math-460-set-complements.tex

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\newcommand{\llnot}{\lnot \,} % is accepted


\subtitle{%Lesson 6 \\ \vspace{1.0in} 
The whole problem with the world is that fools and fanatics are always so certain of themselves, and wiser people so full of doubts. \\ \vspace{0.25in} Bertrand Russel}
\title{\textbf{Set complements}}
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\begin{document}



\frame{\titlepage}

\begin{frame}{The set complement}

\begin{mydef}  Let \( U \) be the universal set. For any set \(A\), we define its complement \(A^c\) by \(A^c = U  \setminus A\).

\end{mydef}


\begin{checklist}

\item We have
\[
     (x \in A^c)  \equiv   (x \in U)    \land  (x \notin A)  .
\]

\item Assisted by Mr.\  DeMorgan, we have
\[
     (x \notin A^c)  \equiv   (x \notin U)    \lor   (x  \in A)  .
\]


\end{checklist}
\end{frame}


\begin{frame}{Every definition deserves a theorem}


\begin{myth} Let \(A\) be a set.  We have
\begin{align*}
    \varnothing^c &= U, \\
    U^c  &= \varnothing, \\
    (A^c)^c & = A, \\
      A \cap A^c &= \varnothing, \\
       A \cup A^c &=  U.
\end{align*}

\end{myth}
\end{frame}

\begin{frame}

\begin{myproof}[ \(U  = \varnothing^c\)] We have
\[
     \varnothing^c = \{ u \in U \mid x \notin \varnothing \} = \{ u \in U \mid  \true \} = U.
\]
\end{myproof}


\begin{checklist}

\vspace{0.2in}

\item We used the fact that   \( (\forall x)(x \notin \varnothing) \equiv \true\).

\vspace{0.2in}
\item The proof is a string of equalities. The conclusion of the proof is to compare the far left to the far right of the string.

\vspace{0.2in}

\item Alternatively, we could show that \(U  \subset  \varnothing^c\) and \(\varnothing^c \subset U\).  But I think this proof is more clear.
\end{checklist}

\end{frame}
\begin{frame}

\begin{myproof}[\((A^c)^c  = A\)] Let \(A\) be a set. We have
\[
     (A^c)^c = \{ u \in U \mid u \notin A^c \} =  \{ u \in U \mid u \in A \} = A .
\]
\end{myproof}

\begin{checklist}

\item Arguably this proof makes too large a jump in logic from  \(u \notin A^c \) to \(u \in A\).

\item Here is a fix:
\begin{align*}
      \{ u \in U \mid u \notin A^c \} &=  \{ u \in U \mid  (u \notin U) \lor  (u  \in A)\}, \\
                                                               &=  \{ u \in U \mid  \false  \lor  (u  \in A)\} , \\
                                                              &=   \{ u \in U \mid  u  \in A\},\\
                                                              &    = A.
 \end{align*}

 \item We have
 \[
     x \in A \setminus B \equiv (x \in A) \land (x \notin B)
 \]


  \item So
 \[
     x \notin A \setminus B \equiv    \lnot ((x \in A) \land (x \notin B)) = (x \notin A) \lor (x \in B).
 \]
\end{checklist}

\end{frame}

\begin{frame}

\begin{myproof}[\(  \varnothing  = U^c\)]
We've already shown that \(\varnothing^c = U\).  Using this fact, we have
\[
  \varnothing = (\varnothing^c)^c  = (U)^c.
\]
\end{myproof}

\begin{checklist}

\item Our proof uses the fact that for all sets \(A\), we have \((A^c)^c = A\).
\vspace{0.2in}
\item With malice aforethought, we proved the statements in the proposition in a different order than they were presented.

\vspace{0.2in}
\item  We switched up the order to make use of \((A^c)^c = A\) in a later proof.
\end{checklist}
\end{frame}

\begin{frame}{Looks like homework}


\begin{myproof}[  \(A \cap A^c = \varnothing\)]

(This looks like it should be homework--it's all up to you. You are certainly allowed to use all the results we have proved so far.)
\end{myproof}

\vspace{0.2in}

\begin{myproof}[  \(A \cup A^c = U \)]
(This looks like it should be homework--it's all up to you. You are certainly allowed to use all the results we have proved so far.)
\end{myproof}
\end{frame}

\end{document}