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\begin{document}

\noindent
{\large \bf Additional notes \#1.}

\bigskip

\noindent
{\bf Two Nice Theorems.}

\bigskip

\noindent
{\bf Theorem} (Cantor-Schroeder-Bernstein)
If $A$ is equivalent to a subset of $B$ and $B$ is
equivalent to a subset of $A$, then $A$ is equivalent 
to $B$.

\medskip

{\it Proof.} We prove this theorem in the case that 
$A$ and $B$ have no elements in common.
This is sufficient, since if they have
elements in common we can replace $B$ by an equivalent
set $B'$ that has no elements in common with $A$,
and then prove that $A$ is equivalent to $B'$
(hence is equivalent to $B$).

Since $A$ is equivalent to a subset of $B$,
then there is a 1-1 function $f:A\rightarrow B$
onto a subset of $B$. Similarly, the hypotheses
of the theorem imply that there is a 1-1
function $g:B\rightarrow A$ onto a subset of $A$.
We will use $f$ and $g$ to define a 1-1
function $h:A\rightarrow B$ onto $B$.

If $a\in A$, then define the {\em parent} of 
$a$ to be $g^{-1}(a)$ if this element exists.
If $b\in B$, then the {\em parent} of
$b$ is $f^{-1}(b)$ if this element exists.
Note that for any $x\in A\cup B$ the parent
of $x$ is unique if it exists, since $f$ and
$g$ are 1-1. Call $y$ an {\em ancestor}
of $x$ if $y=x$ or $y$ 
is the parent of the parent of
the parent $\cdots$ (etc) of the parent of $x$.
A {\em terminal ancestor} of $x$, written
$x_t$, is
any ancestor of $x$ that has no parent.
The element $x$ may not have a terminal ancestor,
but if it does then the terminal ancestor
is unique since parents are unique.

Define a function $h: A\rightarrow B$ by 
$$
h(x) = \left\{\begin{array}{ll}
      g^{-1}(x) & \mbox{\rm if $x_t$ exists and is in $B$}\\
      f(x) & \mbox{\rm otherwise.}
       \end{array}\right.
$$
%That is, $h(x)$ is either the parent of $x$
%or the ``child'' of $x$, with the two cases being
%distinguished by whether or not the terminal
%ancestor of $x$ is in $B$.
Observe that $h$ is a well-defined function
from $A$ to $B$ (check this!), as is its inverse:
$$
h^{-1}(y) = \left\{\begin{array}{ll}
      g(y) & \mbox{\rm if $y_t$ exists and is in $B$}\\
      f^{-1}(y) & \mbox{\rm otherwise.}
       \end{array}\right.
$$
Since $h:A\rightarrow B$ has an inverse, it is a
1-1 function from $A$ onto $B$. \hfill $\Box$


\bigskip\bigskip

\vfill
\pagebreak

\noindent
{\bf Theorem} (Cantor) 
$|X| < |{\mathcal P}(X)|$
for any set $X$.

\medskip

{\it Proof.}
Since $f:X\rightarrow {\mathcal P}(X): x\mapsto \{x\}$
is a 1-1 function from $X$ onto a subset of 
${\mathcal P}(X)$, it follows that 
$|X|\leq |{\mathcal P}(X)|$. To prove the theorem 
we must show only that $|X| \neq |{\mathcal P}(X)|$.
Assume otherwise that there is a cardinal
$\kappa$ such that 
$|X| = \kappa = |{\mathcal P}(X)|$. Then it is
possible to enumerate $X$ with $\kappa$:
$$X = \{x_i\;|\;i < \kappa\},$$
and also to enumerate ${\mathcal P}(X)$ with $\kappa$:
$${\mathcal P}(X) = \{U_i\subseteq X\;|\; i < \kappa\}.$$
Define $U = \{x_i\in X\;|\;x_i\not\in U_i\}$.
Clearly $U\subseteq X$, so $U\in {\mathcal P}(X)$,
so $U = U_{i_0}$ for some $i_0<\kappa$.

\smallskip

\noindent
{\bf Claim.} $x_{i_0}\in U$ if and only if $x_{i_0}\not\in U_{i_0}$.

\smallskip

{\em Proof of Claim.} This follows from the definition of $U$.

\medskip

But $U=U_{i_0}$!
So the Claim shows that 
$x_{i_0}$ is neither in nor out of $U$.
This is a contradiction, which 
implies that we made an incorrect assumption somewhere 
in this proof. The only assumption that we did not
justify was the assumption that there exists
a cardinal that enumerates both $X$ and 
${\mathcal P}(X)$, so we conclude that
$|X|\neq |{\mathcal P}(X)|$.
\hfill $\Box$
\end{document}

