Claim: Cantor's Proof on infinite arithmetic is wrong!
G. Chen
-------------------------------------------------------------------
Following is an article by my friend who is, self-sponsored, doing
research in mathematics in China. Some people say he is a crazy guy with
such a crazy idea(in his paper) that he claims that the cantor's
infinite arithmetic is a "dead-body" in mathematics, and even more,
the basis for mathematics should be changed somehow...
The whole article is in LaTeX format, please read it and
send your comments and suggestions to
thanks in advance...
--Guoying Chen
%=================================cut it here =============================
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\title{On Infinite-Arithmetic}
\author{ Han-qing Liu
\thanks{
The original manuscript was written in Chinese, this English
version is translated by Guoying Chen, a Ph.D student of
Computer Science Department
at Courant Institute, New York University.
Readers can contact the author with the address:
Han-qing Liu, c/o Guoying Chen,
Courant Inst, New York University, New York, NY 10012.
Email: che...@cs.nyu.edu }}
\date{ March 7, 1989 \\
(Translated from Chinese, September 10,1989) \\
(Revised November 8, 1989) \\}
\newpage
\maketitle
\begin{document}
\begin{titlepage}
\begin{abstract}
With proof by contradiction,
Cantor showed that there exist distinguished
cardinalities for infinite sets.
In this paper, we first analyze the proof, and
point out the incorrectness of the proof,
then as a result of its
inference, we construct a paradox, which is not only
very hard to interpret in the point of
view of formalism but also of a new paradox in set theory of
axioms(ZFC system).
Thus, we claim that Cantor's infinite-arithmetic
is a dead-body in mathematics.
As a matter of fact, as for intuitionist,
the criticism on classic set theory, philosophically or
logically, has been developed as an excuse
for supporting Cantor's theory.
No way of destruction could be better than find
out the implicit(hidden) false from the objects of system
and from the procedure
of logical reasoning when following the theory proper.
With this idea in mind, we find out that it is incorrect
to use the method of proof by
contradiction to show that there are distinguished
cardinalities in infinites sets. We conclude that cardinalities
of any two infinite sets
are incomparable.
Thus, we can completely throw out the classic set theory.
Consequently,
the continuum hypothesis becomes false and the
selection axiom has to be
discarded. More seriously,
since the G\"{o}del's
incomplete theorem was proved on the basis of the existence
of distinguished cardinalities, as a result, it is
a logical mistake.
\end{abstract}
\end {titlepage}
\pagenumbering{arabic}
\setcounter{page}{1}
\section{Introduction}
There has been continuous
arguments about the concept of
``infinite'' throughout the history.
Since the idea of ``real infinite'' was introduced by
Cantor when he
investigated
on infinite series, it has been accepted by most mathematicians
(rational or perhaps irrational), even though it was ``attacked'' by
Kroureker and is still refused by intuitionist. Cantor's idea
has been developed into one of
mathematical concepts, turning the infinite set
theory into the foundation of mathematics and finding its applications in
every branches of applied mathematics.
One of the key part of Cantor's idea was
{\em one-one correspondence}.
As a result,
Cantor introduced(under erroneous logic) the concept of
the cardinality and the order
of infinite set and then constructed the infinite-arithmetic theory.
Furthermore,
he proposed the famous ``continuum hypothesis''
(the first of 23 Hilbert problems).
Later, G\"{o}del and Cohen showed that the continuum
hypothesis
plays the same role in mathematics as the 5th axiom hypothesis does in
Geometry. In other words,
we can build any mathematical system and in which
the continuum hypothesis can be negated, specifically, the continuum hypothesis
cannot be decidable in ZF Axioms System. It is really very odd
that a proposition from one theory cannot even be
decidable in the theory itself but can be higher than the theory.
In fact, as we can
see later, the continuum hypothesis doesn't exist at all.
Because of the discovery of fully-logical paradox(Russell paradox) in
primitive infinite set theory, the ZFC axiom system emerged to
eliminate the paradox and provide the set theory for all classic analysis.
After that, there
are several different approaches of mathematical philosophy coming out.
The most radical one is intuitionism. It is entirely different from the classic
mathematics. In the view of intuitionism, many mathematical
results are invalid.
As for the infinite, intuitionists
insist on Aristotle's potential infinite and they disagree
to think of an
infinite as a closed entity
(the Cantor's real infinite that most mathematicians
believe in). They explain that's why it is more
difficult than the Russell's paradox in mathematics. For instance,
Poincare once predicted
that people in future would be free from the real infinite
(as the result from our research, it's very true);
Brouwer was against using logic, especially ``middle exclusion'',
in infinite set without limitation,
As a result of their philosophy, the study of
Cantor's infinite set theory proper would be negated, since the only
way in constructing
Cantor's infinite-arithmetic theory is by ``middle exclusion''.
However, the idea of intuitionists didn't get the right point to refute
it. In this
paper, we admit that ``middle exclusion'' can be used in infinite set.
However, after
extensive analysis of operations of ``middle exclusion'' in infinite and
finite set,
We found that there are fundamental errors in Cantor's proof, especially the
incorrectness of showing that there are distinguished
cardinalities in infinite sets with {\em proof by contradiction}.
That is to say, applying contradiction cannot make it sure that different
cardinalities exists. Consequently,
its logical application, the infinite set theory is not correct.
It is surely believed that the consequence of the above is of mass
misconception
of mathematicians, when applying proof by
contradiction in the research of infinite
set theory. Most importantly, the G\"{o}del's
proof of incomplete theory, being based
on the existence of infinite-set with different cardinalities,
cannot hold either. Put in another way,
for any formal system, given any one of it's statements, we
cannot decide whether this statement is true or not, nor decide whether it
can be proved or not. As a matter of fact, whether
or not a system is of contradictory property is the research scope of
mathematical philosophy.
In addition, selection theory would be mathematically negated
since it cannot be decidable which is larger between
two different cardinalities of infinite-set. These results are very surprising,
but they are not so scaring. I'll explain why it is in subsequent
sections.
\section{One-One Correspondence}
As stated above, one would ask: is the ``middle exclusion principle''
applicable to the problem of cardinality comparison in infinite set?
Given a finite-set $A$:
\{1,2,3 ... ,n\}. $A$ constitutes an one-one relationship
with itself, while there can be
any correspondence between elements in $A$. In
other words, the correspondence of any elements in two sets($A$ and
itself) constitutes an one-one correspondence.
More importantly, for any two finite-sets,
the correspondence between any elements in both sets either
may constitute an one-one correspondence or may not. Moreover,
whether the
correspondence between any elements of them constitutes
an one-one correspondence
is independent of correspondence form of selected elements, i.e., the one-one
correspondence is unique for {\em finite} sets.
Hence, the conventional logic abstracted from
finite-set makes it fully necessary and efficient to be applied to finite
set and cannot produce any ambiguous results. Applying proof by contradiction
method, we can obtain
{\em one and only one}
negation because of the uniqueness
of one-one correspondence in finite-set.
Given an infinite set $A$:
\{1,2,...,n\},
then $A$ would have an one-one correspondence with itself.
The form can be
\begin{center}
\( 1 \longleftrightarrow 1\) \\
\( 2 \longleftrightarrow 2 \) \\
\( 3 \longleftrightarrow 3 \) \\
. \\
. \\
. \\
\( n \longleftrightarrow n \) \\
. \\
. \\
.
\end{center}
or any other forms such that the one-one correspondence is fixed but their
positions can be put in any places.
But $A$ and its subset can also constitute an one-one
correspondence relationship
(it is, in fact, the definition given by Cantor),
let $B$ be a subset of $A$, and $B$ = \{1,2,...n\},
then we have
\begin{center}
\( 1 \longleftrightarrow 2\) \\
\( 2 \longleftrightarrow 3 \) \\
\( 3 \longleftrightarrow 4 \) \\
. \\
. \\
. \\
\( n \longleftrightarrow n+1 \) \\
. \\
. \\
.
\end{center}
That is to say, for an infinite set $A$, in some cases, it might be one-one
correspondence with itself;
but in the other cases, it might not. It only corresponds to its subset
(For correspondence relationship, $A$ is not of identity).
This is the fundamental difference in the correspondence form
between an infinite set and a finite set. That is,
the finite set is {\em independent}
of correspondence, but the infinite set
{\em depends} on it.
The answer to the question -- whether there is a correspondence -- is that
the correspondence is unique for finite set, but not always unique for
infinite set.
Therefore, when proving by contradiction ,
we must take it into consideration.
Take the natural set $A$:
\{1,2, ... ,n\}
for example again. The instance
that it can be one-one corresponded to itself is
\begin{center}
\( 1 \longleftrightarrow 1\) \\
\( 2 \longleftrightarrow 2 \) \\
\( 3 \longleftrightarrow 3 \) \\
. \\
. \\
. \\
\( n \longleftrightarrow n \) \\
. \\
. \\
.
\end{center}
the instance that cannot be one-one corresponded to itself can be
\begin{center}
\( 1 \longleftrightarrow \space \) \\
\( 2 \longleftrightarrow 1 \) \\
\( 3 \longleftrightarrow 2 \) \\
. \\
. \\
. \\
\( n+1 \longleftrightarrow n \) \\
. \\
. \\
.
\end{center}
Hence, set $A$ can be one-one corresponded with itself($ A \sim A $)
unless in which each
element corresponds to each, and one cannot be one-one correspondent with
itself( $A \not\sim A$) unless in which each element doesnot
corresponds to each.
\section{Analysis of Cantor's Proof}
Let's analyze the Cantor's proof. It shows that
the cardinality of a real
number set $B$: (0,1) is greater than that of the natural number set $A$.
Proof by contradiction is used(the famous diagonal method).
The proof can be briefly sketched as follows:
Assume that there is an one-one correspondence between
the natural number set $A$ and the real number set $B$. Then
we can write
each real number as an infinitesimal number, and each real number
can correspond to a natural number n:
\begin{center}
Table 1
\\
1 $\longleftrightarrow 0.a_{1,1}a_{1,2}a_{1,3} \ldots$ \\
2 $\longleftrightarrow 0.a_{2,1}a_{2,2}a_{2,3} \ldots$ \\
$\ldots$ $\ldots$ \\
$\ldots$ $\ldots$ \\
$\ldots$ $\ldots$ \\
n $\longleftrightarrow 0.a_{n,1}a_{n,2}a_{n,3} \ldots$
\end{center}
Define another real number $b = 0.b_{1}b_{2}b_{3} \ldots b_{n} \ldots$ ,
where if $a_{r,r}$ = 1,then $b_{r}$ = 9,
otherwise $b_{r}$ = 1. Then b is different from
every real number listed in Table 1, and it contradicts the assumption
that Table 1 contains all real number in (0,1), therefore,
the cardinality of the real number set(0,1) is
greater than that of the natural
number set. $\Box$
{\tt We claim that the proof above cannot hold}.
We first
study with the assumption that two different infinite sets are
one-one corresponded, and see what inference we can obtain,
then we analyse, in detail, the above proof procedure of
$\| B \| > \| A \|$
(where $\| A \|, \| B \|$ are the notations for cardinality of $A$ and
$B$ respectively).
Suppose $A$ and $B$ are different infinite sets and $A \sim B$.
That is to say,
\begin{enumerate}
\item we assumed that there is a set
(let us call it $B'$) of some elements in $B$ such
that each element in $A$ can be designated to be ono-one
correspondent with each element in $B'$, while this $B'$ contains
all elements in $B$;
\item Meanwhile, we also assumed
that each element in $A$ can be designated to be ono-one
correspondent with each element in $B''$, while this $B''$
does not contains
all elements in $B$, i.e., it contains only a subset of $B$
(call it $B_{sub}$).
\end{enumerate}
These two points are direct corollaries from the assumption
of $A \sim B$.
Therefore, we can say now that there exist a set of some elements in
$B$ such that each element in $A$ can be designated to be one-one
correspondent with each element in that set. Consider the
case that the set
is $B'$, which
contains all elements of $B$(let it be $\{ a, b,\cdots\}$).
From this assumption, if we get a contradiction with
the method of proof by contradiction,
then we could obtain two negations. One is the negation of the
assumption of $A \sim B$, the other is
the negation of the assumption of $B' = \{ a, b,\cdots\}$ containing
all $B$'s elements. Since now $\{a, b,\cdots\}$ are not all elements
of $B$, they must be all elements of a subset of $B$(i.e., $B_{sub}$).
Because of the assumption of $A \sim B$, all elements in $B_{sub}$
will be able to constitute an one-one correspondence with all
elements in $A$. Thus it doesn't negate the assumption of $A \sim B$.
Hence, there are two cases, the first is to negate $A \sim B$;
the second is not to negate $A \sim B$ but only negate the assumption
that $\{a, b,\cdots\}$ contains all elements in $B$. Therefore,
it cannot be decidable whether or not we can negate the
assumption of $A \sim B$ with the method of proof by
contradiction.
On the other hand, suppose $A \sim B$, we can obtain the inferences
as follows:
all elements in $A$ is one-one correspondent with elements in $B$,
while it is impossible to decide whether these elements in $B$
corresponded with $A$ are
all elements in $B$ or a subset of $B$'s elements unless
it is given that these elements are all elements of $B$ or
a subset of $B$, or unless
a set of uniform rules is given to decide it
(but, unfortunately, such rules do not exist).
Therefore, in order for $A \sim B$ to have a definite inference,
we have
\begin{enumerate}
\item
$\{a, b,\cdots\}$
must be all elements of $B$. However, if we
can derive a contradiction with proof by contradiction, we can
only negate that $\{a, b,\cdots\}$ are all given numbers.
\item
Assume $\{a, b,\cdots\}$ are all numbers of $B$, if we derive a
contradiction, then it would negate the assumption that
$\{a, b,\cdots\}$ are all numbers in $B$, i.e., $\{a, b,\cdots\}$
is a subset of $B$.
\end{enumerate}
Now, let's look at the Cantor's proof on $\| B \| > \| A \|$.
Suppose $A \sim B$. Since that each element in $B$ is represented
by an infinitesimal number, then for every infinitesimal
number there must be a designated natural number
$n$ that corresponds to it so that the assumption can be satisfied.
Consider the finite set. From the assumption, for any number
in $B$ will be one-one correspondent with all numbers in $A$, while
these numbers in $B$ are all its numbers. However, we cannot
say here, `` $A \sim B$ simply because
any number in $B$ is one-one correspondent to
that of $A$''. The reason is that for any number, it could always
be a subset(or in a subset), while it holds that a subset is
one-one correspondent with $A$, which does not negate the assumption.
Moreover, for any number in $B$, we can designate a natural number
$n$, but we cannot determine
whether these numbers are all elements in $B$
or all elments in $B_{sub}$.
Then, for the assumption $A \sim B$, one would
ask: what are those numbers in $B$ such that
we can obtain all number in $B$ after designating a natural number
$n$? In Table 1, Cantor didn't imply or provide what those numbers in
$B$ are in his proof by contradiction.
So, in Table 1, it is uncertain whether or
not the numbers right hand side
are $B$'s every infinitesimal numbers.
Thus, the contradiction between Table 1's containing all real numbers
in (0,1) and finding a new number $b$ implies two negations:
1). negation of the assumption of $A \sim B$.
2). negation of the assumption that Table 1 contains all real
numbers in (0,1), put in another way,
the numbers right hand side are in a subset
of $B$. While a subset of $B$ can be one-one correspondent to
$A$, it is therefore not negating the assumption of $A \sim B$.
We now can see it clearly that the conclusion derived from proof by
contradiction is that, on one hand, it can negate the assumption
of $A \sim B$, on the other hand, it cannot. Thus we cannot
conclude that $\| B \| > \| A \|$, i.e., the cardinalities of
$A$ and $B$ is incomparable. The assumption of $A \sim B$ is just
an imaginary abstraction.
In order for the inference result in Table 1 to be uniquely definite,
we require predefining a set of uniform rules such that the numbers
that are one-one correspondent to every numbers in A are none other
than all numbers. However, we will see that such a set of rule doesn't
exist at all.
In addition, the numbers given in Table 1
only satisfy the sufficiency of numerability.
In order for the result to be unique,
the important is that, for real number set $B$,
we must have already
constructed every real number so that we can
inference from the assumption of $A \sim B$.
We can also think in this way,
since the real numbers in Table 1 are arbitrary
(but undecidable whether they are all numbers in $B$ or
all numbers in a subset of $B$),
the defined real number $b$ therefore
just makes it sure that the real
numbers in Table 1 are of a subset. As such, $b$
can be added into Table 1, satisfying
the efficiency of enumerability. Therefore,
we have following correspondences:
\begin{center}
Table 2
\\
1 \( \longleftrightarrow 0.b_{1}b_{2}b_{3} \ldots b_{n} \ldots \) \\
2 \( \longleftrightarrow 0.a_{1,1}a_{1,2} \ldots .. \ldots \) \\
3 \( \longleftrightarrow 0.a_{2,1}b_{2,2} \ldots .. \ldots \) \\
\( \ldots \ldots \ldots \) \\
\( \ldots \ldots \ldots \) \\
\( \ldots \ldots \ldots \) \\
n \( \longleftrightarrow 0.a_{n-1,1}a_{n-1,2} \ldots \)
\end{center}
In the same way as we did before,
we can define a new real number $b'$
which is not in Table 2,
we can also add it into Table 2 and construct a new table -- Table 3:
\begin{center}
Table 3
\\
1 \( \longleftrightarrow 0.b'_{1}b'_{2}b'_{3} \ldots \) \\
2 \( \longleftrightarrow 0.b_{1}b_{2}b_{3} \ldots \) \\
3 \( \longleftrightarrow 0.a_{1,1}a_{1,2} \ldots \) \\
4 \( \longleftrightarrow 0.a_{2,1}a_{2,2} \ldots \) \\
\( \cdots \cdots \cdots \)
\end{center}
$b'$ also satisfies the sufficiency of enumerability. Continue this
way repeatly,
as long as we can add $b'$ into a new table,
the new number can always be found.
In order for the assumption of $A \sim B$ to be able to
inference uniquely, we may make an assumption of $B$'s
configuration. For example, we may assume that \\
\begin{center}
$0.a_{1,1}a_{1,2}
\cdots a_{1,n} \cdots, 0.a_{2,1}a_{2,2} \cdots a_{2,n} \cdots, \cdots \cdots
0.a_{n,1}a_{n,2} \cdots a_{n,n}, \cdots \cdots$ \\
\end{center}
are $B$'s all numbers,
$B$ can constitute an one-one correspondence with $A$. But
we can find a new number $b$, which contradicts the assumption
that \\
\begin{center}
$0.a_{1,1}a_{1,2}
\cdots a_{1,n} \cdots, 0.a_{2,1}a_{2,2} \cdots a_{2,n} \cdots, \cdots \cdots
0.a_{n,1}a_{n,2} \cdots a_{n,n} \cdots \cdots \cdots $ \\
\end{center}
are $B$'s all numbers.
Hence, we cannot negate the assumption of $A \sim B$.
Therefore, after we assume that two infinite sets are of one-one
correspondence,{\em with proof by contradiction},
we cannot derive the unique conclusion to negate
the assumption, because we don't have the required configuration
(when we need the configuration, we have to have construct it).
Thus, it is similarly not correct that, {\em with proof by contradiction},
we could conclude that an infinite set is {\em not}
one-one correspondent with the infinite constructed from all
its subsets.
Furthermore,
we cannot determine if the cardinality of real numbers is greater than that of
natural numbers.
Let's briefly summarize the above our ideas.
For any two infinite sets $A$ and $B$, assume that $A \sim B$,
we can only have following inferences: all elements in $A$ is one-one
correspondent with some elements in $B$(we call a set of these
elements $B'$); We cannot tell whether $B'$
is $B$ or a subset of $B(B_{sub})$ unless we are able to
construct all $B$'s elements or we can provide a method to
determine whether all $A$'s elements are one-one correspondent
to $B$'s elements and the set of these elements is exactly set $B$.
But as we know, it is impossible to achieve any one of them.
In addition, if we apply the Cantor's inference method, we
may assume that $A \not\sim B$ and
$\| B \| > \| A \|$, then
we can derive that $A$ only corresponds with the subset $B_{sub}$,
and at least there is
one element $\theta \in B$ while $A$ cannot correspond
to $B_{sub} \bigcup \theta$.
But, by the definition of one-one correspondence,
$A \sim B_{sub} \bigcup \theta$ surely holds.
A Contradiction! Therefore it is also mistaken that $A \sim B$.
\section{Paradox}
As we see above, it verifies again that
any rule must be an abstract repeatly
correct for concrete objects and can be applied to bijective
mapping objects.
Beyond it,
the rules have to be analyzed case by case.
Moreover, any rule must
observe strict logic inferences and the most primitive facts.
For one-one correspondence among infinite sets, indirect proof
can only show that there exit some rules but cannot make definite
rules. Meanwhile, rules themselves are reflected in one-one
correspondence relationship.
In the view
of formalism, $n \leftrightarrow 0.\eta_{n1}\eta_{n2}\eta_{n3}\ldots$ is
predeterminedly understood as every real number in $B$ corresponds to every
natural number
in $A$ and they are completely correspondent. No doubt, it's incorrect
because it does not have
the sufficient rights for rules.
Assume $A \sim B$, same as before we have done,
we can only make
following inferences:
all elements in $A$ is one-one
correspondent with some elements in $B$(we call a set of these
elements $B'$); We cannot tell whether $B'$
is $B$ or a subset of $B(B_{sub})$ unless
$B$ has already been constructed or $B'$ is known as
a subset of $B$.
Therefore, as a result of inference,
there exists only
the assumption, but no predetermination.
Moreover,
if it were predetermined, we could have the
assumption $A \sim B$,
then the set $B'$
is surely $B$. This clearly contradicts the
Cantor's view of real infinite,
since for $A$, we have $A \sim A$, we also
have $A \not\sim A$.
In addition, if
$n \leftrightarrow 0.\eta_{n1}\eta_{n2}\eta_{n3}\ldots \eta_{nn}
\cdots$ is
predetermined as $B$'s all elements, then
we can find a new $\eta'$ which is not in them.
But $\eta \in B$, therefore, all $B$'s elements
must include $\eta$ so that
$0.\eta_{n1}\eta_{n2}\eta_{n3}\ldots \eta_{nn} \cdots$
{\em cannot} be predetermined
to be thought as
all $B$'s elements.
In fact, suppose $A \sim B$, the correct inference should
make another assumption that
$n \leftrightarrow 0.\eta_{n1}\eta_{n2}\eta_{n3}\ldots \eta_{nn}$
are $A$'s every number which one-one corresponds to $B$'s every
number.
The ignorance to make this assumption is the reason why
the result $\| B \| > \| A \|$ is incorrectly derived
with the proof by contradiction.
Next, let's construct a proposition which
is undecidable in ZF systems.
By Cantor's method, given an infinite set $A$,
we have $A \sim A$,
which one belongs to itself.
We also have $A \not\sim A$, which one doesn't
belong to itself.
This provides us the background to construct following
propositions.
\newtheorem{prop}{Proposition}
\begin{prop}
we can have a relation which is composed of all those relations,
each of which
can not only belong to itself(unless it already belonged to itself)
but also not belong to itself(unless it already didn't belong to
itself).
\end{prop}
We may ask: does this relation(the new constructed relation)
belong to itself?. Suppose it does, then it may not belong to itself(unless
it belong to itself), and may belong to itself(unless it doesn't belong
to itself); vice versa if we suppose it doesn't.
This must have close relationship
with one-one correspondence of infinite set
unless it already could belong
to itself and not belong to itself(otherwise there is no inference
available). In this
case it will solve whether-or-not-negation-itself by itself.
If we remove
\it
unless it already belonged to itself
\rm
and
\it
unless it already didn't belong to itself
\rm
from Proposition 1, we would have
a new ``Paradox'' in the real ZF Axiom Systems.
Here, by ``Paradox'', we don't regard it as
its ordinary meanings.
\begin{prop}
If we have a relation $R$ which is composed of all those relations, each
of which can not only
belong to itself but also not belong to itself,
then there always exists an object such that
it is not in $R$.
\end{prop}
If this relation belongs to itself, then, from the proposition 2,
it may belong to itself and may not.
This proposition doesn't go beyond the ZF Axiom Systems, but
it cannot be decidable. It embodies 2-value logic. The reason
why it happens
lies in the fact that the ``unclear'' is thought
to be ``clear''.
It seems that this proposition
cannot be stated clearly in logic
with axioms. This proposition surely has close relationship
with the one-one correspondence of infinite sets. In fact, the
proposition is derived from it.
For the proposition, unless it is positive about itself or
not positive about itself, otherwise,
it is even much harder than the Russell's paradox.
The reason is that it is undecidable.
\section{Summary}
The ``middle exclusion'' failed in problem of cardinality comparison
among infinite sets. For any two infinite sets, with
{\em proof by contradiction} we cannot tell
whose cardinality is larger than the other's.
As the result, all mathematical branches associated with infinite sets
should be examined carefully again. Those relied on the result of
infinite-arithmetic and selection axiom must be removed. Thus, there remain
two questions: one is on what basis the mathematical should be founded,
the other is what the infinite is about and how we should understand it.
As author's belief, it can be
predicted that these two questions will stimulate to produce
many new branches of mathematics and a large amount of new mathematics
knowledge. For the infinite, we must
face this situation, i.e, we can regard
infinite as a closed ``intelligent''
entity, then we rebuild a non-Cantor infinite
set theory, but we should make sure
it is definitely correct in the procedure of logic
inference. It could be a
possible way. However, it seems that there is no
hope applying classical logic, we can only return to ``potential
infinite'' for help, the mathematics built on the basis of it ( as
intuitionists did) will cut off many fruitful results. Our belief(predefined)
is that all infinites are equally ``large'', i.e, all of them are one-one
correspondent to the natural number set . With this in mind, all of
them can be reduced
to natural number set, thus mathematical induction will become
a general rule and
there is a terminal point in the sequence of natural numbers. With this
understanding of infinite, it is good for classical analysis and,
as we can see,
what the infinite is about
is reflected in the finite set. Also, the difference
between finite and infinite is just that they are
of same form. Hence, for infinite,
it will be understandable both for our rationalists and intuitionists.
Mathematics results from rational work and repeat of
continuous experiences.
Originated from extensive experiences of human being
(including long-historically accumulated rationalism), mathematics must
return back to the human being experiences. Perhaps, it is only the most
extensive source and basis of mathematics. Man-made basis could be
with idealist meanings and could limit our creativity and practicability.
Mathematics is neither from intuition, nor from formal symbols; It will always
opens its door to experiences and enrich itself from experience. There is
no doubt that it is rooted profoundly in experience. $\Box$
\end{document}
% =================end of article ==============================
Gerald Edgar
>In article <1287...@acf5.NYU.EDU> gxc...@acf5.NYU.EDU (G. Chen) writes:
>>Following is an article by my friend who is, self-sponsored, doing
>
>I haven't seen any replies to this. Doubtless this means everyone has
>accepted his argument as true, obvious, and trivial? :-)
No, we thought it was kinder just to write directly to the poster
to communicate the problems in the document. It does not refute
Cantor's set theory.
--
Gerald A. Edgar
Department of Mathematics Bitnet: EDGAR@OHSTPY
The Ohio State University Internet: ed...@mps.ohio-state.edu
Columbus, OH 43210 ...!{att,pyramid}!osu-cis!shape.mps.ohio-state.edu!edgar
Doug Merritt
I haven't seen any replies to this. Doubtless this means everyone has
accepted his argument as true, obvious, and trivial? :-)
Doug
--
Doug Merritt {pyramid,apple}!xdos!doug
Member, Crusaders for a Better Tomorrow Professional Wildeyed Visionary