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\begin{document}
\lhead{Set Theory \\ Assignment 1} 
\rhead{Solution Key \\ September 13}

\noindent
\begin{enumerate}
\item[(1)]  (Exercise 1.3.1.)
  Show that the set of all $x$ such that $x\in A$ and $x\notin B$ exists.
\end{enumerate}

\vfill
\color{blue}
The set of the problem, which is typically denoted by $A-B$ or $A\setminus B$,
may be constructed by the Axiom of Separation as follows:
$$
A-B = \{x\in A\;|\; x\notin B\}.
$$
\color{black}
\vfill

\begin{enumerate}
\item[(2)]  (Exercise 1.3.6.)
  Show that ${\mathcal P}(X)\subseteq X$
    is false for any $X$. In particular, ${\mathcal P}(X)\neq X$ for
    any $X$. This proves again that a ``set of all sets'' does not exist.
    [Hint:
Let $Y = \{u\in X\;|\; u\notin u\}$; $Y\in {\mathcal P}(X)$ but $Y\notin X$.]
\end{enumerate}

\vfill
\color{blue}
Assume that ${\mathcal P}(X)\subseteq X$ and let
$Y = \{u\in X\;|\; u\notin u\}$. We must have $Y\subseteq X$, so
$Y\in {\mathcal P}(X)$, so we should have $Y\in X$.
Let us show that the assumption that $Y\in X$
leads to a Russell-type paradox.
\medskip

Case 1: ($Y\in X$ and $Y\notin Y$)
Then $u:=Y$ satisfies the property that defines membership
in $Y$, so $Y=u\in Y$. The properties $Y\notin Y$ and $Y\in Y$
contradict one another.
\medskip

Case 2: ($Y\in X$ and $Y\in Y$)
Then $u:=Y$ fails the property that defines membership
in $Y$, so $Y=u\notin Y$. The properties $Y\in Y$ and $Y\notin Y$
contradict one another.
\color{black}
\vfill

\begin{enumerate}
\item[(3)]
\begin{enumerate}
\item[(a)]  (Exercise 2.2.8(a))
  $A\times B=\emptyset$ if and only if $A=\emptyset$ or $B=\emptyset$.
\end{enumerate}
\end{enumerate}

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\color{blue}
$[\Rightarrow]:$ Arguing the contrapositive,
assume that $A\neq \emptyset\neq B$.
There must exist $a\in A$ and $b\in B$,
in which case $(a,b)\in A\times B$, so $A\times B\neq\emptyset$.

$[\Leftarrow]:$ Again, arguing the contrapositive, assume
that $A\times B\neq\emptyset$. There must exist $(a,b)\in A\times B$,
in which case $a\in A$ and $b\in B$,
so $A\neq\emptyset$ and $B\neq \emptyset$.
\color{black}
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\begin{enumerate}
\item[(3)] 
\begin{enumerate}
\item[(b)]  (First part of Exercise 2.2.8(b))
$(A_1 \cup A_2) \times B =(A_1 \times B) \cup (A_2 \times B)$.
\end{enumerate}
\end{enumerate}

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\color{blue}
Choose an element $(a,b)$ from the left hand side:
$(a,b)\in (A_1 \cup A_2) \times B$.
By the definitions of $\times$ and $\cup$, we have
$b\in B$ and either $a\in A_1$ (Case~1)
or $a\in A_2$ (Case~2). If we are in Case~1,
then $(a,b)\in A_1\times B$; 
if we are in Case~1,
then $(a,b)\in A_2\times B$. In either case we have
$(a,b)\in (A_1 \times B) \cup (A_2 \times B)$, according
to the definition of $\cup$.

Now choose an element $(a,b)$ from the right hand side:
$(a,b)\in (A_1 \times B) \cup (A_2 \times B)$.
By the definition of $\cup$, we have
$(a,b)\in A_1\times B$ (Case 1) or 
$(a,b)\in A_2\times B$ (Case 2).
This yields $b\in B$ in either case and either
$a\in A_1$ (in Case 1) or $a\in A_2$ (in Case 2).
Therefore, in either case we have $a\in A_1\cup A_2$ and $b\in B$,
so $(a,b)\in (A_1 \cup A_2) \times B$.
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