\magnification=\magstep1
\centerline{MATH 242 -- Principles of Analysis}
\centerline{Solutions for Problem Set 1 -- due:  Jan. 31}
\baselineskip=15pt
\bigskip
\noindent
{\it `A` Section}
\bigskip
\item{1.} Assume that $A,B$ are sets of integers.
\itemitem{a.}  What is the contrapositive of the statement: ``If $x$ is even then $x \in A \cup B$''?  Express without
using {\it not}.
\itemitem{} {\it Solution:}  The contrapositive of ``if $p$ then $q$''  is ``if not $q$ then not $p$.''
Here, by the DeMorgan Law, $x \notin A \cup B$ is equivalent to $x \in A^c$ and $x \in B^c$.  Also, ``not even''
is equivalent to ``odd.''  So, without
using {\it not}, we can state the contrapositive as ``If $x \in A^c$ and $x \in B^c$, then $x$ is odd.'' 
\itemitem{b.}  What is the converse of the statement in part a?
\itemitem{} {\it Solution:}  The converse is: ``If $x\in A\cup B$, then $x$ is even.''  
\item{2.}  Let $A = \{x \in {\bf R} \mid x^2 - 5x + 4 = 0\}$, $B = (0,1) = \{x \in {\bf R} \mid 0 < x < 1\}$ and $C = \{{x\over x^2 + 9} \mid x \in {\bf R}\}$  (Note:  $C$ is the range of the function $f$ defined by  $f(x) = {x\over x^2 + 9}$.)
\itemitem{a.}  Express the set $C$ as a union of one or more closed intervals $[a,b]$ in ${\bf R}$. (Note:  You should use facts from calculus to solve this.  Don't worry that we have not justified them yet.)
\itemitem{} {\it Solution:}  The function $f(x) = {x\over x^2 + 9}$ has $f'(x) = {9 - x^2\over (x^2 + 9)^2}$.
This is $ = 0$ at $x = \pm 3$.  Moreover $f'(x) < 0$ for $x < -3$, $f'(x) > 0$ for $-3 < x < 3$ and
$f'(x) < 0$ for $x > 3$.  Therefore, at $x = -3$, $f$ has a local minimum with $f(-3) = -1/6$.  Similarly,
at $x = 3$, $f$ has a local maximum with $f(3) = 1/6$.  We also see $\lim_{x \to \pm \infty} f(x) = 0$.
Hence $f(-3) = -1/6$ is also an absolute minimum, and $f(3) = 1/6$ is also an absolute maximum.  We will prove
a general theorem later in the course that shows that every $y$ with $-1/6 < y < 1/6$ must also be in the range,
but this can also be checked directly here since the equation 
$$y = {x\over x^2 + 9}$$
can be rearranged to $y x^2 - x + 9y = 0$. If $y = 0$, then $x = 0$.  Otherwise, by the quadratic formula this has roots
$$x = {1\pm \sqrt{1 - 36y^2}\over 2y}.$$
The expression in the square root is nonnegative exactly when $-1/6 \le y \le 1/6$ and we get $x$ with $f(x) = y$ (two of them in fact for $y \ne 0, -1/6, 1/6$).  Hence $C = [-1/6,1/6]$.
\itemitem{b.}  Find the sets $B \cap A$ and $B\cap C$.
\itemitem{} {\it Solution:}  Since $A = \{1,4\}$, we see that $B \cap A = \emptyset$ and
$B \cap C = (0,1/6]$.
\itemitem{c.}  Find the sets $B\cup A$ and $B\cup C$ and express using set notation.
\itemitem{} {\it Solution:}  We have $B \cup A = (0,1] \cup \{4\} = B$.  Then by part a,
$B \cup C = (0,1) \cup [-1/6,1/6] = [-1/6,1)$.
\item{3.}     For $n$ a general natural number, let $B_n = \{0,2n\}$.
 What are $\cap_{n=1}^\infty B_n$ and $\cup_{n=1}^\infty B_n$?
\item{} {\it Solution:}  The union, $\cup_{n=1}^{\infty} B_n$, is the set
$$\{0,2,4,\cdots\} = \{2n \mid n \ge 0\},$$
or the set of nonnegative even integers.
The intersection, $\cap_{n=1}^{\infty} B_n$, is the set $\{0\}$, since that is the
only element in $B_n$ for all $n \ge 1$.
\item{4.}  Let $I_n = [-1/n,1/n]$ for any $n \ge 1$.
 What are $\cap_{n=1}^\infty I_n$ and $\cup_{n=1}^\infty I_n$.
(Explain your reasoning intuitively.)
\item{} {\it Solution:}  Note first that $I_m \subset I_n$ whenever $m > n$.  This shows that the union
is the same as $I_1 = [-1,1]$.  The intersection contains only $0$.  We will see in about
a week how to justify the claim that for any real $a > 0$, there is some $n \ge 1$ such
that $1/n < a$.  Hence $a$ is not in the intersection.  The same is true on the negative
side:  for any $b < 0$, there exists some $n \ge 1$ such that $b < -1/n$.  Hence $b$ is
not in the intersection either.  This leaves only $0$ which does satisfy $-1/n < 0 < 1/n$
for all $n \ge 1$.
\item{5.}  Let $f : {\bf R} \to {\bf R}$ be the function defined by $f(x) = \tan^{-1}(x)$.
\itemitem{a.}  Is $f$ one-to-one?  Why or why not?
\itemitem{}  {\it Solution:}  Yes, the inverse tangent of $x$ is defined as the unique
angle $\theta$ in the interval $(-\pi/2,\pi/2)$ such that $\tan(\theta) = x$.  
So $\tan^{-1}(x) = \theta = \tan^{-1}(x')$ implies that $x = x'$.   
\itemitem{b.}  Is $f$ onto?  Why or why not?
\itemitem{} {\it Solution:}  No, since the range is just the interval 
$(-\pi/2,\pi/2)$.  
\itemitem{c.}  If $I = (0,\sqrt{3})$, what is the set $f(I)$?  Explain.
\itemitem{} {\it Solution:}  $f(I) = (0,\pi/3)$ since $\tan(0) = 0$ and $\tan(\pi/3) = \sqrt{3}$. 
\itemitem{d.}  If $J = (-\pi/4,\pi/4)$, what is the set $f^{-1}(J)$.  Explain.
\itemitem{} {\it Solution:}  $f^{-1}(J) = \{x \mid -\pi/4 < \tan^{-1}(x) < \pi/4\}$, which is the 
same as $\tan(-\pi/4) < x < \tan(\pi/4)$, so $-1 < x < 1$.  Hence $f^{-1}(J)$ is the open interval $(-1,1)$.   
\bigskip
\noindent
{\it `B' Section}
\bigskip
\item{1.}  Prove part (f) of Theorem 1.1.3 in the text.  These are the {\it De Morgan Laws} for
complements.
\item{} {\it Solution:}  We show $(A \cap B)^c = A^c \cup B^c$.  Let $x \in (A \cap B)^c$, then
$x \notin A\cap B$, which says $x \notin A$ {\it or} $x \notin B$.  But then $x \in A^c \cup B^c$,
and it follows that $(A \cap B)^c \subset A^c \cup B^c$.
Conversely, if $x \in A^c \cup B^c$, then $x \notin A$ or $x \notin B$.  This shows $x \notin A\cap B$,
so $x \in (A\cap B)^c$, and it follows that $A^c \cup B^c \subset (A \cap B)^c$.  Since we have both
inclusions, $(A \cap B)^c = A^c \cup B^c$.  The second statement $(A \cup B)^c = A^c \cap B^c$ is
proved similarly.
\item{2.}  Let $A$ and $B$ be arbitrary sets.  Does $A = A - (B - B)$, as we might expect
if we looked at the formula through the lens of ordinary algebra?  If this is always true,
prove it; if it is not, give both a counterexample (an example where the formula is not true),
and a correct statement with proof.
\item{} {\it Solution:}   This is true since for any set $B$, we have $B - B = \emptyset$.
This follows from part g of Theorem 1.1.3, for instance: $B - B = B \cap B^c = \emptyset$.
But then $A - \emptyset = A$, since $A - \emptyset = A \cap \emptyset^c = A \cap U = A$ (where
$U$ denotes the universal set).
\item{3.}  Let $f : A \to B$ be a function.
\itemitem{a.} Let $C, D$ be subsets of $A$.  Is it always
true that $f(C\cap D) = f(C) \cap f(D)$?  If this is always true prove it; if it is not,
give a counterexample.
\itemitem{} {\it Solution:}  This is not true.  For instance, let $f : {\bf R} \to {\bf R}$
be defined by $f(x) = x^2$.  Let $C = (-1,0)$ and $D = (0,1)$.  Then $f(C) = f(D) = (0,1)$, so
$f(C) \cap f(D) = (0,1)$.  But $C \cap D = \emptyset$, so $f(C\cap D) = \emptyset$ as well.
Note that other similar examples can be constructed any time that $f$ is {\it not one-to-one}.  
\itemitem{b.}  Show that $f$ is one-to-one if and only if $f^{-1}(f(C)) = C$ for all subsets $C$ of $A$.
\itemitem{} {\it Solution:}  First note that $C \subseteq f^{-1}(f(C))$ for all $f$ and all $C$
since if $x \in C$, then $f(x) \in f(C)$, so $x \in f^{-1}(f(C))$ and hence $C \subseteq f^{-1}(f(C))$. 
So what we need to show here can be restated as follows:  (1) if $f$ is one-to-one, then we need to show 
$f^{-1}(f(C)) \subseteq C$ for all $C$.
And conversely (2) if $f^{-1}(f(C)) \subseteq C$ for all $C$, then we need to show that $f$ is one-to-one.

\itemitem{} To prove (1), let $f$ be one-to-one.  For each $y \in f(C)$, there is some $x\in C$ such
that $f(x) = y$.  But if $f(x') = y$, then $f$ being one-to-one implies that $x = x'$.  Hence
the only elements of $A$ that map to $f(C)$ are the elements of $C$, so $f^{-1}(f(C)) \subseteq C$.  

\itemitem{} To prove (2), let $f^{-1}(f(C)) \subseteq C$ for all subsets $C$ of $A$.  In particular, 
let $C = \{x\}$ for some particular element $x \in A$.  Suppose that $f(x') = f(x)$.  Then
by definition, $x$ and $x'$ are both elements of $f^{-1}(f(C))$.  But by assumption $f^{-1}(f(C)) = \{x\}$, 
so $x = x'$.  This shows that $f$ must be one-to-one.
\bigskip
\item{4.}  Let $f : A \to B$ and $g : B \to C$.
\itemitem{a.}  Show that if $f$ and $g$ are both onto, then $g \circ f : A \to C$ is
also onto.
\itemitem{} {\it Solution:} Let $z \in C$.  Since $g$ is onto, there exists
$y \in B$ such that $g(y) = z$.  But then since $f$ is onto, there exists
$x \in A$ such that $f(x) = y$.  Combining these statements, we see that $g(f(x)) = (g\circ f)(x) = z$.
Since $z$ was arbitrary, this shows that $g\circ f$ is onto.  
\itemitem{b.}  Is the converse of the statement in part a true?  That is, if you know
that $g \circ f$ is onto, does it follow that $f$ and $g$ are onto?  Prove
or find a counterexample.
\itemitem{} {\it Solution:}  This statement is {\it not true}.  Let $A = B = {\bf R}$
and $C = [0,\infty)$, and let $f : A \to B$ be defined by $f(x) = x^2$ and $g(y) = \sqrt{|y|}$.
Then for all $z \in C$ we have $z = g(f(z))$, so $g\circ f$ is onto.  However,
$f$ is not onto since its range contains no negative numbers.  (It does follow in general that $g$
must be onto, but as in the counterexample, if $g$ is not one-to-one, the range of $f$
only needs to contain one inverse image of each element $z \in C$.)
\bye
