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\homework{8: More Set Theory}{Due Friday, Week 4}

Do \textbf{one} of the \textbf{three} problems below!

\begin{enumerate}
\item In class on Monday, for any two sets $x, y$ we defined the \textbf{ordered pair} $(x,y)$ to be the two-element set $\{\{x\}, \{x, y\}\}$.  We picked this definition because we could uniquely translate any ordered pair $(x,y)$ into such a two-element set, and always decode any such two-element set uniquely into an ordered pair.

You might hope that a similar construction would work for \textbf{ordered triples}: i.e. one might hope that for any sets $x,y,z$, the definition
\begin{align*}
(x,y,z) := \big\{ \{x\}, \{x,y\}, \{x,y,z\}\big\}
\end{align*}
would give us a way to describe ordered triples.  Show that this definition is ``bad,'' in the following sense: find objects $a,b,c,d,e,f$ such that $(a,b,c) \neq (d,e,f)$ but $\big\{ \{a\}, \{a,b\}, \{a,b,c\}\big\} = \big\{ \{d\}, \{d,e\}, \{d,e,f\}\big\}$.  (In other words, this definition classifies different ordered triples as the same thing, which is not what we'd want!)
\item Show that there is no set $A$ that contains every ordered pair.  
\item Suppose that $A, B, C$ are all sets.  Prove or disprove the following statements:
\begin{enumerate}
\item  $A \times (B \cup C) = (A \times B) \cup (A \times C)$.
\item If $A \neq \emptyset$ and $A \times B = A \times C$, then $B = C$.
\item If $D$ is the set $\{A \times X ~|~ X \in B\}$, then\footnote{On our last problem set, we defined $\bigcup A$ as the union $\bigcup_{A' \in A} A'$ of all of the elements of $A$. } $A \times \left( \bigcup B\right) = \bigcup D$.
\end{enumerate}
\end{enumerate}
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