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\centerline{MATH 242 -- Principles of Analysis}
\centerline{Solutions for Exam 2 -- April 1, 2011}
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\noindent
I.  
\item{A)} (20) State and prove the Monotone Convergence Theorem
for sequences.  (You may give the proof in the case that the 
sequence is monotone increasing.)
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\item{} {\it Solution:}  The Theorem says that every bounded monotone
sequence in ${\bf R}$ converges to some real number $a$.   If $\{x_n\}$
is monotone increasing, let $a = {\rm lub}(\{x_n \mid n \ge 1\})$.
then for all $\varepsilon > 0$, $a - \varepsilon$ is not an
upper bound for $\{x_n \mid n \ge 1\}$.  Hence there exists
some $n_0$ such that $a - \varepsilon < x_{n_0} \le a$.  
Since the sequence is monotone increasing, for all $n \ge n_0$,
we have $a - \varepsilon < x_{n_0} \le x_n \le a$.  But this
implies that $|x_n - a| < \varepsilon$ for all $n \ge n_0$.  Hence $x_n \to a$.
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\item{B)} (10)  Suppose that $x_n$ is a sequence of strictly 
{\it negative} numbers and $x_{n+1}/x_n \le 1$ for all $n \ge 1$. 
Show that $\lim_{n\to\infty} x_n = a$ exists in ${\bf R}$
and satisfies $a \le 0$.
\bigskip
\item{} {\it Solution:}  Since the $x_n < 0$, from the inequality
$x_{n+1}/x_n \le 1$, we get $x_{n+1} \ge x_n$ for all $n \ge 1$.
(The inequality reverses when we multiply by the negative number
$x_n$.)
This implies the sequence is {\it monotone increasing}.  Since it
is also bounded above by $0$, part (A) shows that $x_n \to a$
for some $a \le 0$.  
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\noindent
II. (20) Suppose $\{x_n\}$ is a sequence such that 
$|x_n - 10| < 100$ for all $n \ge 1$.  Show that there exists
some number $a \in [-90,110]$ and a subsequence 
$\{x_{n_k}\}$ such that $x_{n_k} \to a$.  State any ``big theorems''
you are using.
\bigskip
\item{} {\it Solution:}  Since $|x_n - 10| < 100$ for all $n$, it follows
that $-100 < x_n - 10 < 100$, so $-90 < x_n < 110$.  Hence 
$|x_n| \le 110$ for all $n$, so this is a bounded
sequence.  By the Bolzano-Weierstrass theorem, there must be
a convergent subsequence $x_{n_k} \to a$ for some $a \in {\bf R}$.
The limit must also lie in the interval $[-90,110]$ because of the 
``order limit theorem'' (Corollary 2.2.8 in our text).  This result says:
If $a \le x_n \le b$ for all $n$, and $x_n \to c$, then $a \le c \le b$
also.
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\noindent
III.  (20) Show using the $\varepsilon,\delta$ definition that 
$\lim_{x\to 2} x^2 - x = 2$. 
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\item{}{\it Solution:}  (``prep'':  $|x^2 - x - 2| = |x - 2||x + 1|$.
If $|x - 2| < 1$, then $1 < x < 3$, so $|x + 1|$ satisfies
$2 < |x+1| < 4$.)
\medskip
\item{} {\it proof:}  Given $\varepsilon > 0$, let 
$\delta = \min\left(1,{\varepsilon\over 4}\right)$.
Then for all $x$ with $0 < |x - 2| < \delta$, $1 < x < 3$, so 
$|x + 1| < 4$.  Therefore, for all such $x$,
$$|x^2 - x - 2| = |x - 2| |x + 1| < {\varepsilon\over 4}\cdot 4 = \varepsilon.$$
This shows $\lim_{x\to 2} x^2 - x = 2$.//
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\noindent
IV. True/False.  For each true statement, give a short proof.
For each false statement, give a counterexample or reason.  
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\item{A)} (10) If $\{x_n\}$ diverges, then for all $a\in {\bf R}$,
there exist $\varepsilon_0 > 0$ and $n_0 \in {\bf N}$ such that
$|x_n - a| \ge \varepsilon_0$ for all $n \ge n_0$.
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\item{} {\it Solution 1:}  This is FALSE.  A counterexample is the sequence
$x_n = (-1)^n$.  If $a = 1$ (or $a = -1$), note there is no $n_0$ and no $\varepsilon_0 > 0$
such that $|x_n - a| \ge \varepsilon_0$ for {\it all} $n \ge n_0$ because
$|(-1)^n - 1| = 0$ for all even $n$.  (Similarly, $|(-1)^n + 1| = 0$ for all odd $n$.)
\medskip
\item{} {\it Solution 2:} We can also see that this is FALSE if we
think of the meaning of the statement:  For all $a \in {\bf R}$, 
$x_n \not\to a$ (that is, $x_n$ {\it does not converge to $a$}.).
In symbolic form, by negating the definition of the statement
$x_n \to a$, this says 
$$\forall a, \exists \varepsilon_0 > 0, \forall n_0, \exists n \ge n_0,\ s.t.\ |x_n - a| \ge \varepsilon_0.$$
(Note: the quantifier on the $n$ is $\exists$, not $\forall$.  The given statement
would certainly imply that the sequence does not converge to $a$, but it
is not necessarily true for divergent sequences, as the example in
the first solution shows.)
\bigskip   
\item{B)} (10) It is possible to find a function $f: {\bf R} \to {\bf R}$ that 
is continuous at $c = \sqrt{2}$, with $f(\sqrt{2}) = 0$, but
satisfying $f(x) = 1$ for all rational $x$.
\bigskip
\item{} {\it Solution 1:}  This is FALSE.  If $\{x_n\}$ is any sequence
of rational numbers with $x_n \to \sqrt{2}$, then by a standard property 
of continuous functions, $f(x_n) \to f(\sqrt{2})$.  But $f(x_n) = 1$
for all $n$, and $f(\sqrt{2}) = 0$.  So this is not possible.  
\medskip 
\item{} {\it Solution 2:}  We can also see that this is FALSE if we
apply the definition of continuity directly.  Aiming for a contradiction,
suppose there were such a function. 
Then for all $\varepsilon > 0$, there would exist $\delta > 0$
such that $|f(x) - f(\sqrt{2})| < \varepsilon$ for all $x$
with $|x - \sqrt{2}| < \delta$.  If $f(\sqrt{2}) = 0$ and 
we take $\varepsilon < 1$, then the interval defined by $|y - 0| < \varepsilon$
does not contain $1$.  But every interval defined by $|x - \sqrt{2}| < \delta$ for
$\delta > 0$ contains rational numbers $x$ and for those $x$, $f(x) = 1$. 
This is a contradiction which shows no such $f$ exist.   
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\item{C)}  (10)  The sequence defined by $x_1 = 1$
and $x_n = \sqrt{3x_{n-1} + 1}$ for $n\ge 2$ is monotone increasing. 
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\item{} {\it Solution:}  This is TRUE.  We have $x_1 = 1$ and $x_2 = \sqrt{4} = 2$.
so $x_2 > x_1$.  This is the base case for an induction.  Assume $x_{k+1} \ge x_k$
for a given $k$.  Then by the induction hypothesis,
$$x_{k+2} = \sqrt{3x_{k+1} + 1} \ge \sqrt{3x_k + 1} = x_{k+1}.$$
Therefore $\{x_n\}$ is increasing by induction.  
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\noindent
{\it Extra Credit.}  (10) Assume that $\{x_n\}$ is a sequence that 
converges to $a$.  Construct a new sequence $\{y_n\}$
by making $y_n$ the average of the first $n$ terms in $\{x_n\}$:
$y_n = {x_1 + x_2 + \cdots + x_n\over n}.$
Show that $y_n$ also converges to $a$.  
\bigskip
\item{} {\it Solution:}  Here's the idea.  By the triangle inequality, note that 
$$\eqalign{
|y_n - a| &= \left|{x_1 + x_2 + \cdots + x_n\over n} - a\right|\cr
 &= {1\over n} | x_1 + x_2 + \cdots + x_n - na|\cr
&\le {1\over n} \left(|x_1 - a| + \cdots + |x_n - a|\right).\cr}$$
Since $\{x_n\}$ converges to $a$, recall that we know the sequence is
bounded.  This implies that there exists $M$ such that
$|x_n - a| \le M$ for all $n$ as well.  Now, given $\varepsilon > 0$, 
there exists $N$ such that $|x_n - a| < \varepsilon/2$ for all $n \ge N$.
So if $n \ge N$, continuing from the last line above,
$$\eqalign{
&\le {1\over n}(|x_1 - a| + \cdots + |x_{N-1} - a|) + {n - N + 1\over n} {\varepsilon\over 2}\cr
&< {N M\over n} + {\varepsilon\over 2}.\cr}$$
Since we can think of the $N$ that works here as fixed, but $n$ is still 
allowed to grow, note that we can now find $n$ large enough
so that ${N M\over n} < {\varepsilon\over 2}$ as well by taking $n > {2NM\over \varepsilon}$.
In other words, given $\varepsilon > 0$, we take 
$n_0 = \max(N,{2NM\over \varepsilon})$.  Then $n \ge n_0$ implies $|y_n - a| < \varepsilon$.      
\bye
