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\fancyhead[L]{\Large \textbf{Math 157 Fall 2013}}
\fancyhead[R]{\large \textbf{Homework 1 - selected answers}}



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\bp 
\ben
\item Prove that one of the following two formulas about sets is always
right and the other is sometimes wrong:
\begin{gather*}
A-(B-C)=(A-B)\cup C\\
A-(B\cup C)=(A-B)-C
\end{gather*}
\ba Here is a proof that $(A-B)-C=A-(B\cup C)$ for any sets $A,B$, and $C$.

\begin{proof}Let $a\in A-(B\cup C)$.  This means $a\in A$ and $a\notin B\cup C$.  Since
$a\notin B\cup C$,  $a\notin B$, so $a\in A-B$.  Since $a\notin B\cup
C$, $a\notin C$.  So, $a\in (A-B)-C$.  This proves $A-(B\cup
C)\subseteq (A-B)-C.$

Now let $a\in (A-B) -C$.  This means that $a\in A-B$ and $a\notin C$.
Since $a\in A-B$, $a\in A$ and $a\notin B$.  It follows from $a\notin C$
and $a\not in B$ that $a\notin B\cup C$.  The fact that $a\in A$ and
$a\notin B\cup C$ implies that $a\in A-(B\cup C)$.  This shows that
$(A-B)-C\subseteq A-(B\cup C).$

Since $A-(B\cup
C)\subseteq (A-B)-C$ and $(A-B)-C\subseteq A-(B\cup C)$ together imply
that $(A-B)-C=A-(B\cup C).$
\end{proof}

Sometimes, it's not true that $A-(B-C)=(A-B)\cup C.$  
For example, let $A=\{1,2,3,4\}$, $B=\{1,2,5,6\}$, $C=\{2,3,6,7\}$.  Then $B-C=\{1,5\}$
and $$A-(B-C)=\{2,3,4\}.$$   One the other hand $A-B=\{3,4\}$ and
$$(A-B)\cup C=\{2,3,4,6,7\}.$$


\ea
\item State some additional necessary and sufficient condition for the
  formula which is sometimes incorrect to be always right.
\ba First note that for all sets $A,B$, and $C$, we have 
$$A-(B-C)\subset (A-B)\cup C.$$
To prove this, let $a\in A-(B-C)$.  Then $a\in A$ and $A\notin B-C$.
If $A\notin B-C$, we have either $a\notin B$ or $a\in C$.  If $a\in
C$, then $a\in (A-B)\cup C$.  If $a\notin B$, then $a\in A$ and
$a\notin B$ imply that $a\in A-B$, hence $a\in (A-B)\cup C$.

Now, we prove that if $C-A\neq \emptyset$ then $$A-(B-C)=(A-B)\cup
C.$$
\begin{proof}
Since it's already been shown that we always have $A-(B-C)\subset
(A-B)\cup C$, it remains to prove that if $C-A= \emptyset$ then
$(A-B)\cup C\subseteq A-(B-C).$  So, assume that $C-A = \emptyset$ and
let $a\in (A-B)\cup C$.  This means that $a\in A-B$ or $a\in C$.  If
$a\in A-B$, we have $a\in A$ and $a\notin B$.  If $a\notin B$, we have
$a\notin B-C$, so we have $a\in A-(B-C)$.  On the other hand if $a\in
C$, then $a\in A$ (since $C-A=\emptyset$) and $a\notin B-C$, so $a\in A-(B-C)$.
\end{proof}

Finally, note that the statement 
\begin{quote}
$C-A\neq \emptyset$ if and only if $A-(B-C)=(A-B)\cup C$.  
\end{quote}
We've already shown that if $C-A=\emptyset$, then $A-(B-C)=(A-B)\cup
C$.  On the other hand if $C-A \neq \emptyset$ then there is an
element $c\in C-A$.  Then $c\in (A-B)\cup C$.  But since 
$c\in C-A\Rightarrow c\notin A$, it follows that $c \notin A-(B-C)$.
Therefore, $(A-B)\cup C\neq A-(B-C)$.
\ea
\een
\ep

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