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\title{Math 2001 \\ Quiz 6}
\date{\today}

\begin{document}
\maketitle

\begin{enumerate}[label={\arabic*.}]
\item (1 pt) Write your name in the top \emph{right} corner of the page.

\item (2 pts) You proved: if $C \neq \varnothing$ and $B \times C \subseteq C \times D$, then $B \subseteq D$. Why is it necessary that $C \neq \varnothing$? Fill in the blanks with a concrete example that illustrates why the theorem is false when $C = \varnothing$.

\begin{solutions}
Any choice of $B$ and $D$ where $B \not\subseteq D$ is a correct answer for this problem.
\end{solutions}

\begin{quote}
If $B =$ 
\ifthenelse{\boolean{print_solutions}}{{\color{blue} $\{1\}$}}{\underline{\hspace{1in}}},
$D =$ 
\ifthenelse{\boolean{print_solutions}}{{\color{blue} $\{2\}$}}{\underline{\hspace{1in}}}
, and $C = \varnothing$. Then $B \times C =$ 
\ifthenelse{\boolean{print_solutions}}{{\color{blue} $\varnothing$}}{\underline{\hspace{1in}}},
and $C \times D$ = 
\ifthenelse{\boolean{print_solutions}}{{\color{blue} $\varnothing$}}{\underline{\hspace{1in}}},
but $B \not\subseteq D$.
\end{quote}

\item Sketch a proof for the statement: if $B \subseteq C$, then $A \times B \subseteq A \times C$.

\begin{enumerate}
\item (2 pts) Write a one or two sentence introduction for the proof of this statement.

\ifthenelse{\boolean{print_solutions}}{}{\vspace{30mm}}

\begin{solutions}
Suppose $A$, $B$, and $C$ are sets and that $B \subseteq C$. We prove that $A \times B \subseteq A \times C$ by showing that if $(x,y) \in A \times B$, then $(x,y) \in A \times C$.
\end{solutions}

\item (4 pts) Arrange the statements to give an outline for the body of the proof. Justify each implication in the space after each line (e.g. cite a definition).

\begin{minipage}{.55\textwidth}
\begin{align*}
\ifthenelse{\boolean{print_solutions}}{{\color{blue} a}}{\underline{\hspace{10mm}}}
&\implies 
\ifthenelse{\boolean{print_solutions}}{{\color{blue} c}}{\underline{\hspace{10mm}}}
\qquad \text{(
\ifthenelse{\boolean{print_solutions}}{{\color{blue} definition of Cartesian product}}{\hspace{2.7in}}
)}\\
&\implies 
\ifthenelse{\boolean{print_solutions}}{{\color{blue} d}}{\underline{\hspace{10mm}}}
\qquad \text{(
\ifthenelse{\boolean{print_solutions}}{{\color{blue} definition of subset, and $B \subseteq C$ (given)}}{\hspace{2.7in}}
)}\\
&\implies 
\ifthenelse{\boolean{print_solutions}}{{\color{blue} b}}{\underline{\hspace{10mm}}}
\qquad \text{(
\ifthenelse{\boolean{print_solutions}}{{\color{blue} definition of Cartesian product}}{\hspace{2.7in}}
)}\\
\end{align*}
\end{minipage}
\begin{minipage}{.4\textwidth}
\begin{enumerate}[label={\alph*.}]
\item $(x,y) \in A \times B$
\item $(x,y) \in A \times C$
\item $x \in A$ and $y \in B$
\item $x \in A$ and $y \in C$
\end{enumerate}
\end{minipage}
\end{enumerate}

\item (0.5 pts per blank) Fill in the blanks to complete a proof of the following statement: if $C \neq \varnothing$ and $A \times C \subseteq B \times C$, then $A \subseteq B$.
\end{enumerate}


\doublespace

\begin{proof}
Let \ifthenelse{\boolean{print_solutions}}{{\color{blue} $A$, $B$, and $C$ be sets}}{\underline{\hspace{3in}}},
where $A \times C \subseteq B \times C$ and $C \neq \varnothing$. We prove that \ifthenelse{\boolean{print_solutions}}{{\color{blue} $A \subseteq B$}}{\underline{\hspace{1.5in}}}
by showing that 
\ifthenelse{\boolean{print_solutions}}{{\color{blue} if $x \in A$, then $x \in B$}}{\underline{\hspace{3.75in}}}.


Suppose $x \in $ 
\ifthenelse{\boolean{print_solutions}}{{\color{blue} $A$}}{\underline{\hspace{1in}}}.
Since $C \neq \varnothing$, the set $C$ contains at least one element; call that element $y$. Therefore, since $x \in A$ and $y \in C$, we know that $(x,y) \in$ 
\ifthenelse{\boolean{print_solutions}}{{\color{blue} $A \times C$}}{\underline{\hspace{1.5in}}}
by the definition of 
\ifthenelse{\boolean{print_solutions}}{{\color{blue} Cartesian product}}{\underline{\hspace{3in}}}.
So, 
\ifthenelse{\boolean{print_solutions}}{{\color{blue} $(x,y)$}}{\underline{\hspace{1in}}}
$\in B \times C$ since 
\ifthenelse{\boolean{print_solutions}}{{\color{blue} $A \times C$}}{\underline{\hspace{1in}}}
$\subseteq B \times C$. Hence, by the definition of 
\ifthenelse{\boolean{print_solutions}}{{\color{blue} Cartesian product}}{\underline{\hspace{1.5in}}}, we see that $x \in$ 
\ifthenelse{\boolean{print_solutions}}{{\color{blue} $B$}}{\underline{\hspace{1in}}}
and $y \in $ 
\ifthenelse{\boolean{print_solutions}}{{\color{blue} $C$}}{\underline{\hspace{1in}}}.
Thus we have shown that if $x \in $ 
\ifthenelse{\boolean{print_solutions}}{{\color{blue} $A$}}{\underline{\hspace{1in}}},
then $x \in $ 
\ifthenelse{\boolean{print_solutions}}{{\color{blue} $B$}}{\underline{\hspace{1in}}},
and therefore $A \subseteq B$ by the definition of 
\ifthenelse{\boolean{print_solutions}}{{\color{blue} subset}}{\underline{\hspace{2in}}}.
\end{proof}
\end{document}