\magnification=\magstep1
\nopagenumbers
\overfullrule=0pt
\centerline{Mathematics 242 -- Principles of Analysis}
\centerline{Problem Set 1 Solutions}
\centerline{September 9, 2005}
\bigskip
\noindent
1.2.1. a) As in the proof of Theorem 1.1.1, we argue by contradiction.
Suppose $\sqrt{3}$ is rational, so $\sqrt{3} = {m\over n}$, where
the fraction is in lowest terms.  This says $m^2 = 3n^2$.  So $3$
divides $m^2$, which implies $3$ divides $m$: $m = 3 k$ for some
integer $k$.  Hence $9k^2 = 3n^2$, so $3k^2 = n^2$.  Hence
$3$ divides $n$ by the same reasoning.  This is a contradiction
because we assumed that ${m\over n}$ was in lowest terms, so
$m$ and $n$ have no common (integer) factors.    The same
reasoning does work to show that $\sqrt{6}$ is irrational since
if $6$ divides $n^2$ for some integer, then $6$ must divide $n$.

b)  If $4$ divides $n^2$, then it is not necessarily true that
$4$ divides $n$.  For instance $4$ divides $6^2 = 36$, but $4$
does not divide $6$.
\bigskip
\noindent
1.2.2. a) {\it False}:  Consider the sets $A_n = [0,1/n)$.
Then each $A_n$ contains infinitely many real numbers,
but $\cap_{n=1}^\infty A_n = \{0\}$ is not infinite.

b) {\it True}:  The intuitive reason is: Since each $A_n$, including $A_1$, is 
finite, there can be only a finite number of strict inclusions in
the chain.  In other words, after a finite number of steps, the chain 
``stabilizes,'' and we have all the $A_n$'s the same for $n \ge N$:
$A_N = A_{N+1} = \cdots$. The intersection is the set $A_N$.

c) {\it False}:  Let $A = \{1,3\}, B = \{1\}, C = \{2\}$.
Then $A \cap (B \cup C) = \{1\}$, but $(A\cap B) \cup C = \{1,2\}$.

d) {\it True}: This can be seen with a Venn diagram, for instance.

e) {\it True}: This can be seen with a Venn diagram, for instance.

\bigskip
\noindent
1.2.3c.  Let $x \in (A\cup B)^c$, then $x$ is not an element of either
$B$ or $C$ (since the elements of $A\cup B$ are the $x$ that
are elements of $A$, or of $B$, or of both sets).  Saying this another
way, $x$ is not an element of $A$ and not an element of $B$.  Hence
$x \in A^c \cap B^c$.  This shows $(A \cup B)^c \subseteq A^c \cap B^c$.
Now we show the other inclusion.  Let $x \in A^c \cap B^c$.  This 
says $x$ is not an element of $A$ and also not an element of $B$.
Hence $x$ is not an element of $A \cup B$ (see above).  This shows
$x \in (A\cup B)^c$, so $A^c \cap B^c \subseteq (A\cup B)^c$.
Since both inclusions are true, the sets are equal.

\bigskip
\noindent
1.2.6. a)  $f(A) = [0,4]$ and $f(B) = [1,16]$.  $f(A) \cap f(B) = [1,4] = 
f(A\cap B)$ since $A\cap B = [1,2]$.  $f(A) \cup f(B) = [0,16] = f(A \cup B)$.

b)  The key observation here is that $f(x) = f(-x)$, so $f(A)$ and $f(B)$
can have equal elements coming from unequal (opposite sign) $x$'s.
An ``extreme'' example:
Let $A = [1,2]$ and $B = [-2,-1]$.  Then $f(A) = f(B) = [1,4]$, so
$f(A) \cap f(B) = [1,4]$, but $A\cap B = \emptyset$, so $f(A\cap B) = \emptyset$.

c)  If $y \in g(A\cap B)$, then $y = g(x)$ for some $x\in A\cap B$.
This $x$ is in both $A$ and $B$.
This means $y = g(x)$ for some $x\in A$ and also $y = g(x)$ for some
$x \in B$.  Hence 
$y \in g(A) \cap g(B)$ and $g(A\cap B) \subseteq g(A) \cap g(B)$.

d)  The general statement we can make here is that
$g(A\cup B) = g(A) \cup g(B)$.  Proof:  Let $y \in g(A\cup B)$.
Then $y = g(x)$ for some $x \in A \cup B$.  This means $x \in A$
or $x \in B$.  Hence $y = g(x)$ for some $x\in A$ or some $x\in B$,
which implies $y \in g(A) \cup g(B)$.  Hence $g(A\cup B) \subseteq g(A) \cup g(B)$.
For the other inclusion, let $y \in g(A) \cup g(B)$.  Then either
$y = g(x)$ for some $x \in A$ or some $x \in B$ (or both -- but note
that it might be $g(x) = g(x')$ for some $x\ne x'$ with $x\in A$ 
and $x' \in B$).  Either
way, $y = g(x)$ for some $x \in A\cup B$.  So the other inclusion holds
as well.

\bigskip
\noindent
1.2.7. a) $f^{-1}(A) = [-2,2]$ and $f^{-1}(B) = [-1,1]$ (note that only 
the subinterval $[0,1] \subset [-1,1]$ is ``hit'' by $f$ though).
$f^{-1}(A \cup B) = f^{-1}( [-1,4]) = [-2,2] = f^{-1}(A) \cup f^{-1}(B)$.
$f^{-1}(A \cap B) = f^{-1}( [0,1]) = [-1,1] = f^{-1}(A) \cap f^{-1}(B)$.

b) To show $f^{-1}(A \cup B) = f^{-1}(A) \cup f^{-1}(B)$, first let
$x \in f^{-1}(A \cup B)$, then $f(x) \in A \cup B$, so $f(x) \in A$
or $f(x) \in B$.  This shows $x\in f^{-1}(A)$ or $x\in f^{-1}(B)$, 
so $x\in f^{-1}(A) \cup f^{-1}(B)$, showing that 
$f^{-1}(A \cup B) \subseteq f^{-1}(A) \cup f^{-1}(B)$.
For the opposite inclusion, we can just reverse the reasoning here.

To show $f^{-1}(A \cap B) = f^{-1}(A) \cap f^{-1}(B)$, first let
$x \in f^{-1}(A \cap B)$, then $f(x) \in A \cap B$, so $f(x) \in A$
and $f(x) \in B$.  This shows $x\in f^{-1}(A)$ and $x\in f^{-1}(B)$, 
so $x\in f^{-1}(A) \cap f^{-1}(B)$, showing that 
$f^{-1}(A \cap B) \subseteq f^{-1}(A) \cap f^{-1}(B)$.
For the opposite inclusion, we can just reverse the reasoning here.


\bigskip
\noindent
1.2.10. a) The base case is $y_1 < 4$, which is true since $y_1 = 1$.
Now assume that $y_k < 4$ and consider $y_{k+1} = (3y_k + 4)/4$.
Since $y_k < 4$, $y_{k+1} < (3\cdot 4 + 4)/4 = 16/4 = 4$.  Hence
$y_n < 4$ for all $n$.

b)  We must show $y_n \le y_{n+1}$ for all $n$.  The base case
is the statement $y_1 \le y_2$, which is true since $y_1 = 1$,
and $y_2 = (3 + 4)/4 = 7/4$.  Now assume that $y_k \le y_{k+1}$.
We will show $y_{k+1} \le y_{k+2}$.  From the inequality
$y_k \le y_{k+1}$, it follows that $(3y_k + 4)/4 \le (3y_{k+1}+4)/4$,
since adding the same quantity to both sides and multiplying
by positive numbers preserve the inequality.  But now we
see $(3y_k + 4)/4 = y_{k+1}$ and $(3y_{k+1}+4)/4 = y_{k+2}$.
So we are done by induction.

\bigskip
\noindent
1.2.11. We argue by induction on $n$.  The base case here is $n=0$
(that is, $A = \emptyset$).  There is exactly 1 subset of $A$,
namely $\emptyset$.  So $A$ has $2^0 = 1$ subsets.  Now
assume that every set with $k$ elements has exactly $2^k$ different
subsets.  Consider $A$ with $k+1$ elements, say $A =\{x_1,\ldots,x_{k+1}\}$.
Every subset of $A$ either contains the ``last element'' $x_{k+1}$,
or not.  The ones that don't are the same as the subsets
of $A' = \{x_1,\ldots,x_k\}$.  So by induction, there are $2^k$ of these
subsets.  Next consider the subsets of $A$ that {\it do} contain
$x_{k+1}$.  If we {\it remove} $x_{k+1}$ from one of these subsets,
what is left is a subset of $A'$.  Putting $x_{k+1}$ ``back in''
never gives the same subset of $A$ twice if the subsets of $A'$
were different.  So by induction again there are also $2^k$ subsets
of $A$ containing $x_{k+1}$.  Adding we find $2^k + 2^k = 2\cdot 2^k = 2^{k+1}$
different subsets of $A$.

\vfill\eject
\centerline{Mathematics 242 -- Principles of Analysis}
\centerline{Problem Set 2 Solutions}
\centerline{September 16, 2005}
\bigskip
\noindent
1.2.8. a)  For some real numbers satisfying $a < b$ and for all 
$n \in {\bf N}$, $a + 1/n \ge b$.

b) Most natural form of negation:
There exist two real numbers such that there is no rational number 
between them.

c)  For some natural number $n\in {\bf N}$, $\sqrt{n}$ is not
a natural number and $\sqrt{n}$ is rational.

d)  For some real number $x\in {\bf R}$,
$x \ge n$ for all $n\in {\bf N}$.   
(Or:  There is some real number greater than every natural number.)

\bigskip
\noindent
1.3.2. a)  A real number $t$ is the {\it greatest lower bound} 
of a set $A$, written $t = \inf(A)$ if 
\item{(i)} $t$ is a lower bound for $A$ (that is, $t \le x$ for all $x \in A$), and
\item{(ii)} If $c$ is any lower bound for $A$, then $c \le t$.

b) Statement:  Let $t$ be a lower bound for $A \subseteq {\bf R}$.  Then
$t = \inf(A)$ if and only if for every $\varepsilon > 0$, 
there exists $a \in A$ with $t \le a < t + \varepsilon$.
(Note: in intuitive terms, this is another way of 
saying that no number greater than
$t$ is a lower bound for $A$.)

\bigskip
Proof:  $\Rightarrow$:  Let $t = \inf(A)$.  The contrapositive
form of (ii) in the definition of part a says: if $c > t$, then
$c$ is not a lower bound for $A$.  In other words, for any $\varepsilon > 0$,
taking $c = t + \varepsilon > t$, there exists some $a\in A$ 
with $t \le a < t + \varepsilon$. 

$\Leftarrow$:  Assume that $t$ is a lower bound for $A$ and
that for every $\varepsilon > 0$, 
there exists $a \in A$ with $t \le a < t + \varepsilon$.
This says that no number greater than $t$ is a lower bound for $A$.
So if $c$ is any lower bound for $A$, then 
$c \le t$.  Hence by part a, $t = \inf(A)$.

\bigskip
\noindent
1.3.3.  a) 
Assuming $A$ is nonempty (not stated explicitly!), the 
set $B$ is bounded above.  Hence the Axiom of Completeness 
implies that $s = \sup(B)$ exists in  ${\bf R}$.  To show 
$s = \inf(A)$, we use the definition of $\inf(A)$ given in 1.3.2a above.  

\bigskip
First we will show $s \le a$ for all $a\in A$.  We can argue by 
contradiction.  If not, there would be some $a\in A$ with $a < s$.  
But then by Lemma 1.3.7, there would be some $b \in B$ with 
$a < b \le s$.  This is impossible since $b\in B$ is supposed to be a
lower bound for $A$.  Therefore, $s \le a$ for all $a\in A$.
Hence (i) of the definition of $\inf(A)$ must hold. 

\bigskip
Next, let $c$ be any lower bound of $A$.  By definition this
says $c \in B$.  Hence by the definition of $\sup(B)$,
$c \le s$.  Hence (ii) of the definition of $\inf(A)$ holds
for $s$ as well.

\bigskip
\noindent
b)  We do not need to assert that greatest lower bounds exist 
separately because the proof in part a here shows that their
existence follows from the existence of least upper bounds.

\bigskip
\noindent
c)  An alternate way to use the Axiom of Completeness to prove
that sets bounded below have greatest lower bounds is to note
that taking negatives reverses inequalities, turns upper bounds
to lower bounds, etc.  In fact if $A$ is bounded below, then
$-A = \{-x : x \in A\}$ is bounded above, and $\sup(-A) = -\inf(A)$.

\bigskip
\noindent
1.3.4.  Since $B\subset A$, any upper bound for $A$ is also
an upper bound for $B$.  By definition, $\sup(A)$ is an upper
bound for $A$ and also for $B$.  Therefore, $\sup(A) \ge \sup(B)$,
since if $t$ is any upper bound for $B$, then  $g \ge \sup(B)$.

\bigskip
\noindent
1.3.6. a) $A = \{n\in {\bf N} : n^2 < 10\} = \{1,2,3\}$.  
$\inf(A) = 1$ and $\sup(B) = 3$.

\bigskip
b)  $B = \{n/(m+n) : m,n \in {\bf N}\}$.  Note that $0 < n/(m+n) < 1$
since $m,n \ge 1$ are natural numbers).  $\inf(B) = 0$ (think of 
fixing $n$ and letting $m \to \infty$). $\sup(B) = 1$ (think
of fixing $m = 1$ and letting $n\to \infty$.  

\bigskip
c)  $C = \{n/(2n+1) : n\in {\bf N}\}$.  Note that $n/(2n+1)$
increases with $n$. So  $\inf(C) = 1/3$ ($n=1$) and $\sup(C) = 1/2$ 
(think of what happens as $n\to\infty$).

\bigskip
d)  $D = \{n/m : m,n \in {\bf N}, m + n \le 10\}$.  This is 
a finite set:  $\inf(D) = 1/9$, $\sup(D) = 9/1 = 9$.

\bigskip
\noindent
1.3.7. The first condition in the definition of $\sup(A)$ is
true by hypothesis.  So we only need to show that if $b$
is any upper bound for $A$, then $b\ge a$.  But this follows
immediately because $a \in A$.

\bigskip
\noindent
1.3.8. Assume $\sup(A) < \sup(B)$.  Then by Lemma 1.3.7, there
exists some $b\in B$ such that $\sup(A) < b \le \sup(B)$.  This $b$
is an upper bound for $A$ since $b > \sup(A) \ge a$ for all $a \in A$.

\bigskip
\noindent
1.3.9. a) {\it True} (finite means the set has finitely many elements;
the situation is like that in 1.3.6 above).
\bigskip
b) {\it False}: Take $B$ from 1.3.6 b,
$C$ from 1.3.6 c,  or any similar example.
\bigskip
c) {\it False}:
An example:  $A = \{1 - 1/n : n \in {\bf N}\}$ and 
$B = \{1 + 1/n : n \in {\bf N}\}$.  Note $\sup(A) = 1 = \inf(B)$.
\bigskip
d) {\it True}.  The intuitive idea is that $s+t$ is an upper
bound for the set of sums.  Moreover, $A$ contains elements
arbitrarily close to $s = \sup(A)$ and $B$ contains
elements arbitrarily close to $t=\sup(B)$. Hence the set of sums
$A+B$ contains elements arbitrarily close to $s+t$.
\bigskip
e) {\it False}:  Take $A = \{1\}$ and 
$B = \{1 - 1/n: n\in {\bf N}\}$.  $\sup(A) = \sup(B) = 1$, but
no element of $B$ is an upper bound for $A$.  



\vfill\eject
\centerline{Mathematics 242 -- Principles of Analysis}
\centerline{Problem Set 3 Solutions}
\centerline{September 23, 2005}
\bigskip
\noindent
1.4.4.  The first statement of the
Archimedean Property is that if $x$ is any real
number then there exists a natural number $n$ with $n > x$.
So let $\varepsilon > 0$ be any positive number (as close to zero
as we like).  Applying the Archimedean Property, there exists
some $n > 1/\varepsilon$, so $1/n < \varepsilon$.  This shows
that no positive number is an lower bound for the set 
$\{1/n : n \in {\bf N}\}$.  But the set is obviously bounded below 
($0$ is a lower bound since $1/n > 0$ for all $n$), so the $\inf$ exists
by the Axiom of Completeness.  
It follows that $0 = \inf(\{1/n : n \in {\bf N}\})$.

\bigskip
\noindent
1.4.5.  If there were some $x \in \cap_{n=1}^\infty (0,1/n)$, then 
$0 < x < 1/n$ for all $n\in {\bf N}$.  This would say 
the positive number $x$ would be a lower bound for
the set $A$ from the previous problem 1.4.4.  But by that problem, 
$0 = \inf(\{1/n : n \in {\bf N}\})$.  So there can be no such $x$.

\bigskip
\noindent
2.2.1. a)  Let $\varepsilon > 0$ and take $N > 1/\sqrt{6\varepsilon}$
in ${\bf N}$.  If $n \ge N$, then
$$\eqalign{ \left|{1\over 6n^2 + 1} - 0\right| 
          & = {1\over 6n^2 + 1}\cr
          &\le {1\over 6n^2}\cr
          &\le {1\over 6N^2}\cr
          &< \varepsilon\cr}$$
This shows that $\lim {1\over 6n^2 + 1} = 0$.

b)  Let $\varepsilon > 0$ and take $N > 13/(4\varepsilon)$
in ${\bf N}$.  If $n \ge N$, then
$$\eqalign{ \left|{3n + 1\over 2n + 5} - {3\over 2}\right| 
          & = {13\over 4n + 10}\cr
          &\le {13\over 4n}\cr
          &\le {13\over 4N}\cr
          &< \varepsilon \cr}$$
This shows that $\lim {3n+1\over 2n+5} = {3\over 2}$.

c)   Let $\varepsilon > 0$ and take $N > 4/\varepsilon^2$
in ${\bf N}$.  If $n \ge N$, then
$$\eqalign{ \left|{2\over \sqrt{n+3}} - 0\right| 
          & = {2\over \sqrt{n+3}}\cr
          &\le {2\over{\sqrt n}}\cr
          &\le {2\over \sqrt{N}}\cr
          &< \varepsilon \cr}$$
This shows that $\lim {2\over \sqrt{n+3}} = 0$.

\bigskip
\noindent
2.2.2.  An example:  Let $x_n = (-1)^n$ and $x = 0$.  $(x_n)$
``verconges'' to $x$ since for $\varepsilon = 2$ (for instance),
for all $N$, $n\ge N$ implies $|x_n - x| = 1 < 2$.
That inequality is true for all $n\in {\bf N}$, in fact.

This definition says that all the terms in the sequence $(x_n)$
lie in  some interval $(x - \varepsilon, x + \varepsilon)$.  
In other words, saying $(x_n)$ ``verconges'' to $x$ just says
that all the terms in the sequence lie in a bounded interval
containing $x$.  

\bigskip
\noindent
2.2.3.  a)  Must show that at some US college, no student is at 
least seven feet tall (equivalently, all students are less than 
seven feet tall).

b)  Must show that at some US college, every professor gives grades other
than A or B to some students.

c)  Must show that at all US colleges, some students are less than six feet 
tall.

\bigskip
\noindent
2.2.4.  Let $\varepsilon < 1$.  Then for all $N\in {\bf N}$,
there exist $n \ge N$ such that $|x_n - 0| = 1 > \varepsilon$.
Hence the sequence does not converge to $0$.  The rest of the
question is asking about the set of $\varepsilon$ for which
it would be true that there exists some $N\in {\bf N}$ such
that $|x_n - 0| < \varepsilon$ for all $n\ge N$.  As long
as $\varepsilon \ge 1$, there is no problem finding such 
an $N$.  You can take $N = 1$, in fact!  But as above, for 
$\varepsilon < 1$, there is no such $N$.

\bigskip
\noindent
2.2.6. a)  {\it larger}.  Reason:  If $N$ ``works'' for $\varepsilon$
in the definition of convergence, and $N' > N$, then 
$n\ge N'$ implies $n\ge N$ also, so $|x_n - a| < \varepsilon$
for all $n\ge N'$.    b)  {\it larger}.  If $|x_n - a| < \varepsilon$
for all $n\ge N$, and $\varepsilon' > \varepsilon$, then
we have $|x_n - a| < \varepsilon'$  too.

\bigskip
\noindent
2.2.7. a)  Definition:  $\lim x_n = +\infty$ if for every positive
number $B$ (no matter how {\it big}), there exists an $N$ such that 
whenever $n \ge N$, $x_n > B$.  

Consider $x_n = \sqrt{n}$.  Given $B > 0$, let $N > B^2$ in ${\bf N}$.
Then for all $n\ge N$, $\sqrt{n} \ge \sqrt{N} > B$.  Hence
$\lim \sqrt{n} = +\infty$.

b)  The given sequence 
does not converge to $+\infty$, because if $B > 0$, there
is no $N$ such that $x_n > B$ for all $n \ge N$ (the zero terms 
always fail this inequality).

\bigskip
\noindent
2.2.8. a)  frequently, not eventually.  

b)  eventually implies frequently,
so eventually is stronger.  

c)  We can say:  $(x_n)$ converges to $x$
if for every $\varepsilon > 0$, $(x_n)$ is eventually in the 
interval $(x-\varepsilon, x+\varepsilon)$.  

d)  We can say the sequence is frequently in 
the interval, but not necessarily eventually in the interval.  For
instance, consider $x_n = \cases{2 & if $n$ even\cr
                                 3 & if $n$ odd\cr}$.  
This sequence is not eventually in the interval $(1.9,2.1)$
because of the infinitely many terms equal to 3, but
it is frequently in that interval.                    
                                                    
\vfill\eject
\centerline{Mathematics 242 -- Principles of Analysis}
\centerline{Problem Set 4 Solutions}
\centerline{October 17, 2005}
\bigskip
\noindent      
2.3.8.  a)  Let $x_n = (-1)^n$ and $y_n = (-1)^{n+1}$.  Then
$(x_n)$ and $(y_n)$ diverge.  But $x_n + y_n = 0$ for all $n$, so the 
sum series converges.

b)  Impossible by Theorem 2.3.3 (ii) and (i):  Note that 
$y_n = -x_n + (x_n + y_n)$.  If $(x_n)$ and $(x_n + y_n)$ converge, 
so does $(-x_n)$, and hence so does $y_n$.

c)  Example:  $b_n = 1/n$.  

d)  Impossible by Theorem 2.3.2.  If $(b_n)$ is convergent, then
$(b_n)$ is bounded:  $|b_n| \le B$ for all $n$.  If $(a_n - b_n)$
is also bounded, $|a_n - b_n| \le A$, 
then 
$$|a_n| = |a_n - b_n + b_n| \le |a_n - b_n| + |b_n| \le A + B$$
for all $n$.  So $|a_n|$ would also have to be bounded.

e)  Example:  $a_n = 1/n$, $b_n = n$.  Note $\lim a_n = 0$, $\lim a_nb_n = 1$,
but $(b_n)$ is not convergent.

\bigskip
\noindent
2.4.2.  a) First note that if $x < 4$, then $1/(4-x) > 0$.
We have $x_1 = 3$ and $x_2 = 1$.  Moreover if
$x_k > x_{k+1}$, then $4 - x_k < 4 - x_{k+1}$, and 
$1/(4-x_k) > 1/(4 - x_{k+1})$.  This shows $x_{k+1} > x_{k+2}$.  
By induction, therefore, $(x_n)$ is a decreasing sequence bounded below
by zero.  By Theorem 2.4.2 (the Monotone Convergence Theorem), $(x_n)$
is convergent.  

b)  This should be intuitively clear, since $(x_{n+1})$ is 
just the same sequence, ``shifted by one'' in the index.  Here
is a formal proof:  If $(x_n)$ converges to $a$, then for all $\varepsilon >0$,
there exist $N\in {\bf N}$ such that $|x_n - a| < \varepsilon$
for all $n \ge N$.  If $n \ge N$, then $n+1 > n \ge N$ too, so
$|x_{n+1} - a| < \varepsilon$ whenever $n \ge N$.  Hence
$x_{n+1}$ also converges to $a$. 

c)  Say $L = \lim x_n = \lim x_{n+1}$.  Then we have $L = 1/(4 - L)$, 
so $L^2 - 4L + 1= 0$.  The smaller positive root is $L = {4 - \sqrt{12}\over 2}
= 2 - \sqrt{3}$.  (This is the limit of the sequence since the
other root $2 + \sqrt{3} > 3$.  Since $x_1 = 3$ and the sequence is decreasing
$2 + \sqrt{3}$ cannot be the limit.)

\bigskip
\noindent
2.4.5. a)  By the method of Lagrange Multipliers from MATH 241, the 
function $(x+y)/2$ is minimized at the points on the constraint curve
$xy = 2$ with $x,y > 0$ where $x = y = \sqrt{2}$ and the minimum value is
$(\sqrt{2} + 2/\sqrt{2})/2 = \sqrt{2}$.  It follows that
if $x > \sqrt{2}$, then $(x + 2/x)/2 > \sqrt{2}$.  Now we argue by 
induction.  The base case is $n = 1$ and $x_1 = 2 > \sqrt{2}$.   
By the above, if $x_k > \sqrt{2}$, then $x_{k+1} = (x_k + 2/x_k)/2 > \sqrt{2}$
also.  Therefore $x_n > \sqrt{2}$ for all $n$, so $x_n^2 > 2$.   

Then 
$$x_n - x_{n+1} = {1\over 2}\left(x_n - {2\over x_n}\right) = {1\over 2}{x_n^2 - 2\over x_n} > 0$$
since $x_n^2 > 2$ for all $n$.  It follows that $(x_n)$ is a monotone
decreasing sequence.  By the Monotone Convergence Theorem (2.4.2), it
converges to some $L$ in ${\bf R}$.  As in 2.4.2 and 2.4.4 above, $L$ satisfies
$$L = {1\over 2}\left(L + {2\over L}\right) \Rightarrow 2L^2 = L^2 + 2,$$
so $L^2 = 2$.  It follows that $L = +\sqrt{2}$, since all the terms
of the sequence are positive.  

b)  The sequence defined by $x_{n+1} = {1\over 2}\left(x_n + {c\over x_n}\right)$
converges to $\sqrt{c}$ by the same kind of argument given for part a.

\bigskip
\noindent
2.5.3. a)  An example: $x_n = \cases{1/n & if $n$ is odd\cr
                                     1+1/n & if $n$ is even.\cr}$
                                     
b)  Impossible:  Let $x_{n_k}$ be the convergent subsequence, say
$\lim_{k\to \infty} x_{n_k} = L$.  Let $\varepsilon > 0$.
Then there exists a $K$ such that whenever $k \ge K$, 
$|x_{n_k} - L| < \varepsilon$.  Since the whole sequence is 
monotone this implies that for all $n \ge N = n_K$, 
$|x_n - L| < \varepsilon$ also.  Thus, $(x_n)$ converges to $L$ also.

c)  One way:  Let $(x_m)$ the sequence produced as follows:
$$1,1,1/2,1,1/2,1/3,1,1/2,1/3,1/4,1,1/2,1/3,1/4,1/5,1, \ldots$$
(repeating the same pattern of starting over from $1$ and adding
one additional $1/n$ on each iteration).
Then $(x_m)$ has subsequences converging to $1/n$ for each $n\in {\bf N}$
because each $1/n$ appears infinitely many times in the $x_m$ sequence.
\bigskip
\noindent
Alternate Method:  Take a sequence consisting of all rational numbers in the
interval $(0,1)$, listed in some order.  Then by the Density 
Theorem for ${\bf Q}$, there are subsequences
converging to every real number in $(0,1)$, including the ones given(!)

d)  An example:  $x_n = \cases{n & if $n$ is even\cr  
                               1/n & if $n$ is odd,\cr}$
$(x_n)$ is unbounded because of the odd-numbered terms, but the 
even-numbered subsequence converges to zero.

e)  Impossible:  The bounded subsequence is itself a bounded
sequence, so the Bolzano-Weierstrass Theorem says it has 
a convergent subsequence.  That subsequence is also a subsequence
of the original sequence.                                   
         
\bigskip
\noindent
2.5.4.  Aiming for a contradiction, suppose that $(a_n)$
is a bounded sequence such that every convergent
subsequence converges to $a$, but the whole sequence does
not converge to $a$.  Then by the definition, there exists
$\varepsilon > 0$ such that for all $N \in {\bf N}$,
 $|a_n - a| \ge \varepsilon$ for some $n \ge N$.  
For the rest of this solution we will pick 
a particular $\varepsilon > 0$ as above.  
For $N = 1$,  let $n_1\in {\bf N}$ be such that
$a_{n_1}$ satisfies $|a_{n_1} - a| \ge \varepsilon$.
Then with $N = n_1+1$, let $n_2 \ge n_1 + 1$ satisfy
$|a_{n_1} - a| \ge \varepsilon$.  In the same way, we
produce a strictly increasing sequence $n_k$ of natural
numbers, and a subsequence $a_{n_k}$ such that 
$|a_{n_k} - a| \ge \varepsilon$ for all $k \ge 1$.  
The $(a_{n_k})$ is a subsequence of the original $(a_n)$ sequence. 
Hence it is also bounded.  Then the Bolzano-Weierstrass theorem 
implies that $(a_{n_k})$ has a convergent subsequence.  This 
is also a subsequence of the original sequence, so by hypothesis 
it must be converging to $a$.  But that is a contradiction because
we have $|a_{n_k} - a| \ge \varepsilon$ as above for all $k$.
        
        
            
\bigskip
\noindent
2.5.6.  Since $(a_n)$ is a bounded sequence, the set $S$ is 
bounded above.  So $\sup(S)$ exists in ${\bf R}$.  By Lemma
1.3.7, given any $\varepsilon > 0$, there is some 
$x \in S$ with $s - \varepsilon < x \le s$.  By the definition 
of $S$, $x\in S$ means there are infinitely many $n$ such
that $s - \varepsilon < x < a_n \le s$.  
So we argue as follows: For each $k \in {\bf N}$
in turn, taking $\varepsilon = {1\over k}$ in the above, we
let $n_k$ be the smallest integer $> n_{k-1}$
with $s - {1\over k} < a_{n_k}$.  This gives a subsequence
$a_{n_k}$ converging to $s$, since we have 
$s - {1\over k} < a_{n_k} \le s$ for all $k$.  
We have  $\lim_{k\to\infty} s - {1\over k} = s$, so
by the Limit Squeeze Theorem,
$\lim_{k\to\infty} a_{n_k} = s$.  
         
\vfill\eject          
\bigskip
\noindent
2.6.1. a)  We just need to exhibit a non-monotone convergent
sequence.  An example:  $x_n = 1 + {(-1)^n\over n}$.  

b)  An example $x_n = n$.

c)  Impossible:  Theorem 2.6.4 (the Cauchy Criterion) says
a Cauchy sequence coverges.  By Theorem 2.5.2, every subsequence
of a convergent sequence converges to the same limit as the whole
sequence.

d) An example $x_n = \cases{n & if $n$ is odd\cr
                            0 & if $n$ is even\cr}$.
The whole sequence is unbounded because of the odd-numbered
terms, but the even-numbered subsequence is Cauchy (constant, in fact!)

\bigskip
\noindent
2.6.3.  a)  The difference is the condition $|s_{n+1} - s_n| < \varepsilon$
rather than $|s_m - s_n| < \varepsilon$.  This condition only deals
with {\it pairs of consecutive terms} in the sequence, not with arbitrary
pairs ``far enough out''.

b)  The sequence $s_n = \sum_{k=1}^n {1\over n}$ (partial sums of the 
harmonic series) is pseudo-Cauchy, since $s_{n+1} - s_n = {1\over n+1}$.
Given any $\varepsilon > 0$, we can make ${1\over n+1} < \varepsilon$ by taking $n \ge N$ for 
suitably large $N$.
However, it is not convergent, because the harmonic series diverges
(see Example 2.4.5 and the class notes).  

\vfill\eject
\centerline{Mathematics 242 -- Principles of Analysis}
\centerline{Problem Set 5 Solutions}
\centerline{October 14, 2005}
\bigskip
\noindent
I.  A)  We know $\sum_{k=1}^\infty {1\over \sqrt{k}}$ diverges
by an example we did in class.  Hence $\sum_{k=1}^\infty {1\over \sqrt{3k}}$
diverges too, since the terms are just a constant
times the terms in the first series.   For all $k\ge 1$, 
$\sqrt{2k + 1} \le \sqrt{2k + k} = \sqrt{3k}$.  Hence
${1\over \sqrt{2k + 1}} \ge {1\over \sqrt{3k}}$.  By
the Comparison Test, $\sum_{k=1}^\infty {1\over \sqrt{2k+1}}$
diverges, since the terms are greater than the terms in 
a divergent series of positive terms.

\bigskip
\noindent
B)  We know $\sum_{k=1}^\infty {1\over k^2}$ converges
by an example we did in class. Hence $\sum_{k=1}^\infty {1\over 4k^2}$
converges too, since the terms are just a constant
times the terms in the first series.   For all $k\ge 1$, 
$(2k+1)^2 > 4k^2$.  Hence
${1\over (2k + 1)^2} \le {1\over 4k^2}$.  By
the Comparison Test, $\sum_{k=1}^\infty {1\over (2k+1)^2}$
converges, since the terms are less than the terms in 
a convergent series of positive terms.

\bigskip
\noindent
C) This is a geometric series with ratio $-2/9$. 
Since $|-2/9| < 1$, it converges.

\bigskip
\noindent
D) Diverges, since the terms do not go to zero:
$$\lim_{n\to\infty} {n^3+ n\over 2n^3 + n + 1} = {1\over 2}.$$
(Recall, if $\sum a_n$ converges, then $\lim_{n\to\infty} a_n = 0$
by 2.7.3 in text.)
\bigskip
\noindent
II.  A)  This is a geometric series with ratio $\sin(x)$.
Since $|\sin(x)| \le 1$ for all real $x$,
it converges for all $x$ such that $\sin(x) \ne -1,1$.  For  
$x\in [0,2\pi]$, this is all $x$ except $x = \pi/2, 3\pi/2$.  The sum is 
${\sin{x} \over 1 - \sin{x}}$.

B)  This is also a geometric series with ratio $e^x$.  It
Converges for all $x$ such that $0 < e^x < 1$, so for $x < 0$.
The sum is ${e^x \over 1 - e^x}$.
\bigskip
\noindent
2.7.3.  a)  (By contraposition) If $\sum p_n$ and $\sum q_n$
both converge, then $\sum p_n + q_n$ also converges by Theorem 2.7.1
(the Algebraic Limit Theorem for series).
But $p_n + q_n = a_n$ by definition, so $\sum a_n$ converges too.

b)  $\sum a_n$ conditionally convergent means $\sum a_n$ converges
but $\sum |a_n|$ diverges.  Now by definition, $a_n = p_n + q_n$
and $|a_n| = p_n - q_n$.  We will use the Algebraic Limit
Theorem for series to derive
a contradiction if either $\sum p_n$ or $\sum q_n$ is convergent.
Suppose $\sum p_n$ converges.  Then 
from $\sum a_n$ convergent, we get $\sum q_n = \sum (a_n - p_n)$
is also convergent.  Hence $\sum |a_n| = \sum (p_n - q_n)$ is
convergent too.  But that contradicts the hypothesis, so $\sum p_n$
cannot be convergent.  The same argument shows $\sum q_n$ cannot
be convergent either.

\bigskip
\noindent
2.7.4. Take $x_n = y_n = 1/n$.  Then $\sum x_n$ and $\sum y_n$
are harmonic series (divergent).  But $\sum x_n y_n$ is the 
series $\sum {1\over n^2}$, which we know converges.  

\bigskip
\noindent
2.7.9.  a) Given any $r < 1$, there exist $r'$ with $r < r' < 1$, since
for instance we can take $r' = (1 + r)/2$.  The hypothesis
$\lim \left|{a_{n+1}\over a_n}\right| = r < 1$ says that 
if we take $\varepsilon = r' - r > 0$ in the 
definition of convergence of a sequence, then there exists some $N$
such that 
$$\left|\left|{a_{n+1}\over a_n}\right| - r\right| < r' - r$$
whenever $n \ge N$.  But this last inequality implies
$$ \left|{a_{n+1}\over a_n}\right| < r' $$
so $|a_{n+1}| < |a_n| r'$ for all $n\ge N$.  


For future reference, we note that starting from 
$n = N$, it follows by induction that 
$|a_{N+k}| \le |a_N| (r')^k$ for all $k \ge 0$.  The base
case is $k = 0$ which is clear.  If the statement 
$|a_{N+\ell}| \le |a_N| (r')^\ell$ is true for some $\ell$, then
$$|a_{N+\ell+1}| \le |a_{N+\ell}| r' \le |a_N| (r')^\ell\cdot r' = |a_N| (r')^{\ell+1}.$$
Hence $|a_{N+k}| \le |a_N| (r')^k$ for all $k \ge 0$.

b)  Because $\sum_{k=0}^\infty (r')^k$ is {\it a geometric series} 
with ratio $0 < r' < 1$, this series converges.

c) The series $\sum |a_n|$ converges for the following reason: The finitely
many terms with $n < N$ are irrelevant for convergence.  Once $n \ge N$,
say $n = N + k$ with $k \ge 0$, we have from the last part of part a:
$|a_n| \le |a_N| (r')^k$.  Hence the ``tail'' of the 
series: 
$$\sum_{n = N}^\infty |a_n| = \sum_{k=0}^\infty |a_{N+k}|$$
converges by comparison with the convergent geometric series 
$|a_N| \sum_{k=0}^\infty (r')^k$ from part b.  

\bigskip
\noindent
III.  Take $a_n = (x-1)^n/3^n$ in the Ratio Test. We have
$\lim \left|{a_{n+1}\over a_n}\right| = {1\over 3} |x - 1|$.
The Ratio Test says the series converges absolutely for all $x$
with $|x-1| < 3$ or $-2 < x < 4$.  At $x = -2$, the series
is $\sum (-1)^n$ which diverges (terms do not go to zero).  
The series also diverges at $x = 4$.  The interval of convergence
is $(-2,4)$ in this example.

\vfill\eject
\centerline{Mathematics 242 -- Principles of Analysis}
\centerline{Problem Set 6 Solutions}
\centerline{October 21, 2005}
\bigskip
\noindent
3.2.1.  a)  The hypothesis is used when dealing with 
$\varepsilon = \min\{\varepsilon_i\}$.  Since we are taking
the minimum of a finite set of positive numbers, we get one of the
elements of the set, which is still strictly positive.  If there
were an infinite collection of open sets the intersection could
be empty (for instance if $\inf(\{\varepsilon_i\}) = 0$), or
the intersection could be non-open (see the next part).  

b)  An example:  Let $O_n = (0 - 1/n, 1 + 1/n)$, so 
$O_1 \supset O_2 \supset O_3 \supset \cdots$.
As $n\to \infty$, $\lim -1/n = 0$ and $\lim 1 + 1/n = 1$
so the intersection $\cap_{n=1}^\infty$ is $[0,1]$.  

\bigskip
\noindent
3.2.2.  This is similar to Example 3.2.9 (i)

a) Limit points $\{-1,1\}$.  

b)  No, since for instance $1 \ne B$.

c)  No, since for instance $4/5 \in B$ (take $n = 4$) but there is no open 
interval $V_\varepsilon(4/5)$ contained in $B$.  (The closest
points to $4/5$ in $B$ are $2/3$ and $6/7$ and none of the 
points in $(2/3,4/5)$ or $(4/5,6/7)$ are in $B$.

d)  Yes.  By what we said in part c, $4/5$ is an isolated point in $B$.
All the other points in $B$ are isolated as well. 

e) $\overline{B} = B \cup \{-1,1\}$.  
\bigskip
\noindent
3.2.3.
a)  Neither.  Let $a$ be any rational number and $\varepsilon > 0$
be any positive real number.  The set 
$V_\varepsilon(a) = (a - \varepsilon, a+\varepsilon)$
contains both rational and irrational numbers.  Every real number is
a limit of a sequence of rational numbers so any irrational number is
a limit point of ${\bf Q}$ not contained in ${\bf Q}$.

b)  Closed, not open.  If $\varepsilon > 0$, then
$V_\varepsilon(1)$ never consists only of natural numbers.

c)  Open, not closed.  $0$ is a limit point of $\{x \in {\bf R}: x > 0\}$
not contained in the set, since $\lim 1/n = 0$.

d)  Neither.  There is no $V_\varepsilon(1)$ contained in $(0,1]$,
since any such set contains numbers $> 1$ also.  $0$ is a limit
point of $(0,1]$ not contained in the set.

e)  Neither.  This set is the set of partial sums of the convergent
$p$-series with $p = 2$.  The sum of the series, $\pi^2/6$, is a limit
point not contained in the set, so it is not 
closed.  Every point of the set is isolated point, so the set is not open.

\bigskip
\noindent
3.3.1.  By the Heine-Borel Theorem (3.3.4), $K$ is closed and bounded.  
Since $K$ is bounded, $\inf(K)$ and $\sup(K)$ exist in ${\bf R}$.  
Moreover, as we know, there are sequences of elements of $K$
converging to $\inf(K)$ and to $\sup(K)$.  Since $K$ is closed,
this implies $\inf(K)$ and $\sup(K)$ are elements of $K$.

\bigskip
\noindent
3.3.5. a) Not compact -- take any sequence of rational numbers converging
to an irrational limit.  b)  Not compact -- as in a, but take the irrational
number in the interval $[0,1]$. (Note this set is bounded, but not closed).
c)  Not compact -- take any unbounded sequence like $x_n = n$.  (This set is closed,
but not bounded.) d)  Compact (any finite set is compact).  e)  Not compact --
0 is a limit point not contained in the set, since $\lim 1/n = 0$.  
f)  Compact -- $\lim {n \over n+1} = 1$, which is in the set.  
\vfill\eject
\centerline{Mathematics 242 -- Principles of Analysis}
\centerline{Solution for Discussion 2}
\centerline{October 31, 2005}
\bigskip
\noindent
C)  1)  We use the following notation and facts.  For sets
$A,B \subset {\bf R}$ and $c\in {\bf R}$, write $A + B = \{x+y: x\in A, y\in B\}$
and similarly $cA = \{cx : x \in A\}$.  If $B = \{b\}$ is a 
single point, then we write $A + b$, which is just the translation (shift)
of $A$ by $b$ along the number line.  If $A = [r,s]$ and $B = [u,v]$
are closed intervals, then note $A + B = [r+u, s+v]$ and $cA = [cr,cs]$.
The ordinary distributive law in ${\bf R}$ implies that
$r(A+B) = rA + rB$ and $(r+s)A = rA + sA$.

We show that $C_n + C_n = [0,2]$ for all $n \ge 1$ by induction.
The base case is $n = 1$, where $C_1 = [0,1/3] \cup [2/3,1]$.
Hence 
$$\eqalign{
C_1 + C_1 &= ([0,1/3] + [0,1/3]) \cup ([0,1/3]+[2/3,1]) \cup ([2/3,1]+[2/3,1])\cr
&= [0,2/3]\cup [2/3,4/3] \cup [4/3,2]\cr
&=[0,2]\cr}$$
Now, assume as induction hypothesis that $C_k + C_k = [0,2]$, and 
consider $C_{k+1}$.  The key observation we need here is that 
$C_{k+1}$ is the union of two scaled and translated copies of $C_k$:
$$C_{k+1} = (1/3) C_k \cup ((1/3)C_k + 2/3).$$
Hence, using the induction hypothesis:
$$\eqalign{ C_{k+1} + C_{k+1} &= ((1/3) C_k + (1/3) C_k) \cup 
((1/3) C_k + (1/3)C_k + 2/3) \cr
&\qquad\quad \cup ((1/3)C_k + 2/3 + (1/3)C_k + 2/3)\cr
&= (1/3)[0,2] \cup ((1/3)[0,2] + 2/3) \cup ((1/3)[0,2] + 4/3)\cr
&= [0,2/3] \cup [2/3,4/3] \cup [4/3,2]\cr
&= [0,2].\cr}$$
Hence $C_n + C_n = [0,2]$ for all $n\ge 1$.

2)  By part 1, we know that given $s\in [0,2]$, 
for each $n$ there are $x_n,y_n \in C_n$ such that $s = x_n + y_n$.
We want to show that $s = x + y$ for some $x,y\in C = \cap_{n=1}^\infty C_n$.
The sequences $(x_n)$ and $(y_n)$ live in the bounded set $[0,1]$.
Hence by the Bolzano-Weierstrass Theorem, there exists a convergent
subsequence $(x_{n_k})$.  By Bolzano-Weierstrass again, from the sequence
$(y_{n_k})$ we can select a convergent subsequence $(y_{n_{k_\ell}})$ and
$(x_{n_{k_\ell}})$ is a subsequence of a convergent sequence so 
it converges too.  Let $x = \lim_{\ell\to\infty} x_{n_{k_\ell}}$
and $y = \lim_{\ell\to\infty} y_{n_{k_\ell}}$.
Since $s = x_n + y_n$ for all $n$, we still have $s = x + y$
by the Algebraic Limit Theorem.  

Then to finish, we must show that $x,y \in C = \cap_{n=1}^\infty C_n$. 
To do this in a ``nice'' way, we will use some of the ideas
from class on Friday, 10/28.
First for each $\ell$, we have the set $C_{n_{k_\ell}}$, which
is a finite union of closed intervals, hence closed.   We
claim that $x,y \in C_{n_{k_\ell}}$.  This follows because
of the fact that the $C_n$ are nested:
$C_1 \supset C_2 \supset C_3 \supset \cdots$.  For all $\ell' > \ell$,
$x_{n_{k_{\ell'}}} \in C_{n_{k_\ell}}$ too.  
This means $x,y$ are limit points of convergent sequences 
in the closed set $C_{n_{k_\ell}}$.  Hence $x,y$ 
are in $C_{n_{k_\ell}}$.  

Next, we claim that 
this implies that $x,y$ are in $C_n$ for all $n$.  The reason
is that because $n_{k_\ell}$ is an increasing sequence
in ${\bf N}$, given any $n$ there exists an $\ell$ such that 
$n_{k_\ell} > n$ and hence $x,y \in C_{n_{k_\ell}} \subset C_n$).
It follows that $x,y\in C$, since they are in $C_n$ for all $n$.

\vfill\eject
\centerline{Mathematics 242 -- Principles of Analysis}
\centerline{Problem Set 6 Solutions}
\centerline{November 11, 2005}
\bigskip
\noindent
4.2.1.  a)  Given $\varepsilon > 0$, let $\delta = \varepsilon/2$.
Then for all $x$ with $0 < |x - 2| < \delta$, we have
$$\eqalign{
|(2x + 4) - 8| &= 2|x - 2| \cr
             &< 2(\varepsilon/2) \cr
             &= \varepsilon.\cr}$$
This shows $\lim_{x\to 2} 2x + 4 = 8$.

b)  Given $\varepsilon > 0$, let $\delta = {\root 3\of \varepsilon}$.
Then for all $x$ with $0 < |x| < \delta$, we have
$$\eqalign{
|x^3 - 0| &= |x|^3 \cr
             &< ({\root 3\of \varepsilon})^3 \cr
             &= \varepsilon.\cr}$$
This shows $\lim_{x\to 0} x^3 = 0$.

c)  Given $\varepsilon > 0$, let $\delta = \min(1,\varepsilon/19)$.
Then for all $x$ with $0 < |x - 2| < \delta$, we have $1 < x < 3$ so 
$|x^2 + 2x + 4| < 3^2 + 2\cdot 3 + 4 = 19$ (because $x^2 + 2x + 4$
is increasing for all $x > -1$).  Therefore
$$\eqalign{
|x^3 - 8| &= |x - 2| |x^2 + 2x + 4|\cr
             &< (\varepsilon/19)\cdot 19 \cr
             &= \varepsilon.\cr}$$
This shows $\lim_{x\to 2} x^3 = 8$.

d)  Given $\varepsilon > 0$, let $\delta = \pi - 3 \doteq .14\cdots$.
Then for all $x$ with $0 < |x-\pi| < \delta$, we have 
$3 < x < 2\pi - 3 \doteq 3.28\cdots$, so $[[x]] = 3$ for all such $x$ and
$$|[[x]] - 3| = 0 < \varepsilon.$$
This shows $\lim_{x\to \pi} [[x]] = 3$.

\bigskip
\noindent
4.2.3. a)  Let $x_n = 1/n$.  Then $|x_n|/x_n = 1$ for all $n\ge 1$,
and $\lim |x_n|/x_n = 1$.  On the other hand, let $y_n = -1/n$.  
Then $|y_n|/y_n = -1$ for all $n\ge 1$, and $\lim |y_n|/y_n = -1$.
Since we have two sequences converging to the same limit
$\lim x_n = 0 = \lim y_n$, but $\lim f(x_n) \ne \lim f(y_n)$, 
Corollary 4.2.5 implies that $\lim_{x\to 0} |x|/x$ does not exist.

b)  Let $x_n$ be any sequence of rational numbers converging to 
$1$ (for instance $x_n = 1 + 1/n$).  Then $\lim g(x_n) = 1$, since
$g(x_n) = 1$ for all $n\ge 1$.  On the other hand, let $y_n$
be any sequence of irrational numbers converging to 1 (for
instance $y_n = 1 + \sqrt{2}/n$.  Then $\lim g(y_n) = 0$, since
$g(y_n) = 0$ for all $n\ge 1$.  Corollary 4.2.5 implies
that $\lim_{x\to 1} g(x)$ does not exist.

\bigskip
\noindent
4.3.1. a)  Given $\varepsilon > 0$, let $\delta = \varepsilon^3$.
Then for all $x$ with $|x| < \delta$, we have
$$\eqalign{
|{\root 3 \of x} - 0| &= {\root 3\of {|x|}}\cr 
&< {\root 3\of {\varepsilon^3}}\cr
                      &= \varepsilon.\cr}$$
This shows $\lim_{x\to 0} {\root 3 \of x} = 0 = {\root 3\of 0}$.
So $g(x) = {\root 3\of x}$ is continuous at $x = 0$.

b)   Given $\varepsilon > 0$, let $\delta = \min(|c|, c^{2/3} \varepsilon)$.
Then for all $x$ with $|x - c| < \delta$, first since $\delta < |c|$, 
$c - |c| < x < c + |c|$.  So,
it follows that $x$ and $c$ have the {\it same sign} (both $+$ or
both $-$).  Then by
the identity $a^3 - b^3 = (a-b)(a^2 + ab + b^2)$ and the fact 
that all terms in the denominator on the right of the first line
are positive:
$$\eqalign{
|{\root 3 \of x} - {\root 3\of c}| &= \left|{ (x - c) \over 
x^{2/3} + {\root 3\of c} {\root 3 \of x} + c^{2/3} }\right|\cr
&\le { |x - c| \over c^{2/3} }\cr
&< {c^{2/3} \varepsilon\over c^{2/3}}\cr
&= \varepsilon}$$
This shows $\lim_{x\to c} {\root 3 \of x} = {\root 3\of c}$.
So $g(x) = {\root 3\of x}$ is continuous at $x = c$.

\bigskip
\noindent
4.3.2. a)  Let $\varepsilon > 0$.  Since $g$ is continuous at
$f(c)$, there exists a $\delta > 0$ such that if $|y - f(c)| < \delta$,
then $|g(y) - g(f(c))| < \varepsilon$.   But then since $f$
is continuous at $c$, for this $\delta$, there exists a
$\delta_1 > 0$ such that if $|x - c| < \delta_1$, then
$|f(x) - f(c)| < \delta$.  For any such $x$, taking $y = f(x)$ in the first 
set of inequalities, $|f(x) - f(c)| < \delta$ implies
$|g(f(x)) - g(f(c))| < \varepsilon$.  This shows that 
$g(f(x)) = (g\circ f)(x)$ is continuous at $c$.

b)  Let $x_n$ be any sequence converging to $c$ in the domain
of $g\circ f$.  Then since $f$ is continuous at $c$, part (iv) of 
Theorem 4.3.2 implies that $f(x_n)$ is a sequence converging
to $f(c)$ in the range of $f$.  Using the same part of the 
theorem, since $g$ is continuous at $f(c)$, $g(f(x_n))$ is a
sequence converging to $g(f(c))$.  This shows that for
any sequence $x_n$ converging to $c$, $(g\circ f)(x_n)$
is a sequence converging to $(g\circ f)(c)$.  By the 
equivalence of parts (i) and (iv) in the theorem, this
implies that $g\circ f$ is continuous at $c$.
%
%\bigskip
%\noindent
%4.3.3. If $a = 0$, then $f = b$ is constant and is clearly 
%continuous.  So we assume $a \ne 0$ in the following.
%Let $c\in {\bf R}$ be arbitrary
%Given $\varepsilon > 0$, let $\delta = \varepsilon/a$.
%Then for all $x$ with $0 < |x - c| < \delta$, we have
%$$\eqalign{
%|(ax + b) - (ac + b)| &= a|x - c| \cr
%             &< a(\varepsilon/a) \cr
%             &= \varepsilon.\cr}$$
%This shows $\lim_{x\to c} ax + b = ac + b = f(c)$.
%Hence $f(x) = ax + b$ is continuous at all $c \in {\bf R}$.
%
\bigskip
\noindent
4.3.9.  a)  Let $\varepsilon > 0$ and take $\delta = \varepsilon/c$.
Then for any $a\in {\bf R}$, for all $x$ with $|x - a| < \delta$,
$$|f(x) - f(a)| \le c |x - a| < c( \varepsilon/c) = \varepsilon.$$
Hence $f$ is continuous at all $a \in {\bf R}$.  (Note: this would be
true even if $c > 1$!)

b) (Note: for notational convenience, I'm starting the sequence from 
$n = 0$ rather than $n = 1$ as in the text.  Also, instead
of writing $|y_m - y_n|$ as in the usual definition of 
Cauchy sequences, it will be easier to take $m > n$ and 
write $m = n + k$.)
 Define the sequence $(y_n)$ by picking $y_0$ arbitrarily, 
then iterating: $y_{n+1} = f(y_n)$ for all $n \ge 0$.  Let $n$ be
arbitrary and $k \ge 1$.  Then by an easy induction argument
$|y_{n+k} - y_n| =$
$$ |f(y_{n+k-1}) - f(y_{n-1})| \le c|y_{n+k-1} - y_{n-1}|
\le c^2|y_{n+k-2} - y_{n-2}| \le \cdots \le c^n |y_k - y_0|$$
We might be tempted to say that $\lim c^n = 0$, so this tends to zero,
but we must still deal with the fact that $|y_k - y_0|$ depends on $k$.
We can deal with $|y_k - y_0|$ as follows.  By the triangle inequality,
and similar reasoning to the above:
$$\eqalign{
|y_k - y_0| &\le |y_k - y_{k-1}| + |y_{k-1} - y_{k-2}| + \cdots + |y_1 - y_0|\cr
&\le (c^{k-1} + c^{k-2} + \cdots + 1)|y_1 - y_0|\cr
&\le {1\over 1 - c}|y_1 - y_0|\cr}$$
(where in the last line we have used the sum formula for the convergent
geometric series $\sum c^n$ and eliminated the dependence on $k$).
Putting everything together we have
$$|y_{n+k} - y_n| \le {c^n \over 1 - c}|y_1 - y_0|.$$
Now (since the factors $|y_1 - y_0|\cdot {1\over 1-c}$ are fixed)
we can argue that for any $\varepsilon > 0$, there exists some
$N$ such that $n \ge N$ (hence $n + k \ge N$ too)
implies ${c^n \over 1 - c}|y_1 - y_0| < \varepsilon$,
so the sequence $y_n$ is Cauchy.

c)  Let $y$ be the limit of the sequence $y_n$ from part b.  Since
$f$ is continuous (part a), we have by part (iv) of Theorem 4.3.2
$$f(y) = f(\lim y_n) = \lim f(y_n) = \lim y_{n+1} = y.$$
In other words, $y$ is a {\it fixed point of $f$}.

For the last part of this question, we must show that there is only
one fixed point of $f$.  So suppose there are two:  $y,y'$ with
$f(y) = y$ and $f(y') = y'$.  By hypothesis on $f$:
$$|y - y'| = |f(y) - f(y')| \le c |y - y'|$$
with $0 < c < 1$.  This is only possible if $|y - y'| = 0$.  So 
$y = y'$ and the fixed point is unique.

d) This follows from parts b and c.  No matter where we start $x = y_0$,
part b says we get a Cauchy, hence convergent sequence by iterating $f$.  
Part c says the limit is a fixed point of $f$, but $f$ has only one 
fixed point.  

\bigskip
\noindent
4.3.12. a)  Let $c \in C$.  We claim that there exist sequences $x_n, y_n$
with $\lim x_n = c = \lim y_n$ with $x_n \in C$ and $y_n \notin C$.
The easiest way to see this is to recall one of the facts we 
proved in Discussion 2:  {\it $c$ is an element of the Cantor 
set if and only it has a ternary expansion
$$c = (.d_1d_2\cdots)_3 = {d_1\over 3} + {d_2 \over 3^2} + \cdots$$ 
containing ternary 
digits $d_i$ that are all either $0$ or $2$.}.  
We can get a sequence $x_n\in C$ converging to $c$ by 
taking 
$$x_n = {d_1\over 3} + {d_2 \over 3^2} + \cdots + {d_n\over 3^n}$$
with 0's the rest of the way.   Similarly we can get a sequence
$y_n\notin C$ by replacing the $n$th digit
in the ternary expansion of $c$ by a $1$.  
Then $y_n \to c$ also since the difference between $y_n$
and $c$ is $\pm {1\over 3^n}$ and goes to zero as $n\to \infty$.  
This construction gives $\lim g(x_n) = 1 \ne 0 = \lim g(y_n)$.  Hence
$g$ is not continuous at $c$ by Corollary 4.2.5.

b)  The complement of $C$ is open, since $C$ is closed.  Hence
for every $c$ in the complement of $C$, there is an open interval
$(c-\delta,c+\delta)$
centered at $c$ contained in the complement.  The value of $g$
at all points in that interval is $0$ so $g$ is continuous at $c$
(reason: $|g(x) - g(c)| = |0 - 0| < \varepsilon$ whenever
$|x - c| < \delta$.)

\vfill\eject
\centerline{Mathematics 242 -- Principles of Analysis}
\centerline{Problem Set 7 Solutions}
\centerline{November 18, 2005}
\bigskip
\noindent
I.

\bigskip
\noindent

%4.4.2.  Let $x,y \in [1,\infty]$, then since $x,y \ge 1$
%$$\left| {1\over x} - {1\over y}\right| = {|x - y| \over xy} < |x - y|.$$
%Hence if $|x - y| < \delta = \varepsilon$, then $|1/x - 1/y| < \varepsilon$
%(note $\delta$ doesn't depend on where $x,y$ are, just how far apart
%they are!)
%
%On the other hand for $x,y$ in the set $(0,1]$, the difference
%is that ${|x - y| \over xy}$  cannot be bounded by any constant multiple 
%of $|x - y|$.  
%To prove nonuniform continuity, though, Theorem 4.4.6 is definitely
%the easiest way:  Let $x_n = 1/n$ and $y_n = 1/(2n)$.
%Then $\lim |x_n - y_n| = \lim 1/(2n) = 0$, but
%$|f(x_n) - f(y_n)|  = |n - 2n| = n$, which is not even bounded
%as $n \to \infty$.  We can take $\varepsilon_0 = 1$ in Theorem 4.4.6.
%
%\bigskip
%\noindent
4.5.2. a)  {\it False:}  Let $f(x) = x^2$ (continous for all $x$). 
$f((-1,1)) = [0,1)$, which is not open (but is bounded of course!)

b) {\it False:} See part a.

c)  {\it True:}  Follows from the Extreme and Intermediate
Value Theorems:  Assume $f$ is continuous on $[a,b]$.  
By the Extreme Value Theorem, $f$ attains a maximum $s$ and
a minimum $t$ on $[a,b]$.  Say $f(x_1) = s$ and $f(x_2) = t$.
Then the Intermediate Value Theorem implies that for all real 
$r$ with $t < r < s$, there exist $x$ in the interval
with endpoints $x_1$ and $x_2$ with $f(x) = r$.
Hence $f([a,b])$ is the closed interval $[t,s]$. 

\bigskip
\noindent
4.5.3.  There are no such functions.  Assume that there was
a continuous $f : {\bf R} \to {\bf R}$ with $f({\bf R}) = {\bf Q}$.
Let $r_1 < r_2$ be any two rational numbers.  Then there
are $x_1$ and $x_2$ with $f(x_1) = r_1$ and $f(x_2) = r_2$.  
But in the interval $(r_1,r_2)$, by the Density Theorems
there are irrationals too:  there always exist irrational
$s$ with $r_1 < s < r_2$.  By the Intermediate Value Theorem,
applied to $f$ on the interval with endpoints $x_1,x_2$, there
exists some $x$ with $f(x) = s$.  This is a contradiction
because we assumed $f$ took only rational values. 
 
\bigskip
\noindent
4.5.7.  If $f(0) = 0$ or $f(1) = 1$ then there
is nothing to prove.  So we will assume $0 < f(0) \le 1$ and 
$0\le f(1) < 1$.  Consider the function $g(x) = f(x) - x$.  
We know $g$ is continuous on $[0,1]$ by the Algebraic Continuity 
Theorem (Theorem 4.3.4).  
We have $g(0) = f(0) - 0 > 0$, but $g(1) = f(1) - 1 < 0$.
Hence by the Intermediate Value Theorem, there must
be some $c$ in $(0,1)$ where $g(c) = f(c) - c = 0$.
This means $f(c) = c$, so $c$ is a fixed point of $f$.

\bigskip
\noindent
5.2.3.  Let $$f(x) = \cases{ x^2 & if $x$ is rational\cr
                            0 & if $x$ is irrational\cr}$$
We claim $f$ is differentiable at $0$ ($f'(0) = 0$), but 
$f$ is not differentiable at any other point.  To see
$f'(0) = 0$, note that 
$${f(x) - f(0)\over x - 0} =  \cases{x & if $x$ is rational\cr
                               0 & if $x$ is irrational\cr}$$
Hence $\lim_{x\to 0} {f(x) - f(0)\over x - 0} = 0$, and $f$ is
differentiable at $0$ with $f'(0) = 0$.  

On the other hand, $f$ is {\it not differentiable} at any other 
$x\in {\bf R}$ because it is not even continuous there.
Note that if $c \ne 0$, then there is a sequence
$x_n \to c$ with $x_n \in {\bf Q}$ for all $n$, 
and another sequence $y_n \to c$ with $y_n \notin {\bf Q}$ for 
all $n$.  Then $\lim_{n\to\infty} f(x_n) = \lim_{n\to\infty} x_n^2 = c^2$, 
but $\lim_{n\to\infty} f(y_n) = 0$.
Since $c\ne 0$, this says $\lim_{x\to c} g(x)$ does not exist.  

\bigskip
\noindent
5.2.4.  Note that the two ``halves'' of $f_a$ are continuous
separately for all $a$ (but $f_a(0)$ is undefined as stated if $a < 0$).  

a)  $\lim_{x \to 0^+} x^a = 0$ for all $a > 0$, so $f_a(x)$ is 
continuous at $x = 0$ for all $a > 0$.

b)  $f_a$ is differentiable at $x = 0$ as long as $a > 1$, because
$$\lim_{x\to 0} {f_a(x) - f_a(0) \over x - 0} = 
\lim_{x\to 0} \cases{x^{a-1} & if $x$ rational \cr 0 & if $x$ irrational}
= 0.$$
The derivative
function is $f_a'(x) = \cases{ ax^{a-1} & if $x \ge 0$\cr 0 & if $x < 0$\cr}$,
so the derivative function is continuous for all $a > 1$ by the same
reasoning as in part a (it's a constant times the function $f_{a-1}(x)$).

c)  By the last observation in part b,
$f_a'$ is differentiable at $x = 0$ as long as $a - 1 > 1$, or $a > 2$.

\bigskip
\noindent
II.  
\item{A)}  We will use the fact that if $f$ is a polynomial of odd degree, then
as $x\to \pm \infty$, one of the limits is $+\infty$ and the other is $-\infty$
(which is which depends on the sign of the leading coefficient).  Hence if $B$
is sufficiently large $f(B)f(-B) < 0$.  So by the Intermediate Value Theorem 
$f$ has a root in the interval $[-B,B]$.  
\item{B)}  If $f$ has even degree, then as $x\to \pm \infty$, both of the limits are
either $+\infty$ or $-\infty$
(which it is again depends on the sign of the leading coefficient). 
Hence if $f(a) < 0 < f(b)$, the Intermediate Value Theorem implies there is 
a root of $f$ in $[a,b]$.  But then $f$ must change sign again on $(-\infty,a)$
if $\lim_{x\to -\infty} f(x) = +\infty$ (leading coefficient positive), 
or on $(b,+\infty)$ if  $\lim_{x\to +\infty} f(x) = -\infty$ (leading coefficient negative).
It follows that in any case, $f$ has at least two distinct real roots.
\item{C)}  {\it False:}  There could be a single double 
root, as in the case of $f(x) = x^2$.

\bigskip
\noindent
III.  Consider $f(x) = x^2 - 3^x$. $f$ is continuous on all of ${\bf R}$
since it is a sum of continuous functions.
Note that $f(-1) = 1 - 1/3 = 2/3 > 0$
while $f(0) = 0 - 1 = -1 < 0$.  Hence by the Intermediate Value
Theorem, there is some $c \in (-1,0)$ such that $f(c) = 0$, 
and hence $c^2 = 3^c$.  

\bigskip
\noindent
IV.  
\item{A)}  This is like the proof of 4.5.7 above.  Consider
$g(x) = f(x) - x^2$.  We have $g(0) = f(0) - 0 < 0$ and
$g(1) = f(1) - 1 > 0$.  Hence by the Intermediate Value
Theorem, there is some $c\in (0,1)$ where $g(c) = 0$, so
$f(c) = c^2$.  
\item{B)}  Consider  $h(x) = f(x) - g(x)$.  We have $h(0) = f(0) - g(0) < 0$ and
$h(1) = f(1) - g(1) > 0$.  Hence by the Intermediate Value
Theorem, there is some $c\in (0,1)$ where $h(c) = 0$, so
$f(c) = g(c)$.  

\vfill\eject
\centerline{Mathematics 242 -- Principles of Analysis}
\centerline{Problem Set 9 Solutions}
\centerline{November 18, 2005}
\bigskip
\noindent
5.3.2.  Call the closed interval $A$.
By the Mean Value Theorem, we have for all $x,y \in A$
$f(x) - f(y) = f'(c)(x - y)$ for some $c\in (x,y)$.  Taking absolute
values $|f(x) - f(y)| = |f'(c)| |x - y|$.  By hypothesis, 
$f'$ is continuous on $A$, hence by the Extreme Value Theorem, 
it attains a maximum $s = \max |f'| < 1$.
Therefore,  $|f(x) - f(y)| \le s |x - y|$ for all $x,y \in A$.
It follows that $f$ is contractive on $A$.

\vfill\eject
\centerline{Mathematics 242 -- Principles of Analysis}
\centerline{Solutions -- Problem Set 8}
\centerline{December 7, 2005}
\bigskip
\noindent
5.3.3.  a)  $h$ is differentiable, therefore continuous on $[0,3]$.  Hence
the function $g(x) = h(x) - x$ is also continuous on $[0,3]$.  We 
have $g(0) = h(0) = 1$, and $g(3) = h(3) - 3 = -1$.  So $h$ changes
sign on $[0,3]$.  By the Intermediate Value Theorem, there is some
$d \in (0,3)$ where $g(d) = 0$, so $h(d) = d$.

b)  By the Mean Value Theorem, there is some $c \in (0,3)$ where
$1 = h(3) - h(0) = h'(c)(3 - 0)$, so $h'(c) = 1/3$.

c)  By an argument similar to that for part b, there is some
$b\in (1,3)$ where $0 = h(3) - h(1) = h'(b)(3 - 1)$, so $h'(b) = 0$.
Now by the intermediate value property for derivatives (the Darboux theorem
5.2.7), since $0 < 1/4 < 1/3$, there is some $x$ in the interval 
where $h'(x) = 1/4$.

\bigskip
\noindent
5.3.4.  a)  We are given that $f,g$ are continuous on $[a,b]$ and
differentiable on $(a,b)$.  Consider 
$h(x) = (f(b) - f(a))g(x) - (g(b) - g(a))f(x)$.  $h$ is also
continuous on $[a,b]$ and differentiable on $(a,b)$ by the 
usual algebraic properties of limits and derivatives.  
By the Mean Value Theorem, there exists some $c\in (a,b)$
where $h(b) - h(a) = h'(c)(b - a)$.  But $h(b) = g(a)f(b) - f(a)g(b)$
and $h(a) = f(b)g(a) - f(a)g(b) = h(b)$. Hence $h'(c) = 0$. But
 $h'(c) = (f(b) - f(a))g'(c) - (g(b) - g(a))f'(c)$, which implies
$${f(b) - f(a) \over g(b) - g(a)} = {f'(c)\over g'(c)},\leqno(1)$$
which is what we wanted to show.

b) Consider the parametric curve defined by $x = f(t), y = g(t)$
for $t \in [a,b]$.  Then part a says that there is some $c$ 
where the slope of the tangent line:
$${dy \over dx} = {dy/dt \over dx/dt} = {g'(c) \over f'(c)} $$
equals the slope of the secant line through the endpoints
$(f(a),g(a))$ and $(f(b),g(b))$:  ${g(b) - g(a) \over f(b) - f(a)}$
(of course, the equation in (1) is obtained by taking 
reciprocals on both sides).

%\bigskip
%\noindent
%5.3.5.  Suppose we have two different fixed points in the interval, say
%$f(s) = s$ and $f(t) = t$.  By the Mean Value Theorem, 
%$s - t = f(s) - f(t) = f'(c) (s - t)$ for some $c$ between $s$ and $t$.
%But this implies $f'(c) = 1$.  If $f'(x) \ne 1$ for all $x$ in the
%interval this gives a contradiction.  Hence there is at most one
%fixed point in the interval.

%\bigskip
%\noindent
%5.3.7. a) $\Leftarrow$:  If $f$ is increasing and differentiable on $(a,b)$, 
%let $x \in (a,b)$ be arbitrary.
%For $y < x$ we have ${f(y) - f(x) \over y - x} \ge 0$ and if 
%$x < y$ we have ${f(y) - f(x) \over y - x} \ge 0$.  Hence the limit
%of the difference quotient must satisfy
%$f'(x) = \lim_{y\to x} {f(y) - f(x) \over y - x} \ge 0$. 
%
%$\Rightarrow$:  By the Mean Value Theorem, if $f'(x) \ge 0$ for all
%$x\in (a,b)$, then for all $y > x$, there is some $c \in (a,b)$ such that
%$f(y) - f(x) = f'(c) (y - x)$.  Since $f'(c) \ge 0$ and $y > x$, this shows
%$f(y) - f(x) \ge 0$.  Hence $f(y) \ge f(x)$, so $f$
%is increasing on $(a,b)$.
%
%b)  If $x\ne 0$, we can use the usual derivative
%rules to find $g'(x) = 1/2 + 2x\sin(1/x) - \cos(1/x)$.
%If $x = 0$, then we need to use the definition and compute
%$\lim_{h\to 0} {g(h) - g(0) \over h} = \lim_{h\to 0} 1/2 + h \sin(1/h) = 1/2$.
%Hence $g$ is differentiable at $0$ also and $g'(0) > 0$. 
%
%We claim that $g$ is not increasing on any open interval
%containing $x = 0$.  For each $n\in {\bf N}$, let $x_n = {1\over 2n\pi} $.  Then
%by the first part of the above, 
%$$g'(x_n) = 1/2 + {2\over 2n\pi}\sin(2n\pi ) - \cos(2n\pi)
%= 1/2 - 1 = -1/2.$$
%By part a, this shows that $g$ cannot be increasing on any 
%interval containing ${1\over 2n\pi}$.
%Since $\lim {1\over 2n\pi} = 0$, every interval containing zero contains
%one (in fact infinitely many) of these points, so $g$ cannot be increasing 
%on any interval containing zero.  
%
\bigskip
\noindent
7.2.2. a)  $L(f,P) = (3\cdot 1/2) + (4\cdot 1/2) + (5\cdot 1) = 17/2$.
$U(f,P) =  (4\cdot 1/2) + (5\cdot 1/2) + (7\cdot 1) = 23/2$, and 
$U(f,P) - L(f,P) = 3$.

b) We get $U(f,P) - L(f,P) = 22/2 - 18/2 = 2$.  

c) The regular partition with any $n \ge 5$ is one.

\bigskip
\noindent
7.2.3.  If $f(x) = c$ is constant, then for each subinterval in the
partition $m_k = M_k = c$.  Hence 
$L(f,P) = \sum_k c(x_k - x_{k-1})$.  This sum telescopes to 
$L(f,P) = c(b - a)$.  Similarly $U(f,P) = c(b-a)$.  Since
$U(f,P) - L(f,P) = 0$ for all $P$,  the criterion from Theorem 7.2.8
is certainly met, and $f$ is integrable on $[a,b]$.  The integral
is $\int_a^b f = c(b-a)$.

\bigskip
\noindent
7.2.4. a) This follows directly from Theorem 7.2.8.  If
$\lim_{n\to\infty} U(f,P_n) - L(f,P_n) = 0$, then 
for any $\varepsilon > 0$, there exists $N$ such that whenever
$n \ge N$, $U(f,P_n) - L(f,P_n) < \varepsilon$.

b) $P_n = \{i/n: 0\le i \le n\}$, so 
$$U(f,P_n) = \sum_{i=1}^n {i\over n}\cdot {1\over n} = {1\over n^2}\sum_{i=1}^n i = {n^2 + n\over 2n^2}
= {1\over 2} + {1\over 2n}$$
Similarly:
b) $P_n = \{i/n: 0\le i \le n\}$, so 
$$L(f,P_n) = \sum_{i=0}^{n-1} {i\over n}\cdot {1\over n} = {1\over n^2}\sum_{i=0}^{n-1} i = {n^2 - n\over 2n^2}
= {1\over 2} - {1\over 2n}$$

c)  From part b, we have $U(f,P_n) - L(f,P_n) = {1\over n}$.
Hence $\lim_{n\to\infty} U(f,P_n) - L(f,P_n) = 0$ and part a implies
that $f(x) = x$ is integrable on $[0,1]$.  The value of $\int_0^1 x$ is 
$$\lim_{n\to\infty} U(f,P_n) = \lim_{n\to\infty} L(f,P_n) = 1/2.$$

%\bigskip
%\noindent
%7.2.6.  Consider the regular partition $P_n$ 
%of $[a,b]$ into $n$ equal subintervals: $P_n = \{x_i : 0 \le i \le n\}$
%with $x_i - x_{i-1} = (b-a)/n$ for all $i$.  Then since $f$ is increasing
%$M_i = f(x_i)$ for all $i$ and 
%$$U(f,P_n) = {(b - a)\over n} \sum_{i = 1}^n f(x_i)$$
%Similarly, $m_i = f(x_{i-1})$ for each $i$ and 
%$$U(f,P_n) = {(b - a)\over n} \sum_{i = 1}^{n} f(x_{i-1}) = {(b - a)\over n} \sum_{i = 0}^{n-1} f(x_i).$$
%In the difference $U(f,P_n) - L(f,P_n)$, all the terms
%except the last one in $U$ and the first one in $L$ cancel, leaving
%$$U(f,P_n) - L(f,P_n) = (b-a)(f(b) - f(a))\cdot {1\over n}$$
%(note $x_n = b$ and $x_0 = a$).  Since $(b - a)(f(b) - f(a))$ is constant,
%we have $\lim_{n\to\infty} U(f,P_n) - L(f,P_n) = 0$.  Therefore by part a
%of Exercise 7.2.4 (or Theorem 7.2.8), $f$ is integrable on $[a,b]$. 

%\vfill\eject
%\centerline{Mathematics 242 -- Principles of Analysis}
%\centerline{Problem Set 10 Solutions}
%\centerline{November 28, 2005}

\bigskip
\noindent
7.3.1. a)  No matter which partition we take, every subinterval
will contain points where $f(x) = 1$ so, $m_i = \inf \{f(x) : x_{i-1} \le x \le x_i\} = 1$ for all $i$.
Therefore $L(f,P) = \sum_{i=1}^n (x_i - x_{i-1}) = x_n - x_0 = 1$.

b)  We can take any partition where the subinterval containing $x = 1$ has width
$< 1/10$.  A ``lazy'' example is the partition $P = \{0,10/11,1\}$
Then $U(f,P) = 1\cdot 10/11 + 2\cdot 1/11 = 12/11 = 1 + 1/11 < 1 + 1/10$.

c)  Along the lines of part b, given $\varepsilon > 0$, let
$\delta < \varepsilon$ and consider $P = \{0,1-\delta,1\}$.
Then $U(f,P) = 1\cdot (1-\delta) + 2\cdot \delta = 1 + \delta < 1 + \varepsilon$.

\bigskip
\noindent
7.3.4. a)  This can either be proved by an argument following the 
ideas of the proof of Theorem 7.3.2, or by using that theorem
and some of the results in the next section.  Here is the second approach:
Say $g$ has the same value as $f$ except at $c \in [a,b]$.  Then
$g$ and $f$ are equal on $[d,b]$ for all $d$ with $b > d > c$.  This says $g$ is integrable
on $[c,b]$ by Theorem 7.3.2.  Similarly $g$ is integrable on $[a,c]$.  
Finally Theorem 7.4.1 implies $g$ is integrable on $[a,b]$.  

Next, we will show $\int_a^b f = \int_a^b g$.  
Say $f(c) - g(c) > 0$ first for simplicity.
The difference $f(x) - g(x)$ is zero except at $x = c$ and integrable by 
Theorem 7.4.2.  If we take a partition $P=\{a,c-\delta,c+\delta,b\}$
with $\delta < {\varepsilon \over 2(f(c) - g(c))}$ then
$U(f-g,P) = (f(c) - g(c)) \cdot 2\delta < \varepsilon$.  Since $\varepsilon$
can be any positive number, this shows $\int_a^b f-g = 0$, so $\int_a^b f = \int_a^b g$.  
The other case $f(c) - g(c) < 0$ follows by an argument parallel to this but using $L(f-g,P)$,
or $U(g-f,P)$.  

b)  Suppose $f(x)$ equals $g(x)$ except at $x = c_1, \ldots, c_n$.  
Then $\int_a^b f = \int_a^b g$ follows by an induction argument in which we
alter one point at a time.  The base case is part a.  The induction step
uses exactly the same reasoning as the base case.  

c)  The Dirichlet function $f(x) = \cases{1 & if $x$ rational\cr 0 & if $x$ irrational\cr}$
is an example of this.  The function $g(x) = 0$ is integrable on $[0,1]$ since
it is constant, hence continuous.  The set of rational numbers in $[0,1]$
is countable, hence we can get $f$ from $g$ by changing a countable number of values.
But we know $f$ is not integrable according to our definition. 

\bigskip
\noindent 
7.3.5.  Let $0 < c < 1$.  Then in the interval $[c,1]$ there are only finitely
many numbers of the form $1/n$ for $n$ a natural number.  This implies that
$f$ is integrable on $[c,1]$ by part b of Exercise 7.3.4.  But then 
$f$ is also integrable on $[0,1]$ by Theorem 7.3.2.  (This can also be proved
directly of course, using a proof like the proof of 7.3.2.)  The value is 
$\int_0^1 f = 0$.  For any $0 < \varepsilon < 1$, note that
there are at most $2/\varepsilon$ integers $n$ with $1/n > \varepsilon/2$.
Let $P$ be the partition
containing $0,\varepsilon/2$ and points $1/n - \varepsilon^2/8, 1/n + \varepsilon^2/8$
for each of the $n$ with $1/n > \varepsilon/2$.  Then 
$$U(f,P) = 1\cdot \varepsilon/2 + \sum_{\{n : 1/n > \varepsilon/2\}} 1\cdot \varepsilon^2/4
< \varepsilon/2 + (2/\varepsilon)\cdot \varepsilon^2/4 = \varepsilon.$$
This shows that $\int_0^1 f = 0$ since there are partitions $P$
with $U(f,P) < \varepsilon$ for all $\varepsilon > 0$.

\bigskip
\noindent
7.4.2.  By the ``usual case'' of this additivity result, if $c \le a \le b$, then
$\int_c^b f = \int_c^a f + \int_a^b f$.  But then Definition 7.4.3 says
$\int_a^c f = -\int_c^a f$, so $\int_a^b = \int_a^c f + \int_c^b f$ works here too.

\bigskip
\noindent
7.4.4. a)  {\it False:}  Consider the function 
$f(x) = \cases{1 & if $x$ rational\cr -1 & if $x$ irrational\cr}$.
$|f|$ is the constant function with value 1 for all $x$ and hence is integrable on $[0,1]$.  
But $f$ is not integrable on $[0,1]$ for the same reason the Dirichlet function is not:
$U(f,P) = 1$ for all partitions $P$ since every subinterval contains
rational numbers, while $L(f,P) = -1$ for all partitions since every subinterval
contains irrational numbers.

b) {\it False:}  The function from Exercise 7.3.5 is a counterexample.

c) {\it True:}  If $g$ is continuous on $[a,b]$, $g(x) \ge 0$, and $g(x_0) > 0$, then
take $\varepsilon = g(x_0)/2$ in the 
definition of continuity.  There exists $\delta > 0$ such that for all $x$
with $|x - x_0| < \delta$, $|g(x) - g(x_0)| < \varepsilon/2$.  This implies
$g(x) > \varepsilon/2 > 0$ for all $x$ with $|x - x_0| < \delta$.
By an ``area under the curve'' estimate, we get
$$\int_a^b g \ge (\varepsilon/2)(2\delta) = \varepsilon\delta > 0.$$

d) {\it True:}  Since we are given $\int_a^b f > 0$, we know $f$ is integrable,
and $\sup \{L(f,P) : P \}$ is positive.  This means that for some
partitition $P$, $L(f,P) > 0$.  Hence for some subinterval $[c,d]$ 
in that partition we have $m_i = \inf \{f(x) : x\in [c,d]\} > 0$.  
Take $\delta = m_i$.  Then $f(x) \ge \delta$ for all $x\in [c,d]$.

%\bigskip
%\noindent
%7.4.5.  a)  Let $I = [x_{i-1},x_i]$ be any one of the intervals in 
%the partition $P$.  The statement to be proved follows since 
%$$M_i(f+g) = \sup_{x\in I} (f(x) + g(x)) \le \sup_{x\in I} f(x) + \sup_{x\in I} g(x) = M_i(f) + M_i(g).$$
%Hence
%$$\eqalign{
%U(f+g,P) &= \sum_i M_i(f+g)(x_i - x_{i-1})\cr
%         &\le \sum_i (M_i(f)+M_i(g))(x_i - x_{i-1})\cr
%         &= \sum_i M_i(f)(x_i - x_{i-1}) + \sum_i M_i(g)(x_i - x_{i-1})\cr
%         &= U(f,P) + U(g,P)\cr}$$
%
%An example where the inequality is strict:  Let $f(x) = x$, $g(x) = 1 - x$
%and $P = \{0,1/2,1\}$ on $[0,1]$.  Then since $f + g = 1$, $U(f+g,P) = 1$. But
%$U(f,P) = 1/4 + 1/2 = 3/4$ and $U(g,P) = 1/2 + 1/4 = 3/4$.  So $U(f+g,P) < U(f,P) + U(g,P)$.
%
%The corresponding statement for the lower sums is 
%$L(f+g,P) \ge L(f,P) + L(g,P)$.
%
%b) We use Exercise 7.2.4.  For all $P$ we have by part a
%$$\eqalign{U(f+g,P) - L(f+g,P) &\le (U(f,P) + U(g,P)) - (L(f,P)+L(g,P))\cr 
%&= (U(f,P) - L(f,P)) + (U(g,P) - L(g,P)).\cr}$$
%Since $f,g$ are integrable there is a sequence of partitions $P_n$ such that
%$\lim_{n\to\infty} U(f,P_n) - L(f,P_n) = 0$ and 
%$\lim_{n\to\infty} U(g,P_n) - L(g,P_n) = 0$.
%(We find sequences $Q_n,R_n$ for $f$,$g$ separately,then
%take the common refinement $P_n = Q_n \cup R_n$ and use Lemma 7.2.3.) 
%So then $\lim_{n\to\infty} U(f+g,P_n) - L(f+g,P_n) 
%= 0$ as well.  This shows that $f+g$ is 
%integrable. 
%
%We still need to see that $\int_a^b f+g = \int_a^b f + \int_a^b g$.  To see
%this note that from part a, with the sequence $P_n$ as above,
%$$\int_a^b f+g = \lim_{n\to\infty} U(f+g,P_n) \le \lim_{n\to\infty} U(f,P_n) + \lim_{n\to\infty} U(g,P_n)
%=\int_a^b f + \int_a^b g.$$
%But on the other hand we also have
%$$\int_a^b f+g = \lim_{n\to\infty} L(f+g,P_n) \ge \lim_{n\to\infty} L(f,P_n) + \lim_{n\to\infty} L(g,P_n)
%=\int_a^b f + \int_a^b g.$$
%Therefore $\int_a^b f+g = \int_a^b f + \int_a^b g$.
%


  


\bye
