math-460-sets-as-ordered-pairs.tex

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%\subtitle{Lesson 4 \\
%\vspace{1.0in}
 %\tiny Barton Willis, Attribution 4.0 International (CC BY 4.0), 2020 \normalsize}
\title{\textbf{Sets as ordered pairs}}
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\begin{document}

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\begin{frame}{Ordered pairs}

An ordered pair is a familiar object--the Cartesian coordinates of a point in a plane, 
for example, is an ordered pair of real numbers.

\begin{myex}
Examples of ordered pairs of real numbers:
\[
    (0,0), \quad (2,-6),  \quad  (2, \sqrt{42}), \quad   (107, 28).
\]
\end{myex}

\begin{checklist}

\item We say that \(a\) is the \emph{first coordinate} of the ordered pair \((a,b)\), and \(b\) is its \emph{second coordinate}.

\item  We efine equality of ordered pairs using   \([ (a,b) = (a^\prime, b^\prime)]   \equiv  [a = a^\prime] \land  [b = b^\prime] \).

\item Of course, the symbols \(a\) through \(b^\prime\) need to be objects for which 
equality is defined.
\end{checklist}

\end{frame}
\begin{frame}{As nice this may be}

\textbf{Question} Ordered pairs are somewhat like sets, but the order matters.  
Can we define an ordered pair as a set?

\textbf{Answer}   Sure.  To an ordered pair \((a,b)\) we associate it with the 
set  \(\{\{a\}, \{a,b\}\} \).

\vspace{0.2in}

\begin{checklist}

\item Since \(\{\{a\}, \{a,b\}\} \) is  a set of sets, we can form the intersection 
of its members. Define  \(I = \{\{a\}, \{a,b\}\} \). Then
\[
   \underset{x \in I}{\cap} x =  \{a\} \cap \{a,b\} = \{a\}.
\]
So the intersection of the member of \(I\) gives the first coordinate of the 
ordered pair \((a,b)\).

\item How do we extract the second coordinate?
\[
    \underset{x \in I}{\cup} x   \setminus \underset{x \in I}{\cap} x  = \{a,b\} \setminus \{a\} = \{b\}.
\]



\end{checklist}


\end{frame}
\begin{frame}

\begin{myth}  If  \(  \{\{a\}, \{a,b\} \} =   \{\{a^\prime\}, \{a^\prime,b^\prime\} \}\), 
  then \(a = a^\prime\) and  \(b = b^\prime\).
\end{myth}

\begin{checklist}

\item A proof uses the ingredients:  If \( \{a \} = \{a^\prime \}\), 
then \(a = a^\prime\). It also uses
\[
   \underset{x \in I}{\cap} x =  \{a\} \cap \{a,b\} = \{a\},
\]
and
\[
    \underset{x \in I}{\cup} x \setminus \underset{x \in I}{\cap} x  = \{a,b\} \setminus \{a\} = \{b\}.
\]
where \(I = \{\{a\}, \{a,b\}\} \).

\end{checklist}
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