m| < \epsilon$
for all $n,m\geq \N$.  Then if $m \geq n\geq \N$, we also have
$$
|s_m - s_n| = |a_{n+1} +\dots a_m|
	\leq |a_{n+1}| + \dots |a_m| 
	= |t_m - t_n| < \epsilon.
$$
Hence $\{s_n\}$ is Cauchy, and we're on to the next problem.\qed
\vspace*{.2in}

\item Let $X$ be a metric space and $E\subset X$.  Prove that
the interior of $E$ is open.
\vspace*{.1in}

\proof
Let $U$ be the interior of $E$.  Given a point $p\in U$, we must show that
$p$ is an interior point of $U$.  By definition of $U$, we know that there
is $r>0$ such that $N_r(p)\subset E$.  But then if $q\in N_r(p)$, we
also have
$$
N_{r-d(p,q)}(q) \subset N_r(p) \subset E.
$$
Hence $q\in U$, too.  But $q\in N_r(p)$ was arbitrary, so $N_r(p)\subset U$.
We conclude that $p$ is an interior point of $U$, as desired.
\qed
\vspace*{.2in}

\item Prove that there is no $x\in\Q$ such that $x^3 = 24$.
\vspace*{.1in}

\proof Suppose there is such an $x$.  We can write $x = p/q$ where
$p$ and $q$ have no common divisors.  Then $p^3 = 24 q^3$. In particular,
all factors of 24 divide $p^3$.  Therefore, the prime factors, such as
3 divide $p$ itself.  So we can write $p = 3k$ for some $k\in\N$.  This
gives us that $9 k^3 = 8 q^3$.  Now we see that $3$ divides $q^3$ 
and therefore also $q$.  We conclude that 3 is a common factor of $p$
and $q$, contradicting our initial assumption that $p$ and $q$ had no
common divisors.  This shows that $x$ cannot exist.
\qed
\vspace*{.2in}

\item Prove that a closed subset of a compact set is compact.
\vspace*{.1in}

\proof
Let $E$ be a closed subset of a compact set $K$ in a metric space $X$.
Let $\mathcal{U}$ be an open cover of $E$.  Then 
$\mathcal{U}\cup\{X-E\}$ is an open cover of $K$.  So by hypothesis,
we have elements $U_1,\dots, U_n\subset\mathcal{U}$ such that
$$
K\subset U_1\cup\dots \cup U_n\cup X-E.
$$
It follows immediately that $X\subset U_1\cup \dots\cup U_n$.  Therefore,
$X$ is compact, and this proof is closed.  Get it, \emph{closed}?  
\qed
\end{subproblem}

\newpage
\item Take home problem(s), due Monday 10/14 by class time.  Do one of 
two---note that the second problem is both longer and worth more points 
than the first one.  If you turn in solutions to both, I will grade only 
the first.  The only resources you are to use in solving these problems are 
your textbook and yourself.  I will answer questions about the problems only 
insofar as they clarify what is written here.

\begin{subproblem}
\item Let $\{a_n\}\subset X$ be a sequence in a complete metric
space such that $\sum_{n=0}^\infty d(a_{n+1},a_n)$ converges.
Show that $\{a_n\}$ converges. (15 points)

\vspace*{.1in}
\proof
Since $X$ is complete it is enough to show that $\{a_n\}$ is Cauchy.
Let $s_n = d(a_2,a_1) + \dots d(a_{n+1},a_n)$ be the partial
sums of the series above.  Then we know that $\{s_n\}$ is Cauchy
by hypothesis.  Given $\epsilon > 0$ choose $N\in\N$ such that
$$
|s_n - s_m| < \epsilon
$$
for all $n,m\geq N$.  Then if $m\geq n\geq N+1$ we have from the triangle
inequality that
$$
d(a_n,a_m) \leq d(a_{n+1},a_n) + \dots d(a_m,a_{m-1})
	= |s_{n-1} - s_m| < \epsilon.
$$
So $\{a_n\}$ is Cauchy, like we wanted it to be.
\qed
\vspace*{.25in}


\item Let $U\subset\R$ be an open set.  Complete the following outline to
show that $U$ is a finite or countable union of mutually disjoint intervals.
(20 points total)
\begin{itemize}
\item For $x,y\in U$, let us say that $x\sim y$ if and only if there is an 
      open interval $(a,b)\subset U$ containing both $x$ and $y$.  Show that 
      $\sim$ is an equivalence relation.
\vspace*{.1in}

\proof
Fix $x\in U$.  Since $U$ is open, we have $r>0$ such that
$N_r(x) = (x-r,x+r) \subset U$.  Therefore $x\sim x$, and $\sim$
is reflexive.  Moreover, if $x\sim y$, then there exists $(a,b)\subset U$
containing both $x$ and $y$, so $y\sim x$, too.  Therefore $\sim$ is
symmetric.  Finally, if $x\sim y$ and $y\sim z$, then we have open
intervals $(a,b),(c,d)\subset U$ such that $x,y\in (a,b)$ and
$y,z \in (c,d)$  In particular, $(a,b)\cap (c,d) \neq \emptyset$,
so $(a,b)\cup (c,d)$ is again an open interval.  Therefore, $x\sim z$,
and $\sim$ is transitive.
\qed
\vspace*{.2in}

\item For each $x\in U$, let $U(x)$ be the equivalence class of $x$.
      Show that $U(x)$ is an open interval.
\vspace*{.1in}

\proof
Fix $x\in U$ and consider any point $y\in U(x)$.  Then 
$x,y\in (a,b)\subset U$ for some $(a,b)$.  This implies that
$(a,b)\subset U(x)$.  So $y$ is an interior point of $U(x)$, and $U(x)$
is open.  

Suppose now that $y,z\in U(x)$ and $y<t<z$.  Then by transitivity, $y\sim z$.
That is, there is an interval $(a,b)\in U$ such that $y,z\in (a,b)$.
Again by transitivity, we see that $(a,b)\subset U(x)$.  In particular
$t\in U(x)$.  So $U(x)$ is an interval.
\qed
\vspace*{.2in}

\item Show that there are at most countably many distinct 
      equivalence classes $U(x)\subset U$.
\vspace*{.1in}

\proof
Fix $x\in X$.  Since $U(x)$ is open, we can choose a point
$q = q(x) \in \Q\cap U(x)$.  Now if $y\in X$ is another point,
we have either $U(x) = U(y)$, in which case, we can take $q(y) = q(x)$;
or $U(x)\cap U(y) = \emptyset$, in which case, we definitely ahve
$q(x)\neq q(y)$.

So if we define $f(U(x)) = q(x)$, we see that $f$ is an injective function
from the set of equivalence classes of $\sim$ into $\Q$.  In particular,
the cardinality of the set of equivalence classes is the same as that
of some subset of $\Q$, so it is therefore finite or countable.
\qed
\vspace*{.2in}

\item Conclude.  Then draw a little filled in square at the end of your
      proof, taking scrupulous care to ensure that all four sides have
      the same length and that all four angles measure $\pi/2$. 

\vspace*{.1in}
There's not really much left to say.  The equivalence classes of $\sim$
form a partition of $U$.  As there are countably many of them, and
each is an open interval, we have proven that $U$ is a union of 
finitely or countably many mutually disjoint open intervals.
\qed
\end{itemize}
\end{subproblem}
\end{problems}

\end{document}
\end
