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\lhead{Set Theory \\ Assignment 7} 
\rhead{Solution Key \\ November 3}

\noindent
\begin{enumerate}
\item[(1)]
Working in ZF, show that an infinite well-orderable set is Dedekind infinite. Conclude that an amorphous set is not well-orderable. (Interpret ``well-orderable'' to mean ``equipotent with an ordinal''.)
\end{enumerate}

\smallskip
\color{blue}
Assume that the bijection $f\colon \alpha\to X$
establishes that $X$ is equipotent with the ordinal $\alpha$.
Since $X$ is infinite, $\alpha$ must be infinite.
The ordinals are totally ordered by $\in$, so
$\alpha$ must be $\in$-comparable with $\omega$.
We do not have $\alpha\in\omega$, since $\alpha$ is infinite,
so we must have $\omega=\alpha$
or $\omega \in \alpha$. In either case, $\omega\subseteq \alpha$.
The inclusion map $\iota \colon \omega \to \alpha$
composed with $f$ is an injective function
$f\circ \iota\colon \omega\to X$, establishing
that $|\omega|\leq |X|$, so $X$ is Dedekind infinite.

For the last line of the problem, an amorphous set $A$ is
infinite and Dedekind finite, so there can be no
bijection $f\colon \alpha\to A$ for any ordinal $\alpha$.
\color{black}

\smallskip

\begin{enumerate}
\item[(2)] Do Exercise 6.1.3. (There exist $2^{\aleph_0}$
  well-orderings of the set of all natural numbers.)
\end{enumerate}

\smallskip
\color{blue}
We have already shown that there exist
$2^{\aleph_0}$-many distinct bijections from $\omega$ to $\omega$.
Each such bijection $\pi\colon \omega\to \omega$
may be used to define a 
binary relation $<_{\pi}$ on $\omega$:
\[
m <_{\pi} n \Leftrightarrow \pi^{-1}(m) \in \pi^{-1}(n).
\]
With this definition, the bijection
$\pi\colon \omega\to \omega$
becomes an order-isomorphism
\[
\pi\colon \langle \omega; \in \rangle\to \langle \omega; <_{\pi} \rangle.
\]
Since $\langle \omega; <_{\pi} \rangle$ is order-isomorphic to
$\langle \omega; \in \rangle$,
$\langle \omega; <_{\pi} \rangle$ is a well-ordered set, so
$<_{\pi}$ is a well-ordering of $\omega$.

I claim that distinct bijections $\pi, \pi'$
yield distinct order relations $<_{\pi}, <_{\pi'}$
on $\omega$. Assume instead that
$\pi \neq \pi'$, but $<_{\pi} =  <_{\pi'}$.
Then $\pi$ and $\pi'$ must be distinct isomorphisms
from the well-ordered set
$\langle \omega; \in \rangle$ to the well-ordered set
to $\langle \omega; <_{\pi} \rangle = \langle \omega; <_{\pi'} \rangle$.
This contradicts the fact that isomorphisms between
well-ordered sets are unique when they exist.

The set
\[
\{<_{\pi}\in {\mathcal P}(\omega\times \omega)\;|\;\pi\textrm{ a permutation of } \omega\}
\]
contains $2^{\aleph_0}$-many distinct
well-orderings of $\omega$. There
cannot be more, since there are only $2^{\aleph_0}$-many
binary relations on $\omega$.
(The total number of binary relations is
$|{\mathcal P}(\omega\times\omega)|=|{\mathcal P}(\omega)|=|2^{\omega}|=
2^{\aleph_0}$.)
\color{black}

\smallskip

\begin{enumerate}
\item[(3)]
  Do Exercise 6.2.8. (If $X$ is a nonempty set of ordinals,
  then $\bigcap X$ is an ordinal. Moreover,
  $\bigcap X$ is the least element of $X$.)
\end{enumerate}  

\smallskip

\color{blue}
The arguments for this were presented in class
duing the October 30 and November 1 lectures.

On October 30, in Lemma 1, Part 4, we showed that the intersection of a class
of ordinals is an ordinal. The idea was to show that if each element
of $X$ is a transitive set of transitive sets, then
$\bigcap X$ is also a transitive set of transitive sets.

On November 1, when showing that ON is well-ordered
as a class, we showed that if $X$ is a nonempty subset
of ordinals, then the ordinal $\bigcap X$ is the least
element $X$. The idea was this: we know that
$\bigcap X\subseteq X_i$ for every $X_i\in X$. Hence
either  (i) $\bigcap X\subsetneq X_i$ for every $X_i\in X$
or (ii) $\bigcap X = X_i$ for some $X_i\in X$
and $\bigcap X\subsetneq X_j$ for every $X_j\in X, j\neq i$.
Using Lemma~2 from October~30, 
Case (i) leads to $\bigcap X\in X_i$ for every $X_i\in X$,
hence to $\bigcap X\in \bigcap X$, which contradicts the Axiom of
Foundation.
Case (ii) leads to
the conclusion that $\bigcap X = X_i$ is the $\in$-least 
element of $X$, as desired.
\color{black}

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