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\lhead{Set Theory \\ Assignment 9} 
\rhead{Solution Key \\ December 1}

\noindent
\begin{enumerate}
\item[1.] (Exercise 8.1.9)
Prove (using AC):  
Every infinite set is equipotent to some
of its proper subsets. Equivalently, Dedekind
finite sets are precisely the finite sets.
\end{enumerate}

\bigskip
\color{blue}
This was proved on Slide 6 of the slides
from the October 23 lecture, but I will include a brief argument here.

We must explain why, under ZFC, Dedekind
finite sets are finite. Contrapositively, 
we must explain why an infinite set $A$
is Dedekind infinite. Using the Well-ordering Principle,
enumerate $A$ by some ordinal $\alpha$.
Since $\omega$ is infinite and its elements
are finite, $\omega$ is the least infinite ordinal.
Therefore, $\omega\subseteq \alpha$, so $|\omega|\leq |\alpha|=|A|$,
so $A$ is Dedekind infinite. (Here I am using a ZF-theorem
that we proved: a set $D$ is Dedekind infinite iff $|\omega|\leq |D|$.)
\color{black}
\bigskip

  
\begin{enumerate}
\item[2.] (Exercise 8.1.16.)
  Prove:
  If, for any sets $A$ and $B$, either
$|A|\leq |B|$ or $|B|\leq |A|$, then the Axiom of
Choice holds.
\end{enumerate}

\bigskip

\color{blue}
It suffices to argue that if,
for any sets $A$ and $B$, either
$|A|\leq |B|$ or $|B|\leq |A|$, then every set can be well-ordered.

For this, choose $A$ arbitrarily and let $B = h(A)$ (= the Hartogs number
of $A$). Then $|h(A)|\leq |A|$ is impossible, so $|A|\leq |h(A)|$.
This yields an injective function $f\colon A\to h(A)$.
Well-order $A$ by pulling back the order on the ordinal $h(A)$ to $A$:
$x<y$ in $A$ if $f(x)<f(y)$ in $h(A)$.
\color{black}
\bigskip

\begin{enumerate}
\item[3.]
Suppose you are working in a universe of sets satisfying ZF in which all proper classes have the same size in the following sense: whenever $C$ and $D$ are proper classes, then there is a class bijection $F\colon C\to D$. Show that the Axiom of Choice holds in your universe.  
\end{enumerate}
  
\bigskip

\color{blue}
Assume that $F\colon V\to \textrm{ON}$ is a class bijection
from the class of all sets to the class of ordinal numbers.
We can use $F$ to define a choice function $c_A$ for
any set $A$, namely if $U\subseteq A$ and $U\neq \emptyset$,
choose $c_A(U) = u_0\in U$ where $F(u_0)$ is the $\in$-least
element of $F(U)$.
\color{black}

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