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\begin{document}
\lhead{Set Theory \\ Assignment 6} 
\rhead{Solution Key \\ October 27}

\noindent
\begin{enumerate}
\item[(1)]
Let $A$ be an amorphous set.
    Show that $A\times A$ is not amorphous.
    Conclude that $A$ is an infinite set
    such that $|A\times A|\neq |A|$.
\end{enumerate}

\smallskip
\color{blue}
The fact that $A$ is amorphous means
that $A$ is infinite, but it cannot
be partitioned into two infinite cells.

Assume that $A$ is amorphous and
choose $a\in A$. We intend to argue that (i)~$A\times \{a\}$
is an infinite subset of $A\times A$ and (ii)~the complementary
set $A\times (A-\{a\})$ is also an infinite subset of $A\times A$.
This will show that $\Pi = \{A\times \{a\}, A\times (A-\{a\})\}$
is a partition of $A\times A$ into two infinite cells,
preventing $A\times A$ from being amorphous.

We establish Claim (i) by observing that
$A\times \{a\}$ is equipotent with $A$,
and $A$ is infinite, so $A\times \{a\}$ must be infinite.
(One may establish equipotence by observing that
the first projection function
$\pi_1\colon (A\times \{a\})\to A\colon (x,a)\mapsto x$ is a bijection.)

We can establish Claim (ii) with a slight modification
of the argument in the previous paragraph.
The first projection function
$\pi_1\colon (A\times (A-\{a\}))\to A\colon (x,a)\mapsto x$ is a
\underline{surjection} from $A\times (A-\{a\})$ onto an infinite
set $A$. By the contrapositive
of Theorem~2.5 of HJ\footnote{This result states that
if the domain a function is finite, then the image
is finite. The contrapositive
asserts that if the image is infinite, then the domain must be infinite.}
this is enough to conclude that 
$A\times (A-\{a\})$ is infinite.

One may establish Claim (ii) a different way based on
Theorem~2.4 of HJ.\footnote{This theorem states that a subset
of a finite set is finite.}
Argue as follows:
$A$ is infinite. If $a\in A$, then $A-\{a\}$ is also infinite
according to Question~1 of Quiz~5. Choose $b\in A-\{a\}$.
The set $A\times \{b\}$ is infinite by the argument for Claim (i).
Since $A\times \{b\}$ is an infinite subset of $A\times (A-\{a\})$,
it follows from the contrapositive form of Theorem~2.4 that
$A\times (A-\{a\})$ is infinite.
\color{purple}
\bigskip

For the final sentence of the problem, we observe that
the property of being amorphous is an equipotence invariant:
If $A$ is amorphous and $|A|=|B|$, then $B$ is amorphous.
To see this, assume instead that $f\colon A\to B$ is a bijection.
If $\{B_0, B_1\}$ is a partition of $B$
into two infinite cells, then
$\{f^{-1}(B_0), f^{-1}(B_1)\}$
is a partition of $A$ into two infinite cells.
\color{black}

\smallskip

\begin{enumerate}
\item[(2)] Do Exercise 5.1.7. (Use Cantor's Theorem to show that the ``set of all sets'' does not exist.)
\end{enumerate}

\smallskip
\color{blue}
Assume that the class $\mathcal S$ of all sets is a set.
Since ${\mathcal P}({\mathcal S})$ consists of sets,
${\mathcal P}({\mathcal S})\subseteq \mathcal S$ and therefore
the inclusion map witnesses that
(i) $|{\mathcal P}({\mathcal S})|\leq |\mathcal S|$ holds.
By Cantor's Theorem,
(ii)~$|\mathcal S|<|{\mathcal P}({\mathcal S})|$ holds.
Apply the Canter-Bernstein-Schr\"{o}der Theorem to (i)
and (ii) to obtain
$|\mathcal S|=|{\mathcal P}({\mathcal S})|$, which contradicts
Cantor's Theorem.
\color{black}

\smallskip

\begin{enumerate}
\item[(3)]
\begin{enumerate}
\item[(a)] Give an example of a set that is transitive, but not well-ordered by epsilon.
  \bigskip

  \color{blue}
\[
T={\mathcal P}({\mathcal P}({\mathcal P}(\emptyset))) = \{\emptyset, \{\emptyset\}, \{\{\emptyset\}\}, \{\emptyset, \{\emptyset\}\}\}
\]

$T$ is transitive, since $\emptyset, \{\emptyset\}\in T$.
and the only instances of sets $R, S, T$ satisfying $R\in S\in T$ are
when $R=\emptyset$ or $R=\{\emptyset\}$.

$T$ is not well-ordered by $\in$, since
$\{\{\emptyset\}\}$ and $\{\emptyset, \{\emptyset\}\}$
are incomparable under $\in$.  
\color{black}

\item[(b)] Give an example of a set that is well-ordered by epsilon, but is not transitive.
  \bigskip

\color{blue}
\[W=\{\{\emptyset\}\}.\]

$W$ is not transitive, since $\emptyset\in \{\emptyset\}\in W$,
but $\emptyset\notin W$.

$W$ is well-ordered by $\in$, since $W$ has only one
element and any one-element ordered set is
well-ordered by $\in$.
\color{black}
\end{enumerate}  
\end{enumerate}  

\smallskip


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