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\lhead{Set Theory \\ Assignment 8} 
\rhead{Solution Key \\ November 17}

\noindent
\begin{enumerate}
\item[1.] (Exercise 7.1.3)
  For any set $A$,
  there is a mapping of ${\mathcal P} (A\times A)$ onto $h(A)$.
\end{enumerate}


\bigskip
\color{blue}
Define $f: {\mathcal P}(A\times A)\to h(A)$ by
\[
f(R) =
\begin{cases}
  \alpha& \textrm{if $R$ is a well-ordering of type $\alpha$ of a subset of $A$}\\
0  &\textrm{otherwise}.
\end{cases}
\]
There exists an $R\subseteq A\times A$ that is a well-ordering
of type $\alpha$ of a subset of $A$ iff $\alpha$ is embeddable in $A$
iff $\alpha<h(A)$. Thus, $f$ maps onto $h(A)$.
\color{black}
\bigskip

  
\begin{enumerate}
\item[2.] (Exercise 7.1.4.)
  $|A|<|A|+h(A)$.  (It suffices to explain why $|A|<|A\cup h(A)|$.)
\end{enumerate}

\bigskip

\color{blue}
Let's prove the second version of the problem.
The inclusion function $\iota: A\to A\cup h(A)$ is 1-1,
so $|A|\leq |A\cup h(A)|$.
If we had equality, $|A|=|A\cup h(A)|$, then there would exist
a bijection $f: A\cup h(A)\to A$. The restriction of $f$
to $h(A)$ would be an embedding of $h(A)$ into $A$.
This is impossible, so $|A|\neq |A\cup h(A)|$, and therefore
$|A|<|A\cup h(A)|$.

The original version of the problem follows from the
second version 
as follows. Let $A'$ be a set equipotent with $A$
and which contains no ordinals. Then $|A'|=|A|$ and $h(A')=h(A)$,
but we have arranged that the sets $A'$ and $h(A')$ are disjoint.
Now
\[|A|=|A'| < |A'\cup h(A')| = |A'|+|h(A')|=|A'|+h(A') = |A|+h(A).\]
\color{black}
\bigskip

\begin{enumerate}
\item[3.] (Exercise 7.1.5.)
$|h(A)| < |{\mathcal P}({\mathcal P}(A \times A))|$
  for all $A$. 
\end{enumerate}
  
\bigskip

\color{blue}
It is a general fact that if $f: X\twoheadrightarrow Y$ is surjective, then
$
f^{-1}\colon Y\rightarrowtail {\mathcal P}(X): y\mapsto f^{-1}(y)
$
is injective.
Applying this to
$f: {\mathcal P}(A\times A)\twoheadrightarrow h(A)$
from Problem~7.1.3 we get that
$f^{-1}: h(A)\rightarrowtail
{\mathcal P}({\mathcal P}(A\times A))$
is 1-1.
\color{black}

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