ContinuumHypothesis

The Continuum Hypothesis

H. Vic Dannon

[email protected] September 2007

Abstract

We prove that the Continuum Hypothesis is equivalent to the Axiom

of Choice. Thus, the Negation of the Continuum Hypothesis, is

equivalent to the Negation of the Axiom of Choice.

The Non-Cantorian Axioms impose a Non-Cantorian definition of

cardinality, that is different from Cantor’s cardinality imposed by

the Cantorian Axioms.

The Non-Cantorian Theory is the Zermelo-Fraenkel Theory with the

Negation of the Axiom of Choice, and with the Negation of the

Continuum Hypothesis. This Theory has distinct infinities.

Keywords: Continuum Hypothesis, Axiom of Choice, Cardinal,

Ordinal, Non-Cantorian, Countability, Infinity.

2000 Mathematics Subject Classification 03E04; 03E10; 03E17; 03E50; 03E25; 03E35; 03E55.

Contents

Preface………... 4

1 Hilbert’s 1st _{problem: The Continuum Hypothesis...6}

Preface to2………...14

2 Non-Cantorian Cardinal Numbers………...16

Preface to3………38

3 Rationals Countability and Cantor’s Proof...39

Preface to4………50

4 Cantor’s Set and the Cardinality of the Reals... 52

Preface to5………75

5 Non-Cantorian Set Theory………...76

Preface to6………....86

6 Cardinality, Measure, Category………....87

Preface to7………...100

Non-Preface

The Continuum Hypothesis says that there is no infinity between the

infinity of the natural numbers, and the infinity of the real numbers.

The account here, follows my attempts to understand the

Hypothesis.

In 2004, I thought that I found a proof for the Hypothesis. That

turned out to be an equivalent statement to the Continuum

Hypothesis.

That condition is the key to the Non-Cantorian Theory, but it took

me until 2007 to comprehend its meaning, and to apply it.

The first hurdle is to comprehend that the condition holds in

Non-Cantorian Cardinality.

The second hurdle is to realize that the Non-Cantorian cardinality is

imposed by an Axiom of the Non-Cantorian Theory.

The Non-Cantorian Theory is essential because Cantor’s Theory

Theorems, and borrows from the Non-Cantorian Theory Axioms

that do not hold in Cantor’s Theory.

At the end, Cantor’s Theory produces only one infinity.

To obtain distinct infinities, we need the Non-Cantorian Theory.

The equivalence between the Hypothesis and the Axiom of Choice,

renders the consistency result of Godel trivial.

The equivalence between the Negation of the Hypothesis, and the

Negation of the Axiom of Choice says that Cohen’s result must be

wrong. The mixing of the Axiom of Choice with its Negation must

lead to inconsistency.

The Hypothesis is the most illusive statement of the Axiom of

1

Hilbert’s 1

st

_{problem: The Continuum}

Hypothesis

H. Vic Dannon October 2004

Gauge Institute Journal, Volume 1, No 1, February 2005 Abstract: There is no set

X

with

 

₀

cardX



2

0.

Introduction

The continuum hypothesis reflects Cantor’s inability to construct a

set with cardinality between that of the natural numbers and that of

the real numbers. His approach was constructive, But if he was

right, such set cannot be constructed, and he needed a proof by

contradiction.

That contradiction remained out of reach at Cantor’s time. Even

when Hilbert presented Cantor’s Continuum Hypothesis as his first

problem remained open.

In 1963, Cohen proved that if the commonly accepted postulates of

set theory are consistent, then adding the negation of the hypothesis

does not result in inconsistency. [1, p.97]. That left the impression

that Hilbert’s first problem was either solved, or is unsolvable. But

it became commonly accepted that the problem was closed.

Not that the question was settled. According to [2,p.189],

“Mathematicians do not tend to assume the Continuum Hypothesis

as an additional axiom of set theory mostly since they cannot

convince themselves that this statement is “true” as many of them

have done for the axioms of ZFC including the axiom of choice.

However, a mathematician trying to prove a theorem will usually

regard a proof of the theorem from the generalized continuum

hypothesis as a partial success”.

Cohen result was interpreted to mean that there is another set

theory that utilizes the negation of the continuum hypothesis.

under the assumptions of Cantor’s set theory.

1. Proof

To understand the gap between cardinal numbers, it is natural to

examine the convergence

  

n_{0 0}0, and try to pinpoint where the jump from

 

₀

(

₀n

)

to

2 (

0



₀0

)

occurs. This idea1 motivates our proof.

By [3, p.173], the sequence

2

0 0

...

0n n

    

, converges to the series

2 0 0 0 0 1

...

n

...

n n  

     





.

The series has a well-defined sum



, which is a cardinal number that does not depend on the order of the summation.

By [3, p.174],

In particular, 0 0 0 0 1

...

n _{n N}



   _













. By [2, p.106], (or [3, p.183])

0

CardN

₀

n N

  



. Therefore, 0

0

2

CardN



_

_

_

_. That is, 0

2







.

Now, suppose that there is a set

X

with

0 0

cardX

2



 



.

Then, for any finite

n

,

2 3

0 0 0

...

0n

n

CardX

      

.

Tarski ([4], or [3, p.174]) proved that for any cardinal numbers, and

1

2

...

n

m

  

m

m



m

,

imply the inequality

1 2

...

n

....

m

  

m

m

 

m

. Since for any

n

n

cardX

 

,

by Tarski result

cardX



. Combining this with

0

2

cardX



,

yields by transitivity of cardinal inequalities [3, p. 147],

0

2





, which contradicts





2

0.

2. Discussion:

Why did Cantor fail to prove his hypothesis? Tarski result was not

partial sums

2

0 0

...

n0

   

,

that are the cardinality of all the roots of all the polynomials in

integer coefficients of degree



n

. In that context,

n

must be always finite, and the partial sum never exceeds



₀. Only when we take infinite

n

, we obtain terms such as

 

₀0

2

0



₀. The infinite summation over all

n

, is the key to our proof. In spite of the degenerate character of



n₀ , there is a jump to

2

0 that forbids a cardinal number betweenCardN , and CardR.

The gap between cardinal numbers may be traced to the jump

between finite cardinal numbers n, and the first infinite cardinal

0



. That incomprehensible jump may be the reason for the jump from the cardinal number

2 3 0 0 0

...

0n n0 0

      

to the cardinal 0

2

0 n  

  



Perhaps, there is no cardinality between the integers and the real

numbers, because there is no infinite cardinal number between

n

, and



₀. But an infinite cardinality between

n

, and



₀, will contradict the definition of



₀, as the first infinite cardinal.

Was Cohen right? In earlier stages of this work, we wondered

whether our proof for the Continuum Hypothesis proves Cohen

wrong. But our further studies affirm Cohen’s result. Under the

assumptions of Cantor’s set theory, we can prove the Continuum

Hypothesis. But with one assumption changed, we can construct a

Non-Cantorian set theory , where there is a set X with

CardN CardX CardR  . Cohen’s result predicts the vulnerability of Cantor’s set theory, and makes it easier for us to

References

[1] Machover, Moshe, Set Theory, logic, and their limitations. Cambridge Univ. press, 1996.

[2] Levy, Azriel, Basic Set theory. Dover, 2002.

[3] Sierpinski, Waclaw, Cardinal and Ordinal Numbers. Warzawa, 1958. [4] Tarski, A. “Axiomatic and algebraic aspects on two theorems on sums of

cardinals.” Fund. Math. 35 (1948), p. 79-104.

[5] Dannon, H. Vic. “The Continuum Hypothesis is equivalent to

 

2_{0 0}.” www.gauge-institute.org

Preface to 2

How can the Continuum Axiom be proved?

It is equivalent to the claim that the rationals cardinality is

CardN

, which Cantor considered a proven fact.

In7, we’ll see that this proven fact is a Cantorian Axiom.

Therefore, there is no proof that the rationals cardinality is

CardN

. In other words, if Cantor have proved that the rationals cardinality is

CardN

, then we have proved the Hypothesis.

We exhibit in 7 one Axiom that says that the rationals Non-Cantorian cardinality is greater than

CardN

, and another Axiom, that says that the rationals Cantorian Cardinality is

CardN

.

In2, I was not aware of the Non-Cantorian cardinality.

Furthermore, it is not clear in 2 why we must have the Non-Cantorian Set Theory.

In4, I proved that in Cantor’s Set Theory there is only one infinity. The different infinities are all Non-Cantorian.

2

Non-Cantorian Cardinal Numbers

by H. Vic Dannon May, 2005

Abstract:

 

2_{0 0} is an assumption of Cantor’s set theory, and we develop the arithmetic of Non-Cantorian cardinal numbers

Introduction: The Continuum Hypothesis says that there is no set

X

with

 

₀

cardX



2

0 . In 1963, Cohen proved that if the commonly accepted postulates of set theory are consistent, then the

addition of the negation of the hypothesis does not result in

inconsistency [1].

Cohen’s result was interpreted to mean that there is another set

theory that utilizes the negation of the Continuum Hypothesis.

However, sets with cardinality between the natural numbers and the

Recently, we proved the Continuum Hypothesis in Cantor’s set

theory [2]. But according to Cohen, the Continuum Hypothesis is an

assumption. How can an assumption be proved? Close scrutiny of

our proof will reveal that

  

2_{0 0} Continuum Hypothesis. We further show that the converse is also true, and that the Continuum

Hypothesis is equivalent to

 

2_{0 0}.

Thus, Non-Cantorian set theory is founded on the converse

assumption that

cardQ cardN



.

We first show that the continuum Hypothesis and

 

2_{0 0}, are equivalent.

1.

  

2_{0 0} Continuum Hypothesis:

We prove:

Assume

CardN 

2

CardN. Then



_CardN

2

_

_CardN

_

Continuum Hypothesis.

( )



By [3, p.173], the sequence

  2 3

...

 n

CardN



CardN



CardN

 

CardN

sums up1[3] to the series

  2 3

...

CardN



CardN



CardN



. By [3], the series has a well defined sum

   

CardN

CardN

CardN











CardN

 



CardN

 



CardN



...





,

which is a cardinal number.

By [3, p.174],





any component of the series. In particular,



is greater than the infinite product term



CardN

 

CardN

 

CardN



...













CardN

1 1 1 ....  



.

By [3, p.183],

Therefore,



_CardN

CardN





.

Or, using the product notation, by [4, p. 106], we obtain similarly,





CardN n N

CardN

CardN











. Since



CardN

CardN



2

CardN,

by transitivity of cardinal inequalities [3, p. 147], we conclude that

2

CardN





.

Now suppose that the continuum hypothesis does not hold, and there

is a set

X

with

2

CardN

CardN CardX





.

Since we assume 

CardN

2



CardN

, then for any

1,2,3...

n 

,

 2

...

_

_

n

CardN



CardN





CardN



CardX

. Tarski ([5], or [3, p.174]) proved that

,

,

, and

1 2

..

n

m m m

m

are any cardinal numbers so that for any

n 

1,2,3....

,

1

2

...

n

m m

  

m



m

, then,

1

2

...

n

....

m m

    

m

m

. By Tarski result,  2

...





n

...

CardN



CardN





CardN





CardX

. Since we assume

CardX 

2

CardN , by transitivity of cardinal inequalities [3, p. 147], 

_CardN

2





3

.

_...

2

CardN

CardN

CardN

















That is,

2

CardN



,

which contradicts



2

CardN .

This completes the proof that



_CardN

2

_

_CardN

_

Continuum Hypothesis.



( )



By [3, p.155], For any cardinals

m n

, ,

,

m

₁

, and

n

₁

,

m n



, and

m

₁



n

₁



m m

  

₁

n n

₁. Since

2

CardN

CardN 

, we have

2

CardN

2

CardN

CardN CardN







. But

2

CardN



2

CardN



2

CardN CardN



2

CardN. Hence,



CardN 

2

2

CardN. Therefore,

 2  2

₂

CardN

CardN



CardN



CardN



CardN



. Namely, if

CardN





CardN

2, the rationals serve as a set

X

which cardinality is between

CardN

, and

CardR

, and the Continuum Hypothesis does not hold. That is,

 2

CardN



CardN



Hypotesis Negation

Thus, the Hypothesis implies 

CardN

2



CardN

.



2.

 

2_{0 0} is an assumption.

The equivalence of the continuum hypothesis with the countability

of the rationals, indicates that the countability of the rationals is an

assumption, just as the continuum hypothesis is, and as such it

cannot be proved.

Consequently, Cantor’s proof of it, that does not utilize the

Continuum Hypothesis, has to raise doubts.

Indeed, if we could prove that Q is countable, then Cantor’s

Continuum Hypothesis will be a fact, and Cantor’s set theory would

be the only valid set theory, in contradiction to Cohen’s result.

just like the continuum hypothesis.

If the countability of Q is assumed, we obtain Cantor’s set theory,

and Cantor’s rules for cardinal numbers.

The assumption

 

2_{0 0} is an Axiom of cantor’s set theory, that can replace the Continuum Hypothesis, similarly to the

well-ordering Axiom that can replace the Axiom of Choice.

If it seems believable that the rationals can be arranged in a

sequence, note that the Axiom of Choice too seems believable, but

has to be assumed.

Non-Cantorian set theory is founded on the assumption that Q is

uncountable. and

 

₀ 2₀.

Cohen result alludes to a Non-Cantorian set theory where the

negation of Cantor’s Continuum Hypothesis holds. Then, Q must

be uncountable, and

cardN



cardQ cardR



.

That is, the set that eluded Cantor is the set of rational numbers

Q

. The Non-Cantorian set theory distinguishes the cardinality of

Q

from the cardinality of

N

, and from the cardinality of

R

.

3. Cantor’s Zig-Zag Proof Well-Orders Q, but does it arrange Q in a Sequence?

Non-Cantorian set theory was indicated in 1963 by Cohen, but was

born before 1915 when Cantor well-ordered2Q, and believed that he

arranged it in a sequence as well, so that he may conclude

that

 

2_{0 0}.

Given two infinite sequences of numbers

1 2

, ,... ,...

n

a a

a

1 2

, ,... ,...

n

b b b

we can arrange them in one sequence

1 1 2 2

, , , ,... , ,...

n n

a b a b a b

The method applies to n sequences, but not to an infinite number of

Cantor reasoned [5] that we can zigzag through an infinite triangle

of rationals, starting from the vertex of the triangle, and progressing

towards the triangle’s base.

But Scanning the infinite triangle by zigzag lines, forces us

eventually to go through an infinite number of infinite sequences.

The zig-zag starts with finite sequences, but there are infinitely

many infinite sequences above the infinite triangle base.

* * * * * ... ... * ... * ... ... ... ... * * * * ... ... ... * ... * ... ... ... ... * * * ... ... ... * ... * ... ... ... ... * * ... ... .... * ... * ... ... ... ... * ... ... ... * ... * ... ... ... ... * ... * ... * ... * ... * ... *                              

Cantor did not draw the zigzag. Instead, for each rational





,

he defined a variable

  

 

,

that is supposed to remain well defined, but long before the triangles

base is reached,



, and the correspondence between N, and

Q breaks down.

Cantor’s zig-zag well-orders Q. Depending on the direction of the

zig-zag scan, we have either

1

1

1

1



 







 

 





, or

1

1

1

1



 







 

 





.

However, Well-Ordering does not guarantee arranging in a

sequence. For instance, the real numbers, may be well-ordered, but

Specifically, the real numbers between 0, and 1, can be arranged in

an infinite triangle using their binary representation. The first row

will have 0 and 1, the second will have 00, 01, 10, and 11, …. The

nth row has

2

n numbers. The well ordering will be defined by zig-zag from one row to the next, towards the base of the infinite

triangle. * * * * * * * * * ... ... ... ... * ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ...                 

If Cantor’s diagonal proof is correct, this Well-Ordering zig-zag

does not arrange the real numbers in a sequence. But it does not

the rational numbers.

In conclusion, we maintain that Cantor’s zig-zag may not arrange

the rationals in a sequence.

On the other hand, the assumption that Q may be sequenced may

not be easily disproved.

Consequently, we may assume that the rational numbers can be

arranged in a sequence, an assumption that is equivalent to the

Continuum Hypothesis, and implies Cantor’s set theory, with no

inconsistencies or contradictions.

Otherwise, we may assume the negation of the Continuum

Hypothesis, and recognize CardQ as a cardinal number different

from either CardN, or Card R. Then, the rationals may present the

different infinite cardinality that Cantor himself was seeking.

We proceed to develop the arithmetic of non-Cantorian cardinal

numbers.

Non-Cantorian set theory is founded on the assumption that Q is uncountable. Then, 2 0 0

 

, and

cardN



cardQ cardR



.

That is, the set that eluded Cantor is the set of rational numbers

Q

. We proceed to develop the arithmetic of non-Cantorian cardinal

numbers.

Theorem 1

 

₀n ₀n1, for

n 

1,2,3...

.

Proof

For

n 

1

,

 

₀ 2₀,is the non-Cantorian assumption.

By induction on

n

, similarly to our proof that

2 3

0 0

 

, we conclude that for all finite n,

1

0 0

n n 

The



n₀ are newly added cardinalities that in the Cantorian set theory degenerated into



₀.

Theorem 2.

  

₀n ₀0

2

0 .

Proof

For each finite n,



n₀ is the cardinality of all the roots of all the polynomials in integer coefficients of degree n. Therefore,

0

0

2

n

_cardR



 



.

Since



n₀ is monotonically increasing, 0 0 0 n 

 

. Since

 

₀

2

, 0 0 0

2

 

 

. To show that 0 0 0

2

 

 

,

0 2 3

1 2

_{    }

3

...

_n

n

2

 _. By Tarski, [5] 0 2 3

1 2

     

3

...

n

n

... 2

 That is, 0 0

2

cardN n n N

n

 









. Thus, 0 0 0

2

 

 

, and 0 0 0 0

2

n  

  

.



Theorem 3

 

n

cardR

Proof 0 0 0

(2 )

2

2

n  n n 

 





.



Thus, while

 

2_{0 0} is an assumption,

 

2 holds in Non-Cantorian Set Theory too.

Theorem 4

2

0



n

0



cardR

, for all finite

n 

2,3,4...

.

proof 0 0 0 0 0

2



_

n



_{ }



2

_.



Theorem 5

2

0



2

20. Proof If

2

0



2

02, then 2 2 2 2 0 0 0 0 0 0 0 0 0

2



_

(2 )

 

_

(2 )

 

_

(2 )

 

_

(2 )

  3 0 0 0 0

(2 )

 

.... (2 )

 n

...









0 0 2 1 2

(2 )

 

2









,

since by Theorem 2,

 

n₀

2

0. This contradicts

c 

2

c.

Theorem 6

2

n0



2

n0 1.

By the same argument of Theorem 5. Theorem 7

2

n0



2

. Proof 0 0 0 0 2

2

n

_

2



_

2



_

2

_.



The

2

0n are newly added cardinalities, that in the Cantorian set theory degenerated into

2

0.

5. Rational and Irrational Numbers:

Theorem 8

cardR cardQ



, for Cantorian or Non-Cantorian cardinals.

Proof

Based on the preceding theorems we have

0 0 2 0 0

2

cardR

_



_{  }



cardQ

_.



This proof replaces Cantor’s diagonal proof that does not apply to

0

cardQ 

.

Recall that Cantor’s diagonal proof exhibits the diagonal element as

the one not counted for, to obtain a contradiction. It leaves a

lingering doubt why that one missing element cannot be added to

the listing, that will still remain countable. In fact, that listing of the

real numbers will remain countable after countably many such

elements will be exhibited and added to it, while it seems that only

one such element may be produced.

Theorem 9

Card Irrational Numbers

(

)



CardR

, for Cantorian or Non-Cantorian cardinals.

Proof

By [3, p. 417], the Axiom of Choice implies for any cardinal

numbers

m

, and

n

, so that

m n



m n m

 

. Since by Theorem 8

cardR cardQ



6. Algebraic and Transcendental numbers

Theorem 10

Card Al ebraic Numbers

( g

)



CardR

, for Cantorian or Non-Cantorian cardinals.

Proof

The cardinality of the set of all the roots of all the polynomials in

integer coefficients of degree at most n is

2 3

0 0 0

...

0n

    

.

Therefore, for any

n N



,

2 3

0 0 0

...

n0

Card Algebraic Numbers

(

)

     

.

By Tarski [5],

2

0 0

...

n0

...

Card Algebraic Numbers

(

)

     

. Since 0 0 2 0 0

...

n0

...

0

2



      

,

by transitivity of cardinal inequalities,

0

2





Card Algebraic Numbers

(

)

. On the other hand,

0

(

)

2

Card Algebraic Numbers

_

CardR

_

_.

Consequently,

0

(

)

2

Card Algebraic Numbers



CardR



.



Cantor had

     

₀ 2₀

...

n₀

...

₀. But In chapter 4 we show that for Cantor’s Cardinals,

2

0



₀.

Theorem 12

Card transcendental numbers

(

)



CardR

, for Cantorian or Non-Cantorian cardinals.

Proof

By [7], if

a

is non-zero, real algebraic number, then

e

a is a transcendental number. The function

a

a



e

is an injection from the algebraic numbers into the transcendental

numbers.

Therefore,

(

)

(

)

Card Algebraic Numbers

CardR





.

Thus,

(

)

Card transcendental numbers



CardR

.



References

[1] Machover, Moshe, Set theory, logic, and their limitations. Cambridge U. Press, 1996. p. 97.

[2] Dannon, H. Vic, Hilbert’s 1st Problem: Cantor’s Continuum Hypothesis. ABSTRACTS of

Papers Presented to the American Mathematical Society. Vol. 26, Number 2, issue 140, p. 405.

[3] Sierpinski, Waclaw, Cardinal and Ordinal Numbers. Warszawa, 1958 (or 2nd edition)

[4] Levy, Azriel, Basic Set Theory. Dover, 2002.

[5] Tarski, A., Axiomatic and algebraic aspects on two theorems on sums of cardinals. Fund. Math. 35 (1948), p.79-104.

[6] Dannon, H. Vic, Cantor’s Proof of Rationals Countability,www.gauge-institute.org [7] Siegel, Carl, Ludwig, Transcendental Numbers. Princeton University Press, 1949, p.17

Preface to 3

The response to my claim in2 that the rationals cardinality need not be

CardN

, was invariably

“But Cantor’s mapping is a bijection…”

One critique pointed out to me that

“

( , )

i j

1

i j

 

is a one-one mapping….” I found out that Cantor did not exhibit a one-one mapping.

Later generations drew the famous Zig-Zag, and everyone believed

in the Zig-Zag, and in the effective countability of the rationals.

In 3, I supply two proofs for the sequencing of the rationals. The first proof is based on Tarski. The second proof constructs an

injection, that Cantor failed to find, from the rational numbers into

the natural numbers.

Indeed, we can sequence the rationals, but we cannot prove what

In 7 we’ll see that by one Axiom, the Cantorian cardinality of the rationals numbers is

CardN

, and by another Axiom, the Non-Cantorian cardinality of the Rationals is greater than

CardN

. It is plain, that Cantor was not aware of any of that.

3

Rationals Countability and Cantor’s Proof

H. Vic Dannon January, 2006

Gauge Institute Journal, Volume 2, No 1, February 2006

Abstract: Cantor’s proof that the rational numbers are countable

uses a mapping that is not one-one. Thus, the countability of the

rationals was not proved by Cantor.

We prove by cardinal number methods, using a result of Tarski, that

the rationals are countable. We confirm this by exhibiting a one-one

mapping from the rationals into the natural numbers.

1. Cantor’s Mapping is not One-One.

Cantor’s proof appears in [1]. Cantor wrote

“By (6) of

§

3,

 

₀



₀ is the cardinal number of the aggregate of bindings

where



and



are any finite cardinal numbers which are independent of one another. If also



represents any finite cardinal number, so that

{ }



,

{ }



, and

{ }



are only different notations for the same aggregate of all finite numbers, we have to show that

{( , )} { }







.

Let us denote

 



by



; then



takes all the numerical values

2,3,4,...

and there are in all



1

elements

( , )



for which

  

 

, namely:

(1,



1)

,

(2,



2)

,…,

(



1,1)

.

In this sequence imagine first the element

(1,1)

, for which

2



, put , then the two elements for which



3

, then the three elements for which



4

,

and so on. Thus, we get all the elements

( , )



in a simple series:

and here, as we easily see, the element

( , )



comes at the



th place, where (9)

(

1)(

2)

2

 

 

 

 

 

 

.

The variable



takes every numerical value

1,2,3,...

, once. Consequently, by means of (9), a reciprocally univocal relation subsists between the aggregates

{ }



and

{( , )}



.”

Clearly, the variable



takes several times some of numerical value

1,2,3,...

, The mapping is not one-one.

We have,

1



,



1

,



1

, and

(1,1)



1

.

1



,



2

,



2

, and

(1,2)



2

.

2



,



1

,



2

, and

(2,1)



2

.

1



,



3

,



4

, and

(1,3)



4

.

2



,



2

,



4

, and

(2,2)



4

.

1



,



4

,



7

, and

(1,4)



7

.

2



,



3

,



7

, and

(2,3)



7

.

3



,



2

,



7

, and

(3,2)



7

.

4



,



1

,



7

, and

(4,1)



7

.

Apparently, Cantor expected his mapping to be a bijection, did not

check it, and did not see that it is not even one-one.

It is well known [2] that a one-one mapping is required to establish

that

cardQ cardN



. Thus, Cantor’s claim is unfounded, and his use of

 

2_{0 0}, amounts to adding another axiom to his set theory.

2. Proof of Rationals Countability by Tarski result.

We use the graphical interpretation of Cantor’s proof by a zig-zag

through an infinite triangular matrix of rationals. While the zig-zag

by itself does not prove the countability, it is useful to clarify our

1/1.

The second line has two rationals,

1/2, and 2/1.

………

The n-th line has n rationals,

1/n, 2/(n-1),…,(n-1)/2, n/1, 1 1 1 1 1 5 1 2 3 4 2 2 2 2 1 2 3 4 3 3 3 1 2 3 4 4 1 2 5 1 ... ... * ... * ... ... ... ... ... ... ... * ... * ... ... ... ... ... ... ... * ... * ... ... ... ... ... ... .... * ... * ... ... ... ... ... ... ... * ... * ... ... ... ... * ... * ... * ... * .                             .. * ... *  

Summing the number of the rationals along the zig-zag, for

1,2,3....

Thus, 0

2(1 2 3 ....

    

n

)

. That is, 2 0

n n

 

. (2) Tarski ([3], or [4, p.174]) proved that

for any sequence of cardinal numbers,

m

₁

,

m

₂

,

m

₃

,…, and a cardinal

m

, the partial sums inequalities

1

2

...

n

m m

  

m



m

, for

n 

1,2,3....

imply the series inequality

1

2

...

n

....

m m

    

m

m

.

Applying to (1),

0

1 2 3 ....

     

n

...

. Regarding (2), this says

2

0 0 0

  

.

Since

  

2_{0 0} 2₀, by transitivity of cardinal inequalities [3, p. 147],

2 0 0

 

. Since

 

₀ 2₀, 2 0 0

 

.

3. Proof of Rationals Cantorian Countability by injection

Aided by the zig-zag listing of the rationals, we produce a one-one

mapping from the rationals into the natural numbers. We construct

our mapping with numerical examples. Then we give the general

formula.

The first line in the zig-zag has one rational

1/1

which we assign as follows

1

1

_{1 2}

₃

1   

.

The second line in the zig-zag has two rationals,

1/2, and 2/1,

2

1

_{1 2}

₅

2   

, 2

2

_{2 2}

₆

1   

.

The third line in the zig-zag has three rationals

3/1, 2/2, 1/3,

which are assigned as follows

3

3

_{1 2}

₉

1   

, 3

2

_{2 2}

₁₀

2   

, 3

1

_{3 2}

₁₁

3   

.

The fourth line in the zig-zag has four rationals

1/4, 2/3, 3/2, 4/1

which are assigned as follows

4

1

_{1 2}

₁₇

4   

, 4

2

_{2 2}

₁₈

3   

,

4

3

_{3 2}

₁₉

2   

. 4

4

_{4 2}

₂₀

1   

.

The fifth line in the zig-zag has five rationals

5/1, 4/2, 3/3, 2/4, 1/5,

which are assigned as follows

5

5

_{1 2}

₃₃

1   

, 5

2

_{2 2}

₃₄

4   

, 5

3

_{3 2}

₃₅

3   

. 5

2

_{4 2}

₃₆

4   

. 5

1

_{5 2}

₃₇

5   

. If

1

2

m n

  

even



k

,

1 2

m n

  

k

m n

  

1 2

k

rationals

1/n, 2/(n-1),…(n-1)/2, n/1,

which are assigned as follows

2

1

_{1 2}

_k

n  

, 2

2

_{2 2}

1

k

n



 

, ………. 2

1

₂

_{1 2}

2

k

n

_{   }

_k

_, 2

2

2

1

k

n

_

_k

_

. If

1

2

1

m n

  

odd

 

k

,

the

m n

   

1 2

k

1

zig-zag line has the

1 2

1

m n

   

k

rationals

m/1, (m-1)/2,…, 2/(m-1),1/m,

which are assigned as follows

2 1

1 2

k

2 1

1

_{2 2}

2

k

m

  

_, ………. 2 1

2

₂

₂

1

k

k

m







, 2 1

1

₂

_k

_{1 2}

_k

m



 

 .

This defines a one-one function from the rationals into the natural

numbers.

References

[1] Cantor, Georg, Contributions to the founding of the theory of Transfinite Numbers, p.107. Open Court Publishing 1915, Dover 1955.

[2] Lipschutz, Seymour, Theory and problems of Set Theory and Related Topics, McGraw-Hill, 1964.

[3] Tarski, Alfred, Axiomatic and algebraic aspects of two theorems on sums of cardinals, Fundamenta Mathematicae, Volume 35, 1948, p. 79-104. Reprinted in Alfred Tarski Collected papers, edited by Steven R. Givant and Ralph N. McKenzie, Volume 3, 1945-1957, p. 173, Birkhauser, 1986.

[4] Sierpinski, Waclaw, Cardinal and Ordinal Numbers. Warszawa, 2ndedition, 1965, (Also in the 1958, 1st edition).

Preface to 4

The conclusion of3 was that the rationals can be sequenced.

Not knowing that the Cantorian Cardinality of the rationals is

CardN

by an Axiom, I accepted that the rationals cardinality was proven to be

CardN

.

Thus, I had to conclude that the Hypothesis was a proven fact, and

there was neither Non-Cantorian Theory, nor Non-Cantorian

cardinals.

However, Cantor’s claim that his mapping was one-one puzzled me,

and the question on my mind was

“What else did he do wrong…”

Taking a second look at the Cantor set, I found out that its properties

were as advertised. But Cantor did not take full advantage of them.

The construction of the Cantor set produces a set of rational

numbers with cardinality

2

0. These numbers can serve as a range for an injection from the real numbers to conclude that there are no

Consequently, in Cantor’s Theory the only infinity is

CardN

. Cantor’s definition of cardinality that does not distinguish between

the infinities of integers and rationals, does not distinguish between

the infinities of the integers and the reals.

In 7 we show that the existence of unique infinity in Cantor’s Theory is an Axiom equivalent to the Hypothesis.

Therefore, the inequality

CardN



CardR

is an Axiom equivalent to the Negation of the Hypothesis.

Thus, Cantor’s “Diagonal Argument” is a “proof” of a

4

Cantor’s Set and the Cardinality of the Reals

H. Vic Dannon February, 2006

Gauge Institute Journal, Volume 3, No 1, February 2007,

Introduction:

The Cantor set is obtained from the closed unit interval [0,1] by a

sequence of deletions of middle third open intervals. The remaining

set contains no interval, but has cardinality equal to

2

CardN

CardR 

.

If Cantor was attempting to find a cardinality between

CardN

, and

2

CardN _{, that set might have led him to his Continuum Hypothesis}

that there is no set

X

with

CardN



CardX



2

CardN.

But if the cardinality of such a meager set equals

Card

(0,1)

, how large is

Card

(0,1)

?

1 issue: Card(N)<Card(R)

To prove that

CardN



CardR

,

Cantor listed the reals in

(0,1)

, presumed to be countably many, and exhibited the diagonal element as the one not counted for, in the

list.

By [1, p. 57], every real number between 0, and 1, has a unique

infinite decimal representation

( ) ( ) ( ) 1n n n2 3

....

c c c

Missing from that listing is the real number

1 2 3

....

c c c

where ( )

0, if

n

0

n n

c



c



, and

c

_n



1, if

c

( )_nn



0

. What is so crucial about one missing element out of infinitely

many? Why cannot we add the one missing diagonal element to the

additions.

Two infinite sets have the same power, even if one set is “half” of

the other. For instance, the natural numbers, and the odd natural

numbers have the same power. For infinite sets, we can tolerate the

missing of countably many elements long before we conclude a

contradiction. In other words, the diagonal proof does not apply

credibly to infinite sets.

In this paper we disprove Cantor’s claim that

CardN



CardR

.

2nd issue: Card(A)<Card( P(A)).

The inequality

2

CardN

CardN 

,

seems like a wish modeled after the finite case

2

n

n 

. But as

n  

, we have

lim

_{n }

lim2

n _.

2

CardN

CardN 

indeed strict?

By [1, p. 87], Cantor proved that

The set of all subsets of any given set A is of greater power than the set A

The proof aims to show that

( )

( ( ))

Card A



Card P A

,

in order to conclude that for every cardinal number there is a greater

cardinal number, and to establish

( )

( ( ))

( )

Card N



Card P N



Card R

.

But Cantor’s proof uses the concept of set of sets which for infinite

sets is not well-understood, since it may lead to the Russell

paradoxial set.

Russell (1903) defined his set

y

by

x y

  

x x

. Then, in particular,

which is a contradiction.

In fact, as pointed out in [2, p.87], if we apply Cantor’s theorem to

the universal class of all objects V, every subset of V is also a

member of V, and we have

( )

( ( ))

Card V



Card P V

.

Avoiding this fact by claiming that V is not a set, while leaving the

definition of what is a set vague enough to suit other results, does

not establish the credibility of Cantor’s claim. Perhaps, for infinite

sets strict inequality does not exist, and we have only

( )

( ( ))

Card A



Card P A

.

In this paper we prove that

CardN



Card P N

( ( ))

.

3rd issue: Card(Cantor set)=Card(R)

The association of CardR with the “Power of the continuum”, gives

the impression that CardR must be greater than CardN.

But Cantor’s set, which cardinality equals to CardR, is no

Cantor’s set demonstrates that cardinality fails to distinguish

between intervals, and sets with no intervals. The cardinality of

(0,1)

equals that of the meager Cantor set. Thus, it is conceivable that Card

(0,1)

may not be any greater than Card(Q).

We present six different proofs of the equality

(0,1)

( )

Card



Card Q

. Our first proof constructs an injection form the real numbers into the rationals numbers.

1. Proof by injection that Card(0,1)=Card(Q)

We list the real numbers in

[0,1]

, in the dictionary order, using their binary representation.

The first row has the sequence that starts with 0, and the sequence

that starts with 1, each followed by infinitely many zeros.

1,1

(0,0,0,..., 0,0,0...)



x

,

1,2

(1,0, 0,...,0, 0,0...)



x

2,2

(0,1,0,..., 0...) x



, 2,3

(1,0, 0,...,0,...)



x

2,4

(1,1,0,..., 0,...) x



The third row has the

8

sequences

3,1

(0,0,0,0,...)



x

, 3,2

(0,0,1,0,...) x



, 3,3

(0,1,0, 0,...) x



, 3,4

(0,1,1,0,...) x



, 3,5

(1,0, 0,0,...) x



, 3,6

(1,0,1,0,...)



x

, 3,7

(1,1,0,0,...)



x

, 3,8

(1,1,1,0,...)



x

. The n-th row lists the

2

n sequences that start with

,1

(0,0,0,0,...)



x

n , and end with

,2

(1,1,1,1,...1,0,...)



x

_n n.

The



₀ row will list the

2

0sequences that represent the binary expansions of all the real numbers in the interval

(0,1)

.

The sequences in each row are rational numbers, but the infinitely

many rows contain all the real numbers. Denote

,1 ,2 ,3 ,2

{ ,

,

,....

n

}

n n n n n

X



x x x

x

Then, 1 2

...

n n 1

...

X



X

 

X



X





We list the sequences=numbers as follows

0 1

00 01 10 11

000 001 010 011 100 101 110 111 ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ...

We are guaranteed that this listing will enumerate all the real

numbers between 0, and 1.

1

[0,1]

n n

X

 





2 /2

of digits, in one of the rows. But the irrational number

2 /2

is only in the last row.

The last row has

infinitely many sequences with a single 1, and infinitely many zeros,

infinitely many sequences with two 1’s, and infinitely many zeros,

infinitely many sequences with three 1’s, and infinitely many zeros,

………

as well as

infinitely many sequences with one, and infinitely many 1’s,

infinitely many sequences with two zeros, and infinitely many 1’s,

infinitely many sequences with three zeros and infinitely many 1’s

………

We proceed to exhibit an injection from the real numbers into the

rationals:

We want to map each real number one-one to a distinct rational

number. All the rationals in the range of the map, have to be

produced in the generation of the Cantor set [3].

The Cantor set is obtained from the closed unit interval [0,1] by a

sequence of deletions of middle third open intervals.

First, we delete the open interval (1/3,2/3).

The numbers left in the intervals [0,1/3], and [2/3,1]

have either 0, or 2 of the fraction 1/3 in their expansion in base 3.

The two rational endpoints

1,1

1

3 y



, and

2

3



y

1,2

remain in the Cantor set after indefinitely many deletions.

We denote





1 1,1 1,2

,

Y



y y

.

Second, we remove the open intervals (1/9,2/9) and (7/9,8/9).

The numbers left in the intervals

[0,1/9], [2/9,1/3], [2/3,7/9], and [8/9,1],

have either 0, or 2 of the fraction 1/9 in their expansion in base 3.

2,1

1

9



y

,

2

9



y

2,2,

7

9 y



2,3,

8

9 y



2,4,

remain in the Cantor set after indefinitely many deletions.





2 2,1 2,2 2,3 2,4

,

,

,

Y



y y y y

Third, we remove the open intervals

(1/27,2/27), (7/27,8/27), (19/27,20/27), (25/27,26/27).

The numbers left in the 8 closed intervals have either 0, or 2 of the

fraction 1/27 in their expansion in base 3.

The eight rational endpoints

3,1 1 27 y , 272 y3,2, 3,3 7 27 y , 278 y3,4, 3,5 19 27 y , 2027 y3,6, 3,7 25 27 y , 2627 y3,8,

remain in the Cantor set after indefinitely many deletions.





3 3,1 3,2 3,3 3,4 3,5 3,6 3,7 3,8

,

,

,

,

,

,

,

In the n-th step, we remove

2

n1 open intervals leaving

2

n closed intervals, which numbers have either 0, or 2 of the fraction

1/ 3

n in their expansion in base 3.

The

2

n rational endpoints of the removed open intervals, remain in the Cantor set after indefinitely many deletions.

1 n n

Y

 



is a subset of the rationals in [0,1] because the midpoints of the removed intervals, ½, 1/6, 5/6,.. are not in any

Y

_n.

We define a one-one function from the real numbers listed in

dictionary order, onto the rational endpoints of the Cantor set.

The two reals in

X

₁ are assigned by their listing order to each of the two rational endpoints produced in the first deletion 1/3, and 2/3, by

the bijection

1

:

1 1

f X



Y

(0,0,0,..., 0,0,0,....)



1/ 3

,

The four reals in

X

₂ are assigned by their listing order to each of the four endpoints of the two deleted open intervals,

1/9, 2/9, 7/9, 8/9.

by the bijection

2

:

2 2

f X



Y

.

The eight reals in the

X

₃ are assigned by their listing order to each of the eight endpoints of the four deleted open intervals,

1/27, 2/27, 7/27, 8/27, 19,27, 20/27, 25/27, 26/27.

by the bijection

3

:

3 3

f X



Y

The

2

n reals in

X

_n are assigned by their order to each of the

2

n endpoints of the

2

n 1 deleted open intervals, by the bijection

:

n n n

f X



Y

. Now define 1 1

:

n n n n

f



X



Y











so that

f

_Xn



f

_n

The mapping

f

is one-one from the real numbers in 1 n n

X

 



onto the rational numbers in

1 n n

Y

 



. Consequently, 1

[0,1]

n n

CardR Card

Card



X









1 n n

Card Y



CardQ CardR











Hence 2 0

[0,1]

Card



CardQ



.

2. Proof by Tarski result for Cardinals.

The 1st row in the dictionary listing of the reals has

2

1 reals. The 2nd row has

2

2 reals.

The 3rd row has

2

3reals. ……….

………..

Summing the number of the reals along the listing, for

1,2,3....

n 

, 2 3 0 0 0 0 0

2 2

_{         }

2

.... 2

n

...

_. That is, 2 3 2 0

2 2

_{    }

2

.... 2

n

Tarski ([4], or [1, p.174]) proved that for any sequence of cardinal

numbers,

m

1

,

m

2

,

m

3

,…, and a cardinal

m

, the partial sums inequalities

1

2

...

n

m m

  

m



m

, for

n 

1,2,3....

imply the series inequality

1

2

...

n

....

m m

    

m

m

. Applying Tarski result, we obtain

2 3 2 0

2 2

     

2

.... 2

n

...

. Now,





0 2 1 2 2_{    }.... 2n ... lim 2n _{ }2 lim 2n _2_.

Hence,

0 2

0

2





. On the other hand,

0 2₀

2





CardR CardQ







, Consequently, 0 2 0

2



_

_.

_

3. Proof by Cardinals and Ordinals

By [1, p.277], every ordinal number



has a next ordinal number

1



 



,

and no intermediate ordinal number



with



  

1

 

. The ordinal



7

is preceded by



6

, and is classified as 1st kind.

The smallest ordinal number that is not preceded by any ordinal is

classified as 2ndkind, and is denoted by

increasing transfinite sequence of ordinal numbers. In particular,

lim 2

n n 





_ .

By [1, p. 318, Theorem 1], the function

( ) 2

n

f n 

is continuous in

n

, and lim

lim 2

n

2

n n

2

n   



 







. That is,

2





Now, by [2, p. 88, Corollary 2.19]



is a cardinal number. By [2, p.90, Corollary 2.33], 0



 

. Thus,



2

says 0 0

2



 

.

By a Theorem of Schonflies (1913) [2, p. 126, Theorem 2.11], for

ordinal numbers



, and



( ) max(

( ),

( ))

Card







Card



Card



. Therefore,

0

(2 ) max(

(2),

( ))

Card



_

Card

Card

_

_

_.

On the other hand, by [2, p.126, (2.9)], exponentiation is a repeated

multiplication , and for all



, and



  











. Hence,

2

2

n   





. Therefore,

(2 )

2

n

Card



Card

 





.

Now, by [2, p. 106, proposition 4.15], if

a

is a well ordered cardinal,

( ) Card u

a



a

Therefore, 0 ( )

2 2

Card

2

n    







. In conclusion, 0 0

2



 

.

5. Proof by Effective Countability of Ordinals

By the axiom of Choice, the set of ordinals up to the ordinal

2

is well ordered, by magnitude, and is sequenced. Therefore, it is well

known to be effectively countable [5], and its power is



₀.

6. Proof by Tarski result applied to Algebraic Numbers

 

2

CardN

_

_CardN

CardN

...

CardN CardN CardN









  2 3

_...

CardN

CardN

CardN









Now, since

  2 3

_...

then, for any

n 

1,2,3....

 2

_...

 n

CardN



CardN

 

CardN



cardN

. Therefore, By Tarski result,

  2 3

_...

CardN



CardN



CardN

 

CardN

. Thus,

2

CardN

_

_CardN

_.

_

7. Transcendental and Irrational numbers

The rationals are effectively countable, and

 

2_{0 0}. For the algebraic numbers,

0

CardN

Card Algebraic Numbers

(

)

CardR

0

 







.

Hence,

0

(

)

Card Algebraic Numbers 

.

By [6], if

a

is non-zero, real algebraic number, then

e

a is a transcendental number. The mapping

a

a



e

is an injection from the algebraic numbers into the transcendental

numbers. Therefore,

0

CardR Card Transcendental Numbers

(

)

 



0

(

)

Card Algebraic Numbers





.

Thus,

0

(

)

Card Transcendental Numbers 

. Finally,

0

CardR Card Irrational Numbers

(

)

 



0

(

)

Card Transcendental Numbers





.

Hence,

0

(

)

Card Irrational Numbers 

.

8. Non-Cantorian Set Theory

Since Cantor’s set theory leads to a single infinity

 

₀

2

0, we may wish to develop non-Cantorian set theory with more infinities.