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A STUDY OF THE DIAGONALIZATION TECHNIQUE
AND INFINITELY MORPHING OBJECTS
Part I by
Webster Kehr
P.O. Box 7452
Overland Park, KS 66207
INTERNET: webster@tyrell.net
Part II by
Rusty Johnson
Mystech Associates
Falls Church, Virginia
INTERNET: rusty@lyra.mystech.com
Version: 13 July 1994
(c) Copyright April 25, 1994 by R. Webster Kehr, All Rights Reserved.
This paper *may* be freely photocopied and/or electronically
transmitted if and only if it is photocopied/transmitted in its
entirety. No fee or royalty is requested. However, no significant
part of this paper may be published in a magazine, other periodical,
or book without written permission of the copyright owner. This
paper may be retyped as long as this copyright notice is included.
ABSTRACT:
The primary purpose of this paper is to mathematically deal with the
Diagonalization Technique and the properties of the "new number" created
by the Diagonalization Technique. The focus of this paper will be math-
ematical, not logical and not philosophical. Many new definitions,
algorithms, mappings, theorems and concepts will be introduced. The pri-
mary contribution of this paper will be to introduce a series of
multi-linked bijections combined with the concept of "width," which will
be defined near the beginning of this paper. This paper will view the
real numbers from the perspective of permuations rather than the more
common perspective of points on a geometric line.
PART I (Webster Kehr)
SECTION 1) INTRODUCTION AND PRELIMINARY AXIOMS
ASCII notations used in this paper:
N/o Aleph Naught or Aleph Null, the cardinality of the counting
or natural numbers.
[n] The set of all counting number from 1 to n, meaning:
{1, 2, 3, 4, ..., n-1, n}
Other notations will be defined as they are introduced.
In this paper the term "countable" means "infinite and countable."
Sometimes the latter phrase will be used to add emphasis. Also, all ref-
erences to the set of terminating decimals in R(0,1) refer only to sig-
nificant digit positions in the elements of the set. Also, all refer-
ences to "permutations" imply permutations where redundancy is allowed
and order is significant.
The History of Transfinite Set Theory:
Transfinite Set Theory has evolved into three distinct layers of
theorems and concepts. The three layers form an inverted pyramid. The
first, or bottom, layer is the Diagonalization Technique/Theorem and the
"new number" created by this technique, developed by Georg Cantor in
1891. The second, or middle, layer is the well established belief that
the real numbers are uncountable. While there are several proofs that R
is uncountable, the Diagonalization Theorem, and its close cousin, the
Power Set Theorem, both developed by Cantor, are by far the most powerful
and significant.
The third layer consists of two different types of theorems: 1)
theorems regarding the Continuum Hypothesis, and 2) theorems which use
the Diagonalization Theorem or which develop a technique which is the
equivalent of the Diagonalization Technique. Examples of the third layer
include theorems or works by Godel, Church and several others.
Theorems in these three layers should not be confused with each
other. For example, theorems in the third layer are totally independent
of theorems which directly deal with the Diagonalization Technique.
Theorems or works in the third layer directly assume R is uncountable, or
directly or indirectly assume the Diagonalization Theorem/Technique is
valid, or use a technique similar to the Diagonalization Technique.
Proofs in the third layer may be perfectly valid internally, but
they, by definition, rely on an assumption that the first and second lay-
ers are valid or that the Diagonalization Technique is valid. This is
vital to remember. Perfectly valid proofs in the third layer may be
found to be false (and in fact their results may be reversed), not be-
cause of any internal fault, but because the assumptions upon which they
are founded are proven to be false.
Part I of this paper will deal with first and second layer theorems.
Part II of this paper, by Rusty Johnson, will deal with several key third
layer theorems, including Godel's Incompleteness Theorem.
THE AXIOM OF THE COUNTING NUMBERS (AXIOM 1):
Given two sets, Set M and Set N, if both sets consist exclusively of
*consecutive* counting numbers beginning with the number 1, and both sets
are infinite and countable, then there exists a bijection between Sets M
and N, where each counting number is mapped onto itself in both direc-
tions, and both sets are equal in cardinality.
When Cantor proved the Diagonalization Theorem, the assumption which
he used, which will be discussed below, was based on the theory that the
counting numbers can be mapped onto any countable set. In other words,
he assumed there was only one definition of countable.
The above axiom is actually far more powerful that it appears at
first glance. It is necessary in order to deal with several double stan-
dards which frequently emerge in discussions of transfinite cardinality.
AXIOM 2: A well-defined set has a cardinality which is either finite,
countable or uncountable. No well-defined set can have a cardinality
which is in two of these categories, nor can the cardinality of a
well-defined set be in-between these three categories.
This axiom should not be confused with the continuum hypothesis.
There is nothing in this axiom which states *what* the smallest uncount-
able set is. This axiom is simply a formal statement of some well ac-
cepted definitions. A finite set is bounded and an infinite set is un-
bounded. A set cannot therefore be in both conditions at the same time,
by definition. Nor can any set be in-between these two conditions, which
are inclusive of all cases, by definition. Likewise, the very definition
of uncountable means "bigger" than countable, which creates a mutually
exclusive condition for a single set, and, by definition, prohibits a
single set from being in-between these two states.
Axiom #2 basically states that a mapping is not needed to prove a
set is countable. If it can be proven that a set is larger than finite
and smaller than uncountable, then it is countable by Axiom #2. For ex-
ample, a set created by a function which has a domain of an infinite num-
ber of counting numbers, must be countable, whether a mapping is provided
or not. It is not finite because the domain is infinite, and it is not
uncountable, because the domain is a subset of a countable set.
Likewise, Cantor proved that a countable number of countable sets is
countable. Axiom #2 can extend this concept to processes. Suppose a
process with a countable number of steps creates either a finite or
countable number of unique elements in each step. Then the set created
by this process (i.e. the union of the subsets) is countable, by Axiom #2
and Cantor's proof.
Theorem #1: Given a bijection between countable Set M and countable Set
N, and a bijection between countable Set N and countable Set P, then
there exists a bijection between Set M and Set P.
Proof: Given a bijection between Sets M and N and between Sets N and P,
let Set M be a proxy for Set N in the bijection between Sets N and P,
thus creating a bijection between Sets M and P.
Q.E.D.
SECTION 2) THE CONCEPT OF WIDTH
DEFINITION: The "width" of a single number is a set of consecutive
counting numbers, always beginning with the number 1, which map to the
significant digits in that number.
For example, the number: 600.031508 has 9 significant digits in it.
The "width" or "width set" of this number is therefore: {1, 2, 3, 4, 5,
6, 7, 8, 9} or [9]. The "width" can be thought of as a set of indexes,
but they are technically a set of counting numbers, nevertheless. As an-
other example, the width of pi is {1, 2, 3, 4, ...}, the set of all
counting numbers.
DEFINITION: The "width" of a (parent) *set* is a set of consecutive
counting numbers, beginning with the number 1, which map to the sig-
nificant digit positions of the element of the set with the maximum
width.
In other words, it is the set of counting numbers from 1 to the su-
premum of the set of cardinalities of the widths of the elements of the
set. If the supremum of the cardinalities of the elements of a set is n,
the width of the set is [n].
For example, consider the following set (the second column repre-
sents the cardinality of the width of that element):
Cardinality of
Element Width
5,071,100.30000000000000... 8
160,901.30000000000000... 7
101.00000005007700... 15
42,103.66000000000000... 7
The cardinality of the above set is 4 (i.e. it has 4 elements), but
the cardinality of the width (set) of the above set is 15 (i.e. the
supremum of the set of cardinalities of widths). The width of the above
set is therefore [15].
It is important to understand that in this paper theorems regarding
cardinality are based on two different types of sets. The cardinality of
a set is the number of elements in the set. The cardinality of the width
of a set is the cardinality of the subset of counting numbers which make
up the width of the set (later, "height" sets will be introduced which
also have this characteristic). If the meaning is clear from the con-
text, the term "width" of a set will also be used to mean "the cardinali-
ty of the width of a set." This is reasonable because calculating the
width set of a set and calculating the cardinality of the width set of a
set are really the same thing. The term "width" of a set implies the
"cardinality of the width set."
Theorem #2: The width of every countable set of numbers is [N/o]. The
countable set is assumed to be permutation-based and to
have a finite base (e.g. base 10) of characters.
Proof:
Assume an infinite, countable set had a finite width, say n. In
base 10, the summation of 10^w, from w=1 to w=n, is the maximum number of
representations (i.e. representations are generated by permutations) in
this set and would include all possible permutations of from 1 to n digit
positions from the set: {0, 1, ..., 9}. If n is finite, so is the
maximum number of representations. The cardinality of this set cannot
exceed the maximum number of representations, which is finite, but may be
less. But in either case the cardinality of the set is finite, which is
a contradiction to the assumption.
Q.E.D.
The real purpose of this theorem is to prove that the width (set) of
the counting numbers and the width (set) of the set of terminating
decimals in R(0,1) are both [N/o]. Note: Examples of "terminating
decimals" would be: .543, .7634, .60065, etc. Examples of
"non-terminating decimals" would be: pi, e, pi/e, 1/3, 8/9, etc.
SECTION 3) BASE 10 TREES
Base 10 Trees:
Consider the following definitions of a base 10 tree (in this paper
the terminology that is developed is chosen thinking about a base 10 tree
which is displayed from left to right):
The 'trunk' or 'level 0' of a base 10 tree is the first position of
the tree, to which all branches are ultimately attached. From a cardin-
ality standpoint level 0 is ignored because it contains only a "place
holder." 'Level 1' of a base 10 tree has ten 'branches.' 'Level 2' of a
base 10 tree has 100 'branches.' 'Level 3' of a base 10 tree has 1,000
'branches,' etc. where the number of branches in 'level n' is represented
by 10^n and the cardinality of all branches from level 1 to level n
equals the summation of 10^w, for w=1 to w=n.
DEFINITION: The 'height' (set) of a base 10 tree for level n is the set
of counting numbers from 1 to 10^n, which map to the
branches of the base 10 tree for *only* level n.
The 'height' of a base 10 tree for a certain level, n, equals
[10^n], the counting numbers which map to the number of representations
generated by the permutations of exactly n digits, taken from a set of 10
elements: {0, 1, ..., 9}. As with 'width' the definition of 'height'
creates a set of counting numbers. The total cardinality of a base 10
tree (i.e. the number of branches) up to and including level n, is the
summation of the cardinalities of all heights (sets) from level 1 to
level n, inclusive.
The Counting Numbers Base 10 Tree:
Now consider a base 10 tree for the counting numbers, defined as Set
A, where each branch is occupied by a unique counting number. While the
design of this tree might seem strange at this point in the paper, it
will eventually become clear why it is done in this manner. In the trunk
will be placed the number '5.' In level 1 will be placed the numbers
'50' through 59,' from top to bottom. Before proceeding, the '5' at the
beginning of each counting number can be thought of merely as a place
holder, much like what a decimal point is to the terminating decimals.
In terms of representations, level 0 has zero representations (be-
cause the first '5' of each number is merely a place holder), and level 1
has ten representations (ditto): 50 through 59. Level 2 has 100 repre-
sentations: 500 through 599. Level 3 has 1,000 representations: 5000
through 5999. Remember, the first '5' of each number is merely a place
holder, but each element (e.g. 500) is a counting number, nevertheless.
The tree contains all counting numbers which begin with a '5.'
Because the width of the counting number base 10 tree is N/o
(Theorem #2), the number of levels of the base 10 tree is N/o, and the
total cardinality of the counting numbers is the summation of 10^w, for
w=1 to w=N/o. More will be said about this later.
The Terminating Decimals Base 10 Tree:
A base 10 tree for the terminating decimals in R(0,1), defined as
Set R+, can be constructed *exactly* like the base 10 tree for the count-
ing numbers. There are only two differences:
1) The place holder is a decimal point, instead of a '5' and
2) Consecutive terminating decimals are not significant for the Set R+
base 10 tree.
The major difference is the second item. It is easy to show, keep-
ing this item in mind, that the cardinality of Set R+, for any width n,
is 10^n, rather than the summation of 10^w, for w=1 to w=n. This is be-
cause many of the branches are duplicated. For example, the branch .56
and the branch .560000 are considered the same branch, even though they
are located on different levels.
The Non-Terminating Decimals Base 10 Tree:
Let the non-terminating decimals in R(0,1) be defined as Set Rnt.
Thinking of base 10 trees, let a "path" be the sequence of branches which
need to be passed through to get to the number under discussion. For ex-
ample, the path for .876321 would pass through the .8 branch on the first
level, the .87 branch on the second level, the .876 branch on the third
level, and so on, to the .876321 branch on the sixth level.
For the non-terminating numbers, there are no branches, but there
are paths. The positions where the branches would be are called
"non-branches." The paths for non-terminating numbers are paths which do
not stop at any non-branch. In other words, the Set Rnt base 10 tree has
no branches at all, but does have non-branches and paths.
The Difference Between the Set R+ Tree and the Set Rnt Tree:
To understand the difference between Set R+ and Set Rnt in their re-
spective trees, it is best to study them in finite space. Suppose that
all "real" numbers stopped at 1 trillion decimal positions. The Set R+
base 10 tree would have a total cardinality of 10^1 trillion. This would
represent the number of *significant branches* in the first 1 trillion
levels of the tree because branches with consecutive terminating decimals
are not significant. If consecutive terminating decimals were sig-
nificant, the height of this tree, at level 1 trillion *only*, would be
10^1 trillion.
The Set Rnt base 10 tree, on the other hand, has a total number of
possible paths to level 1 trillion, of 10^1 trillion, one for each
non-branch in the 1 trillionth level. In other words, the total number
of possible paths for the first 1 trillion levels of the tree exactly
equals the number of non-branches at the maximum level *only*, if con-
secutive terminating decimals were significant. Remember, the branches
at each level consist of all permutations of n decimals, for level n,
from a set of 10 characters: {0, 1, 2, ..., 9}. This is exactly how many
possible paths there are for level n only (try this on a 4 digit wide
tree). In other words, there is a bijection (a 2-way function mapping)
for the permutations for *only* level n and the possible paths to level n
which do not stop before level n.
Some of these paths, however, consist of paths which have con-
secutive terminating zeros. These paths would not technically represent
non-terminating decimals, so the total cardinality of the technically ac-
curate non-terminating paths is less than 10^1 trillion. The point is
that the number of *paths*, up to any level, n, is equal to the number of
branches or non-branches in that same level, if all zeros were sig-
nificant.
Note that in this case, the number of *significant* branches of the
Set R+ tree at level 1 trillion *only*, is exactly equal to the number of
*significant* non-terminating paths which represent non-terminating
decimals, meaning paths with consecutive terminating zeros are ignored.
Theorem #3: The width of Set Rnt and Set R+ are equal.
Proof:
By Axiom #1 and Theorem #2 this is trivial, however, it is instruc-
tive to pursue this proof in more detail. To prove this, a set known as
K-Cauchy Sequences will be developed. K-Cauchy Sequences should be
thought of as a well defined and special type of Cauchy Sequence.
Definition: A K-Cauchy Sequence is an infinite set of terminating
decimals in R(0,1), which represent the branches on the path of a spe-
cific non-terminating decimal in R(0,1). Another way to define it would
be the set of all initial segments of a non-terminating decimal.
For example, the K-Cauchy Sequence for the non-terminating element of
R(0,1), pi/10, is,
Element # Branch on Path
1 .3
2 .31
3 .314
4 .3141
5 .31415
6 .314159
and so on.
Let us call this Set Kpi (i.e. Kpi/10). Note that for element 5,
which represents level 5, the width of the 5th element of Set Kpi is 5.
Element 1 trillion would have a width of 1 trillion, and so on. At each
level, the element number and the width of the Set Kpi element for that
level are equal. We know that Set R+ is countable (general knowledge) and
we therefore know that the width of Set R+ is countable (Theorem #2).
Suppose Set Kpi were finite. Then it would end on element, t, for ex-
ample. The width of this element would be, t, by construction. The car-
dinality of Set R+ would therefore be less than or equal to 10^(t+1) (see
above comments). This would mean Set R+ was finite, a contradiction.
Set Kpi is not uncountable because it is a subset of Set R+, it is there-
fore countable by Axiom #2.
Because Set Kpi is countable, the *width* of Set Kpi is countable
(Theorem #2 and by construction). Note that both the cardinality of the
K-Cauchy Sequence elements and the *width* of the K-Cauchy Sequence el-
ements are always equal. In infinite space, both the cardinality of the
elements and the cardinality of the width set are equal to N/o, the
cardinality of the counting numbers.
Now let us examine the non-terminating "new number" which is created
using the Diagonalization Theorem. Because it is a non-terminating el-
ement, there exists a K-Cauchy Sequence for this element of R(0,1), call
it Set Knn. In fact, the K-Cauchy Sequence is the set of initial seg-
ments of the "new number" as it is created. The question is, does the
new number ever reach a point where there exists a digit position in the
new number which is not a digit position of an element of the K-Cauchy
Sequence, Set Knn?
Suppose there was. Because for each digit position in the new num-
ber, called Number Nn, there exists an initial string for this digit po-
sition, let us assume the first initial string which was not an element
of the K-Cauchy Sequence was: .555...556, where the last digit position,
occupied by a '6,' was not attained by any element of the K-Cauchy Se-
quence. This would mean that the K-Cauchy Sequence must end before
.555...556.
Now let us link the *set* of countable steps of the Diagonalization
Technique to the *set* of countable steps of the K-Cauchy Sequence. Both
sets are sets of counting numbers and both sets are countable. As with
Set Knn, the width of each initial segment of the new number is always
equal to the step number of the Diagonalization Technique, which is a
counting number. The two sets of counting numbers will be mapped to each
other, element to element.
If there existed a digit position achieved by Number Nn which was
not achieved by the K-Cauchy Sequence, then that would mean there was a
counting number used in the Diagonalization Technique which was not a
counting number used in the K-Cauchy Sequence process. This would be a
contradiction because there is only one definition of counting numbers
(Axiom #1).
Q.E.D.
The question naturally becomes, if there is not an infinitely wide
single element of Set R+, or of a Set K, how can the set of terminating
decimals be as wide as the set of non-terminating decimals? The answer
is that we are comparing *sets*, not *single elements* of sets. If you
compare the width of a non-terminating decimal, which itself is the end
result of an infinite process, to a single element of Set R+ or a Set K,
the element in Set Rnt will be considered to be wider. But the *set* Set
R+ does achieve the same width as the *set* Set Rnt.
The difference between Set R+ and Set Rnt is one of construction.
The elements of Set Rnt are each made infinitely wide *before* the infi-
nitely wide set is created (in order to create limit points). The el-
ements of Set R+, on the other hand, are each created *as* the infinitely
wide set is created. It is the difference between a person walking in
the sand on the sea shore (who leaves footprints) and a bird flying above
the sand (which does not leave footprints). Both the person and the bird
have the same destination in mind, but the bird leaves no footprints.
More will be said about these issues below.
SECTION 4) THE CARDINALITY OF THE REAL NUMBERS
At this point it is necessary to define what I call: "Cantor's First
Assumption."
Cantor's First Assumption: Given two infinite sets, Sets M and N, if it
is known that Set M is countable and it can
be proven there cannot exist a mapping from
Set M onto Set N, then Set N is uncountable.
This is technically not the original assumption Cantor used in the
Diagonalization Theorem, but it is very similar and will be used as a vi-
able substitute in this section.
DEFINITION: A 'calculator' is an object which holds an infinite number
of digits and is totally *void* of digits unless they are
added via a mapping or process.
Let us assume there are an inexhaustible supply of calculators for
use in this section. Let us choose calculators and create the digits in
these calculators by using the following algorithm:
1) We will post digits to these calculators as we move across (i.e.
meaning successive levels) the Set A base 10 tree.
2) The zero digit (i.e. level 0) in each calculator will always be
a decimal point, which will be a place holder, like the number '5' is a
place holder in the level 0 in the Set A base 10 tree.
3) For the first level of the Set A base 10 tree, we will choose 10
calculators and place the numbers: .0 through .9 in them, which come from
the 50 through 59 elements of the Set A base 10 tree.
4) As we get to the second level of the Set A base 10 tree, we will
choose 90 more calculators, bringing the total number of calculators to
100, and place the numbers: .00 through .99 in them, which come from the
Set A base 10 tree elements: 500 through 599, respectively. Note that
the first 10 calculators are used for both levels.
5) For the third level of the Set A base 10 tree, we will choose
900 more calculators, bringing the total number of calculators to 1,000,
and will place the numbers: .000 through .999 in them, which come from
the Set A base 10 tree elements: 5000 through 5999, respectively.
6) We will continue this process a countable number of times, for
the entire width of Set A.
Conclusions:
1) Because the width of Set A is countable, meaning the number of
levels in the Set A base 10 tree is N/o, this is a countable process.
2) A countable number of calculators are adequate for this infinite
process because we add calculators and choose and post numbers to these
calculators based solely on elements of the set of counting numbers. No
calculator is chosen unless a counting number drives the process.
It is very important to remember that each calculator is capable of
containing an infinite number of digits, but each calculator and the dig-
its in each calculator are created by new levels of the Set A base 10
tree, meaning from the counting numbers. In much the same way, the real
number line (geometrically speaking) from 0 to infinity is created by the
units (0,1], (1,2], (2,3], (3,4], etc. The real number line can be
thought of as a calculator and the units of R themselves can be thought
of as the digits in the calculator.
We will now define the set of all of the calculators we have chosen
during this infinite process as Set Rc. In Set Rc, consecutive terminat-
ing zeros *will* be significant, because these calculators were created
from subsets of the counting numbers. Note that for each level, n, the
cardinality of Set Rc, up to that level only, is exactly 10^n, the exact
cardinality of the height of Set A for that level. This is not surpris-
ing because every element of Set Rc has exactly the same width at all
times, namely the width of the elements residing in the nth level (ex-
cluding place holders). Several things are clear about Set Rc, in
totality (i.e. in infinite space):
1) Because a base 10 tree does not change its shape or construction
between finite and infinite space, cardinalities of different levels can
make the jump from finite to infinite space successfully (the issue is
always counting and cardinality),
2) each element of Set Rc is countable in width because Set A is
countable in width, meaning the number of levels is countable.
3) the cardinality of Set Rc is infinite and countable. It is in-
finite because the height of Set A at an infinite level is infinite (i.e.
10^N/o, which will be discussed later) and it is countable because it is
always equal to a subset of Set A for any given level, by creation and by
Axiom #2.
4) Because of the way 'levels' work in a base 10 tree, the repre-
sentations in Set Rc contain all possible permutations of N/o digits,
taken from a set of 10 characters: {0, 1, 2,... 9}, by construction.
Each digit is significant, by definition, even consecutive terminating
zeros are significant.
Now let us remember the set of all non-terminating elements of the
real number line from 0 to 1 (i.e. R(0,1) or [0,1]), Set Rnt. As was
noted above, if all non-terminating permutations of N/o digits from a
pool of 10 characters were created, it would include all terminating and
non-terminating elements of R, because consecutive terminating zeros are
not significant, but they are legitimate permutations. Whether con-
secutive terminating zeros are significant in Set Rnt is irrelevant be-
cause consecutive terminating zeros in the elements of Set Rc are sig-
nificant and this is the set that will be compared to Set Rnt. In
standard mathematics Set Rnt is considered an uncountable set. If, how-
ever, Set Rnt were countable, then R(0,1) would be countable because two
countable sets is a countable (Cantor proved that a countable number of
countable sets is countable).
Before the theorem, a comment should be made about the necessity of
the calculators. The purpose of these calculators is to map to the el-
ements of Set Rnt. Because every element of Set R+ is *defined* to have
an infinite number of zeros behind them, some interpret this to prove
that Set R+ is not as wide as Set Rnt, in terms of significant digits.
It has already been shown that there is no digit position achieved by any
non-terminating decimal which is not also achieved by a significant digit
position of an element of Set R+ (i.e. a Set K) (Theorem #3).
Also because all elements of Set R+ have a finite width, *by defini-
tion*, some think that the width of Set R+ is less than the width of Set
Rnt. It has also been proven that the width of Set R+, referring only to
significant digits, is N/o (Theorem #2). The consecutive terminating ze-
ros behind each terminating decimal is merely a paradoxical observation
caused by a definition, it is not a proof of anything. Nevertheless,
this is why the calculators were defined the way they were.
The next theorem will prove that R is countable. The technique that
is used in this proof is quite different than the technique that will be
used later to prove that the Diagonalization Technique can only create a
countable number of "new numbers." This theorem is placed first in order
to introduce concepts which will be needed later in this paper, so as not
to introduce too many concepts in one theorem.
Theorem #4: R is countable
Proof:
Let us define the *width* set of Set Rc to be Set WRc and let us de-
fine the *width* set of Set Rnt to be Set WRnt. It is vital to remember
that Set WRc and Set WRnt contain *only* counting numbers, because they
are width sets. We have proven above that Set Rc is countable, therefore
Set WRc is countable (Theorem #2). Considering the relationship between
Set WRc and Set WRnt, there are only two possibilities. These will be
called Case 1 and Case 2. In Case 1, there will be an assumed mapping
from Set WRc onto Set WRnt. In Case 2, it will be assumed there is *not*
a mapping from Set WRc onto Set WRnt.
Case #1: There exists a mapping from Set WRc onto Set WRnt.
Step 1: By assumption, this case means that given any element of Set
WRnt, there exists an element of Set WRc mapped to that element from the
countable mapping from Set WRc onto Set WRnt. This basically means that
the elements of Set WRnt and Set WRc can be listed side-by-side, since
both sets contain only counting numbers. This mapping can be considered
a bijection because it will be assumed that Set WRnt can be mapped onto
Set WRc (i.e. there is no need to prove R is countable if Set WRc is
larger than Set WRnt). This is by Axiom #1.
Step 2: Since Sets WRnt and WRc map to digit positions in Sets Rnt and
Rc, respectively, Step 1 means that given any digit position in Set Rnt,
there exists a digit position in Set Rc mapped to that digit position via
a double or proxy mapping through the width sets (Theorem #1). In other
words, the elements of Sets WRnt and WRc are simultaneously mapped to
each other and to the digit positions in Sets Rnt and Rc, respectively,
making a bijection from the digit positions in Set Rc and Set Rnt.
Step 3: It was shown above that for any digit position, n, achieved by
Set A, the cardinality of Set Rc consists of all representations, n dig-
its wide, generated by creating all permutations of n digits taken from a
pool of 10 characters. This remains true even in infinite space because
of the construction of base 10 trees.
Step 4: It was also shown above that Set A achieves N/o digit positions,
meaning Set Rc consists of all representations of N/o digits taken from a
pool of 10 characters. Also remember that every element of Set Rc is
countably wide.
Step 5: By definition, the cardinality of Set Rnt consists of all repre-
sentations (i.e. generated by permutations) of N/o digits taken from a
pool of 10 characters. This is why the "new number" was claimed to be an
element of R(0,1).
Step 6: Note that at this point we have two sets of infinitely wide num-
bers, both of which contain all permutations of N/o digits taken from a
pool of 10 characters. Every element of Set Rnt is *defined* to be
countably wide. Every element of Set Rc is *created* to be countably
wide by the width of Set A. Is there a difference between defining ver-
sus creating countable sets? Obviously not. For example, *defining* the
'even' counting numbers to be the set of all counting numbers which end
in a 0, 2, 4, 6, or 8 and *creating* the 'even' numbers by starting with
the number 2 and then adding 2 to create each successor, both create the
same set. Remember, the Diagonalization Theorem *created* an irrational
number so there is clear precedence for this process.
Step 7: Because of Steps 3 through 5, the number of representations of
*both* sets is totally dependent on the number of permutations of N/o
digits taken from a pool of 10 characters. But both sets contain *all*
permutations to their defined or created widths. This means that the
only possible way for these sets to be different in cardinality is for
their widths to be different!
For example, the cardinality of Set Rc is equal to the number of
representations in Set Rc, but the number of representations in Set Rc is
exactly equal to the number of *all* permutations, from a pool of 10
characters, to whatever maximum width Set Rc achieves, which is an N/o
width.
Also, the cardinality of Set Rnt is equal to the number of represen-
tations in Set Rnt, but the number of representations in Set Rnt is de-
fined to be equal to the number of permutations of N/o digits taken from
a pool of 10 characters.
Step 8: Therefore, because both sets contain *all* permutations to their
respective widths, the only possible difference between the number of
representations in Set Rnt (i.e. its cardinality) and the number of rep-
resentations in Set Rc (i.e. its cardinality) would be a difference in
*widths* between the two sets (i.e. the number of digit positions
achieved by each set).
((Step 8A: As a parenthetical explanation of Step 8, consider this se-
quence of steps: for both Sets Rc and Rnt:
A) Cardinality is equal to representations,
B) Representations are solely a function of permutations,
C) Permutations are solely a function of width (i.e. the number of digit
positions or levels), therefore,
D) The only possible difference in cardinality between Sets Rc and Rnt
would be caused *exclusively* by a difference in width, but
E) By Axiom #1 and Theorem #1, there is a bijection between the digit po-
sitions of Sets Rc and Rnt and therefore they have exactly the same num-
ber of digit positions, therefore,
F) The number of permutations in Sets Rc and Rnt are the same, and
G) The number of representations in Sets Rc and Rnt are the same, there-
fore,
H) The cardinality of Sets Rc and Rnt are the same.))
Step 9: Because, *by assumption* of this case, given any digit position
in Set Rnt, there exists a digit position in Set Rc mapped to that digit
position, the cardinality (i.e. number of representations) of Set Rnt
cannot exceed the cardinality of Set Rc, because they have identical per-
mutation constructions, and for any digit position achieved by Set Rnt
there is a digit position (i.e. a set of representations) achieved by Set
Rc.
Step 10: To summarize Step 9, because Set Rnt can never achieve a digit
position which Set Rc cannot achieve, the cardinality of Set Rnt can
never exceed the cardinality of Set Rc.
Step 11: Before proceeding, the above concepts will be explained from
another viewpoint. By assumption, the width sets of Sets Rc and Rnt are
permanently 'linked' together by a mapping. What it means to be 'linked'
is that there is no digit position achieved by one set, by assumption,
that is not achieved by the other set. In other words, neither set can
'escape' the link or mapping between them.
As such, the width of neither set can escape the link or bond re-
sulting from this mapping. Because of their similar permutation con-
struction, neither set can excel in cardinality (i.e. representations)
unless that set can escape the linkage or bonding of widths, but by
assumption, neither set can escape the linkage of widths and neither set
can therefore excel in cardinality because they are constructed
similarly.
Step 12: Because Set Rc never becomes uncountable, Set Rnt can never be-
come uncountable (!) because, by assumption, its width can never exceed
the width of Set Rc. Therefore no representation in Set Rnt can exist
which does not also exist in Set Rc, because no digit position can exist
in Set Rnt which does not exist in Set Rc.
Step 13: Set Rnt is therefore countable, because it can never become un-
countable.
Step 14: Because Set Rnt is countable, and because a countable number of
countable sets is countable, *all* of R is countable.
Step 15: In Case 1, all of R is countable.
Case #2: There does *not* exist a mapping from Set WRc onto Set WRnt.
Step 1: By Axiom #1, this case does not even need to be discussed, how-
ever, Axiom #1 will be ignored for a moment.
Step 2: By assumption, there exists an element of Set WRnt which is not
an element of Set WRc (because both Sets WRc and WRnt contain only count-
ing numbers, it is logical that each counting number in Set WRc would be
mapped to its identical counting number in Set WRnt, so if there is no
mapping from Set WRc to Set WRnt for a specific element of Set WRnt, then
that element does not exist in Set WRc).
Step 3: Since it was proven above that Set WRc is countable, and because
by assumption there is no mapping from Set WRc to Set WRnt, then by
Cantor's First Assumption, Set WRnt is uncountable.
Step 4: Because Set WRnt is uncountable, the number of digit positions
in each element of Set Rnt is uncountable, by construction.
Step 5: This is a contradiction because the number of digit positions in
each and every non-terminating number is countable, therefore by con-
struction and definition, the width of Set Rnt, Set WRnt, cannot be un-
countable.
Step 6: Case #2 is rejected.
Q.E.D.
A comment could be made that Set Rnt achieves infinite width instan-
taneously, but Set Rc must achieve it digit by digit. Besides being a
blatant violation of Axiom #1 (i.e. the width of both sets are mapped to
by sets of counting numbers), if Set Rnt achieves an infinite width in-
stantaneously, why doesn't Set Rc achieve it instantaneously also? Set
Rc is created from the counting numbers, which achieve their width in-
stantaneously. It is a double standard to say that one countable set is
larger than another countable set or to place a physical restriction or
paradox on one countable set to make it appear smaller than another
countable set. Remember, the Diagonalization Theorem created the "new
number" - a *non-terminating* number - digit by digit, after an examina-
tion of a countable mapping, so how did the new number achieve infinite
width instantaneously? Fair is fair.
Just how big is an uncountable set? By analyzing the Diagonaliza-
tion Theorem one might think that an uncountable set need be only
slightly larger than a countable set. Not even close. Consider the
definition of a "plate." A "plate" is a collection of googolplex
space-disjoint spherical universes, each of which is 100 billion light
years in radius. A plate is much larger, speaking of volume, than the
neutron of a single hydrogen atom. Much larger. But yet this gap is
very finite. An uncountable set is so big that an infinite number of
countable sets can be removed from it and it will still be uncountable.
That is a big gap, much bigger than the gap between a plate and a single
hydrogen neutron!
Thinking of the "balance" (i.e. how each branch in a tree is suc-
ceeded by a tree equal in size to the original tree) of a base 10 tree,
every branch is equal to every other branch in terms of its association
to "non-terminating paths." If there were only a countable number of el-
ements of Set Rnt for each and every calculator of Set Rc, then Set Rnt
would be countable. If R is uncountable, therefore, there must be as-
sociated with *each and every* element of Set Rc (a single neutron) an
uncountable number of elements of Set Rnt (larger than a plate)!
Given two sets which have identical width-wise permutation construc-
tions, and for which there exists a one-to-one mapping and linking from
the width (i.e. digit positions) of one set to the width (i.e. digit po-
sitions) of the other, does it seem logical to think that for each and
every element of Set Rc, there exist an uncountable number of *addi-
tional* elements or representations of Set Rnt which have no equal
representations in Set Rc?
SECTION 5) A DISCUSSION OF THE DIAGONALIZATION THEOREM AND THE POWER
SET THEOREM
Part 1: Introduction
The first question to ask about the Diagonalization Theorem (DT) is
this: does the DT succeed in proving there exist an uncountable number of
"new numbers" which can be found using the Diagonalization Technique (as
opposed to only a countable number of "new numbers")? That an uncount-
able number can be found is assumed based on the combination of Cantor's
First Assumption and the ability of the Diagonalization Technique (DTq)
to find a "new number" which is not in the listing. Indeed, if R(0,1)
were uncountable, then an uncountable number of "new numbers" could be
found.
On the other hand, if only a countable number of new numbers could
be found, then R(0,1) would be countable. This is because two countable
sets is countable. The two countable sets would be: the listing of el-
ements of R(0,1) and the set of new numbers which could be created.
First it is necessary to discuss the DTq in finite space.
Part 2: The DT in Finite Space
The DTq used in the DT to create a new number basically says this:
"Give me a square matrix, where there are n elements and each element has
a width of n (the rows of the matrix are the set elements and the columns
of the matrix are the digit positions of the elements), and I will find
an element that is n digits wide which is not an element of the set."
The DTq is a simple paradox applicable to any square matrix (in infinite
space a "square" matrix will be referred to as an N/o by N/o matrix, to
avoid defining a "square" matrix in infinite space).
Consider the following set of 10 elements, each of which is 10 dig-
its wide, a square matrix of digits. The right column is the "new num-
ber" which is created using the following algorithm: if the nth digit of
the nth element is a 5, the nth digit of the new number will be a 6, oth-
erwise the nth digit will be a 5:
Element # Element New Number Initial Segments
1 .7867930392 .5
2 .3982176253 .55
3 .3052871672 .556
4 .5400000000 .5565
5 .6541590919 .55656
6 .8901928374 .556565
7 .2132980192 .5565655
8 .5565650000 .55656555
9 .4332113244 .556565555
10 .5432543455 .5565655556
It is easy to verify that the new number is not in the set of 10 el-
ements. If one were to construct the set of all permutations of 10 dig-
its taken from a pool of 10 characters, the cardinality of the set would
be 10,000,000,000 (i.e. 10^10), allowing redundancy and where order is
important. It is obvious that the new number would be in the set of all
representations. In finite space, the DT theorem is perfectly valid be-
cause 10^n is always greater than n.
Part 3: The DT in Infinite Space
The Gap Set
For any n x n matrix, let us define the total number of "new num-
bers" which could be found using the DTq as the "Gap Set" for set size n.
For example, take a 10 x 10 matrix versus the number of permutations of
10 objects taken from a set of 10 elements. The second set has a cardin-
ality of 10,000,000,000, the first set has a cardinality of 10. The pro-
cess of finding a "new number" can only be repeated 9,999,999,990 times
and then there are no more "new numbers" that can be found. These new
numbers would be the Gap Set. If the original set has 1,000,000,000 el-
ements, then there can only be found: (10^1 billion) minus 1 billion, new
numbers for the Gap Set.
Consider the following chart for the DTq:
Counting Number Cardinality of the Gap Set
10 (10^10) - 10
100 (10^100) - 100
1000 (10^1000) - 1000
10000 (10^10000) - 10000
100000 (10^100000) - 100000
and so on.
What happens to the Gap Sets when the base set goes from finite to
countable in cardinality? The cardinality of the Gap Sets, according to
the DT, jump directly from finite to uncountable! This is very il-
logical. In finite space the difference (i.e. gap) between: the cardin-
alities of the set of all permutations of n digits minus the set of num-
bers which fill an n x n matrix, is finite. But suddenly, when these
sets become countable, the Gap Set becomes uncountable.
Consider the following chart:
Set Size = n Gap: Permutations minus Set Size
Finite Finite
Countable Uncountable
What set size generates a Gap Set with a countable cardinality? No
counting number based process, with a fixed formula, which consistently
generates finite incremental increases in the cardinality of the function
set, can *grow* from a finite set to an uncountable set in one step of
the process. That is obvious. But yet that is what is claimed to happen
with the sequence of Gap Sets.
In fact, if the incremental increases became countable, the end re-
sult would still be countable, because there are a countable number of
steps in the process (i.e. a countable number of steps, each of which
create a finite or countable *addition* to the set, creates a countable
set because a countable number of countable sets (i.e. steps) is count-
able - see the discussion after Axiom #2). But the DT does not even
prove or mention there are any countable incremental increases.
There must be many counting numbers which generate countable Gap
Sets. *CLEARLY* all finite numbers generate finite Gap Sets. So what
are the counting numbers that generate countable Gap Sets? Is there a
cardinality for the base sets between the finite numbers and N/o? By
Axiom #2, there is not. The justification for this huge jump in gap size
is explained by Cantor's First Assumption. Something so illogical should
not be proven by a single assumption coupled with a paradoxical mapping.
Theorem #5: The Gap Set of R(0,1) is countable, meaning R(0,1) is
countable.
Proof:
Consider the following mapping from the counting number base 10
*height sets* onto the Diagonalization Theorem's *Gap Sets*:
Set Size or Counting Number Tree Cardinality of
Level Level's Height Set the Gap Set
10 10^10 (10^10) - 10
100 10^100 (10^100) - 100
1000 10^1000 (10^1000) - 1000
10000 10^10000 (10^10000) - 10000
100000 10^100000 (10^100000) - 100000
...
N/o 10^N/o (10^N/o) - N/o
It is very clear that the end result of these infinite processes is
countable. In the left column are counting numbers. Because the width
of the counting numbers is N/o, and because the width of the real numbers
is N/o, both the middle and right columns continue forever. Because the
counting numbers are countable, each height set is countable; note that
height sets map to subsets of the counting numbers. The height sets
achieve their cardinality by counting the elements in the various levels
of the Set A tree, which levels achieve a cardinality of N/o. The Gap
Sets also achieve their cardinality by counting, in this case, counting
elements not listed. Both columns achieve their cardinality by counting
processes. The columns are therefore comparable.
Because non-terminating numbers have been added to Set R+ to create
R(0,1), it is necessary to discuss these elements. Is it possible that
these added elements can boost the cardinality of the Gap Sets to an un-
countable level? Note that a *width* of N/o has already been accounted
for! In other words, all counting numbers have already been accounted
for in terms of width. The non-terminating elements of R are not permit-
ted to achieve a width beyond N/o. So what width cardinality can they
add to the above chart in order to increase the cardinality of the Gap
Set?
Let Set Ca be the set of counting numbers which map to the various
levels of the Set A tree. Let Set Cnt be the set of counting numbers
which map to the digit positions of Set Rnt. Each and every *level*, n,
Set Ca maps to, includes *all* permutations of n digits taken from a pool
of 10 characters. n achieves a level of N/o. By definition, Set Rnt
contains all permutations of N/o digits taken from a pool of 10 charac-
ters (ignoring consecutive terminating zeros which would reduce the car-
dinality of Set Rnt if removed).
The only way that Set Rnt can add cardinality to the above process,
which is already countable, is for Set Cnt to be larger than Set Ca (see
the above discussion about linking representations to permutations to
width). If Set Cnt were larger than Set Ca, then Set Cnt would be un-
countable. This is because: 1) Set Ca has already been proven to be
countable and 2) Cantor's First Assumption. This would be a contradic-
tion. Also, Axiom #1 prohibits two countable sets of counting numbers
from having different cardinalities.
Q.E.D.
In mathematics, 10^N/o is considered an uncountable number. But
this belief is based on the DT or Power Set Theorem (PS), which are
highly related. But it is *clear* that the Gap Sets of both the DT and
the PS (to be discussed below) are never uncountable, because they can be
mapped to a subset of the counting numbers for all n from 10 to N/o.
Does placing the elements of Set A on a tree, instead of in a list,
change the cardinality of Set A?
Corollary: 10^N/o is Countable
Let Set A be defined to be the set of all counting numbers in base
10 form. Several facts are obvious for the totality of the Set A base 10
tree (where the first '5' of each number is a place holder only, meaning
level 0 is ignored from the standpoint of cardinality):
1) because the width of Set A is N/o, as proven above,
2) the base 10 tree for Set A achieves a level of N/o, and
3) the height of level N/o is [10^N/o], therefore
4) the total cardinality of Set A cannot be less than 10^N/o, which
is the cardinality of the height set of the Set A base 10 tree
for "level" N/o, and
5) the representations of the Set A base 10 tree at a level of N/o
is a subset of Set A, therefore
6) the cardinality of Set A consists of all cardinalities of all
height sets from 1 to N/o, inclusive.
In summary, the set: [10^N/o], represents the height at the N/o
level of the counting numbers, and because base 10 trees do not change
shape in infinite space, the cardinality of the N/o level height cannot
exceed the total cardinality of all of the counting numbers. To con-
tinue:
1) a base 10 tree is well defined, and
2) a base 10 tree is extremely well behaved because it is has a
constant structure, and
3) the cardinality of the height of level N/o is a process of
counting the branches (i.e. listing the counting numbers for
that level), and
4) 10^N/o is therefore a symbol which represents a *counting*
operation (i.e. counting branches), not a mathematical op-
eration such as addition or multiplication.
Therefore, if 10^N/o were uncountable, then a subset of the counting
numbers would be uncountable. But that cannot be, therefore 10^N/o and
N/o are two symbols for the same cardinality. The number of elements of
Set A, the width of Set A, and the total height of Set A are all count-
able, N/o, N/o and 10^N/o, respectively.
Two other comments need to be made. First, exponential notation is
an abbreviation for multiple multiplications. Multiplication notation is
an abbreviation for multiple additions. Addition notation is an abbre-
viation for counting. 10^N/o is actually an abbreviation for a counting
operation. Second, because a set of counting numbers can map to the lev-
els of the counting number Base 10 tree, the only way to make 10^N/o an
uncountable number would be to come up with a different set of counting
numbers, so additional levels of the tree could be created. This would
be a violation of Axiom 1.
Q.E.D.
A natural question is whether N/o is a specific number (i.e. omega -
Footnote #1) or a concept of "never ending." Whichever is the case, it
is only fair to use this same definition consistently throughout all of
Transfinite Set Theory. To say that it is a never-ending process in one
discussion, but to say that it is a specific number in another discus-
sion, is simply a double standard, not a proof. There is only one
definition of countable, not two, as Cantor's First Assumption implies.
Part 4: The Power Set Theorem
The PS theorem is more complex than the DT theorem. While it uses
combinations instead of permutations, its main complexity is in the way
the "new subset" is created. But what makes the PS theorem believable is
that a set which contains, as a proper subset, another set should be
higher in cardinality than the set which it contains. For example, in
the PS, the subsets of the Power Set which consist of only one counting
number are elements of the Set of All Subsets of the counting numbers,
and are equal to Set A by themselves. This makes it believable that the
Power Set has higher cardinality than the set of counting numbers. Let's
see if this is obviously true when there is no moving target.
Let Set J be the set of all real numbers greater than zero which
have less than 101 significant digit positions to the right of the
decimal. Examples would be: 3,098,243.8765499, 10,000, 43,321.4,
109,435,123,878.7837235099008982346 and so on.
In infinite space Set J is a subset of the terminating decimals and
hence is well known to be countable. But the counting numbers themselves
are a subset of Set J. In infinite space it is not always true that a
set which is a proper subset of another set has a smaller cardinality.
The counting numbers can easily be mapped to Set J, which contains the
counting numbers! So there is little justification for jumping to the
conclusion that the Power Set of Set A is uncountable.
Let the "Gap Subset" be the set of all possible "new subsets" which
could be created by the PS technique for any original set size, n.
Theorem #6: The Gap Subset of the "Set of All Subsets" of a countable
set is countable.
Proof:
In the PS, a "new subset" is created so that it is guaranteed to be
different than each listed set. This is done by listing the elements of
the original set (infinite set in this case) in the left column and the
elements of the Power Set in the right column. The list is gone through
so that *if* the right column subset contains the element in the left
column the "new subset" does not contain this element. If the right col-
umn subset does not contain the element in the left column then the "new
subset" does contain that element. Thus, the new subset is created to be
different than every subset in the list for at least one number, the num-
ber which is mapped to that subset. This technique is very similar to
the DTq.
As with the DT, the PS is assumed to create an uncountable Gap Sub-
set (i.e. the set of all subsets not in the list) because of Cantor's
First Assumption. And as with the DT, the Gap Subsets jump directly from
finite to uncountable as n goes from finite to countable.
Note that the cardinality of the "set of all subsets" in the PS is
far less than the cardinality of the "number of all permutations" in the
DT, for any n. In the above chart, use the same left and middle columns,
but for the right column use the cardinality of the "set of all subsets"
for n, minus n, to determine the total cardinality of the Gap Subsets.
In this case, the right column would be far smaller than the previous
right column. In short, the counting number base 10 height sets can map
to the cardinality of the Gap Subsets and have plenty of elements left
over. The bigger the n, the larger the gap between the two columns.
Q.E.D.
SECTION 6) A MAPPING FROM SET A AND SET Rc TO R(0,1)
In this discussion, a mapping from a subset of Set A onto the set of
all terminating decimals in R(0,1), defined as Set R+, will be shown
first, then a mapping from another countable set, Set Rc, onto the set of
non-terminating decimals, Set Rnt, will be shown. Technically this is
not necessary because the set of all representations generated by all
permutations of N/o digits from a set of 10 characters includes both
sets, because terminating zeros are not significant, and because the ter-
minating decimals are legitimate permutations. Nevertheless, this
trivial fact will be ignored in order to better demonstrate this type of
mapping.
What is important is that Set A and Set Rc are both countable sets,
and the union of two countable sets is countable.
Mapping to the Terminating Decimals:
Let us define the terminating decimals in R(0,1) as Set R+. We will
map to all of them using the counting numbers which begin with a '3' and
the following rule:
1) Each element of the subset of Set A will map to an element of Set R+
by replacing the initial '3' by a decimal point.
Before a listing is shown it should be pointed out that terminating
zeros are significant in Set A but not in Set R+. This is OK because it
would make Set A larger than Set R+, but Set A is countable and all we
are proving is that Set R+ is not uncountable, so there is no harm done.
Let's look at the list:
Element Set A Element Set R+ Element
1 31 .1
2 32 .2
3 33 .3
4 34 .4
and so on.
It will be left to the reader to prove this is a valid bijection,
ignoring elements of Set A with consecutive terminating zeros.
Mapping to the Non-Terminating Decimals:
As might be expected by now, there need to be calculators created
for this mapping. We will use all elements of Set A which begin with a
'4' to create the calculators (the '4' will be replaced by a decimal
point). We will use a base 10 tree to create this set, as was done
above. Because the width of all counting numbers which begin with a '4'
is countable, the width of the calculators is countable. Remembering the
steps of the proof that R is countable, which will not be repeated, leads
one to realize that the calculators, which are created by a subset of the
counting numbers, can be mapped to all elements of Set Rnt (the '4' was
replaced by a decimal point when Set Rc was created). In fact, they are
the same set and have the same elements (which are mapped to each other),
ignoring elements of both sets which end in consecutive terminating
decimals.
SECTION 7) APPLYING THE DTq TO THE ABOVE MAPPING
In 1891, Cantor published his Diagonalization Theorem as an attempt
to improve upon his 1874 proof of R's uncountability (Footnote #2).
While the above definition of Cantor's First Assumption is probably a
common interpretation of his assumption, the actual stated assumption was
this:
CANTOR'S ORIGINAL ASSUMPTION: "If (M is nondenumerable, then if)
E_1, E_2, ... ,E_y, ... is any simply infinite
sequence of elements of the set M, ... there is
always an element E_0 of M which corresponds to
no E_y." (Footnote #3)
There is little if any difference between Cantor's Original Assump-
tion (COA) and CFA, depending on the definition used for a "simple list-
ing." Both COA and CFA look at determining the cardinality of sets *af-
ter* a listing or mapping is shown. In the assumption above, listings
are provided first and then an element not in the listing is created.
It is clear that if COA or CFA were applied against the above map-
ping, that a "new number" could be found which was not in the listing.
But now we have a conflict, because there *is* a mapping above from Set A
and Set Rc onto R(0,1), and the mapping is to all permutations of N/o
digits. Suppose, however, that the "new number" had to be shown
*first*, rather than the above mapping. Would there be a mapping to that
fixed new number, a single element of R(0,1)?
For example, let's suppose the new number were a terminating
decimal, say: .54298. Which element of Set A maps to this element? Ob-
viously, 354298. Now let's suppose the "new number" was pi/10. Which
element of Set Rc maps to this element? The element of Set Rc called
pi/10.
The issue now becomes, who goes first and who goes last?
Suppose the contest were fair ("fairness" is certainly not a math-
ematical concept, but we do have a contradiction and this discussion has
a dual purpose, as will be seen below). Suppose the two contestants
traded turns coming up with their numbers or mappings, and that both
players could produce their number or mapping *after* the other person's
turn was up. If the real numbers were uncountable, the person who cre-
ates the new number should easily win! For the creation of this new num-
ber, we will use the algorithm mentioned above. For the mapping, we will
use the mapping above from Sets A and Rc to R(0,1). Now consider the
fair contest:
Step Whose Turn Set A The Number/Mapping They Chose
1 First 31 -> .1 first element of listing
2 Second .5 "new number" segment
3 First 35 -> .5 element not mapped to yet
4 Second .55 "new number" segment
5 First 355 -> .55 element not mapped to yet
6 Second .555 "new number" segment
7 First 3555 -> .555 element not mapped to yet
8 Second .5555 "new number" segment
9 First 35555 -> .5555 element not mapped to yet
10 Second .55555 "new number" segment
and so on.
Who will win the contest? Neither party will win, because every
time a new initial segment for the "new number" is chosen, it is realized
that this exact initial segment *is* part of the mapping! At each and
every stage of this process, forever, there exists a mapping to the cur-
rent initial segment of the new number. Because the width of Set A is
infinite and countable, this "race" can continue forever with no winner.
The fair contest is a race between the creation of an irrational new
number and the creation of a K-Cauchy Sequence (see Theorem #3)! The
width set of the K-Cauchy Sequence, which represents an infinite process,
is just as large as the width set of an irrational number, which is also
created by an infinite process. The only way for the "new number" to win
would be if it had an uncountable width (set), which would be against the
rules of the contest and Axiom #1.
With this background, let's start over, with the following scenario:
1) The above mapping is displayed from Sets A and Rc to all of R(0,1).
2) A "new number" is created which is *fixed*; and is a single element
of R(0,1); and which is not mapped to.
3) Once this "new number" is shown and becomes fixed, because Set Rc
maps to all permutations of N/o digits from a set of 10 characters, it is
clear that Set Rc can map to this fixed new number which is a single el-
ement of R(0,1).
The problem is that the "new number" is never a *fixed* number, nor
can it be, it is always an "Infinitely Morphing Object." It *never*
stops morphing so that it becomes a *fixed* element of R(0,1) so it could
be mapped to! It succeeds in being different than every element in the
mapping because it never stops changing. Once it stops changing, as is
shown by the chart in the above contest, an element of Set A or Set Rc
immediately maps to it! It cannot escape the above process and become a
unique and independent element of R(0,1) even if it becomes infinitely
wide.
The "Infinitely Morphing Object" Paradox:
In the DT, there was a claim that the "new number" is guaranteed to
be an element of R(0,1) *because* R(0,1) contains all permutations of N/o
digits. In the above mapping, any *fixed* "new number" is guaranteed to
be part of the mapping *because* the above mapping maps to all permuta-
tion of N/o digits! This is a paradox, because the "new number" is both
*out* of the mapping (i.e. because of the Diagonalizaton Technique) and
*in* the mapping (i.e. because of the mapping) at the same time, for the
same reason (i.e. all permutations of N/o digits)!
The above mapping is perfectly valid. The DTq simply adds a paradox
to a valid mapping. The DT cannot prove there is no mapping, because
there is a mapping! All the DTq can do is keep the "new number" alive
(i.e. different) *only* as long as it is morphing. The instant the "new
number" quits morphing, and becomes a *fixed* element of R(0,1), an el-
ement of either Set A or Set Rc can immediately map to it!
To find a "new number" which is an element of R(0,1), and not in the
above mapping, the following things must be done:
1) prove the new number is not in the above mapping,
2) prove the new number is a *fixed*, *unique*, *single* element of
R(0,1),
3) after the new number is *fixed*, it must be shown there cannot then
be a mapping from Set A or Set Rc to that number.
To be fair, *both* the mapping and the new number should be *fixed*
at the beginning of the contest and a third party should declare the win-
ner! Fair is fair. The above mapping works, no *fixed* "new number" can
be found not in the mapping!
To further prove the DTq is false, using Set Rc, a "new number" can
be created which is *not* in any simple listing of Set Rc, but the new
number would be in Set Rc because Set Rc contains all permutations of N/o
digits. This would mean Set Rc was uncountable. Set Rc, however, is
countable because it was created by subsets of the counting numbers
(Axiom #2).
SECTION 8) A Proof (Using a Mapping) That the Set of All Subsets of a
Countable Set is Countable
Kehr/(Rusty) Johnson Theorem: The Set of All Subsets of a Countable Set
is Countable.
The actual Kehr/Johnson Theorem is more comprehensive than the above
statement, but for the purposes of this paper, the above shortened
theorem is adequate. The set of all subsets of Set A will be used in
this proof.
For Finite Subsets of Set A:
A Delineation Mapping in finite space is a bijection which maps a
finite set of elements to and from a single element of Set A. Because
every element of Set A is *defined* to be finite, a set consisting of a
finite number of finite elements will yield a finite element of Set A, as
will now be shown. An example is the best way to explain the bijection:
Set n = {5, 6, 1, 4, 7} is mapped to and from:
Element of Set A = 11112,111112,2,1112,1111112
(note: the commas are added only for convenience)
Note that each element of the set is mapped to a string segment
which consists of (n-1) 'ones' followed by a single 'two.' These string
segments are then concatenated to form a single counting number. The '2'
acts as a delineator. So if there are 6 2s in a number, it came from a
set with 6 elements in it. It is obvious such a mapping can handle all
of the finite subsets of the Power Set and is unique in both directions,
as long as *only* the elements of Set A (i.e. the strings) which contain
only 1s and 2s, and have at least one 2, and terminate in a 2, are used.
For Infinite Subsets of Set A:
It is not necessary to create a bijection mapping in this part, only
a mapping from the counting numbers, and/or a set created by a mapping
from a subset of the counting numbers, to the infinitely wide Power Sets.
Notice that the above Delineation Mapping used elements of Set A
which had only 1s and 2s in them. Now let's consider the elements of Set
A which have only 4s and 5s in them. In fact, let us map the subset of
Set A consisting of all elements of Set A which have only 4s and 5s in
them, and which begin with a 7 (the place holder), to a countable set of
calculators (we will not need the elements which do not have at least a
countable number of 4s and a countable number of 5s in them, but that is
not important because the entire set is countable, so to use only part of
the elements would still make it a countable mapping).
As was mentioned, the place holder is the number '7.' We are es-
sentially creating a *binary* tree, after the place holder, using base 10
elements. Normally, a binary tree, which is constructed like a base 10
tree, would only have 0s and 1s in it, but our binary tree has 4s and 5s
in it. In other words, it is a binary tree with all permutations of an
infinite number of digit positions from a set of two elements {4, 5}, af-
ter the place holder. This is a countable subset of Set A with a count-
able width. This should be obvious by now because of prior discussions
about base 10 trees.
Now we will apply the same delineation technique as above from this
set of calculators to the infinite subsets of Set A. Because every el-
ement of Set A has a finite width (by definition), any infinite set of
elements of Set A will produce a countable (not uncountable and not fi-
nite) width string (i.e. a countable number of finite sets is countable).
The calculators are both countable in width and contain all permutations
of a countable number of digits from a set of two digits: {4, 5}, after
the place holder. It is clear that this delineation mapping will allow
all infinite subsets of Set A to be mapped to by a set which is known to
be countable.
For example, the set of all "even numbers" will be mapped to by:
.343334333334333333343333333334... (the '7' was replaced by a '.')
Since both the finite and countable delineation mappings are from
countable sets, the "set of all subsets" of a countable set is therefore
countable.
As above for the DT, the attempt to create a "new subset" after an
analysis of the above mapping would yield nothing but a contest of: "who
goes last." In other words, as with the DTq, the "new subset" is an "in-
finitely morphing object" and the above mapping stands.
Q.E.D.
SECTION 9) CONCLUDING COMMENTS OF PART I
The essence of this paper is a race. A race between two sets of
counting numbers: the counting numbers which map to the width of Set A
and the counting numbers which map to the width of Set Rnt. Axiom #1
states that in such a race there is no winner, only a tie. This being
the case, the height sets of Set A, which contain all permutations for
each level and are derived from the digit positions of Set A, are equal
to the permutations derived from the digit positions of Set Rnt. That
the width of these sets is equal is also proven in Theorem #3 using Set
R+.
Because both the height set of Set A, represented by Set Rc, con-
tains all permutations of N/o digits, and because Set Rnt also contains
all permutations of N/o digits, there cannot exist a representation in
Set Rnt which is not in Set Rc, unless Axiom #1 is false (which would al-
low one counting number - i.e. digit position - to exist in one countable
set which does not exist in another countable set). If there does not
exist one representation, there do not exist an uncountable number of
representations. Set Rc can therefore map to the union of Sets R+ and
Rnt.
The difference between Sets Rc and Rnt can also be represented by
the difference between a "growth" set and an "existing" set. The cardin-
ality of Set Rc grows from a cardinality of 10^1 to 10^2 to 10^3 to ...
to 10^N/o. The cardinality of Set Rnt "exists" at 10^N/o. But because
there is no "last" counting number, meaning there is no "last" digit po-
sition in Set Rnt, Set Rnt hardly has a "fixed" cardinality either. In
other words, to create a set which consists of all permutations of "n"
digits, n must be fixed first. But if n equals N/o, n is not fixed first
or ever. Therefore, technically speaking, Set Rnt *cannot* begin by de-
fining N/o as a fixed number, from which to create permutations. What
better way is there to create Set Rnt than by creating Set Rc!
The cardinality of permutation based sets can only be fixed if the
width of its elements is fixed first, something which Sets Rc and Set Rnt
find equally impossible at achieving. In other words, neither Sets Rc
nor Set Rnt have a "fixed" cardinality. A full discussion of omega,
which supposedly is fixed, is beyond the scope of this paper (I have
dealt with it in more lengthy papers), but omega was not created by Can-
tor to avoid Axiom #1, which is essentially a restatement of Cantor's
original assumption in the DT. Furthermore, *defining* all counting num-
bers to be finite in width fails as a definition if omega is a fixed num-
ber. No matter which way the definitions roll, there are clearly major
inconsistencies.
What makes these concepts even more obvious is that the gap in vol-
ume between a plate and a single neutron is finite. It makes no sense to
say that there are an uncountable number of *additional* non-terminating
decimal representations for each and every representation of Set Rc, when
both sets have the same width and both sets have the same representation
and permutation construction.
Because the real numbers are countable, it is clear that the under-
lying assumption of the Continuum Hypothesis is false. Whether this
makes the Continuum Hypothesis true or false is irrelevant, the Continuum
Hypothesis is a moot issue because the assumption upon which it rests is
false.
Part II (Rusty Johnson)
This part will be finished in July of 1994
Footnotes:
1) Dauben, Joseph Warren Dauben; GEORG CANTOR - HIS MATHEMATICS AND PHI-
LOSOPHY OF THE INFINITE; Princeton University Press; Princeton, NJ; Pages
96-99.
2) Ibid, Page 165-168
3) Ibid, Pages 166
