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\lhead{Set Theory \\ Assignment 3} 
\rhead{Solution Key \\ September 29}

\noindent
\begin{enumerate}
\item[(1)]
  Do Exercise 2.3.9 (a). (Show that the set $B^A$ exists.)
  (If you use the hint, show that it is true. Definition 2.3.13 is relevant to this problem.)
\end{enumerate}

\smallskip
\color{blue}
$B^A$ is meant to denote the set of all functions
$f\colon A\to B$. Each such function $f$ is a binary relation
from $A$ to $B$, hence is a subset of $A\times B$,
hence is an element of ${\mathcal P}(A\times B)$.
This shows that $B^A\subseteq {\mathcal P}(A\times B)$.

To show that $B^A$ is a set
we explain how to construct it. We could define it
using the Axiom of Separation by
\[
B^A = \{f\in {\mathcal P}(A\times B)\;|\;\varphi_{\textrm{function}}(f,A,B)\}
\]
if we had a formula $\varphi_{\textrm{function}}(x,y,z)$
expressing the property that $x$ is a function from $y$ to $z$.
This formula needs to say that $x\subseteq y\times z$
and that $x$ satisfies the function rule: for every
$u\in y$ there is a unique $v\in z$ such that $(u,v)\in x$.

Assuming that we have already constructed formulas
for the less-complex concepts
\begin{enumerate}
\item $\varphi_{\textrm{binary relation}}(x,y,z)$: ``$x\subseteq y\times z$''
\item $\varphi_{\textrm{pair}}(p,u,v)$: ``$p=(u,v)$''
\end{enumerate}
we could express $\varphi_{\textrm{function}}(x,y,z)$ as
\[
\varphi_{\textrm{binary relation}}(x,y,z)\wedge
(\forall u)((u\in y)\to(\exists ! v)((v\in z)\wedge ((\exists p)(\varphi_{\textrm{pair}}(p,u,v)\wedge (p\in x))))).
\]
Here $\exists !$ means ``there exists a unique''.
We write $(\exists ! s) Q(s)$ to abbreviate
\[
(\exists t)Q(t) \wedge (\forall r)(\forall s)((Q(r)\wedge Q(s))\to (r=s)).
\]
\color{black}

\smallskip

\begin{enumerate}
\item[(2)]
  Do Exercise 3.2.6. (Prove that each natural number
  is the set of all smaller natural numbers,
i.e., $n=\{m\in \mathbb N\;|\;m<n\}$.)
\end{enumerate}

\smallskip
\color{blue}
Let $\varphi(n)$ be the statement that $n\subseteq \mathbb N$.
\begin{enumerate}
\item (Basis of Induction)
  $\varphi(0)$ holds since $0=\emptyset$
  is a subset of every set.
\item (Inductive Step)
  Assume that $\varphi(k)$ holds for some $k\in\mathbb N$,
  i.e., $k\subseteq \mathbb N$. Since
  $k\in\mathbb N$, we have $\{k\}\subseteq \mathbb N$.
  Since both $k$ and $\{k\}$ are subsets of $\mathbb N$,
  so is $k\cup \{k\} = S(k)$. This shows that
  $k\subseteq \mathbb N$ implies that $S(k)\subseteq \mathbb N$.
%  proves
%  that the truth of $\varphi$ is preserved at successor stages.
\end{enumerate}

Now we argue that any natural number $n$
is the set of its $<$-predecessors
(in symbols, this is the assertion $n=\{m\in \mathbb N\;|\;m<n\}$).
We use that we have defined the order on $\mathbb N$
so that for $m,n\in\mathbb N$ we have
$m<n$ iff $m\in n$. (Thus, $<$ on $\mathbb N$
equals the restriction to $\mathbb N$ of the relation $\in$.)
\begin{enumerate}
\item[(a)] If $x\in n$, then $x\in \mathbb N$ by the inductive
  argument above, hence $x<n$ by the definition of $<$ on $\mathbb N$.
\item[(b)] If $x < n$, then according to the definition
  of $<$ we have $x\in n$.
\end{enumerate}
Items (a) and (b) + the Axiom of Extensionality
prove that $n=\{m\in \mathbb N\;|\;m<n\}$.
\color{black}

\smallskip

\begin{enumerate}
\item[(3)]
  Do Exercise 3.2.8. (Prove that there is no function
  $f\colon \mathbb N\to \mathbb N$ such that for all
  $n\in \mathbb N$, $f(n)>f(n+1)$.)
\end{enumerate}
\smallskip

\color{blue}
One may use the Axiom of Foundation to prove the more general fact that,
for any set $A$, 
there is no function
  $f\colon \mathbb N\to A$ such that for all
$n\in \mathbb N$, $f(n+1)\in f(n)$. That argument goes as follows:
if $f$ were such a function, then its image $\textrm{im}(f)$
is a nonempty set that has no $\in$-minimal element.

But we can prove the statement of Exercise 3.2.8
without using the Axiom of Foundation, since the
image of $f$ consists of natural numbers.
We just observe that $\textrm{im}(f)$
is a nonempty subset of $\mathbb N$
that has no $<$-least element,
contradicting the fact that $\mathbb N$
is well-ordered by $<$.
\color{black}
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