The cardinality of the set of real numbers
Jailton C. Ferreira
\begin{center} A proof that the set of real numbers is denumerable is given. \end{center}
math0108119.tex
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\newtheorem{theorem}{Theorem}[section]
\newtheorem{definition}{Definition}[section]
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\title{The cardinality of the set of real numbers}

\author{Jailton C. Ferreira}

\date{ }
\begin{document}
\maketitle
\pagenumbering{arabic}

\begin{abstract}
\begin{center}
A proof that the set of real numbers is denumerable is given.
\end{center}
\end{abstract}


\section{Introduction} \label{sec-1}

\hspace{22pt}Each real number of the interval [0, 1] can be
represented by an infinite path in a given binary tree. In Section
\textbf{~\ref{sec-2}} the binary tree is projected on a grid $N
\times N$ and it is shown that the set of the infinite paths
corresponds one-to-one to the set $N$. The Theorems
~\ref{teorema-1} and ~\ref{teorema-2} give the first proof and the
Theorem ~\ref{teorema-3} provides a second proof.

\hspace{22pt}Section \textbf{~\ref{sec-3}} examines the Cantor's
proof of 1891. The Section shows that (i) if the diagonal method
is correct, then any denumerable list $L$ to which the diagonal
method was applied is incomplete (Theorem ~\ref{teorema-4}), (ii)
if some complete list exists and if the diagonal method is
correct, then a complete list cannot be represented in the form
used in Cantor's proof (Theorem ~\ref{teorema-5}) and (iii) being
$L$ incomplete nothing affirms or denies that $\lvert N \lvert$ is
the cardinality of the set of real numbers of the interval [0, 1]
(Theorem ~\ref{teorema-6}).

\hspace{22pt} Sections \textbf{~\ref{sec-2}} and
\textbf{~\ref{sec-3}} mean that (i) there is the list, denoted by
$L_H$, such that it contains all members of $N$ and all members of
$F$ and to each member of $F$ of the list corresponds one-to-one a
member of $N$ of the list and (ii) if the diagonal method used in
Cantor's 1891 proof ~\cite{Cantor} is valid, then $L_H$ has a
different form of the form of the list $L$ used by Cantor. In
Section \textbf{~\ref{sec-4}} we try to show that $L_H$ has the
same form as $L$.


\section{The proof of   $\lvert F \lvert  =  \lvert N \lvert$}
\label{sec-2}

\begin{theorem} \label{teorema-1}
Let B be a binary tree such that every node has two children and
its depth is equal to $\lvert N \lvert$. Let A be the set of the
nodes of the binary tree B. The cardinality of A is less than or
equal to $\lvert N \lvert$.
\end{theorem}
\textit{Proof:}

\hspace{22pt}Let us now consider the Figure 1. The horizontal
sequence of finite natural numbers, presented in increasing order
from left to right, contains all numbers of $N$; to each natural
number of the sequence corresponds a vertical line. The vertical
sequence of numbers, presented in increasing order from top to
bottom, contains all numbers of $N$; to each natural number of the
sequence corresponds a horizontal line. To each node of the grid
formed by the horizontal and vertical lines corresponds a pair
($m$, $n$), where $m$ belongs to the horizontal sequence of
numbers and $n$ belongs to the vertical sequence of numbers.

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\begin{center}
\textrm{Figure 1}
\end{center}

Let us project the binary tree $B$ on the grid of the Figure 1 in
the following way: to the root of the tree corresponds the pair
(0, 0); to the 2 nodes of the level 1 correspond the pairs (0, 1)
and (1, 0); to the 4 nodes of the level 2 correspond the pairs (0,
3), (1, 2), (2, 1) and (3, 0); to the 8 nodes of the level 3
correspond the pairs (0, 7), (1, 6), (2,5), (3, 4), (4, 3), (5,
2), (6, 1) and (7, 0) and to the $2^k$ nodes of the level $k$
correspond the pairs

\begin{equation}\label{dois-1}
(0, 2^{k}-1), (1, 2^{k}-2), (2, 2^{k}-3),  \ldots ,(2^{k}-3, 2),
(2^{k}-2, 1) \hspace{8pt} \textrm{and} \hspace{8pt} (2^{k}-1, 0)
\end{equation}

The Figure 2 shows the binary tree $B$ up to the depth 4. The
Figure 3 shows the projection of the binary tree up to the depth 3
on the grid of the Figure 1.

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\begin{center}
\textrm{Figure 2}
\end{center}

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\begin{center}
\textrm{Figure 3}
\end{center}

\hspace{22pt}It is known that the bijection  $f:N \rightarrow  N
\times N$, where $N$ is the set of finite natural numbers, can be
defined applying the diagonal method.

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\begin{center}
\textrm{Figure 4}
\end{center}

The Figure 4 shows how starting from the pair (0, 0) and following
the thick line we can establish the one-to-one correspondence
between the set of the pairs ($m$, $n$) and $N$, this is,

\begin{equation}\label{dois-2}
\begin{tabular}{c c c c c c c c}
(0,0) & (1,0) & (0,1) & (0,2) & (1,1) & (2,0) & (3,0) & \ldots \\
 $\updownarrow$ & $\updownarrow$ & $\updownarrow$ & $\updownarrow$ & $\updownarrow$ & $\updownarrow$ & $\updownarrow$ &  \\
0 & 1 & 2 & 3 & 4 & 5 & 6 & \ldots \\
\end{tabular}
\end{equation}

Considering that to any node of the binary tree corresponds a node
of the grid and that there are nodes in the grid without
corresponding nodes in the binary tree, we conclude that

\begin{equation}\label{dois-3}
\lvert A \lvert  \le \lvert N \lvert
\end{equation}

\begin{theorem} \label{teorema-2}
The cardinality of the set of real numbers of the interval [0, 1]
is equal to $\lvert N \lvert$.
\end{theorem}
\textit{Proof:}

\hspace{22pt}Let us denote by $F$ the set of real numbers of the
interval [0, 1]. Any element of $F$ can be represented in the
binary system by

\begin{equation}\label{dois-4}
f_1 \times 2^{-1} + f_2 \times 2^{-2} + f_3 \times 2^{-3} + f_4
\times 2^{-4} + \ldots
\end{equation}

where
\begin{equation}\label{dois-5}
f_i \in \{ 0, 1 \} \qquad \textrm{and} \qquad i \in N \qquad
\textrm{and} \qquad i \ne 0
\end{equation}


The representation \eqref{dois-4} can be simplified to

\begin{equation}\label{dois-6}
.f_1 f_2 f_3 f_4 f_5 \ldots
\end{equation}


where the first character of the sequence is the point. Each $f_i$
of \eqref{dois-6} is substituted by 0 or 1 to represent a given
number. The $i$-th 0 or 1 on the right of ``." corresponds to
$f_i$. The set $F$ can also be represented by a binary tree where
each node has two children. Each infinite path on the binary tree,
$.q_{\tau 1} q_{\tau 2} \ldots  q_{\tau i} \ldots $ with depth
equal to $\lvert N \lvert$ , represents an element of $F$.

\hspace{22pt} Let us denote by $B$ the binary tree above
mentioned; by $B_i$ the binary tree of depth $i$; by $P$ the set
of all paths starting from the root of $B$; by $P_i$ the set of
all paths from the root of $B_i$ to its leaves. Each element of
$P_i$ or $P$ is the set of the nodes (except the root) that form
the path. Let the set

\begin{equation}\label{dois-7}
\{P_{1}, P_{2}, \ldots , P_{i}, \ldots \} \qquad \textrm{such
that} \qquad i \in N
\end{equation}

The set $\{P_{1}, P_{2}, \ldots , P_{i}, \}$ contains all paths
from the root of the tree $B_i$ and the endings of the paths that
exist in $\{P_{1}, P_{2}, \ldots , P_{i} \}$ are all the nodes of
$B_i$, except the root. This indicates that the set in
\eqref{dois-7} - which corresponds to all the tree $B$ - contains
all paths of $B$.

\hspace{22pt} Let us consider the set of all infinite paths from
the root of $B$

\begin{equation}\label{dois-8}
Q=\{q_{\alpha}, q_{\beta}, q_{\gamma}, \ldots, q_{\tau}, \ldots\}
\end{equation}

The infinite path $q_{\tau}$ can be represented by the set

\begin{equation}\label{dois-9}
\{q_{\tau 1}, q_{\tau 2}, \ldots , q_{\tau i}, \ldots \}
\end{equation}

where $q_{\tau i}$ is the $i$-th node of $q_{\tau}$ after the
root. Let

\begin{equation}\label{dois-10}
Q_{\tau} = \{ \{q_{\tau 1} \}, \{q_{\tau 1} , q_{\tau 2}\}, \ldots
, \{q_{\tau 1}, q_{\tau 2}, \ldots , q_{\tau i} \}, \ldots \}
\end{equation}

The union of all elements of $Q_{\tau}$ is equal to $q_{\tau}$,
that is,

\begin{equation}\label{dois-11}
q_{\tau} = \{q_{\tau 1} \} \hspace{4pt} \cup \hspace{4pt}
\{q_{\tau 1} , q_{\tau 2}\} \hspace{4pt} \cup \hspace{4pt} \ldots
\hspace{4pt} \cup \hspace{4pt} \{q_{\tau 1}, q_{\tau 2}, \ldots ,
q_{\tau i} \} \hspace{4pt} \cup \hspace{4pt} \ldots
\end{equation}

Considering \eqref{dois-11} we can obtain from \eqref{dois-8}

\begin{eqnarray}\label{dois-12}
Q = \{ \{q_{\alpha 1} \} \hspace{4pt} \cup \hspace{4pt}
\{q_{\alpha 1} , q_{\alpha 2}\} \hspace{4pt} \cup \hspace{4pt}
\ldots \hspace{4pt} \cup \hspace{4pt} \{q_{\alpha 1}, q_{\alpha
2}, \ldots , q_{\alpha i} \}
\hspace{4pt} \cup \hspace{4pt} \ldots , \nonumber\\
\{q_{\beta 1} \} \hspace{4pt} \cup \hspace{4pt} \{q_{\beta 1} ,
q_{\beta 2}\} \hspace{4pt} \cup \hspace{4pt} \ldots \hspace{4pt}
\cup \hspace{4pt} \{q_{\beta 1}, q_{\beta 2}, \ldots , q_{\beta i}
\} \hspace{4pt} \cup \hspace{4pt} \ldots , \nonumber\\ \textit{.}
\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots
\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \textit{.} \nonumber\\
\{q_{\tau 1} \} \hspace{4pt} \cup \hspace{4pt} \{q_{\tau 1} ,
q_{\tau 2}\} \hspace{4pt} \cup \hspace{4pt} \ldots \hspace{4pt}
\cup \hspace{4pt} \{q_{\tau 1}, q_{\tau 2}, \ldots ,
q_{\tau i} \} \hspace{4pt} \cup \hspace{4pt} \ldots , \nonumber\\
\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots
\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots
\ldots \textit{.} \} \nonumber\\
\end{eqnarray}

From \eqref{dois-12} we have

\begin{eqnarray}\label{dois-13}
\lvert Q \lvert \leq \lvert \{q_{\alpha 1} \} \hspace{4pt} \cup
\hspace{4pt} \{q_{\alpha 1} , q_{\alpha 2}\} \hspace{4pt} \cup
\hspace{4pt} \ldots \hspace{4pt} \cup \hspace{4pt} \{q_{\alpha 1},
q_{\alpha 2}, \ldots , q_{\alpha i} \}
\hspace{4pt} \cup \hspace{4pt} \ldots \nonumber\\
\cup \hspace{4pt} \{q_{\beta 1} \} \hspace{4pt} \cup \hspace{4pt}
\{q_{\beta 1} , q_{\beta 2}\} \hspace{4pt} \cup \hspace{4pt}
\ldots \hspace{4pt} \cup \hspace{4pt} \{q_{\beta 1}, q_{\beta 2},
\ldots , q_{\beta i} \} \hspace{4pt} \cup \hspace{4pt} \ldots \nonumber\\
\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots
\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots
\ldots \nonumber\\
\cup \hspace{4pt} \{q_{\tau 1} \} \hspace{4pt} \cup \hspace{4pt}
\{q_{\tau 1} , q_{\tau 2}\} \hspace{4pt} \cup \hspace{4pt} \ldots
\hspace{4pt} \cup \hspace{4pt} \{q_{\tau 1}, q_{\tau 2}, \ldots ,
q_{\tau i} \} \hspace{4pt} \cup \hspace{4pt} \ldots \nonumber\\
\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots
\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots
\ldots \lvert \nonumber\\
\end{eqnarray}

Since the sets on the right side of \eqref{dois-13} with
cardinality equal to $i$ are the elements of $P_i$, that is,

\begin{equation}\label{dois-14}
P_{i} = \{q_{\alpha 1}, q_{\alpha 2}, \ldots , q_{\alpha i} \}
\hspace{4pt} \cup \hspace{4pt} \{q_{\beta 1}, q_{\beta 2}, \ldots
, q_{\beta i} \} \hspace{4pt} \cup \hspace{4pt} \{q_{\tau 1},
q_{\tau 2}, \ldots , q_{\tau i} \} \hspace{4pt} \cup \ldots
\end{equation}

we conclude that

\begin{equation}\label{dois-15}
\lvert Q \lvert \leq \lvert P_{1} \lvert \hspace{4pt} +
\hspace{4pt} \lvert P_{2} \lvert \hspace{4pt} + \hspace{4pt}
\lvert P_{3} \lvert \hspace{4pt} + \hspace{4pt} \ldots
\hspace{4pt} + \lvert P_{i} \lvert \hspace{4pt} + \ldots
\end{equation}

\hspace{22pt} The set of the nodes of $B$ has cardinality less
than or equal to $\lvert N \lvert$ (Theorem 2.1). Since the number
of nodes in the level $i$ of the tree $B_i$ is equal to $\lvert
P_{i} \lvert$, we have

\begin{equation}\label{dois-16}
\lvert N \lvert \ge \lvert P_{1} \lvert \hspace{4pt} +
\hspace{4pt} \lvert P_{2} \lvert \hspace{4pt} + \hspace{4pt}
\lvert P_{3} \lvert \hspace{4pt} + \hspace{4pt} \ldots
\hspace{4pt} + \lvert P_{i} \lvert \hspace{4pt} + \ldots
\end{equation}

Therefore

\begin{equation}\label{dois-17}
\lvert Q \lvert \leq \lvert N \lvert
\end{equation} \

Considering the bijection $f:F \rightarrow Q$ and $\lvert F \lvert
\geq \lvert N \lvert$, we conclude

\begin{equation}\label{dois-18}
\lvert F \lvert = \lvert N \lvert
\end{equation}


\begin{theorem} \label{teorema-3}
The cardinal number of the set of infinite paths of the binary
tree B is equal to $ \lvert N \lvert $.
\end{theorem}
\textit{Proof:}

\hspace{22pt}Be the infinite path of $Q$

\begin{equation}\label{dois-19}
q_{\tau 1}, q_{\tau 2}, \ldots , q_{\tau i}, \ldots
\end{equation}

where $q_{\tau i} = 1$ for all $i \in N$, projected in the grid as
shows the Figure 3. In the coordinates of the grid, the path
\eqref{dois-19} corresponds to the pairs

\begin{equation}\label{dois-20}
(0, 0), (0, 1), (0, 3), (0, 7), (0, 15), \ldots , (0, 2^{k}-1),
\ldots
\end{equation}

Let us notice that in (0, 0) all infinite paths begin. By the pair
(0, 1) a part of all infinite paths passes, by (0, 3) a part of
the infinite paths that passed by (0, 1) passes, by (0, 7) a part
of the infinite paths that passed by (0, 3) passes and so forth.
When an entire path is accomplished, the path exists. In the
examined case, \eqref{dois-19} is the accomplished path.

\hspace{22pt}Each pair $(0, 2^{k}-1)$ of the sequence
\eqref{dois-20} belongs to the set of pairs given by
\eqref{dois-1}, this is,

\begin{equation}\label{dois-21}
(0, 2^{k}-1), (1, 2^{k}-2), (2, 2^{k}-3),  \ldots , (2^{k}-3, 2),
(2^{k}-2, 1) \hspace{8pt} \textrm{and} \hspace{8pt} (2^{k}-1, 0)
\end{equation}

Let us denote by $G_k$ the set of pairs given by \eqref{dois-21}
for $k$. For any $k$ the cardinal number of the set of infinite
paths that passes by $(0, 2^{k}-1)$ is equal to the cardinal
number of the set of infinite paths that pass by any other pair of
\eqref{dois-21}. For any $k$ the distance from (0, 0) to $(0,
2^{k}-1)$ in the grid is equal to $\lvert G_k \lvert$. When we
examine the pairs of \eqref{dois-20}, from left to right, and we
accomplished the path \eqref{dois-19}, $G_k$ becomes the set of
the infinite path whose cardinality is equal to the cardinality of
the set of the nodes of the path \eqref{dois-19}.

\hspace{22pt}The cardinality of the set of the infinite paths is
$\lvert F \lvert$ and the cardinality of the set of nodes of any
infinite path is $\lvert N \lvert$. Therefore

\begin{equation}\label{dois-22}
\lvert F \lvert = \lvert N \lvert
\end{equation}

\section{The Cantor's Proof} \label{sec-3}

\subsection{The 1891 proof} \label{sec-3.1}

\begin{quote}

\textit{Theorem.} \hspace{8pt} The set $F$ is not countable.


\textit{Proof.}\hspace{8pt} We assume that the set $F$ is
countable. This means, by the definition of countable sets, that
$F$ is finite or denumerable. Let us notice that $ \lvert F \lvert
$ is not less than $ \lvert N \lvert $. Be

\begin{center}
$F = \{a_0, a_1, a_2, \ldots \}$
\end{center}

where the cardinality of $ \{a_0, a_1, a_2, \ldots \} $ is $
\lvert N \lvert $. We can write their decimal expansions as
follows:

\begin{equation}\label{tres-0}
\begin{tabular}{c c c c c c c}
$a_0$ & = & 0. & $d_{1,1}$ & $d_{1,2}$ & $d_{1,3}$ & \ldots \\
$a_1$ & = & 0. & $d_{2,1}$ & $d_{2,2}$ & $d_{2,3}$ & \ldots \\
$a_2$ & = & 0. & $d_{3,1}$ & $d_{3,2}$ & $d_{3,3}$ & \ldots \\
\ldots & & \ldots
\end{tabular}
\end{equation}

where the $d$'s are binary characters 0 and 1. Now we define the
number

\begin{center}
$x = 0.d_1 d_2 d_3 \ldots $
\end{center}

by selecting $d_1 \ne d_{1,1}$,  $d_2 \ne d_{2,2}$,  $d_3 \ne
d_{3,3}$, \ldots . This gives a number not in the set $ \{a_0,
a_1, a_2, \ldots \} $, but $x \in F$. Therefore, $F$ is not
countable.
\end{quote}

When \eqref{tres-0} is represented as \eqref{tres-1} bellow, the
proof can be called the \textit{written list form} of the Cantor's
argument of 1891 ~\cite{Cantor}.

\subsection{Comments} \label{sec-3.2}

\hspace{22pt}Let us consider the set $N$, the set $F$ of the real
numbers of the interval [0, 1] and the hypothesis

\begin{quote}
$(H_1)$  There is a list, denoted by $L_H$, that contains all
members of $N$ and all members of $F$ and to each member of $F$ of
the list corresponds one-to-one a member of $N$ of the list.
\end{quote}

For Cantor's diagonal method to be applied, it is created a list
$L$

\begin{equation}\label{tres-1}
\begin{tabular}{c c c c c c c}
0 & = & 0. & $d_{1,1}$ & $d_{1,2}$ & $d_{1,3}$ & \ldots \\
1 & = & 0. & $d_{2,1}$ & $d_{2,2}$ & $d_{2,3}$ & \ldots \\
2 & = & 0. & $d_{3,1}$ & $d_{3,2}$ & $d_{3,3}$ & \ldots \\
\ldots & & \ldots
\end{tabular}
\end{equation}

The diagonal method finds a member of $F$ which is not in the list
$L$ and shows that the list cannot contain all members of $F$. The
Cantor's proof supposes that the list $L$ is the list of $(H_1)$
and concludes that the list of $(H_1)$ does not exist.

\hspace{22pt} Theorem ~\ref{teorema-4} below affirms that if the
diagonal method is correct, then any denumerable list $L$ to which
the diagonal method has been applied is incomplete, this is, $L$
does not contain all of real numbers of the interval [0, 1]. The
Theorem ~\ref{teorema-5} affirms that if some complete list exists
and if the diagonal method is correct, then a complete list cannot
be represented in the form \eqref{tres-1}. The Theorem
~\ref{teorema-6} affirms that being $L$ incomplete nothing affirms
or denies that $ \lvert F \lvert = \lvert N \lvert$.

\subsection{Theorems} \label{sec-3.3}

\begin{theorem}\label{teorema-4}
If the diagonal method is correct, then any denumerable list L is
incomplete.
\end{theorem}
\textit{Proof:}

\hspace{22pt}Suppose that the diagonal method is correct. Let us
consider the list $L$ in which it is presumed that the real
numbers of the interval [0, 1] are put into one-to-one
correspondence with the even natural numbers (0 included). We use
the binary system for the real numbers. The list $L$, where any
$d_{i,j}$ is either 0 or 1, is the following:

\begin{equation}\label{tres-2}
\begin{tabular}{c c c c c c c c}
0 & = & 0. & $d_{1,1}$ & $d_{1,2}$ & $d_{1,3}$ & $d_{1,4}$ & \ldots \\
2 & = & 0. & $d_{2,1}$ & $d_{2,2}$ & $d_{2,3}$ & $d_{2,4}$ & \ldots \\
4 & = & 0. & $d_{3,1}$ & $d_{3,2}$ & $d_{3,3}$ & $d_{3,4}$ & \ldots \\
6 & = & 0. & $d_{4,1}$ & $d_{4,2}$ & $d_{4,3}$ & $d_{4,4}$ & \ldots \\
\ldots & & \ldots
\end{tabular}
\end{equation}

Be the real number $0.d_{1,1} d_{2,2} d_{3,3} d_{4,4} \ldots $ .
Let us consider the number $0.k_1 k_2 k_3 k_4 \ldots $ , where
$k_i \ne d_{i,i}$ for every $i \in N$. The number $0.k_1 k_2 k_3
k_4 \ldots $ belongs to the interval [0, 1] and it is not in
\eqref{tres-2}. Therefore, the hypothesis that the list contains
all of the real numbers of the interval [0, 1] is false. Let us
include $0.k_1 k_2 k_3 k_4 \ldots $ in the list:

\begin{equation}\label{tres-3}
\begin{tabular}{c c c c c c c c}
0 & = & 0. & $d_{1,1}$ & $d_{1,2}$ & $d_{1,3}$ & $d_{1,4}$ & \ldots \\
1 & = & 0. & $k_1$ & $k_2$ & $k_3$ & $k_4$ & \ldots \\
2 & = & 0. & $d_{2,1}$ & $d_{2,2}$ & $d_{2,3}$ & $d_{2,4}$ & \ldots \\
4 & = & 0. & $d_{3,1}$ & $d_{3,2}$ & $d_{3,3}$ & $d_{3,4}$ & \ldots \\
6 & = & 0. & $d_{4,1}$ & $d_{4,2}$ & $d_{4,3}$ & $d_{4,4}$ & \ldots \\
\ldots & & \ldots
\end{tabular}
\end{equation}

\hspace{22pt}Starting from of the hypothesis that the set of all
real numbers of the interval [0, 1] is in the list \eqref{tres-2}
and it has cardinality $ \lvert N \lvert $, we showed that there
are real numbers of the interval [0, 1] not included in the set of
real numbers of \eqref{tres-2}.

\hspace{22pt}We can conclude that if the diagonal method is
correct, then any denumerable list $L$ is incomplete.

\begin{theorem} \label{teorema-5}
Be the list \eqref{tres-4} such that (i) the column on the left
contains the elements of N put from top to bottom in increasing
order and (ii) the column on the right contains real numbers of
the interval [0, 1], where $d_{i,j} \in \{0, 1\}$. The sets of the
numbers of the column on the right are equivalent to N.

\begin{equation}\label{tres-4}
\begin{tabular}{c c c c c c c c}
0 & = & 0. & $d_{0,1}$ & $d_{0,2}$ & $d_{0,3}$ & $d_{0,4}$ & \ldots \\
1 & = & 0. & $d_{1,1}$ & $d_{1,2}$ & $d_{1,3}$ & $d_{1,4}$ & \ldots \\
2 & = & 0. & $d_{2,1}$ & $d_{2,2}$ & $d_{2,3}$ & $d_{2,4}$ & \ldots \\
3 & = & 0. & $d_{3,1}$ & $d_{3,2}$ & $d_{3,3}$ & $d_{3,4}$ & \ldots \\
\ldots & & \ldots
\end{tabular}
\end{equation}

If some complete list exists and if the diagonal method is
correct, then a complete list cannot be represented in the form
\eqref{tres-4}.
\end{theorem}
\textit{Proof:}

\hspace{22pt}Let us separate the list \eqref{tres-4} in two lists
so that all of the odd lines belong to one of the lists and the
remaining lines belong to the second list.

\begin{equation}\label{tres-5}
\begin{tabular}{c c c c c c c c}
0 & = & 0. & $d_{0,1}$ & $d_{0,2}$ & $d_{0,3}$ & $d_{0,4}$ & \ldots \\
2 & = & 0. & $d_{2,1}$ & $d_{2,2}$ & $d_{2,3}$ & $d_{2,4}$ & \ldots \\
4 & = & 0. & $d_{4,1}$ & $d_{4,2}$ & $d_{4,3}$ & $d_{4,4}$ & \ldots \\
6 & = & 0. & $d_{6,1}$ & $d_{6,2}$ & $d_{6,3}$ & $d_{6,4}$ & \ldots \\
\ldots & & \ldots
\end{tabular}
\end{equation}
and
\begin{equation}\label{tres-6}
\begin{tabular}{c c c c c c c c}
1 & = & 0. & $d_{1,1}$ & $d_{1,2}$ & $d_{1,3}$ & $d_{1,4}$ & \ldots \\
3 & = & 0. & $d_{3,1}$ & $d_{3,2}$ & $d_{3,3}$ & $d_{3,4}$ & \ldots \\
5 & = & 0. & $d_{5,1}$ & $d_{5,2}$ & $d_{5,3}$ & $d_{5,4}$ & \ldots \\
7 & = & 0. & $d_{7,1}$ & $d_{7,2}$ & $d_{7,3}$ & $d_{7,4}$ & \ldots \\
\ldots & & \ldots
\end{tabular}
\end{equation}

Considering that the sets $\{0, 2, 4, 6, \ldots \}$ and $\{1, 3,
5, 7, \ldots \}$ are equivalent to $N$, we can rewrite the two
lists, respectively, as
\begin{equation}\label{tres-7}
\begin{tabular}{c c c c c c c c}
0 & = & 0. & $d_{0,1}$ & $d_{0,2}$ & $d_{0,3}$ & $d_{0,4}$ & \ldots \\
1 & = & 0. & $d_{2,1}$ & $d_{2,2}$ & $d_{2,3}$ & $d_{2,4}$ & \ldots \\
2 & = & 0. & $d_{4,1}$ & $d_{4,2}$ & $d_{4,3}$ & $d_{4,4}$ & \ldots \\
3 & = & 0. & $d_{6,1}$ & $d_{6,2}$ & $d_{6,3}$ & $d_{6,4}$ & \ldots \\
\ldots & & \ldots
\end{tabular}
\end{equation}
and
\begin{equation}\label{tres-8}
\begin{tabular}{c c c c c c c c}
0 & = & 0. & $d_{1,1}$ & $d_{1,2}$ & $d_{1,3}$ & $d_{1,4}$ & \ldots \\
1 & = & 0. & $d_{3,1}$ & $d_{3,2}$ & $d_{3,3}$ & $d_{3,4}$ & \ldots \\
2 & = & 0. & $d_{5,1}$ & $d_{5,2}$ & $d_{5,3}$ & $d_{5,4}$ & \ldots \\
3 & = & 0. & $d_{7,1}$ & $d_{7,2}$ & $d_{7,3}$ & $d_{7,4}$ & \ldots \\
\ldots & & \ldots
\end{tabular}
\end{equation}

The set of real numbers of \eqref{tres-7} and the set of real
numbers of \eqref{tres-8} are disjoint. Therefore, both
\eqref{tres-7} and \eqref{tres-8} are incomplete, this is, any one
of them does not contain all real numbers of the interval [0, 1].

\hspace{22pt}We can apply the diagonal method to \eqref{tres-7},
for instance, and we obtain a real number that is not in
\eqref{tres-7}. That is consistent because \eqref{tres-7} is
incomplete.

\hspace{22pt}There is uncertainty as to the completeness or not of
\eqref{tres-4}. Considering that the representation \eqref{tres-4}
has the same form of \eqref{tres-7} and \eqref{tres-8}, the
application of the diagonal method to \eqref{tres-4} is to say
that \eqref{tres-4} is incomplete. Concluding, if some complete
list exists and if the diagonal method is correct, then a complete
list cannot be represented in the form \eqref{tres-4}.


\begin{theorem}\label{teorema-6}
Being L incomplete nothing affirms or denies that $ \lvert F
\lvert = \lvert N \lvert $.
\end{theorem}
\textit{Proof:}

\hspace{22pt}Suppose that the list $L$ is the list $L_H$ [Section
~\ref{sec-3.2}]. In this case a proof that $L$ does not contain
all elements of $F$ implies in $(H_1)$ false. This also means that
the one-to-one correspondence between $N$ and $F$ in a list cannot
be done. If $(H_1)$ is false, then $L$ does not exist. However,
$L$ exists. Therefore, it is false the hypothesis that $L$ is the
list $L_H$. Since ``$L$ is not $L_H$" nothing affirms the
existence or not of $L_H$, being $L$ incomplete nothing affirms or
denies that $ \lvert F \lvert = \lvert N \lvert $.

\section{The list $L_H$}
\label{sec-4}

\hspace{22pt} Let $L$ be the list

\begin{equation}\label{quatro-1}
\begin{tabular}{c c c c c c c}
0 & = & 0. & $d_{1,1}$ & $d_{1,2}$ & $d_{1,3}$ & \ldots \\
1 & = & 0. & $d_{2,1}$ & $d_{2,2}$ & $d_{2,3}$ & \ldots \\
2 & = & 0. & $d_{3,1}$ & $d_{3,2}$ & $d_{3,3}$ & \ldots \\
\ldots & & \ldots
\end{tabular}
\end{equation}

where the $d$'s are binary characters 0 and 1. If the diagonal
method is correct, then any denumerable list $L$ is incomplete.
This means - if the argumentation of Section \textbf{~\ref{sec-2}}
is correct - that $L_H$ has not the form \eqref{quatro-1}. We
would have then a very unusual form of list.

\hspace{22pt} Let us now consider the Figure 5. The horizontal
sequence of finite natural numbers, presented in increasing order
from left to right, contains all numbers of $N$; to each natural
number of the sequence corresponds a column. The vertical sequence
of numbers, presented in increasing order from top to bottom,
contains all numbers of $N$; to each natural number of the
sequence corresponds a row. Each element of the matrix formed by
the rows and columns is filled with 0 or 1 in accordance with the
procedure: The first row from left to right is filled from top to
bottom with a succession of the pattern 01. The second row is
filled with a succession of the pattern 0011. The $n$-th row from
left to right is filled from top to bottom with a succession of
the pattern constituted of $2^{n-1}$ 0's followed by $2^{n-1}$
1's.

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\end{picture}
\end{center}
\begin{center}
\textrm{Figure 5}
\end{center}

\hspace{22pt} Let us notice that (i) the rows of the sub-matrix
formed by the $2^{i}$ first rows and $i$ first columns are the
elements of $P_i$ and (ii) all paths belonging to $P_i$ are put
into one-to-one correspondence with a number of the horizontal
sequence of natural numbers. This means that the matrix contains
the tree $B$ and, consequently, the set $Q$.

\hspace{22pt} Let us consider that the matrix of Figure 5 contains
all elements of $Q$. Let us apply the diagonal method shown in the
Figure 6. In the figure the numbers of the vertical sequence of
natural numbers follow the procedure: the number correspondent to
$i$-th row is $n-1$, where $n$ is the number of elements of the
matrix on the thick line from its beginning up to the first
element of the $i$-th row. Let us notice that $\lvert Q \lvert$ is
less or equal to the number of elements of the matrix on the thick
line. Since the set \{0, 2, 3, 9, 10, 20, 21, \ldots \} has
cardinality $\lvert N \lvert$, we conclude that the list $L_H$ can
be represented in the form \eqref{quatro-1}.

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\end{picture}
\end{center}
\begin{center}
\textrm{Figure 6}
\end{center}

\begin{thebibliography}{99}
  \addcontentsline{toc}{chapter}{Bibliography}
\bibitem{Cantor}  G. Cantor, \textit{\"Uber eine elementare Frage der
Mannigfaltigkeitslehre}, Jahresbericht der
    Deutschen Mathematiker-Vereinigung, 1 (1891), 75-78.
\end{thebibliography}


\end{document}

math0108119.extracted_bib
\begin{thebibliography}{99}
  \addcontentsline{toc}{chapter}{Bibliography}
\bibitem{Cantor}  G. Cantor, \textit{\"Uber eine elementare Frage der
Mannigfaltigkeitslehre}, Jahresbericht der
    Deutschen Mathematiker-Vereinigung, 1 (1891), 75-78.
\end{thebibliography}