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\centerline{Mathematics 242 -- Principles of Analysis}
\centerline{Problem Set 5}
\centerline{{\bf Due:} March 15, 2013}
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\noindent
{\it `A' Section}
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\noindent
1.  For each of the following sequences, determine three different subsequences,
each converging to a different limit.  For each one, express your three subsequences
as $x_{n_k}$ for a suitably chosen (strictly increasing) index sequence $n_k$, and give an explicit formula
for $n_k$ as a function of $k$:
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\item{(a)}  $x_n = \cos\left({\pi\sqrt{n}\over 2}\right)$
\item{(b)}  $x_n = {n\over 3} - \left[{n\over 3}\right]$  (as usual, $[\ ]$ denotes the greatest integer function)
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\noindent
2.  Let $x_n = n^{1/4}$.  For each of the following sequences, either express that sequence
as a subsequence of
the sequence $x_n$ for some explicit (strictly increasing) index sequence $n_k$, or say why that is impossible:
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\item{(a)}  $\{2,3,4,5,\ldots\}$
\item{(b)}  $\{\sqrt{3},\sqrt{6},\sqrt{9},\sqrt{12},\ldots\}$
\item{(c)}  $\{1,2,4,8,16,32,\ldots\}$
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\noindent
3.  Let $x_n = \sin\left({n\pi\over 4}\right)$ and $y_n = \cos\left({n\pi\over 4}\right)$.
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\item{(a)}  Find a (strictly increasing) index sequence $n_k$ such that both $x_{n_k}$ and $y_{n_k}$ 
converge.
\item{(b)}  Find a second (strictly increasing) index sequence $n_k$ such that both $x_{n_k}$ and $y_{n_k}$ 
diverge.
\item{(c)}  Find a third (strictly increasing) index sequence $n_k$ such that one of $x_{n_k}$ and $y_{n_k}$ 
converges and the other diverges. 
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\noindent
{\it `B' Section}
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\noindent
1.  (True or False) -- If the statement is true give a proof; if it is false give a counterexample.
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\item{(a)}  If $x_n$ is a sequence of strictly negative numbers converging to $0$, then 
$x_n$ has a strictly increasing subsequence $x_{n_k}$.  
\item{(b)}  If $x_n \to 0$, then $x_n$ contains
a strictly increasing subsequence or a strictly decreasing subsequence (or both).  
\item{(c)}  If $x_n$ is a decreasing sequence with a bounded subsequence $x_{n_k}$, then 
$x_n$ converges.
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\noindent
2.  Consider the sequence $x_n = \cos(n)$ (where we think of $n$ as an angle expressed in radians).
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\item{(a)}  Prove that $x_n$ has a convergent subsequence.
\item{(b)}  In this part of the question we will show that $x_n$ is not convergent, though.
Suppose $\lim_{n\to\infty} \cos(n) = a$ for some real number $a$.  Using a trig identity for 
$\cos(n + 1)$ and considering $\lim_{n\to\infty} (\cos(n+1) - \cos(n))$, show that 
$${a (\cos(1) - 1)\over \sin(1)} = \lim_{n\to\infty} \sin(n).$$  But then use the sequence
$\lim_{n\to\infty} (\sin(n+1) - \sin(n))$ to deduce that $a = 0$, so 
$\lim_{n\to\infty} \cos(n) = \lim_{n\to\infty} \sin(n) = 0$.  But this is a contradiction.
Explain why to conclude the proof.  
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\noindent
3.  A {\it cluster point} of a sequence $x_n$ is a limit of a convergent subsequence $x_{n_k}$.
(See question 1 on the A section for examples of sequences with several different cluster points.)
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\item{(a)}  Show that there exists a sequence $x_n$ whose set of cluster points is all of 
${\bf R}_+ = \{x \in {\bf R}\mid x > 0\}$.
(Hint:  Look at the solution for the Extra Credit problem on Exam 1, which shows how to get
a sequence containing all the positive rational numbers.  Think rational approximations to 
decimals to get a subsequence converging to any given positive $a \in {\bf R}$.  However,
note that you will need to be careful to produce an actual subsequence $x_{n_k}$ with a 
strictly increasing index sequence $n_k$.   In fact, this example
can be ``jazzed up'' to get a sequence whose set of cluster points consists is all of ${\bf R}$!)
\item{(b)}  Show that if $a_m$ is a convergent sequence of cluster points of a given sequence $x_n$,
then $a = \lim_{m\to\infty} a_m$ is also a cluster point of the $x_n$ sequence.  
\bye    
