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\centerline{MATH 243 -- Mathematical Structures}
\centerline{Solutions for Quiz 9 -- December 1, 2017}
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\begin{enumerate}
\item[A)]  (15) State the Completeness Axiom (`Axiom C') for the real numbers.
\item[] {\it Solution:}  The statement is -- Let $A \subset \mathbb{R}$ be a nonempty
a subset that is bounded above.  Then $A$ has a least upper bound $b \in \mathbb{R}$.
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\item[] {\it Note:}  
\begin{enumerate} 
\item[(1)] Saying $b = \sup(A)$, the least upper bound, means first that 
$b$ is an upper bound for $A$ -- that is, all $x \in A$,  $x \le b$ -- and moreover, 
\item[(2)] If $x \le b'$ for all $x \in A$, then $b \le b'$.   (Intuitively, 
$b$ is the smallest number that is an upper bound for $A$.)
\item[(3)] The assumption that $A \ne \emptyset$ is necessary because $\emptyset$
is bounded above, but \emph{every} $b \in \mathbb{R}$ satisfies the condition $x \le b$
for all $x \in \emptyset$ (there aren't any such $x$, so the condition is
\emph{vacuously true}).  There is no smallest upper bound for $A$ in that case.
\end{enumerate}  
\item[B)]  (15)  Let $A \subset {\bf R}$ and $2A = \{2x : x \in A\}$.  Show that
if $\sup(A) = c$, then $\sup(2A) = 2c$.
\item[] {\it Solution 1:}  (a direct proof)  First, since $\sup(A) = c$, we
have that $c$ is an upper bound for $A$.  This implies that $x \le c$ for all 
$x \in A$.  Since $2 > 0$, we can multiply both sides of this inequality by 
$2$ to yield $2x \le 2c$ for all $x \in A$.  This shows that $2c$ is an upper
for $2A$ as in point (1) in the solution for part A above.  Now, to show $2c$ is
the least upper bound, we need to show that point (2) also holds for the 
bound $2c$ and the set $2A$.  So let $d$ be 
any other upper bound for the set $2A$.  This means that $2x \le d$ for all $x \in A$, 
so $x \le \frac{d}{2}$ for all $x$ in A.  By point (2) for the upper bound $c$ for $A$,
this implies $c \le \frac{d}{2}$.  But since $2 >0$, that implies $2c \le d$.  Hence $2c = \sup(2A)$.
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\item[] {\it Solution 2:}  The proof of point (1) is the 
same as before -- if $x \le c$, for all $x \in A$, then $2x \le 2c$, so $2c$ is an
upper bound for $2A$.  Now we argue by contradiction for point (2).  Suppose $2c$ is 
not the least upper bound of $2A$.  That means that there is some $d < 2c$ that
is also an upper bound for $2A$:  $2x \le d$ for all $2x \in 2A$.  But multiplying 
by $1/2 > 0$ yields $x \le \frac{d}{2}$ for all $x \in A$.  This implies $\frac{d}{2}$
is an upper bound for $A$.  However $d < 2c$ implies $\frac{d}{2} < c$ and that contradicts
the assumption that $c$ was the least upper bound of $A$.   
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\item[] Some further comments:  It's tempting to say that $c < 2c$ and try to relate
$c$ and $2c$ that way.  However, this is only true if $c > 0$.  If $c < 0$, then in fact
$2c < c$.  
\end{enumerate}
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