Boolean Logic cannot deal with infinitely many objects

I realized that my last thread (named by "permutations") can’t be understood by professional mathematicians.

Because I don’t know how to write my idea in the common formal way, I am going to do it in a non-formal way, but I will do my best to write it in the clearest way.

So here it is:

Let us check these lists.

P(2) = {{},{0},{1},{0,1}} = 2^2 = 4

and also can be represented as:

00
01
10
11


P(3) = {{},{0},{1},{2},{0,1},{0,2},{1,2},{0,1,2}} = 2^3 = 8

and also can be represented as:

000
001
010
011
100
101
110
111

Let us call any full 01 list, combinations list.

Now, let us use Cantor's Diagonalization method on some finitely long combinations list, for example, the combinations list of number 3:

000
001
010
011
100
101
110
111

We can change the order of the rows, and then use Cantor's Diagonalization method, for example:

001
011
010
000
101
100
111
110

The input for Cantor's Diagonalization method in the first example is 000 and the output is 111.

The input for Cantor's Diagonalization method in the second example is 010 and the output is 101.

In both examples we find that the result is already in the combinations list, and this combination, which is already in the list, is one of the combinations that Cantor's Diagonal does not cover.

The number of the combinations, which are out of the range of Cantor's diagonal is:

2^n - n

Every column, which belongs to some combinations list is a sequence of 01 notations, based on some periodic frequency changes, for example:

the right column of number 3 combinations list, is based on 2^0(=1).

Therefore the periodic frequency changes are 1, and the result in this case is:
01010101.

The result of the middle column is based on 2^1(=2), therefore the sequence is:
00110011.

The result of the left column is based on 2^2(=4), therefore the sequence is:
00001111.

and we get the full combinations list of number 3:

000
001
010
011
100
101
110
111

We can get a combinations list of infinitely many places, by using the ZF Axiom of infinity induction, on the left side of our combinations list, by using the induction on the power_value of each column, for example:

2^0, 2^1, 2^2, 2^3, ...

In this stage we have proven, by induction, that Cantor's diagonal cannot cover any full 01 combinations list, finite or infinite.

Therefore its result is not a new combination (that has to be added to the list).

Because Cantor's diagonal cannot cover the full 01 combinations list (of aleph0 places for each combination) we can conclude that 2^aleph0 > aleph0.

But, because no diagonal's result is a new combination (and therefore not added to the list) each infinitely long sequence of 01 notations can be mapped with some natural number, for example:

...000 <--> 1
...001 <--> 2
...010 <--> 3
...011 <--> 4
...100 <--> 5
...101 <--> 6
...110 <--> 7
...111 <--> 8
...

Therefore we can conclude that 2^aleph0 = aleph0, and we come to contradiction.

(2^aleph0 >= aleph0) = {}, and we have a proof saying that Boolean Logic cannot deal with infinitely many objects in infinitely many magnitudes.

One can say that at least the sequence ...111 is not in the list, for example:

...000 <--> 1
...001 <--> 2
...010 <--> 3
...011 <--> 4
...100 <--> 5
...101 <--> 6
...110 <--> 7
...111 <--> 8
...

Let us examine the infinite from another point of view.

When we have ...111 AND ...000 in an ordered combinations list, it means that the list is complete.

But this is the whole point, infinitely many objects cannot be completed, otherwise they are finitely many objects.

Therefore ...111 AND ...000 are not in the list of infinity many objects.

In other words [...000, ...111) XOR (...000, ...111] .

There are 2 possible structural types of infinitely many 01 notations:

(?...0]
(?...1]

We know how some infinitely long combination starts, but its left side is
unknown (can be 0 XOR 1) and this missing information is essential to the existence of the induction.

Therefore we can find a meaningful missing result by Cantor's Diagonalization method, only in a finite combinations list.


For more details please look at:

http://www.geocities.com/complementarytheory/RiemannsBall.pdf


Organic

The input for Cantor's Diagonalization method in the first example is 000 and the output is 111.

The input for Cantor's Diagonalization method in the second example is 010 and the output is 101.

As I said before, Cantor's Diagonalization method requires that you have as many digits in each number as you have numbers (the way it is normaly applied, this is a countable list of numbers each having a countable number of digits).

In addition, the "input" for Cantor's Diagonalization method is the list of numbers not anyone number.

It is not a matter of not knowing how to "write" mathematics. You've never taken the time to understand the mathematics itself.

Dear HallsofIvy,


Now i know that you don't really want to understand what you read.

And to understand what i wrote you have to do 2 basic things:

1) You have to read it from the first word until the last word.

2) After you read all of it, you have to check if you understand it.

3) After you understand it, than and only than, please write a detailed reply.

Thank you.



Organic

Originally posted by Organic
Dear HallsofIvy,


Now i know that you don't really want to understand what you write.

And to understand what i wrote you have to do 2 basic things:

1) You have to read it from the first word until the last word.

2) After you read all of it, you have to check if you understand it.

3) After you understand it, than and only than, please reply.

Thank you.

Organic

Frankly, Organic, the only thing that would keep me from laughing at the mess you made with that last thread would be my respect for you as a person (until proven otherwise, I assume everyone is worthy of my respect). Posts like this one diminish that. You won't get very far here unless you drop your attitude. And you will get even further if you first learn some real math before trying to invent your own new math.

Dear russ_watters,


First you have to show that you understand what i wrote in this thread, and you can do in this way:

1) You have to read it from the first word until its last word.

2) After you read all of it, you have to check if you understand it.

3) After you understand it, than and only than, please write your detailed reply.


If you can't follow these 3 steps, then you did not show to the persons that read your last raply, that you have any meaningful thing to say about what i wrote.


Yours,


Organic

Perhaps you should try to understand Cantor's diagonal argument before you start claiming that it is invalid.

Usually the thing you call 'combinations list' in your post is referred to as [tex]2^A[/tex].

Cantor's argument is that [tex]|2^A| \neq |A|[/tex]. This should painfully clear if [tex]A[/tex] is a finite set.

A particularly choice abuse is:

But each infinitely long sequence of 01 notations can be mapped with some natural number, for example:

...000 <--> 1
...001 <--> 2
...010 <--> 3
...011 <--> 4
...100 <--> 5
...101 <--> 6
...110 <--> 7
...111 <--> 8
...

Your example pairings are either not well defined, or for example, if the ...'s are all zeros will never map any sequence that contains an infinite number of 1's.

Originally posted by Organic
1) You have to read it from the first word until its last word.

I've read all of it.

Originally posted by Organic
2) After you read all of it, you have to check if you understand it.

I've understood what you are trying to say / prove. However, I do not agree with it.

Originally posted by Organic
3) After you understand it, than and only than, please write your detailed reply.

Your reasoning is flawed. Cantor's argument ONLY works for a list of n numbers, each containing exactly n digits. Since your list does not follow this requirement, you can't use Cantor.

and I also didn't understand what do you mean by this...

Originally posted by Organic
In both examples we find that the result is already in the combinations list, and this combination, which is already in the list, is one of the combinations that Cantor's Diagonal does not cover.

The number of the combinations, which are out of the range of Cantor's diagonal is:

2^n - n
Organic [/B]

To NateTG,

Please show in what i wrote how do you come to the conclusion that what you call A, is finite.

To suyver,

You are right, Cantor's diagonal is limited to aleph0^2(=aleph0), and this is exactly the reason why he can find infinitely many new numbers, which are not in this aleph0^2(=aleph0) list.

The complete list of all R numbers is 2^aleph0.

I proved, by using the ZF axiom of infinity induction, that 2^aleph0 > aleph0.

Then i proved that 2^aleph0 = aleph0 by this:

Each infinitely long sequence of 01 notations can be mapped with some natural number, for example:

...000 <--> 1
...001 <--> 2
...010 <--> 3
...011 <--> 4
...100 <--> 5
...101 <--> 6
...110 <--> 7
...111 <--> 8
...

Therefore we can conclude that 2^aleph0 = aleph0, and we come to contradiction.

(2^aleph0 >= aleph0) = {}, and we have a proof saying that Boolean Logic cannot deal with infinitely many objects.

Originally posted by Organic
I proved, by using the ZF axiom of infinity induction, that 2^aleph0 > aleph0.

how?

http://www.mtnmath.com/book/node53.html

To Guybrush Threepwood,

You wrote:

and I also didn't understand what do you mean by this...

Here it is again:

Now, let us use Cantor's Diagonalization method on some finitely long combinations list, for example, the combinations list of number 3:

000
001
010
011
100
101
110
111

We can change the order of the rows, and then use Cantor's Diagonalization method, for example:

001
011
010
000
101
100
111
110

The input for Cantor's Diagonalization method in the first example is 000 and the output is 111.

The input for Cantor's Diagonalization method in the second example is 010 and the output is 101.

In both examples we find that the result is already in the combinations list, and this combination, which is already in the list, is one of the combinations that Cantor's Diagonal does not cover.

The number of the combinations, which are out of the range of Cantor's diagonal is:

2^n - n

Organic, you don't have to repaste the original message.

In both examples we find that the result is already in the combinations list, and this combination, which is already in the list, is one of the combinations that Cantor's Diagonal does not cover.

If the combination is already on the list, what do you mean is not covered?

And also the argument HallsOfIvy made about the incorrect use of the diagonalization method in this case stands.

I proved it for n, i proved it for n+1, therefore i proved it for infinitely many cases.

Please read my answer to To suyver.

I proved it for n, i proved it for n+1, therefore i proved it for infinitely many cases

Now, that's just bad elementary mathematics!

Originally posted by Organic
To suyver,

I proved, by using the ZF axiom of infinity induction, that 2^aleph0 > aleph0.

I ask again how?? I provided a link to the ZF set of axioms earlier. I just don't see it...

Each infinitely long sequence of 01 notations can be mapped with some natural number, for example:

...000 <--> 1
...001 <--> 2
...010 <--> 3
...011 <--> 4
...100 <--> 5
...101 <--> 6
...110 <--> 7
...111 <--> 8
...

no, you forgot at least the all '1' sequence...

Dear HallsofIvy,

Ok, show in detailed reply why it is bad elementary mathematics.

Let us examine the infinite from another point of view.

When we have ...11111 AND ...00000 in an ordered combinations list, it means that the list is complete.

But this is the whole point, infinitely many objects cannot be completed, otherwise they are finite.

Therefore ...11111 AND ...00000 are not in the list of infinity many objects.

In other words [...000, ...111) XOR (...000, ...111] .

There are 2 possible structural types of infinitely many 01 notations:

(?...0]
(?...1]

We know how some infinitely long combination starts, but its opposite side is
unknown (can be 0 XOR 1) and this missing information is essential to the existence of the induction.

Therefore we can find meaningful missing result by Cantor's diagonal method, only in a finite combinations list.

For more details please look at:

http://www.geocities.com/complementarytheory/RiemannsBall.pdf

Dear HallsofIvy,

Ok, show in detailed reply why it is bad elementary mathematics.

The statement "I proved it for n, i proved it for n+1, therefore i proved it for infinitely many cases." doesn't make any sense.
If you mean "n" as a single specific integer, then "I proved it for n, i proved it for n+1" only proves it for those two particular numbers.
If you meant n to be any natural number, then after "I proved it for n" it isn't necessary to prove it for n+1!
If you meant this as "proof by induction", then it is "Assuming the statement is true for n, prove it is true for n+1" and is only part of what is necessary for a proof by induction.

Hi HallsofIvy,

Please go to the frist post of this thead, find the part of my proof by induction, and then please tell me if this part is proof by induction or not, and if not, why not.

Thank you,

Organic

Might be a poorf, definitley not a proof.

That one misses sequences like:
...01010101010101
which contain infinitely many ones and zeros.

Thank you NateTG, I deleted it.

Please show in what i wrote how do you come to the conclusion that what you call A (look in page 1), is finite.




Organic

Okay, I did go back and read it. You look at 2 bit sequences and show that the number you get by applying Cantor's method to the sequence is in the sequence. Then you look at 3 bit sequences and show that the number you get by applying Cantor's method to the sequence is in the sequence. You then assert that you have proved "by induction" that this will be true for any finite or infinite.
The simplest thing that is wrong with that is that you did not show that "if it is true for N then it is true for N+1".

Actually my objection was to your statement "I proved it for n, i proved it for n+1, therefore i proved it for infinitely many cases."
Even assuming you meant "I proved that IF it is true for any positive integer n, then it is true for n+1, therefore I proved it true for infinitely many cases." that would not be correct- you still have to prove it is true for onecase.(And by the way, induction does not just prove "it is true for infinitely many cases", it proves the statement (which depends on the positive integer n) is true for all n. Saying something is true for all positive integers n is far different than saying it is true for infinitely many of them.)

Somewhat less simple is this: induction proves something is true for all positive integers- not for infinity. You referred to "ZF axiom of infinity induction" but you did not use that. A proof by infinite induction would have to appeal to Zorn's lemma or something equivalent.

Finally, you basic concept of Cantor's method is flawed. There are 4 binary numbers with 2 bits. In applying (your version of) Cantor's method, you used only 2 of them. Of course, the number you got was one of the other 2. There are 8 binary numbers with 3 bits. In applying Cantor's method, you used only 3 of them. Of course, the number you got was one of the other 5. In any list of numbers with n bits, there are 2ⁿ numbers, of which you would use the first n to produce a "new" number. It should be no surprise that that number is in the other 2ⁿ- n.

Cantor applied his method to a list of real numbers, each of which was represented by an infinite (countable) number of digits (or bits if you want to use binary notation). In constructing his new number, with an infinite number of bits, he used all of the numbers in the list. Since the nature of the construction guarenteed that the new number could not be any of those used in its construction, and all numbers in the list were so used, it follows that the new number cannot be on the list. That is the argument that does not apply to any finite list of numbers.

Dear HallsofIvy,

Thank you for your detailed reply.

The ZF Axiom of infinity simply says: if n exists than n+1 exists.

I use this built-in induction on the power_value of 2^power_value,
and the result of using the built-in induction of the ZF Axiom of infinity on the power_value, cannot be but 2^aleph0.

Please show me some mistake in what i wrote in this post.

Yours,


Organic

Originally posted by Organic
Thank you NateTG, I deleted it.

Please show in what i wrote how do you come to the conclusion that what you call A (look in page 1), is finite.

Organic

Originally posted by NateTG
Cantor's argument is that [tex]|2^A| \neq A[/tex]. This should be painfully clear if [tex]A[/tex] is finite

In that phrase "[tex]A[/tex] is finite" is intended to be a hypothesis, not a conclusion.

Please do not wipe out posts that people have responded to. It makes the thread difficult to follow.

Because Cantor's diagonal argument is constructive, it's easy to give an example sequence of 1's and 0's that any particular function [tex]f:\mathbb{N}\rightarrow \{0,1\}^{\mathbb{N}}[/tex] does not cover.

Originally posted by Organic
Dear HallsofIvy,

Thank you for your detailed reply.

The ZF Axiom of infinity simply says: if n exists than n+1 exists.

I use this built-in induction on the power_value of 2^power_value,
and the result of using the built-in induction of the ZF Axiom of infinity on the power_value, cannot be but 2^aleph0.

Please show me some mistake in what i wrote in this post.

Is your intent is to claim something resembling
[tex]\lim_{n\rightarrow \aleph_0} 2^n = \aleph_0[/tex]?

Hi NateTG,

Please do not wipe out posts that people have responded to. It makes the thread difficult to follow.

You are right, i am sorry.

By using the built-in induction of the ZF axiom of infinity, on the power_value of base 2, we get two things:

1) Because the axiom says "if n exists then n+1 exists", the rusult of using it cannot be but 2^aleph0.

2) We get an odered list of (2^aleph0)-1 unique combinations of 01 notations, where the diagonal can cover only aleph0 notations.

Therefore we cannot conclude that the diagonal result is a new sequence (that has to be added to the list).

But the important insight is this:

When we have ...111 AND ...000 in an ordered combinations list, it means that the list is complete.

But this is the whole point, infinitely many objects cannot be completed, otherwise they are finitely many objects.

Therefore ...111 AND ...000 are not in the list of infinity many objects.

In other words [...000, ...111) XOR (...000, ...111] .

There are 2 possible structural types of infinitely many 01 notations:

(?...0]
(?...1]

We know how some infinitely long combination starts, but its opposite side is unknown (can be 0 XOR 1) and this missing information is essential to the existence of the induction.

(
By the way, Rational-like infinitely many sequences, has 4 possible structural types:

0...0
0...1
1...0
1...1
)


Therefore we can find a meaningful missing result by Cantor's Diagonalization method, only in a finite combinations list.


For more details please look at:

http://www.geocities.com/complementarytheory/RiemannsBall.pdf


Organic

most interesting topic, Organic

we must read the way Hilbert end his lecture at paris 1900
to see how much this topic in So significant.

Dear Moshek,

Thank you !

the way Hilbert ended his lecture at paris 1900, can be found here:

http://babbage.clarku.edu/~djoyce/hilbert/problems.html


No organic system is a closed system.


The "key word" for any open information's system is 'uncertainty'.

'Completeness' and 'infinitley many objects' are complementary concepts (like waves and particles in Quantum Mechanics).

Therefore, no infinitely many objects can be completed and well known like finitely many objects.

Please read this again:

By using the built-in induction of the ZF axiom of infinity, on the power_value of base 2, we get two things:

1) Because the axiom says "if n exists then n+1 exists", the rusult of using it cannot be but 2^aleph0.

2) We get an odered list of (2^aleph0)-1 unique combinations of 01 notations, where the diagonal can cover only aleph0 notations.

Therefore we cannot conclude that the diagonal result is a new sequence (that has to be added to the list).

And because nothing is added to the list, we can conclude that (2^aleph0)-1 = aleph0.

And also we can conclude that (2^aleph0)-1 > aleph0, because the diagonal can cover at most aleph0 notations.

Therefore ((2^aleph0-1) >= aleph0) = {}.

Again:

When we have ...111 AND ...000 in an ordered combinations list, it means that the list is complete.

But this is the whole point, infinitely many objects cannot be completed, otherwise they are finitely many objects.

Therefore ...111 AND ...000 are not in the list of infinity many objects.

In other words [...000, ...111) XOR (...000, ...111] .

There are 2 possible structural types of infinitely many 01 notations:

(?...0]
(?...1]

We know how some infinitely long combination starts, but its opposite side is unknown (can be 0 XOR 1) and this missing information is essential to the existence of the induction.

(
By the way, Rational-like infinitely many sequences, has 4 possible structural types:

0...0
0...1
1...0
1...1
)


Therefore we can find a meaningful missing result by Cantor's Diagonalization method, only in a finite combinations list.

(For the record, just because I'm not responding to something doesn't mean it's not wrong nor a rampant abuse of language)

There are 2 possible structural types of infinitely many 01 notations:

(?...0]
(?...1]

We know how some infinitely long combination starts, but its opposite side is unknown (can be 0 XOR 1)

Correction: there is no "opposite side". An infinite sequence has at most one endpoint.

Hi dear Hurkyl,


For me you are the best mathematician in this forum.


Please read my first post, and write your detailed remarks on each part of it.


Thank you.

Yours,


Organic

Correction: there is no "opposite side". An infinite sequence has at most one endpoint.

Rational-like sequences:

(...010010] = (0...0]
(...010101] = (0...1]
(...101010] = (1...0]
(...101101] = (1...1]

will you be satisfy if...

Dear Organic.

Thank you very much for the direction to the way Hilbert end his lecture at Paris:


"The organic unity of mathematics is inherent in the nature of this science, for mathematics is the foundation of all exact knowledge of natural phenomena. That it may completely fulfill this high mission, may the new century bring it gifted masters and many zealous and enthusiastic disciples"

(D.Hilbert 1900)



I am Sorry but I don’t think yet that there is a mistake in Cantor diagonal meted. But I want ot ask you:

Will you be really satisfy if you convict us that there is some problem in Cantor argument.

thank you
Moshek

Hi moshek,

I do not convict anyone in anything, all what i want is to share my ideas with other persons.

Please let me show you some interesting connection between redundancy and uncertatinty, when we construct the combinations list, by using the ZF axiom of infinity.

For example, let us look at 2^2:

0 0
0 1
----
1 0
1 1

And now let us look at 2^3:

0 00
0 01
0 10
0 11
-----
1 00
1 01
1 10
1 11

And 2^4:

0 000
0 001
0 010
0 011
0 100
0 101
0 110
0 111
------
1 000
1 001
1 010
1 011
1 100
1 101
1 110
1 111


In all examples the uniquness of each row, depends on the left most 0 XOR 1 notations.

But when we have infinitely many 01 notations in each row, the left most 0 XOR 1 notations cannot be reached by us, thefore it is unknown, and we always have two identical lists, that cannot be distinguished from each other.

Both uncertainty and redundancy values depends on the number of different notations in any combinations list, for example:

2={'0','1'} , 3={'0','1','2'} , 4={'0','1','2','3'} , ...


I think because of this connection between uncertainty an redundancy (when dealing with infinity), Cantor's Diagonalization method cannot work on infinitely many objects.

This is the main idea of my proof.

Uncertainty and redundancy are essential properties of any rigorous argument dealing with infinitely many objects.

'Completeness' and 'Infinitely many objects' are complementary concepts (exactly like waves and particles in Quantum Mechanics).

When you don't internalize it, then there is no connection between our point of views, about the infinity.




Organic

On Thusday

Organic,

I am really glade that you don’t try to convict me that there is new way to look on mathematics, even if you think so. I will think also and replay to you in about two days.

Thank you
Moshek
[zz)]