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\centerline{Mathematics 242 -- Principles of Analysis}
\centerline{Selected Solutions -- Problem Set 1, ``B portion''}
\centerline{February 6, 2011}
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\noindent
\S 1.1/
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\noindent
3. (a) A question like this is asking for a {\it specific counterexample}
to the statement $(A - B) \cup B = A$ -- an explicit example that shows
the statement is not true.  You can find a simple example
where this fails whenever $A-B$ and $B - A$ are nonempty.
For instance, let 
$$A = \{1,2\}, B = \{2,3\}$$
Then $A - B = \{1\}$ so $(A - B) \cup B = \{1,2,3\} \ne A$.  
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(b)  From examples like this (and/or Venn diagrams), we can see that the correct statement should be $(A - B) \cup B = A \cup B$.  Here are two different
``styles'' of proofs of this, equally correct.
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\noindent
Proof 1:  We will show the two inclusions $(A - B) \cup B \subset A \cup B$
and $A \cup B \subset (A - B) \cup B$.  First, let $x \in (A - B) \cup B$.
then $x \in A - B$ or $x \in B$ by the definition of the union.  Hence
by part (d) of problem 2 (part (g) of Theorem 1.1.3), 
$x \in A$ and $x \notin B$, or $x \in B$.
This shows $x \in A$ or $x \in B$, so $x \in A \cup B$.  Hence 
$(A - B) \cup B \subset A \cup B$.  For the other inclusion, let $x \in A \cup B$.
Then $x \in A$ or $x \in B$.  If $x \in B$, then $x \in (A - B) \cup B$ by
definition of the union.  On the other hand if $x \notin B$, then it must be
true that $x \in A$, and hence $x \in A \cap B^c = A - B$ (using part (d)
of problem 2 again).  This shows $A \cup B \subset (A - B) \cup B$ and 
hence the equality we want.  //
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\noindent
Proof 2:  For this proof, we will use several parts of Theorem 1.1.3 
that incorporate the sort of element-by-element arguments in Proof 1.
We have 
$$\eqalign{
(A - B) \cup B &= (A \cap B^c) \cup B\ ({\rm by}\ 1.1.3 (g))\cr
               &= (A \cup B) \cap (B^c \cup B)\ ({\rm by}\ 1.1.3 (e))\cr
               &= (A \cup B) \cap X\ ({\rm by}\ 1.1.3 (b))\cr
               &= A\cup B\ ({\rm by\ def.\ of\ universal\ set})\cr}$$
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\S 1.2/
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\noindent
5. Let $f : A \to B$ be a function and $C,D \subset A$.
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\noindent
(a)  We want to show $f(C\cup D) = f(C) \cup f(D)$ (subsets of $B$), and 
we will do this by showing each set is contained in the other.  First, let
$y \in f(C\cup D)$.  Then $y = f(x)$ for some $x \in C\cup D$.  Hence
$y = f(x)$ for some $x \in C$ or some $x \in D$.  If $x \in C$, then
$y = f(x) \in f(C)$; if $x \in D$, then $y = f(x) \in f(D)$.  Hence
$y \in f(C) \cup f(D)$.  This shows $f(C \cup D) \subset f(C) \cup f(D)$. 
For the reverse inclusion, suppose $y \in f(C) \cup f(D)$.  Then 
$y \in f(C)$ or $y \in f(D)$.  In the first case, $y = f(x)$ for
some $x \in C$; in the second $y = f(x)$ for some $x \in D$.  
Either way, $y = f(x)$ for some $x \in C \cup D$ since then $x \in C$
or $x \in D$.  It follows that $y \in f(C\cup D)$ and 
$f(C) \cup f(D) \subset f(C\cup D)$.  
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\noindent
(b)  We want to show $f(C\cap D) \subset f(C) \cap f(D)$.  Let
$y \in f(C\cap D)$.  Then $y = f(x)$ for some $x \in C\cap D$.  Hence
$y = f(x)$ for some $x \in C$ which also satisfies $x \in D$.  Hence
$y \in f(C)$ and $y \in f(D)$, so
$y \in f(C) \cup f(D)$.  This shows $f(C \cap D) \subset f(C) \cap f(D)$. 
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\noindent
(c)  If we think carefully 
about how the reverse inclusion might {\it fail}, then it should 
be clear that knowing $y \in f(C) \cap f(D)$ just says that 
we have some $x_1 \in C$ and some $x_2 \in D$ such that
$f(x_1) = y = f(x_2)$.  There's no reason why $x_1$ or $x_2$
must come from $C \cap D$.  Here is an explicit example.
Let $f : {\bf R} \to {\bf R}$ be the function defined by $f(x) = x^2$.
Let $C = \{1\}$ and $D = \{-1\}$.  Then $f(C) = f(D) = \{1\}$, so 
$f(C) \cap f(D) = \{1\}$.  But $C \cap D = \emptyset$, so 
$f(C \cap D) = \emptyset \ne f(C) \cap f(D)$.
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\noindent
6.  We did one part of this in class(!)  You should be reading your
class notes in addition to working on the problem sets.
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12.  {\it Comments:}  This problem is really a continuation of 5 (b) and (c) above.  
There will be interconnected problems on many assignments for this
course; be sure you can spot them and can see what steps are necessary for
the later problems, {\it without repeating work you already did!}  Also,
if a problem states an extra hypothesis (like the hypothesis that $f$
is one-to-one in part (a)), you should ask yourself:  Did I use that
anywhere? How did I use it?
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\noindent
(a)  We know from 5 (b) above that $f(C\cap D) \subset f(C) \cap f(D)$
for all functions $f : A \to B$ and all subsets $C,D \subset A$.  
So, what must be proved for this part is that if $f$ is one-to-one, then
$f(C) \cap f(D)\subset f(C \cap D)$ for all $C,D$.  So let $C,D$
be subsets of $A$, and let $y \in f(C) \cap f(D)$.  Then 
by definition we know $y \in f(C)$ and $y \in f(D)$.  Hence,
$y = f(x_1)$ for some $x_1\in C$ and also (the same) $y = f(x_2)$ 
for some $x_2 \in D$.  {\it Since $f$ is assumed to be one-to-one},
and $f(x_1) = y = f(x_2)$, it must be the case that $x_1 = x_2$. 
 But then this single element
is in both $C$ and $D$, hence in $C \cap D$.  It follows that
$y \in f(C\cap D)$, so $f(C) \cap f(D) \subset f(C\cap D)$.  
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\noindent
(b)  {\it Note:}  A statement of the form ``p if and only if q'' 
means ``if p then q, and if q then p.''  To prove a statement of 
this form, you must supply arguments for {\it both} implications.  
(This is like the standard plan for showing $A = B$, since $A \subset B$
is equivalent to ``if $x \in A$, then $x \in B$'' and 
$B \subset A$ is equivalent to ``if $x \in B$, then $x \in A$''). 
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\noindent
Part (a) shows the statement ``If $f$ is one-to-one, then
$f(C \cap D) = f(C) \cap f(D)$ for all $C,D \subset A$.''  So
that is already known, and what we need to show for this part is
the reverse implication ``If $f(C \cap D) = f(C) \cap f(D)$ for all subsets
$C,D$ of $A$, then $f$ is one-to-one.''  
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Let $x_1,x_2 \in A$ and assume $f(x_1) = f(x_2) = y \in B$.  
Consider the subsets $C = \{x_1\}$ and $D=\{x_2\}$ in $A$.
By our hypothesis, we have 
$$f(C\cap D) = f(C) \cap f(D) = \{y\}.$$
But in particular, this says $C \cap D \ne \emptyset$ (since
otherwise $f(C \cap D) = \emptyset$ too).  Therefore, 
it must be true that $x_1 = x_2$, and hence $f$ is one-to-one.
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\noindent
14.  (a) If $ad - bc = 0$ and $a = 0$, then $bc = 0$, so 
$b = 0$ or $c = 0$.  In the first case we have
$f(x) = {0x + 0\over cx + d} = 0$ for all $x$ so $f$ is constant.
In the second case we have $f(x) = {0x + b\over 0x + d}$ is 
constant.  On the other hand, if $a \ne 0$, then we have
$d = {bc\over a}$ and 
$$\eqalign{
f(x) &= {ax + b\over cx + {bc\over a}}\cr
     &= {a(ax + b)\over c(ax + b)}\cr
     &= {a\over c}\cr}$$
is constant again.
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(b)  Now assume $ad - bc \ne 0$.  We can do most of this part
with one algebraic calculation.  Let $y = f(x)$, and try to solve 
for $x$ as a function of $y$.  If we can do that uniquely,
then $f$ is one-to-one, and we can determine the range and 
the formula for the inverse function all at once!   The computations:
$$\eqalign{
y &= {ax + b\over cx + d}\cr
(cx+d)y  &= ax + b\cr
x(cy - a) &= -dy + b\cr
x &= {-dy + b\over cy - a}\cr}$$
This shows that as long as $y \ne a/c$, there is a unique
$x$ such that $f(x) = y$.  If $y = a/c$, then we get
$ad - bc = 0$ which contradicts our hypothesis.  
Hence $f$ is one-to-one.  The inverse
function is also found from the result of this same calculation:
$$f^{-1}(x) = {-dx + b\over cx - a}.$$
The domain of $f$ is 
$${\rm domain}(f) = 
\cases{ {\bf R} - \{-d/c\} & if $c \ne 0$\cr
              {\bf R} & if $c = 0$\cr}
              $$ 
and the domain
of $f^{-1}$ is 
$${\rm domain}\left(f^{-1}\right) =
\cases{ {\bf R} - \{a/c\} & if $c \ne 0$\cr  
                 {\bf R} & if $c = 0$\cr}$$
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(c) Finally, we ask, when is $f(x) = f^{-1}(x)$ for all $x$.
If we set $f(x) = f^{-1}(x)$, we get
$${ax + b \over cx + d} = {-dx + b\over cx - a}$$
If this is true for all $x$ in the domains, then
$$(ax + b)(cx - a) = (cx + d)(-dx + b)$$
which implies
$$c(a+d) x^2 - (a^2-d^2) x - b(a+ d) = 0.$$
The only way this can be satisfied for infinitely many real $x$
is if $c(a + d) = 0$, $a^2 - d^2 = 0$, and $b(a + d) = 0$.
If $b \ne 0$ or $c \ne 0$, this implies $a = -d$ and the
middle equation is also satisfied.  If $b = c = 0$, then
$a = \pm d$.  Hence the $f$ for which $f = f^{-1}$ are 
all linear fractional mappings of the form
$$f(x) = {ax + b\over cx - a}$$
with $b\ne 0$ or $c \ne 0$, and 
$$f(x) = \pm x.$$ 
 

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