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Gauge Institute Journal, Volume 4, No 1, February 2008,

H. Vic Dannon

The Continuum Hypothesis H. Vic Dannon [email protected] September 2007

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Gauge Institute Journal, Volume 4, No 1, February 2008,

H. Vic Dannon

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.

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Gauge Institute Journal, Volume 4, No 1, February 2008,

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Contents Preface……………………………………………………...... 4 1 Hilbert’s 1 st problem: The Continuum Hypothesis...6 Preface to 2…………………………………………………...14 2 Non-Cantorian Cardinal Numbers………………….......16 Preface to 3……………………………………………………38 3 Rationals Countability and Cantor’s Proof....................39 Preface to 4……………………………………………………50 4 Cantor’s Set and the Cardinality of the Reals............ 52 Preface to 5……………………………………………………75 5 Non-Cantorian Set Theory…………………………….....76 Preface to 6…………………………………………………....86 6 Cardinality, Measure, Category………………………....87 Preface to 7…………………………………………………...100 7 Continuum Hypothesis, Axiom of Choice, and NonCantorian Theory………….……………………………....101

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Gauge Institute Journal, Volume 4, No 1, February 2008,

H. Vic Dannon

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 NonCantorian 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 replaces facts with wishes, attempts to prove Axioms as if they were

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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 Choice.

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1 Hilbert’s 1st 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 20 .

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, the tools for the solution did not exist. Tarski obtained the essential lemma only in 1948. But it was not utilized and the

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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. However, the alternative set theory was never developed, and we shall show here that the Continuum Hypothesis can be proved

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under the assumptions of Cantor’s set theory.

1. Proof To understand the gap between cardinal numbers, it is natural to 0 examine the convergence n0 0 , and try to pinpoint where

n 0 0 1 the jump from 0( 0 ) to 2 ( 0 ) occurs. This idea

motivates our proof. By [3, p.173], the sequence

n 0 02 ...n0 , converges to the series

0 20

...

n0

... n0 . n 1

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],

is any component of the series. 1

We actually use sequences of partial sums of series of cardinal numbers

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In particular,

0 0 ... 0 n 1

0 .

n N

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

0 CardN . 0

n N Therefore,

0 . CardN 2 0 That is,

20 . Now, suppose that there is a set X with

0 cardX 20 . Then, for any finite n ,

n 0 02 03 ...0n CardX . Tarski ([4], or [3, p.174]) proved that for any cardinal numbers, and for n 1,2,...n ... the inequalities

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m1 m2 ... mn m , imply the inequality

m1 m2 ... mn .... m . Since for any n

n cardX , by Tarski result

cardX . Combining this with

cardX 20 , yields by transitivity of cardinal inequalities [3, p. 147],

20 , which contradicts 2 0 .

2. Discussion: Why did Cantor fail to prove his hypothesis? Tarski result was not available to Cantor. Furthermore, Cantor was familiar with the

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partial sums

0 20 ... 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 0 0 0 . 0 2

we take infinite n , we obtain terms such as

The infinite summation over all n , is the key to our proof. In spite n 0 of the degenerate character of that 0 , there is a jump to 2

forbids a cardinal number between CardN , 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

0 20 03 ...n0 n0 0 to the cardinal

0 0 n0 0 2 n N

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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

CardN CardX CardR .

with

Cohen’s result predicts the

vulnerability of Cantor’s set theory, and makes it easier for us to present the NonCantorian set theory [5].

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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 www.gauge-institute.org

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20 0 .”

Gauge Institute Journal, Volume 4, No 1, February 2008,

H. Vic Dannon

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. In 7, 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 NonCantorian cardinality is greater than CardN , and another Axiom, that says that the rationals Cantorian Cardinality is CardN . In 2, I was not aware of the Non-Cantorian cardinality. Furthermore, it is not clear in 2 why we must have the NonCantorian Set Theory. In 4, I proved that in Cantor’s Set Theory there is only one infinity. The different infinities are all Non-Cantorian.

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2 Non-Cantorian Cardinal Numbers by H. Vic Dannon May, 2005 Abstract: 20 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 real numbers were not found, and the alternative set theory was never developed.

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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 20 0 Continuum Hypothesis. We further show that the converse is also true, and that the Continuum Hypothesis is equivalent to 20 0 . Thus, Non-Cantorian set theory is founded on the converse assumption that cardQ cardN . We first show that the continuum Hypothesis and 20 0 , are equivalent.

1. 20 0 Continuum Hypothesis: We prove: Assume CardN 2CardN . Then 2 CardN CardN Continuum Hypothesis.

Proof

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() 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 ... 1 1 1 .... CardN .

By [3, p.183], 1 1 1 .... CardN 1

CardN CardN .

This is “convergence” of infinite series of cardinal numbers precisely as defined in Sierpinski.

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Therefore, CardN CardN .

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

n N

Since CardN

CardN

2CardN ,

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

2CardN . Now suppose that the continuum hypothesis does not hold, and there is a set X with

CardN CardX 2CardN . Since

we

assume

2

CardN CardN ,

then

for

any

n 1,2,3... , 2 CardN CardN ... CardN CardX . n

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

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m1, m2,..mn ,

and

H. Vic Dannon

m

are any cardinal numbers so that for any n 1,2,3....,

m1 m2 ... mn m , then,

m1 m2 ... mn .... m . By Tarski result, 2 CardN CardN ... CardN ... CardX . n

Since we assume CardX

2CardN , by transitivity of cardinal

inequalities [3, p. 147], 3 2 CardN CardN CardN . ... 2CardN

That is,

2CardN , which contradicts 2CardN . Therefore, there is no set X with cardinality between CardN , and

2CardN , and the continuum Hypothesis holds.

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This completes the proof that 2 CardN CardN Continuum Hypothesis.

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

m n , and m1 n1

m, n,, m1,

and

n1 ,

m m1 n n1 .

Since

CardN 2CardN , we have

CardN CardN 2CardN 2CardN . But

2CardN 2CardN 2CardN CardN 2CardN . Hence, 2 CardN 2CardN .

Therefore, 2 2 CardN CardN CardN CardN 2CardN . 2 Namely, if CardN CardN , the rationals serve as a set X

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which cardinality is between CardN , and CardR , and the Continuum Hypothesis does not hold. That is, 2 CardN CardN Hypotesis Negation 2 Thus, the Hypothesis implies CardN CardN .

2. 20 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. Cohen’s proof of the existence of non-Cantorian set theory, mandates that cardQ=cardN is an assumption that cannot be proved

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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 20 0 is an Axiom of cantor’s set theory, that can replace the Continuum Hypothesis, similarly to the wellordering 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

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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-ordered2 Q, and believed that he arranged it in a sequence as well, so that he may conclude that 20 0 . Given two infinite sequences of numbers

a1, a2,...an ,... b1, b2,...bn ,... we can arrange them in one sequence

a1, b1, a2, b2,...an, bn,... The method applies to n sequences, but not to an infinite number of sequences.

2

The Well-Ordering Axiom was conjectured by Cantor in 1883. Its equivalence to the Axiom of choice was proved by Zermelo in 1904.

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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’s zigzag through infinitely many infinite sequences of rational numbers.

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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 not arranged in a sequence.

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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 2n numbers. The well ordering will be defined by zigzag from one row to the next, towards the base of the infinite triangle. *

*

*

*

*

*

*

*

*

...

...

... ...

*

... ... ... ... ...

...

...

...

...

...

...

... ... ...

...

... ... ... ... ... ... ... ... ... ... ... ...

... ...

... ...

... ...

... ...

... ... ... ... ... ... ... ... ...

...

A Well-Ordering zig-zag through infinitely many infinite sequences of real numbers

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 sequence the real numbers any less than Cantor’s zig-zag through

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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.

4. Non-Cantorian Cardinal Numbers

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

0 20 , 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

n0 n0 1 ,

for n 1,2,3... .

Proof For n 1, 0 02 , is the non-Cantorian assumption. By induction on n , similarly to our proof that

20 03 , we conclude that for all finite n,

n0 n0 1 .

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n The 0 are newly added cardinalities that in the Cantorian set

theory degenerated into 0 .

n0 00 20 .

Theorem 2. Proof

n For each finite n, 0 is the cardinality of all the roots of all the

polynomials in integer coefficients of degree n. Therefore,

n0 cardR 20 . n Since 0 is monotonically increasing, 0 n0 0 .

Since 0 2 , 0 0 0 2 .

To show that 0 0 0 2 ,

note that for n 1,2,...

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1 22 33 ... n n 20 . By Tarski, [5]

1 22 33 ... nn ... 20 That is,

cardN n n 20 . 0 n N

Thus, 0 0 0 2 ,

and 0 0 n0 0 2 .

Theorem 3

n cardR

Proof

n (20 )n 2n0 20 . Thus, while 20 0 is an assumption, 2 holds in NonCantorian Set Theory too.

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20 n 0 cardR ,

Theorem 4

for all finite n 2,3,4... . proof 0 0 20 n 0 0 2 .

2

20 20 .

Theorem 5 Proof If 20 20 , then 2

2

2

2

2

20 (20 )0 (20 )0 (20 )0 (20 )0 n

3

(20 )0 .... (20 )0 ... 0

(20 )2

21 2 ,

n 0 c since by Theorem 2, 0 2 . This contradicts c 2 .

n0

2

Theorem 6 Proof

31

1 n 0

2

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By the same argument of Theorem 5.

20 2. n

Theorem 7 Proof 0

0

20 20 22 n

The

n

20

2.

are newly added cardinalities, that in the Cantorian set

theory degenerated into 20 .

5. Rational and Irrational Numbers: Theorem 8

cardR cardQ ,

for Cantorian or Non-Cantorian cardinals. Proof Based on the preceding theorems we have 2 0 cardR 20 0 cardQ . 0

This proof replaces Cantor’s diagonal proof that does not apply to non-Cantorian cardinals because in Non-Cantorian set theory, the rationals are uncountable with

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cardQ 0 . 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

CardR CardQ CardR .

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6. Algebraic and Transcendental numbers Theorem 10

Card(Algebraic Numbers) 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

0 02 03 ... 0n . Therefore, for any n N ,

0 20 30 ... n0 Card(Algebraic Numbers) . By Tarski [5],

0 20 ... n0 ... Card (Algebraic Numbers ) . Since 0 0 0 20 ... n0 ... 0 2 ,

by transitivity of cardinal inequalities,

20 Card(Algebraic Numbers) . On the other hand,

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Card (Algebraic Numbers ) CardR 20 . Consequently,

Card (Algebraic Numbers ) CardR 20 . Cantor had 0 20 ... n0 ... 0 . But In chapter 4 we show that for Cantor’s Cardinals, 20 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 ea is an injection from the algebraic numbers into the transcendental numbers. Therefore,

CardR Card (Transcendental Numbers)

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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

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Preface to 3 The response to my claim in 2 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 their cardinality is.

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In 7 we’ll see that by one Axiom, the Cantorian cardinality of the rationals numbers is CardN , and by another Axiom, the NonCantorian cardinality of the Rationals is greater than CardN . It is plain, that Cantor was not aware of any of that.

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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

{( , )} ,

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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:

(1,1); (1,2) , (2,1); (1,3) , (2,2), (3,1); (1,4) , (2,3),…,

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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 . 3 , 1 , 4 , and (3,1) 4 .

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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 20 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 argument: The first line in the zig-zag has one rational

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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 2

2 1

2 2

3 1

3 2

1 3 2 3 3 3

1 4 2 4

1 5

... ...

*

... *

... ... ... ... ... ... ...

*

...

*

... ... ... ... ... ... ...

*

...

*

... ... ... ... ... ... ....

4 1

4 2

5 1

... ... ...

*

...

*

*

...

*

*

...

*

... ... ... ... *

...

*

... ... ... ... ...

...

*

... *

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

n 1,2,3.... , 1 2 3 .... n 0 . (1)

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Thus,

2(1 2 3 .... n) 0 . That is,

n n 2 0 .

(2)

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

m1 , m2 , m3 ,…, and a

cardinal m , the partial sums inequalities

m1 m2 ... mn m ,

for n 1,2,3....

imply the series inequality

m1 m2 ... mn .... m . Applying to (1),

1 2 3 .... n ... 0 . Regarding (2), this says

0 20 0 . Since 20 0 20 , by transitivity of cardinal inequalities [3, p. 147],

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20 0 . Since 0 02 ,

20 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 21 3 . 1 The second line in the zig-zag has two rationals, 1/2, and 2/1, which we assign as follows

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1 1 22 5 , 2 2 2 22 6 . 1 The third line in the zig-zag has three rationals 3/1, 2/2, 1/3, which are assigned as follows

3 1 23 9 , 1 2 2 23 10 , 2 1 3 2 3 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

1 1 2 4 17 , 4 2 2 24 18 , 3

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Gauge Institute Journal, Volume 4, No 1, February 2008,

H. Vic Dannon

3 3 2 4 19 . 2 4 4 24 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 1 2 5 33 , 1 2 2 2 5 34 , 4 3 3 25 35 . 3 2 4 25 36 . 4 1 5 25 37 . 5 If

m n 1 even 2k , the m n 1 2k zig-zag line has the m n 1 2k

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rationals 1/n, 2/(n-1),…(n-1)/2, n/1, which are assigned as follows

1 1 22k , n 2 2 22k , n 1 …………………….

n 1 2k 1 22k , 2 n 2k 22k . 1 If

m n 1 o dd 2k 1 , the

m n 1 2k 1

zig-zag

line

m n 1 2k 1 rationals m/1, (m-1)/2,…, 2/(m-1),1/m, which are assigned as follows

m 1 22k 1 , 1

48

has

the

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m 1 2 22k 1 , 2 …………………….

2 2k 22k 1 , m 1 1 2k 1 22k 1 . 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, 2nd edition, 1965, (Also in the 1958, 1st edition).

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Preface to 4 The conclusion of 3 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 20 . These numbers can serve as a range for an injection from the real numbers to conclude that there are no more reals than rationals. Hence, the reals too are countable.

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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 NonCantorian Axiom.

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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

CardR 2CardN . If Cantor was attempting to find a cardinality between CardN , and

2CardN , that set might have led him to his Continuum Hypothesis that there is no set X with CardN CardX 2CardN . But if the cardinality of such a meager set equals Card (0,1), how large is Card (0,1)? To motivate our discussion, we point out some unsettled issues regarding infinite cardinalities:

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1st issue: Card(N)

View more...
H. Vic Dannon

The Continuum Hypothesis H. Vic Dannon [email protected] September 2007

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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.

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Contents Preface……………………………………………………...... 4 1 Hilbert’s 1 st problem: The Continuum Hypothesis...6 Preface to 2…………………………………………………...14 2 Non-Cantorian Cardinal Numbers………………….......16 Preface to 3……………………………………………………38 3 Rationals Countability and Cantor’s Proof....................39 Preface to 4……………………………………………………50 4 Cantor’s Set and the Cardinality of the Reals............ 52 Preface to 5……………………………………………………75 5 Non-Cantorian Set Theory…………………………….....76 Preface to 6…………………………………………………....86 6 Cardinality, Measure, Category………………………....87 Preface to 7…………………………………………………...100 7 Continuum Hypothesis, Axiom of Choice, and NonCantorian Theory………….……………………………....101

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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 NonCantorian 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 replaces facts with wishes, attempts to prove Axioms as if they were

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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 Choice.

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1 Hilbert’s 1st 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 20 .

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, the tools for the solution did not exist. Tarski obtained the essential lemma only in 1948. But it was not utilized and the

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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. However, the alternative set theory was never developed, and we shall show here that the Continuum Hypothesis can be proved

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under the assumptions of Cantor’s set theory.

1. Proof To understand the gap between cardinal numbers, it is natural to 0 examine the convergence n0 0 , and try to pinpoint where

n 0 0 1 the jump from 0( 0 ) to 2 ( 0 ) occurs. This idea

motivates our proof. By [3, p.173], the sequence

n 0 02 ...n0 , converges to the series

0 20

...

n0

... n0 . n 1

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],

is any component of the series. 1

We actually use sequences of partial sums of series of cardinal numbers

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In particular,

0 0 ... 0 n 1

0 .

n N

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

0 CardN . 0

n N Therefore,

0 . CardN 2 0 That is,

20 . Now, suppose that there is a set X with

0 cardX 20 . Then, for any finite n ,

n 0 02 03 ...0n CardX . Tarski ([4], or [3, p.174]) proved that for any cardinal numbers, and for n 1,2,...n ... the inequalities

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m1 m2 ... mn m , imply the inequality

m1 m2 ... mn .... m . Since for any n

n cardX , by Tarski result

cardX . Combining this with

cardX 20 , yields by transitivity of cardinal inequalities [3, p. 147],

20 , which contradicts 2 0 .

2. Discussion: Why did Cantor fail to prove his hypothesis? Tarski result was not available to Cantor. Furthermore, Cantor was familiar with the

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partial sums

0 20 ... 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 0 0 0 . 0 2

we take infinite n , we obtain terms such as

The infinite summation over all n , is the key to our proof. In spite n 0 of the degenerate character of that 0 , there is a jump to 2

forbids a cardinal number between CardN , 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

0 20 03 ...n0 n0 0 to the cardinal

0 0 n0 0 2 n N

11

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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

CardN CardX CardR .

with

Cohen’s result predicts the

vulnerability of Cantor’s set theory, and makes it easier for us to present the NonCantorian set theory [5].

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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 www.gauge-institute.org

13

20 0 .”

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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. In 7, 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 NonCantorian cardinality is greater than CardN , and another Axiom, that says that the rationals Cantorian Cardinality is CardN . In 2, I was not aware of the Non-Cantorian cardinality. Furthermore, it is not clear in 2 why we must have the NonCantorian Set Theory. In 4, I proved that in Cantor’s Set Theory there is only one infinity. The different infinities are all Non-Cantorian.

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2 Non-Cantorian Cardinal Numbers by H. Vic Dannon May, 2005 Abstract: 20 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 real numbers were not found, and the alternative set theory was never developed.

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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 20 0 Continuum Hypothesis. We further show that the converse is also true, and that the Continuum Hypothesis is equivalent to 20 0 . Thus, Non-Cantorian set theory is founded on the converse assumption that cardQ cardN . We first show that the continuum Hypothesis and 20 0 , are equivalent.

1. 20 0 Continuum Hypothesis: We prove: Assume CardN 2CardN . Then 2 CardN CardN Continuum Hypothesis.

Proof

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() 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 ... 1 1 1 .... CardN .

By [3, p.183], 1 1 1 .... CardN 1

CardN CardN .

This is “convergence” of infinite series of cardinal numbers precisely as defined in Sierpinski.

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Therefore, CardN CardN .

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

n N

Since CardN

CardN

2CardN ,

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

2CardN . Now suppose that the continuum hypothesis does not hold, and there is a set X with

CardN CardX 2CardN . Since

we

assume

2

CardN CardN ,

then

for

any

n 1,2,3... , 2 CardN CardN ... CardN CardX . n

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

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m1, m2,..mn ,

and

H. Vic Dannon

m

are any cardinal numbers so that for any n 1,2,3....,

m1 m2 ... mn m , then,

m1 m2 ... mn .... m . By Tarski result, 2 CardN CardN ... CardN ... CardX . n

Since we assume CardX

2CardN , by transitivity of cardinal

inequalities [3, p. 147], 3 2 CardN CardN CardN . ... 2CardN

That is,

2CardN , which contradicts 2CardN . Therefore, there is no set X with cardinality between CardN , and

2CardN , and the continuum Hypothesis holds.

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This completes the proof that 2 CardN CardN Continuum Hypothesis.

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

m n , and m1 n1

m, n,, m1,

and

n1 ,

m m1 n n1 .

Since

CardN 2CardN , we have

CardN CardN 2CardN 2CardN . But

2CardN 2CardN 2CardN CardN 2CardN . Hence, 2 CardN 2CardN .

Therefore, 2 2 CardN CardN CardN CardN 2CardN . 2 Namely, if CardN CardN , the rationals serve as a set X

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which cardinality is between CardN , and CardR , and the Continuum Hypothesis does not hold. That is, 2 CardN CardN Hypotesis Negation 2 Thus, the Hypothesis implies CardN CardN .

2. 20 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. Cohen’s proof of the existence of non-Cantorian set theory, mandates that cardQ=cardN is an assumption that cannot be proved

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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 20 0 is an Axiom of cantor’s set theory, that can replace the Continuum Hypothesis, similarly to the wellordering 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

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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-ordered2 Q, and believed that he arranged it in a sequence as well, so that he may conclude that 20 0 . Given two infinite sequences of numbers

a1, a2,...an ,... b1, b2,...bn ,... we can arrange them in one sequence

a1, b1, a2, b2,...an, bn,... The method applies to n sequences, but not to an infinite number of sequences.

2

The Well-Ordering Axiom was conjectured by Cantor in 1883. Its equivalence to the Axiom of choice was proved by Zermelo in 1904.

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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’s zigzag through infinitely many infinite sequences of rational numbers.

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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 not arranged in a sequence.

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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 2n numbers. The well ordering will be defined by zigzag from one row to the next, towards the base of the infinite triangle. *

*

*

*

*

*

*

*

*

...

...

... ...

*

... ... ... ... ...

...

...

...

...

...

...

... ... ...

...

... ... ... ... ... ... ... ... ... ... ... ...

... ...

... ...

... ...

... ...

... ... ... ... ... ... ... ... ...

...

A Well-Ordering zig-zag through infinitely many infinite sequences of real numbers

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 sequence the real numbers any less than Cantor’s zig-zag through

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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.

4. Non-Cantorian Cardinal Numbers

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

0 20 , 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

n0 n0 1 ,

for n 1,2,3... .

Proof For n 1, 0 02 , is the non-Cantorian assumption. By induction on n , similarly to our proof that

20 03 , we conclude that for all finite n,

n0 n0 1 .

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n The 0 are newly added cardinalities that in the Cantorian set

theory degenerated into 0 .

n0 00 20 .

Theorem 2. Proof

n For each finite n, 0 is the cardinality of all the roots of all the

polynomials in integer coefficients of degree n. Therefore,

n0 cardR 20 . n Since 0 is monotonically increasing, 0 n0 0 .

Since 0 2 , 0 0 0 2 .

To show that 0 0 0 2 ,

note that for n 1,2,...

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1 22 33 ... n n 20 . By Tarski, [5]

1 22 33 ... nn ... 20 That is,

cardN n n 20 . 0 n N

Thus, 0 0 0 2 ,

and 0 0 n0 0 2 .

Theorem 3

n cardR

Proof

n (20 )n 2n0 20 . Thus, while 20 0 is an assumption, 2 holds in NonCantorian Set Theory too.

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20 n 0 cardR ,

Theorem 4

for all finite n 2,3,4... . proof 0 0 20 n 0 0 2 .

2

20 20 .

Theorem 5 Proof If 20 20 , then 2

2

2

2

2

20 (20 )0 (20 )0 (20 )0 (20 )0 n

3

(20 )0 .... (20 )0 ... 0

(20 )2

21 2 ,

n 0 c since by Theorem 2, 0 2 . This contradicts c 2 .

n0

2

Theorem 6 Proof

31

1 n 0

2

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By the same argument of Theorem 5.

20 2. n

Theorem 7 Proof 0

0

20 20 22 n

The

n

20

2.

are newly added cardinalities, that in the Cantorian set

theory degenerated into 20 .

5. Rational and Irrational Numbers: Theorem 8

cardR cardQ ,

for Cantorian or Non-Cantorian cardinals. Proof Based on the preceding theorems we have 2 0 cardR 20 0 cardQ . 0

This proof replaces Cantor’s diagonal proof that does not apply to non-Cantorian cardinals because in Non-Cantorian set theory, the rationals are uncountable with

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cardQ 0 . 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

CardR CardQ CardR .

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6. Algebraic and Transcendental numbers Theorem 10

Card(Algebraic Numbers) 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

0 02 03 ... 0n . Therefore, for any n N ,

0 20 30 ... n0 Card(Algebraic Numbers) . By Tarski [5],

0 20 ... n0 ... Card (Algebraic Numbers ) . Since 0 0 0 20 ... n0 ... 0 2 ,

by transitivity of cardinal inequalities,

20 Card(Algebraic Numbers) . On the other hand,

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Card (Algebraic Numbers ) CardR 20 . Consequently,

Card (Algebraic Numbers ) CardR 20 . Cantor had 0 20 ... n0 ... 0 . But In chapter 4 we show that for Cantor’s Cardinals, 20 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 ea is an injection from the algebraic numbers into the transcendental numbers. Therefore,

CardR Card (Transcendental Numbers)

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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

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Preface to 3 The response to my claim in 2 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 their cardinality is.

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In 7 we’ll see that by one Axiom, the Cantorian cardinality of the rationals numbers is CardN , and by another Axiom, the NonCantorian cardinality of the Rationals is greater than CardN . It is plain, that Cantor was not aware of any of that.

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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

{( , )} ,

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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:

(1,1); (1,2) , (2,1); (1,3) , (2,2), (3,1); (1,4) , (2,3),…,

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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 . 3 , 1 , 4 , and (3,1) 4 .

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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 20 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 argument: The first line in the zig-zag has one rational

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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 2

2 1

2 2

3 1

3 2

1 3 2 3 3 3

1 4 2 4

1 5

... ...

*

... *

... ... ... ... ... ... ...

*

...

*

... ... ... ... ... ... ...

*

...

*

... ... ... ... ... ... ....

4 1

4 2

5 1

... ... ...

*

...

*

*

...

*

*

...

*

... ... ... ... *

...

*

... ... ... ... ...

...

*

... *

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

n 1,2,3.... , 1 2 3 .... n 0 . (1)

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Thus,

2(1 2 3 .... n) 0 . That is,

n n 2 0 .

(2)

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

m1 , m2 , m3 ,…, and a

cardinal m , the partial sums inequalities

m1 m2 ... mn m ,

for n 1,2,3....

imply the series inequality

m1 m2 ... mn .... m . Applying to (1),

1 2 3 .... n ... 0 . Regarding (2), this says

0 20 0 . Since 20 0 20 , by transitivity of cardinal inequalities [3, p. 147],

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20 0 . Since 0 02 ,

20 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 21 3 . 1 The second line in the zig-zag has two rationals, 1/2, and 2/1, which we assign as follows

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Gauge Institute Journal, Volume 4, No 1, February 2008,

1 1 22 5 , 2 2 2 22 6 . 1 The third line in the zig-zag has three rationals 3/1, 2/2, 1/3, which are assigned as follows

3 1 23 9 , 1 2 2 23 10 , 2 1 3 2 3 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

1 1 2 4 17 , 4 2 2 24 18 , 3

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H. Vic Dannon

3 3 2 4 19 . 2 4 4 24 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 1 2 5 33 , 1 2 2 2 5 34 , 4 3 3 25 35 . 3 2 4 25 36 . 4 1 5 25 37 . 5 If

m n 1 even 2k , the m n 1 2k zig-zag line has the m n 1 2k

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rationals 1/n, 2/(n-1),…(n-1)/2, n/1, which are assigned as follows

1 1 22k , n 2 2 22k , n 1 …………………….

n 1 2k 1 22k , 2 n 2k 22k . 1 If

m n 1 o dd 2k 1 , the

m n 1 2k 1

zig-zag

line

m n 1 2k 1 rationals m/1, (m-1)/2,…, 2/(m-1),1/m, which are assigned as follows

m 1 22k 1 , 1

48

has

the

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H. Vic Dannon

m 1 2 22k 1 , 2 …………………….

2 2k 22k 1 , m 1 1 2k 1 22k 1 . 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, 2nd edition, 1965, (Also in the 1958, 1st edition).

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Preface to 4 The conclusion of 3 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 20 . These numbers can serve as a range for an injection from the real numbers to conclude that there are no more reals than rationals. Hence, the reals too are countable.

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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 NonCantorian Axiom.

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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

CardR 2CardN . If Cantor was attempting to find a cardinality between CardN , and

2CardN , that set might have led him to his Continuum Hypothesis that there is no set X with CardN CardX 2CardN . But if the cardinality of such a meager set equals Card (0,1), how large is Card (0,1)? To motivate our discussion, we point out some unsettled issues regarding infinite cardinalities:

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1st issue: Card(N)

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