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\begin{document}


\noindent
{\bf Solutions to HW 2.}

\bigskip
  
\begin{enumerate}
\item[1.]
 Is the class of all 1-element sets a set? Explain. 
\end{enumerate}
  
  \bigskip
  
\color{blue}
No.

The class $Z = \{x\;|\;x\;\textrm{has 1 element}\}$ is
describable by a formula, hence it is a class. We now
argue that it is not a set.

If instead $Z$ were a set, then $\bigcup Z$
would also be a set. We shall argue that $\bigcup Z$ is the
class of all sets, which is not a set.

Choose any set $S$. Since
$S\in \{S\}\in Z$ we get $S\in \bigcup Z$,
so $Z$ contains every set.
\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}
