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\noindent
{\bf Solutions to HW 9.}

\bigskip

  
\begin{enumerate}
\item[1.]
  Show that ``$A\subseteq B$ and $B\subseteq A$ implies $A=B$''
  in each of the following two ways.
\end{enumerate}
\bigskip


\begin{enumerate}
\item[(i)] With a direct proof.
\end{enumerate}
\bigskip
\color{blue}
Assume that $A\subseteq B$ and that $B\subseteq A$.

Choose any $a\in A$.  Since $A\subseteq B$, it follows that $a\in B$.
Now choose any $b\in B$.  Since $B\subseteq A$, it follows that $b\in A$.
We have shown that $A$ and $B$ have the same elements, so $A=B$
by the Axiom of Extentionality.
\color{black}

\bigskip

\begin{enumerate}
\item[(ii)] With a proof by contradiction.
\end{enumerate}
\bigskip

  \color{blue}
  Assume that $A\subseteq B$ and that $B\subseteq A$, but $A\neq B$.

  Since $A\neq B$, then by the Axiom of Extentionality
  it is not true that $\forall z((z\in A)\leftrightarrow (z\in B))$,
  so there is some $z\in A\setminus B$ or some $z\in B\setminus A$.
  The two cases are similar, so assume that there is some
  $z\in A\setminus B$. This $z$ witnesses that $A\not\subseteq B$,
  contrary to our assumption.
 \color{black}

 \vfill

\begin{enumerate}
\item[2.]
  Show that ``any nonconstant, real, linear function
  $f(x)=ax+b$ has a unique root''
  in each of the following two ways.
\end{enumerate}
\bigskip

\begin{enumerate}
\item[(i)] With a direct proof.
\end{enumerate}
\bigskip

\color{blue}
Assume that $f(x)=ax+b$ is a nonconstant function.

If $f(x)=ax+b$ is nonconstant, then is must be that $a\neq 0$
(since, when $a=0$, $f(x)=0x+b=b$ IS constant).
Now, using the arithmetic of the real numbers,
$f(x)=0$ holds exactly when $ax+b=0$, or $ax=-b$, or $x=-b/a$.

The fact that the equation $ax+b=0$ is solvable shows that a root
exists, the fact that the equation is uniquely solvable
shows that the root is unique.
\color{black}

\bigskip

\begin{enumerate}
\item[(ii)] With a proof by contradiction.
\end{enumerate}
\bigskip

  \color{blue}
  Assume that  $f(x)=ax+b$ is nonconstant,
  but does not have a unique root.
  If $a=0$, then $f(x)=ax+b = 0x+b = b$ is constant
  which is not true, so we may assume that $a\neq 0$.
%  The fact that
%  this function is nonconstant means $a\neq 0$.

If $f(x)=ax+b$ does not have a unique root, then
either (1) it has no root at all, or (2)
it has multiple roots. 

\noindent
    {\bf Case 1.}
    $f(x)=ax+b$ has no root at all.

By substitution we see that $f(-b/a)=0$, so
$f$ has at least one root when $a\neq 0$. Thus
our assumptions place us in 
  
\noindent
    {\bf Case 2.}
    $f$ has multiple roots.

If $r_1\neq r_2$ are roots of $f$, then $f(r_1)=ar_1+b = 0$ and
$f(r_2)=ar_2+b = 0$.  Subtracting these equations yields
$0 = (ar_1+b)-(ar_2+b) = a(r_r-r_2)$. Dividing through by
$r_1-r_2$ yields $0=a$, contrary to an earlier conclusion.
This contradiction completes the proof.
 \color{black}

\vfill
  
\begin{enumerate}
\item[3.]
  The goal of this problem is to prove that the
  composition of two surjective functions is surjective.
  The type of structure involved looks like
  $\mathbb X=\langle A,B,C;f,g\rangle$
  where $f:A\to B$ and $g:B\to C$ are functions.
  Let the variable $a$ range over the set $A$,
  the variable $b$ range over the set $B$,
  and the variable $c$ range over the set $C$.

  The functions (i) $f$, (ii) $g$, (iii) $g\circ f$
  are surjective if the following sentences hold in $\mathbb X$:
  \begin{enumerate}
\item[(i)] $(\forall b)(\exists a)(f(a)=b)$.
\item[(ii)] $(\forall c)(\exists b)(g(b)=c)$.
\item[(iii)] $(\forall c)(\exists a)(g\ \circ f(a)=c)$.
\end{enumerate}

  To prove that the composition of surjective functions
  is surjective, you must give a winning strategy for
  $\exists$ in the sentence in (iii).
  YOU ARE ALLOWED TO USE the fact that there exist winning
  strategies for $\exists$ in the sentences in (i) and (ii).
  Write a proof that indicates the winning strategy for $\exists$
  in (iii), which accesses the information of the strategies
  for $\exists$ in (i) and (ii). 
\end{enumerate}
\bigskip

\color{blue}
We must provide a winning strategy for $\exists$ for Game (iii). We
are allowed to use that there exist winning strategies for
$\exists$ in Games (i) and (ii).

How do we access the strategies for $\exists$ in Games (i) and (ii)
when developing a strategy for $\exists$ in Game (iii)?

Imagine that Games (i) and (ii) are played by the ``earlier self''
of $\exists$. When $\exists$ needs information about those
games, she will ask her earlier self to
\emph{reveal those winning strategies}.
She does this by pretending to be $\forall$, and playing
Games (i) and (ii) against her earlier self, and watching
how her earlier self responds to each move. Then she uses
that information while developing a strategy for the current game,
Game (iii).

Let's start developing the winning strategy for
$\exists$ for Game (iii): $(\forall c)(\exists a)(g \circ f(a)=c)$.

\begin{itemize}
\item $\forall$ chooses some $c=c_0\in C$.
\item   $\exists$ chooses some $a=a_0\in A$
  using the following strategy:
\begin{itemize}
\item $\exists$ pretends to be the earlier self of
  $\forall$ in Game (ii), and plays
  $c=c_0$.
\item the earlier self of $\exists$ in Game (ii) has a winning strategy
  for Game (ii), so has a way to respond with some $b=b_0\in B$
  such that $g(b_0)=c_0$.
\item Now $\exists$ pretends to be the earlier self of
  $\forall$ in Game (i), and plays
  $b=b_0$.
\item the earlier self of $\exists$ in Game (i) has a winning strategy
  for Game (i), so has a way to respond with some $a=a_0\in B$
  such that $f(a_0)=b_0$.
\end{itemize}
\end{itemize}

\color{red}
This choice of $a=a_0\in A$
is a winning response for $\exists$
against $\forall$'s choice of $c=c_0\in C$,
since if $\exists$ chooses
$b=b_0\in B$ and $a=a_0\in A$ in these ways,
then according to the strategies of Games (i) and (ii) 
we will have $f(a_0)=b_0$ and $g(b_0)=c_0$, so
$c = c_0 = g(f(a_0)) = g\circ f(a)$.

\color{black}

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