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\lhead{Set Theory \\ Assignment 5} 
\rhead{Solution Key \\ October 18}

\noindent
\begin{enumerate}
\item[(1)]
Define a binary operation $\circ$ on $\omega$ as follows.
  Given $m, n\in\omega$, choose sets $A, B$ with $|A|=m$, $|B|=n$
  and define $m\circ n = |A\times B|$.
\begin{enumerate}
\item[(a)] Show that $\circ $ is well defined.
\item[(b)] Show that $m\circ 0=0$ and $m\circ S(n)=(m\circ n)+m$.
  (That is, $\circ$ satisfies the recursion that defines multiplication.)
\item[(c)] Conclude that $m\circ n = mn$. (This shows that
  $|A\times B|=|A|\cdot |B|$ for finite sets.)
\end{enumerate}
\end{enumerate}

\smallskip
\color{blue}
\begin{enumerate}
\item[(a)] We must show that if
  $|A|=|A'|$ and 
  $|B|=|B'|$, then
  $|A\times B|=|A'\times B'|$. For this,
  assume that $f\colon A\to A'$ is a bijection with inverse $f^{-1}$
  and $g\colon B\to B'$ is a bijection with inverse $g^{-1}$.
  I claim that
  \[
  f\times g\colon A\times B\to A'\times B'\colon (a,b)\mapsto (f(a),g(b))
  \]
  is a bijection with inverse $f^{-1}\times g^{-1}$.
  To verify this, observe that
  \[
\begin{array}{rl}  
((f^{-1}\times g^{-1})\circ (f\times g))(a,b)
&= (f^{-1}\times g^{-1})(f(a),g(b))\\
&= ((f^{-1}\circ f)(a),(g^{-1}\circ g)(b))\\
&= (a,b).
\end{array}
\]
A similar calculation shows that
$((f\times g)\circ (f^{-1}\times g^{-1}))(a',b')=(a',b')$,
so $(f\times g)^{-1} = (f^{-1}\times g^{-1})$.
Since $(f\times g)$ is invertible, it is a bijection.
\item[(b)]
  The fact that $m\circ 0=0$ follows from the fact
  t5hat $m\times \emptyset = \emptyset$.
  To show that $m\circ S(n)=(m\circ n)+m$ holds we should
  explain why there is a bijection between
  $m\times S(n) = m\times (n\sqcup \{n\}) = (m\times n)\sqcup (m\times \{n\})$
  and $(m\times n)\sqcup m$. The bijection
  \[F\colon (m\times n)\sqcup (m\times \{n\})\to (m\times n)\sqcup m
  \]
  can be taken to be $\textrm{id}_{m\times n}\cup \pi$
  where $\pi\colon (m\times \{n\})\to m$ is the function
  defined by $(i,n)\mapsto i$.
\item[(c)] Since the circle operation satisfies the recursion
  that defines multiplication of natural numbers,
  the circle operation must equal the operation
  of multiplication of natural numbers according to the uniqueness
  statement of the Recursion Theorem.
\end{enumerate}
\color{black}

\smallskip

\begin{enumerate}
\item[(2)]
Show that the real line has the same cardinality as the real plane.
(Hint: By the CBS Theorem you only need to find
1-1 functions in each direction. For a 1-1 function from the plane
to the line, try mapping a point $(x,y)$ in the plane
to the real number obtained by interlacing the digits
of $x$ and $y$. Be careful to explain exactly what you
mean, noting that some real numbers
have more than one decimal representation.).
\end{enumerate}

\smallskip
\color{blue}
The function $f\colon \mathbb R\to \mathbb R^2\colon r\mapsto (r,0)$
is injective, so $|\mathbb R|\leq |\mathbb R^2|$.

For the other comparison,
following the hint, we will
consider the interlacing function
$g\colon \mathbb R^2\to \mathbb R$.
Here I will write a real number as
\[
\mathbf{a} = \ldots a_2 a_1 a_0 \bullet a_{-1} a_{-2} \ldots,
\]
but this will be done under the assumption
that only finitely many of the digits left of the point
are nonzero. The map is:
\[(\mathbf{a},\mathbf{b}) = 
(\ldots a_1 a_0 \bullet a_{-1} a_{-2} \ldots \textbf{\large{,}}
\ldots b_1 b_0 \bullet b_{-1} b_{-2} \ldots) \mapsto 
\ldots a_1 b_1 a_0 b_0 \bullet a_{-1} b_{-1} a_{-2} b_{-2} \ldots =
\mathbf{a}^{\widehat{\hphantom{a}}}\mathbf{b}.
\]
In order to avoid the type of conflict mentioned in
the problem hint, I will assume that the input
real numbers $\mathbf{a}$ and $\mathbf{b}$ are written
in base 2, so only the digits $0$ and $1$ occur
in their expansions. But I will assume that the output
number $\mathbf{a}^{\widehat{\hphantom{a}}}\mathbf{b}$
is considered to be written in base 10.
Now, if $(\mathbf{a},\mathbf{b}) \neq (\mathbf{a'},\mathbf{b'})$,
then $\mathbf{a}^{\widehat{\hphantom{a}}}\mathbf{b}$
and $\mathbf{a'}^{\widehat{\hphantom{a}}}\mathbf{b'}$
are different decimal numbers.
They cannot represent the same real number,
since if they did they would be two
different decimal expansions, {\it neither
containing the digit 9}, which represent
the same real number. But, when a real number
has more than one decimal expansion, at most one
of the expansions omits the digit $9$.
\color{black}

\smallskip

\begin{enumerate}
\item[(3)]
Show that $|(A^B)^C|=|A^{B\times C}|$.
\end{enumerate}
\smallskip

\color{blue}
There is a bijection 
from $A^{B\times C}$ to $(A^B)^C$ called ``currying'', which assigns
to a function $f\colon B\times C\to A$ the function
$g\colon C\to A^B\colon c\mapsto f(x,c)$.
The inverse of the function $f\mapsto g$ (which is called ``uncurrying'')
is the function that assigns to any $G\colon C\to A^B$
the function $F\colon B\times C\to A\colon (b,c)\mapsto G(c)(b)$.

One should check that
(uncurrying $\circ$ currying) equals the identity and that 
(currying $\circ$ uncurrying) equals the identity.

Suppose we are given $f\colon B\times C\to A$.
Currying $f$ yields $g\colon C\to A^B\colon c\mapsto f(x,c)$.
Uncurrying $g$ yields
\[F\colon B\times C\to A\colon (b,c)\mapsto g(c)(b)  = f(x,c)(b)=f(b,c),\]
so $F(b,c)=f(b,c)$ for all $(b,c)\in B\times C$. This proves that $F=f$.

Now suppose that we are given
$G\colon C\to A^B$ and we uncurry $G$ to
$F\colon B\times C\to A\colon (b,c)\mapsto G(c)(b)$.
Currying $F$ now yields
\[g\colon C\to A^B\colon c\mapsto F(x,c) = G(c)(x),\]
so $g(c)(x) = G(c)(x)$ for all $x\in B$, so $g(c)=G(c)$
for all $c\in C$. This proves that $g=G$.
\color{black}
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