\magnification=\magstep1
\centerline{Mathematics 242 -- Principles of Analysis}
\centerline{Selected Solutions -- Problem Set 1}
\centerline{September 13, 2000}
\bigskip
\noindent
\S 1.3/12 c.  We want to find the smallest $c$ such that 
$$C = \{x | |x - 1| + |x - 13| + |x - 31| + |x - 131| = c\}$$
is not empty, and determine the set of $x$ satisfying
$|x - 1| + |x - 13| + |x - 31| + |x - 131| = c$ for this
smallest $c$ value.  Thinking of $|x - a|$ as the distance
from $x$ to $a$ along the real number line, we can see
that if $x \in [13,31]$, then 
$|x - 1| + |x - 13| + |x - 31| + |x - 131|$ 
equals the total distance between 1 and 131, plus the 
total distance between 13 and 31, or $130 + 18 = 148$.
We claim that $c = 148$ is the smallest value for which
$C \ne \emptyset$.  To see this, suppose first that
$x < 13$.  Then the distance from $x$ to 13, plus
the distance from $x$ to 31 is $|x - 13| + |x - 31| > 31 - 13 = 18$.
Similarly,  $|x - 1| + |x - 131| \ge 131 - 1 = 130$.  Hence
$|x - 1| + |x - 13| + |x - 31| + |x - 131| > 130 + 18 = 148$.
Similarly, if $x > 31$, we get
$|x - 1| + |x - 13| + |x - 31| + |x - 131| > 148$.  Hence
$c = 148$, and the set $C$ for this $c$ value is
the closed interval $[13,31]$.
\bigskip
\noindent
\S 1.5/5.  Suppose that $A$ is bounded above in ${\bf R}$, and let
$\alpha = {\rm lub}(A)$.  Let $B = \{x + k | x \in A\}$.  We
want to show that ${\rm lub}(B) = \alpha + k$.  To do this we
need to show first that $\alpha + k$ is an upper bound for $B$,
and second that any other upper bound for $B$ is greater than
or equal to $\alpha + k$.  

Suppose $x \in A$.  Then by definition $\alpha$ is an upper bound
for $A$, so $\alpha\ge x$ for all $x \in A$.  If we add $k$ to 
both sides of this inequality, we obtain another true inequality:
$\alpha + k \ge x + k$, valid for all $x \in A$.  Hence $\alpha + k$
is an upper bound for $B$.

To show that every other upper bound of $B$ is greater than or equal
to $\alpha + k$, we can argue by contradiction.  Namely suppose
that $\beta\ge x + k$ for all $x \in A$, but $\beta < \alpha + k$.
This implies $\beta - k < \alpha$.  Since $\alpha = {\rm lub}(A)$,
this implies that $\beta - k$ is {\it not} an upper bound for $A$,
so there exists $x\in A$ with $x > \beta - k$.  But this shows
$x + k > \beta$.  This is a contradiction since we assumed
that $\beta$ was an upper bound for the set $B$.  Hence every
upper bound of $B$ is greater than or equal to $\alpha + k$, and 
${\rm lub}(B) = \alpha + k = {\rm lub}(A) + k$.

The proof for the greatest lower bound is similar, reversing all
inequalities.

\bigskip
\noindent
\S 1.5/6.  a)  Suppose that $A$ is bounded above in ${\bf R}$, and let
$\alpha = {\rm lub}(A)$.  Let $k > 0$, and let
$B = \{kx | x \in A\}$.  We
want to show that ${\rm lub}(B) = k\alpha$.  To do this we
need to show first that $k\alpha$ is an upper bound for $B$,
and second that any other upper bound for $B$ is greater than
or equal to $k\alpha$.  

Suppose $x \in A$.  Then by definition $\alpha$ is an upper bound
for $A$, so $\alpha\ge x$ for all $x \in A$.  Since we assume
$k > 0$, when  we multiply $k$ on 
both sides of this inequality, we obtain another true inequality:
$k\alpha \ge kx$, valid for all $x \in A$.  Hence $k\alpha$
is an upper bound for $B$.

To show that every other upper bound of $B$ is greater than or equal
to $k\alpha$, we can argue by contradiction.  Namely suppose
that $\beta\ge kx$ for all $x \in A$, but $\beta < k\alpha$.
This implies $\beta/k < \alpha$ (again, $k > 0$, so $1/k > 0$, so
multiplying both sides by $1/k$ preserves the inequality).  
Since $\alpha = {\rm lub}(A)$,
this implies that $\beta/k$ is {\it not} an upper bound for $A$,
so there exists $x\in A$ with $x > \beta/k$.  But this shows
$kx > \beta$.  This is a contradiction since we assumed
that $\beta$ was an upper bound for the set $B$.  Hence every
upper bound of $B$ is greater than or equal to $k\alpha$, and 
${\rm lub}(B) = k\alpha = k\cdot {\rm lub}(A)$.

The proof for the greatest lower bound is similar, reversing all
inequalities.

\bigskip
\noindent
b) If $k < 0$, then ${\rm lub}(B) = k\cdot {\rm glb}(A)$
and ${\rm glb}(B) = k\cdot {\rm lub}(A)$.

\bigskip
\noindent
\S 1.5/8 b.  We are given that $A$ and $B$ are bounded
subsets of ${\bf R}$ such that for all $x \in A$ and all
$y \in B$, $x \le y$.  Let $\alpha = {\rm lub}(A)$ and
$\beta = {\rm glb}(B)$.   We want to show that
$\alpha \le \beta$.  Suppose on the other hand that
$\beta < \alpha$.  Since $\alpha = {\rm lub}(A)$, this
implies that $\beta$ is not an upper bound for $A$.  Hence
there exists $x\in A$ such that $\beta < x \le \alpha$.
But then since $\beta = {\rm glb}(B)$, $x$ is not a
lower bound for $B$, so there exists some $y \in B$
such that $\beta \le y < x \le \alpha$.  But this
is a contradiction of the hypothesis that $x\le y$
for all $x \in A$ and all $y \in B$.  Hence $\alpha \le \beta$.
\bye
