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\newcommand{\ovs}{\overline{\overline{S}}}
\newcommand{\ovt}{\overline{\overline{T}}}
\newcommand{\ovx}{\overline{\overline{X}}}
\newcommand{\ovy}{\overline{\overline{Y}}}
\newcommand{\ovu}{\overline{\overline{U}}}
\newcommand{\ovv}{\overline{\overline{V}}}
\newcommand{\ovz}{\overline{\overline{Z}}}

\begin{center}

REAL ANALYSIS

CARDINAL NUMBERS

\end{center}

We use $\ovs$ for the cardinal number of a set $S$.

\begin{description}

\item[I]
$\ovs\leq\ovt$ (or $\ovt\geq\ovs$) is to mean \lq\lq $\exists$ a
1-1 correspondence between $S$ and a subset of $T$'' (not
necessarily a proper subset).

\item[II]
$\ovs=\ovt$ is to mean \lq\lq$\exists$ a 1-1 correspondence
between $S$ and $T$.

[$&lt;$ is to mean $\leq$ but not =]

\end{description}

We have that:

\begin{description}

\item[(i)]
The definitions are reasonable when applied to finite sets.

\item[(ii)]

\begin{description}

\item[(a)]
$\leq$ is transitive, i.e.

$$\ovx\leq\ovy\ \ \ovy\leq\ovz\Rightarrow\ovx\leq\ovz$$

\item[(b)]
$=$ is transitive

$$\ovx=\ovy\ \ \ovy=\ovz\Rightarrow\ovx=\ovz$$

= is symmetric

$$\ovs=\ovt\Leftrightarrow\ovt=\ovs$$

= is reflexive

$\ovs=\ovs$

\end{description}

\item[(iii)]
(Bernstein's Lemma) $\ovs\leq\ovt\
\ovt\leq\ovs\Rightarrow\ovs=\ovt$

\item[(iv)]
For any two sets either $\ovs\leq\ovt$ or $\ovt\leq\ovs$.

\end{description}

A set $S$ is said to be enumerable (denumerable, countable)
$\Leftrightarrow\exists$ a 1-1 correspondence between $S$ and the
set of all natural numbers.

$\chi_0$ is called the cardinal number of the set of all natural
numbers.

\begin{enumerate}

\item
If $\ovs\leq\chi_0$ either $S$ is finite or $\ovs=\chi_0$

\item
If $\ovs=\chi_0\ S$ can be put in 1-1 correspondence with proper
subset of itself.

\item
Any infinite subset contains an enumerable subset.

\item
If $\ovs=\chi_0$ and $T$ is infinite then
$\overline{\overline{S\cup T}}=\ovt$.

\item
A set is infinite $\Leftrightarrow$ it can be put into 1-1
correspondence with a proper subset of itself.

\end{enumerate}

\begin{description}

\item[Proof of A]
Suppose $U$ and $V$ are such that $\ovu=\ovv=\chi_0$.

Then $U=u_1\ u_2\ u_3\ldots$

$V=v_1\ v_2\ v_3\ldots$

$U\cup V=u_1\ v_2\ u_2\ v_2\ldots=W=w_1\ w_2\ w_3\ w_4\ldots$
therefore $\overline{\overline{U\cup V}}=\chi_0$.

We define a 1-1 correspondence between $S$ and $T$ thus $T$
contains a subset $T'|\ovt'=\chi_0$.

We map $S\cup T'$ onto $T'$(1-1) and map all the elements of $T$
not in $T'$ onto themselves therefore $\ovs\cup\ovt=\ovt$.

\end{description}

\begin{description}

\item[I]
The set of all pairs $(m,n)$ of all natural numbers is enumerable.

Set up the 1-1 correspondence $(m,n)\Leftrightarrow 2^m3^n$, for
by the theorem of uniqueness of prime factorisation
$2^{m_1}3^{n_1}=2^{m_2}3^{n_2}\Leftrightarrow m_1=m_2\ n_1=n_2$.

Therefore we have mapped the set onto an infinite subset of the
natural numbers which is enumerable.

\item[II]
$S_1,S_2,S_3,\ldots$ enumerable $\displaystyle\Rightarrow
U_{r=1}^\infty S_r$ enumerable.

$S_1=a_{11}\ a_{12}\ a_{13}\ldots$

$S_2=a_{21}\ a_{22}\ a_{23}\ldots$

We assign to $a_{mn}$ the number given by :

$f(mn)$ is a 1-1 correspondence between the set of pairs of
natural numbers and the natural numbers. Assign to an element $x$
of $Us_r$ the least natural number of $f(mn)$ for which
$a_{mn}=x$. Then $Us_r$ is enumerable.

\item[III]
The set of integers $h$ is enumerable.

The set of natural numbers $q$ is enumerable.

The set of all pairs $(h,q)$ is enumerable by (2) $\exists f(h,q)$
mapping the pairs $(h,q)$ in the natural numbers. Now to any
rational $r$ assign the least $f(h,q)$ for which $r=\frac{h}{q}$.

\item[IV]
Suppose $X^1\ X^2\ldots X^n$ are enumerable sets. Then the set of
$(x^{(1)}x^{(2)}\ldots x^{(n)}$ where each $x^{(r)}$ runs
independently through the elements of $X_r$, is enumerable.

$\exists$ a 1-1 correspondence between the $x^{(r)}$ and the
natural numbers $U_u$.

We use induction. Suppose true for $n=m$. $\exists$ a 1-1
correspondence between $(U_1\ldots U_m\ U_{m+1})$ and
$(V,U_{m+1})$.

This is enumerable by III.

\item[V]
The result of $IV$ remains true if we have an enumerable system
$X^{(1)}X^{(2)}\ldots$ and consider all $(x^{(1)}x^{(2)}\ldots
x^{(n)})$ with $n$ variables but finite.

We have $S:(x^{(1)}x^{(2)}\ldots
x^{(n)})$

We have $S_n:(x^{(1)}x^{(2)}\ldots x^{(n)}$, $n$ fixed. This is
enumerable by IV.

$\displaystyle S=\cup_{r=1}^\infty S_n$ and is enumerable by II.

\item[VI]
Consider the set of all polynomials.

$S=b_0x^n+b_1x^{n-1}+\ldots+b_n$ where the $b_i$ are integers.
This set is enumerable by V, but to each $P_n(x)$ corresponds at
most $n$ algebraic numbers. Hence the set of all algebraic numbers
is enumerable.

\begin{description}

\item[Example]
The set of discontinuities of a given monotone function is
enumerable.

\end{description}

\item[VII]
The set of all real numbers is not enumerable.

\begin{description}

\item[(i)]
Consider all real numbers $0&lt;\alpha\leq 1$. Each $\alpha$ has a
unique decimal expansion $$.x_1x_2x)3\ldots$$ providing we insist
that the number of non-zero $x$'s is not finite.

Suppose the set of $alpha$ in $0&lt;\alpha\leq1$ is enumerable.
Enumerate them

\begin{eqnarray*}
\alpha_1&amp;+&amp;.x_{11}x_{12}x_{13}x_{14}\ldots\\
\alpha_2&amp;=&amp;.x_{21}x_{22}x_{23}x_{24}\ldots\\
\alpha_3&amp;=&amp;.x_{31}x_{32}x_{33}x_{34}\ldots\\
\end{eqnarray*}

Let $\beta=.y_1y_2y_3\ldots\ 0&lt;\beta\leq1$ where $y_r=1$ if
$x_{rr}\neq1$ and $y_r=2$ if $x_{rr}=1$.

This $\beta$ is not to be found in the sequence
$\alpha_1\alpha_2\ldots$. $\beta\neq\alpha_n$ since they differ in
the $n$th place.

\item[(ii)]
Let $0\leq\alpha\leq1$ and suppose that $\alpha_1\alpha_2$ is an
enumeration of this set.

Trisect the interval $\left[01\right]$ by 3 closed intervals. At
least one does not contain $\alpha_1$. Choose $J_1$, the interval
nearest the left not containing $\alpha_1$. $\exists J_2$ nearest
the left not containing $\alpha_2$. These intervals tend to a
limit point $l\ 0\leq L\leq1$. For if $J_1=[a_n\ b_n]\ a_n\to l\
b_n\to l$ as $n\to\infty$.

$a_1\not\in J_1,\ \alpha_2\not\in J_2\ldots\ \ J_1\supset
J_2\supset J_3\ldots$ therefore

$l\in J_n$ for $n=1,2,\ldots$ therefore

$l\neq\alpha_n\ldots$

Contradiction- for $0\leq l\leq 1.$

\end{description}

\item[VIII]
For any $a&lt;b$ the points of $(ab)$ have the same cardinal number
as the points in (0\ 1) $\exists$ a 1-1 correspondence between the
points of (0\ 1) and the points of $(a\ b)$ given by $0&lt;t&lt;1\
t\leftrightarrow a+(b-a)t$.

\item[IX]
The cardinal number of the set of real numbers is the same as the
cardinal number of the set of points (0\ 1).

A (1-1) correspondence is $y\leftrightarrow\tanh(x)\ \
-\infty&lt;x&lt;+\infty\ -1&lt;y&lt;1$

We denote the cardinal number of the set of real numbers by $C$.

\item[X]
The set of points in $R_2$ has cardinal $C\ (x\
y)\leftrightarrow\left(\frac{1+\tanh x}{2},\frac{1+\tanh
y}{2}\right)$ maps the points of $R_2$ onto the open square
$0&lt;x&lt;1\ 0&lt;y&lt;1$.

Consider the point $(U\ V)$

$U=.x_1x_2\ldots,\ v=.y_1y_2\ldots$, unique if we exclude
terminating decimals.

$$(UV)\leftrightarrow\alpha=.x_1y_1x_2y_2\ldots$$

If $(U_1V_1)\ (U_2V_2)$ are different

$$(U_1V_1)\leftrightarrow\alpha_1\
(U_2V_2)\leftrightarrow\alpha_2\ \ alpha_1\neq\alpha_2.$$

We have set up a 1-1 correspondence between the set $T$ of points
of the square, and a subset $S$ of the points of (0 1). We can map
the points of $S$ onto a subset of $T$ by
$U\leftrightarrow(U,\frac{1}{2})$. Therefore $\ovt\leq C\ \
c\leq\ovt$ therefore $\ovt=c$

\item[XI]
We can extend the result of X to $R_n$ by induction.

\begin{description}

\item[Example]
The set of all sequences $x_1x_2x_3\ldots$ of real numbers has
cardinal $C$.

\end{description}

\end{description}

\end{document}
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