Is CAT the Key to Understanding Numbers?

[Hurkyl]

And now, in your last message you contradict yourself by writing:

Yes, I did. I generally have been having to guess at the precise meaning of your statements. When you said

Let be R'1 = Cantor's diagonal.

I presumed that you meant that R'1 is the number selected by Cantor's diagonalization method. Specifically, the number who's n-th digit is different from the n-th digit of the real number corresponding to the integer n.

We are in an endless game, because after each Cantor's function operation,
I can return its results to the bijection list.

Since you call it an "endless game", I presume you agree that there does not exist a list such that Cantor's diagonal method fails to produce a real number not contained in that list. Is that correct?


That's the whole point to the proof; no list is enough. The definition of a bijection requries that the list to cover every real number, but if no list can cover everything, then no list can be a bijection.


Hurkyl

 

[Doron Shadmi]

Dear Hurkyl !

Lets put it this way.

If R set is mathematically well-defined then there is no before-time or after-time but now-time so, when Cantor's diagonal exists,
it is in R' set and in R set in no-time, and it is countable.

If you accept Cantor's diagonal argument, then R set is mathematically not
well-defined, I mean there is before-time and after-time.


My endless game is a no-time game.


Yours,

Doron

 

[Hurkyl]

How does time enter the picture?

 

[Doron Shadmi]

This is what I mean, there is no time in my game.

And what about your game ?


Yours,

Doron

 

[Hurkyl]

Of course there's no time in my "game". Time doesn't enter at all into mathematical logic. That's why I'm perplexed why you would even suggest such an idea.

Hurkyl

 

[Doron Shadmi]

Dear Hurkyl !


So, you have to accept this:

... when Cantor's diagonal exists, it is in R' set and in R set
in no-time, and it is countable.

Because:

Theorem: Cantor's diagonal is countable.


Before proof:

First, let's write an example:

Every R set member is a unique member in the aleph0^2 matrix so,
if we take this aleph0^2 R set, we can build another aleph0^2 R' set
matrix with R' set members ,by using the data of Cantor's diagonal,
instead of the data of the original diagonal, which belongs to
aleph0^2 matrix of R set, and then we can do this bijection:

1 <--> R'1 = Cantor's diagonal (*)
2 <--> R1
3 <--> R2
4 <--> R3
5 <--> R4
6 <--> ...


the proof:


Cantor's diagonal is not in R set.

Member R'1 in R' set, can not be equal to member R1 in R set.

Let be member R'1 in R' set = Cantor's diagonal.

Let be R'1 in R set.

R'1 is countable.


QED

(*) Reference:

http://home.ican.net/~arandall/abelard/math12/Cantor.html


You now, there are some problems that can be solved if you see
them from another dimesion so, try look at this "game" from
a 3D point of view instead of aleph0^2 matrix.



Yours,

Doron

 

[Hurkyl]

So, you have to accept this:

... when Cantor's diagonal exists, it is in R' set and in R set
in no-time, and it is countable.

No I don't! As I mentioned, time plays no role in mathematical logic, so it really doesn't make sense to accept any statement involving some concept of time.


I do accept that cantor's diagonal is countable. It consists of 1 real number which is a countable sequence of digits.

But that has no bearing on whether the set of real numbers is countable.


As for your assertion I should think about it in 3-d, I would be glad to if you provided a 3-d object that was provably relevant to the problem.

The proof that the reals are uncountable goes, casting it into the form with your matrices:
Assume there exists a bijection between the natural numbers and the reals.
Write the bijection as a matrix.
Construct a real number via Cantor's method.
That real number is not in the matrix.
Therefore the bijection isn't a bijection.
Our assumption led to a contradiction therefore it cannot be true.
Thus, there is not a bijection between the natural numbers and the reals.

If you wanted to prove that two sets have the same size, you would need to exhibit a single bijection between them.


Either way, we are only concerned with a single matrix. Considering several matrices (or an infinite stack!) is just irrelevant.


However, it's a fun exercise!


Let's mix Cantor's diagonal argument with the other diagonal argument to prove that there exists a real number that does not appear in your 3-d stack of matrices.


Let A(m, n, p) mean the m-th digit of the n-th number in the p-th matrix.

Construct a sequence &theta as (&theta(x) means the x-th digit of &theta):

&theta(1) = 1 - A(1, 1, 1)

&theta(2) = 1 - A(2, 2, 1)
&theta(3) = 1 - A(3, 1, 2)

&theta(4) = 1 - A(4, 3, 1)
&theta(5) = 1 - A(5, 2, 2)
&theta(6) = 1 - A(6, 1, 3)

&theta(7) = 1 - A(7, 4, 1)
&theta(8) = 1 - A(8, 3, 2)
&theta(9) = 1 - A(9, 2, 3)
&theta(10) = 1 - A(10, 1, 4)

...

(I believe that's sufficient to demonstrate the pattern)

&theta is guaranteed to differ from every number in every matrix by at least one digit, therefore &theta is not contained in any of your matrices!

 

[Doron Shadmi]

Hi Hurkyl !


Please read carefully, it is for n-dimesions.




Cantor's diagonal is countable
------------------------------


Theorem: Cantor's diagonal is countable.


Before proof:

First, let's write an example:

Every R set member is a unique member in the aleph0^2 matrix so,
if we take this aleph0^2 R set, we can build another aleph0^2 R' set
matrix with R' set members ,by using the data of Cantor's diagonal,
instead of the data of the original diagonal, which belongs to
aleph0^2 matrix of R set, and then we have this bijection map:

1 <--> R'1 = Cantor's diagonal (*)
2 <--> R1
3 <--> R2
4 <--> R3
5 <--> R4
6 <--> ...


the proof:


Cantor's diagonal is not in R set.

Member R'1 in R' set, can not be equal to member R1 in R set.

Let be member R'1 in R' set = Cantor's diagonal.

Let be R'1 in the bijection map.

R'1 is countable.


QED

It is a collection in n dimensions, which is doesn't hold.
But it is not a collection in n+1 dimensions so, n+1 dimensions bijection map holds.



(*) Reference:

http://home.ican.net/~arandall/abelard/math12/Cantor.html



Yours,

Doron

 

[Hurkyl]

n dimensions it is.

Let A(m, a, b, c, d, e, ...) be the m-th digit of the real number who, in the array system, has first coordinate a, second coordinate b, third coordinate c, et cetera. If a coordiante is requested with more than n dimensions, then it doesn't matter what A gives because it's not in your array, so for simplicity I'll define A to be 0 in those situations.

Pick &theta as:

&theta(1) = A(1, 0)

&theta(2) = A(2, 1)

&theta(4) = A(4, 2)
&theta(5) = A(5, 0, 1)

&theta(6) = A(6, 3)
&theta(7) = A(7, 1, 1)
&theta(8) = A(8, 0, 0, 1)

&theta(7) = A(7, 4)
&theta(8) = A(8, 2, 1)
&theta(9) = A(9, 0, 2)
&theta(10) = A(10, 1, 0, 1)
&theta(11) = A(11, 0, 0, 0, 1)

&theta(12) = A(12, 5)
&theta(13) = A(13, 3, 1)
&theta(14) = A(14, 1, 2)
&theta(15) = A(15, 2, 0, 1)
&theta(16) = A(16, 0, 1, 1)
&theta(17) = A(17, 1, 0, 0, 1)
&theta(18) = A(18, 0, 0, 0, 0, 1)

...


The pattern here is that the k-th grouping (k starts at 0) contains all coordinates so that:

a + 2b + 3c + 4d + 5e + ... = k


Every possible finite coordinate (a, b, c, d, e, ...) (finite meaning it terminates in an infinite string of 0's) will be enumerated by this scheme, and &theta will differ from the element at that coordinate. In particular, for any n, every coordinate for an n-dimensional array appears.


There is an easier way, though. Any countable list of countable lists is countable, so the entire array can be rewritten as a single list. (My scheme above effectively gives an explicit enumeration)

No matter how complicated the countable collection you use, it can be rewritten as a single list to which we can directly apply Cantor's diagonal method.



Anyways, that aside, you still don't understand that you are proving nothing about bijections. Not once in this entire thread have you said anything remotely like "Let r be any real number ... therefore r is in the mapping, thus it is a bijection". You haven't used the definition of a bijection, and you haven't used any theorems about bijections, so how can you possibly think you've proven the existence of a bijection?

If you want to assert the existence of a bijection, then you must assert the existence of a countable list that contains every real number; it follows directly from the definition. If a list contains every real number, it doesn't have to play an "endless game" of applying cantor's diagonal method and then creating new lists to contain the new numbers; it contains every real number so cantor's diagonal method can't possibly choose a real number it is missing.

The fact that your scheme needs to keep creating new lists to include more real numbers proves that it is not a bijection. A bijection cannot exist unless this process can stop, leaving you with a list that really does contain everything without needing any modification.

 

[Doron Shadmi]

Dear Hurkyl !

The fact that your scheme needs to keep creating new lists to include more real numbers proves that it is not a bijection. A bijection cannot exist unless this process can stop, leaving you with a list that really does contain everything without needing any modification.

It is a collection in n dimensions, which is doesn't hold.
But it is not a collection in n+1 dimensions so, n+1 dimensions bijection map holds.

So, my ansewr is:

n dimension and n+1 dimension exists in a NO-TIME mathematical universe.
So, I don't have to keep creating anything, because R set aleph0^2 matrix
which exists in n dimension, and R' set aleph0^2 matrix which exists
in n+1 dimension are both exists without any gap of time, in a NO-TIME mathematical universe.

For example:

Let X be a set of dimensions so, if y is in X then y+1 is in X.

(The above is the FZ axiom of infinity)


So, the bijection map is in n dimension and in n+1 dimension.


Yours,

Doron

 

[Hurkyl]

It is a collection in n dimensions, which is doesn't hold.

What doesn't hold, and why doesn't it?

But it is not a collection in n+1 dimensions so, n+1 dimensions bijection map holds.

What holds and why are you calling it a bijection map?

n dimension and n+1 dimension exists in a NO-TIME mathematical universe.
So, I don't have to keep creating anything, because R set aleph0^2 matrix
which exists in n dimension, and R' set aleph0^2 matrix which exists
in n+1 dimension are both exists without any gap of time, in a NO-TIME mathematical universe.

I think I understand why you think I'm talking about time.

When I say "You have a process that keeps creating new lists", the precise meaning is something like:

You are recursively defining a sequence of lists by providing a base case and an algorithm that computes the n+1-th term from the n-th term.

When I say "This process stops", that means that there is a proposition (which we might call the "stopping condition") such that for some n, the n-th term satisfies the stopping condition.


Time is not involved, though one usual way of speaking about recursions may sound like it refers to time.

Let X be a set of dimensions so, if y is in X then y+1 is in X.

(The above is the FZ axiom of infinity)

The axiom of infinity says there is a set of natural numbers. How is it related to this?

So, the bijection map is in n dimension and in n+1 dimension.

What are you calling a bijection map, why are you callnig it a bijection map, and why is it related to dimension at all?


Yes I am being pedantic. I'm hoping to get you to make at least one somewhat precise mathematical statement sometime in this thread.


Hurkyl

 

[Doron Shadmi]

Dear Hurkyl !

Thank you very much for being pedantic.

I mixed between R set representation, which is countable
and R set which is uncoutable.

So, now my theory is about number-system representation
and it is not compared to ZF theory.

I have changed my article to be a theory of number-system
representation, and I'll be glad to get your comments on it.

Some Graphics of my number-system representation of ALs 1 to 7 ,
you can find here:

http://cyborg2000.xpert.com/ctheory/

It is written in java by a friend of mine, and it is based on
Cartesian Product so, there are some left-right permutations
that can be ignored.

Thank you for your patience.

I put this message under quotes, as a part of my first message,
so anyone can read it.

Yours,

Doron.