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\begin{document}


\noindent
{\bf Solutions to HW 2.}

\bigskip
  
\begin{enumerate}
\item[1.]
We have explained why the Russell class ${\mathcal R}=\{x\;|\;x\notin x\}$
is not a set. Explain why the following classes are also not sets.
\begin{enumerate}
\item[(a)] The class $\mathcal C$ of all sets.
\item[(b)] The class $\mathcal D$ of all 1-element sets.  
\end{enumerate}
(Hints: For (a), show that the assumption that $\mathcal C$ is a set allows you to construct $\mathcal R$ as a set. For (b), show that the assumption that $\mathcal D$ is a set allows you to construct $\mathcal C$ as a set.) 

\medskip

\begin{enumerate}
\item[(a)] The class $\mathcal C$ of all sets.
\end{enumerate}

\medskip

\color{blue}
If $\mathcal C$ was a set, then we could construct
${\mathcal R}=\{x\in{\mathcal C}\;|\;x\notin x\}$
with the Axiom of
Separation, which is a legal set-theoretic construction.
Thus, ``$\mathcal C$ is a set'' implies ``$\mathcal R$ is a set''.
Since we know ``$\mathcal R$ is not a set'', then we
must conclude that ``$\mathcal C$ is not a set''.
\medskip
\color{black}
\begin{enumerate}
\item[(b)] The class $\mathcal D$ of all 1-element sets.
\end{enumerate}

\medskip
\color{blue}
If $\mathcal D = \{\{A\}, \{B\}, \ldots\}$ were a set, then
$\bigcup {\mathcal D} = \{A, B, \ldots\} = {\mathcal C}$
would be a set. We saw in Part (a), that $\mathcal C$
is not a set, so we conclude that $\mathcal D$
cannot be a set, either.
\end{enumerate}
  
\color{black}

\bigskip\bigskip
  
\begin{enumerate}
\item[2.] Your friend offers a wager that,
  under the Kuratowski encoding,
  the ordered pair $(0,1)$
  equals the natural number three.
  Should you take the wager? Explain.
\end{enumerate}
  
\bigskip

\color{blue}
Take the wager! (By offering the wager, the friend
bets that the statement is true. We argue that it is false,
so your friend will lose and you will win.)

$0 = \{{}\}, 1 = \{0\}, 2 = \{0,1\}$ and $3=\{0,1,2\}$.
Thus $(0,1) = \{\{0\},\{0,1\}\} = \{1, 2\}\neq \{0,1,2\}=3$.
(We know that $\{1, 2\}\neq \{0,1,2\}$ since only
one of these two sets has an element equal to the empty set.)


\bigskip

\color{red}
(Here is a second solution: any ordered pair $(x,y)$, considered as a set
$\{\{x\},\{x,y\}\}$, contains
one or two distinct elements. But $3$ contains three distinct elements,
so $(x,y)\neq 3$ for any $x$ and $y$.)
\color{black}
  
\bigskip\bigskip

\begin{enumerate}
\item[3.]
    Show that $\emptyset\times A=\emptyset$.
\end{enumerate}

  
\bigskip

\color{blue}
We must show that $\emptyset\times A$ has no elements.
But this is clear, since if $(x,y)\in\emptyset\times A$,
then $x\in \emptyset$, which is impossible ($\emptyset$ has no elements).
\color{black}
\end{document}
