A new point of view on Cantor's diagonalization arguments

In summary, the conversation is discussing a new perspective on Cantor's diagonalization arguments and thanking individuals for their contributions. The conversation also delves into the topic of alephs and the differences between conventional mathematics and the speaker's own system. The speaker's system claims to be more expansive than Cantor's transfinite universes, with different relationships between aleph0 and 2^aleph0. The conversation also touches on the concept of magnitude in relation to the binary tree representation.
  • #71
phoenixthoth,

You know what, please write in your way how my list is constructed,
without using ... .

Can you do this for me?

Thank you.

Organic
 
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  • #72
Hurkyl,
Let's say I want to count how many photons you have. What use is knowing their energy?
Because we are talking about the magnitude of this quantity.
 
  • #73
Originally posted by Organic
phoenixthoth,

You know what, please write in your way how my list is constructed,
without using ... .

Can you do this for me?

Thank you.

Organic

i will seriously consider trying but this is really your job as a math researcher.

show me what you get to a rigorous statement and i'll look it over for you. we are here to try to help you, you know.
 
  • #74
Hurkyl,

You comparing between notations , I comparing between the results behind these notations.
 
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  • #75
phoenixthoth ,

How we can represent this idea in standard notations?
Code:
 {...,3,2,1,0}=Z*
     2 2 2 2
     ^ ^ ^ ^
     | | | |
     v v v v
{...,1,1,1,1}<--> 1
 ...,1,1,1,0 <--> 2
 ...,1,1,0,1 <--> 3 
 ...,1,1,0,0 <--> 4 
 ...,1,0,1,1 <--> 5 
 ...,1,0,1,0 <--> 6 
 ...,1,0,0,1 <--> 7 
 ...,1,0,0,0 <--> 8 
 ...,0,1,1,1 <--> 9 
 ...,0,1,1,0 <--> 10
 ...,0,1,0,1 <--> 11
 ...,0,1,0,0 <--> 12
 ...,0,0,1,1 <--> 13
 ...,0,0,1,0 <--> 14
 ...,0,0,0,1 <--> 15
 ...,0,0,0,0 <--> 16
 
  • #76
i've done this already in my paper. please read it. I've shown exactly where cantor's argument would fail if the rules are expanded. and yes, expanded, and not contracted. *if the rules stay the same, there is nothing wrong with cantor's argument.* so what you're asking me to do is impossible.
 
  • #77
Dear phoenixthoth,

You asked me what are ",..." trough my point of view.

My answer is: No collection of infinitely many elements is a complete collection, for example:

Code:
 {...,3,2,1,0}=Z*
     2 2 2 2
     ^ ^ ^ ^
     | | | |
     v v v v
          /1 <--> 1
         1 
        / \0 <--> 2
       1   
       /\ /1 <--> 3 
      /  0
     /    \0 <--> 4 
 ... [b]1[/b]    
     \    /1 <--> 5 
      \  [b]1[/b] 
       \/ \[b]0[/b] <--> 6
       [b]0[/b]  
        \ /1 <--> 7
         0
          \0 <--> 8
          
          /1 <--> 9 
         1
        / \0 <--> 10
       [b]1[/b]  
       /\ /[b]1[/b] <--> 11
      /  [b]0[/b] 
     /    \0 <--> 12
 ... [b]0[/b]    
     \    /1 <--> 13
      \  1
       \/ \0 <--> 14
       0  
        \ /1 <--> 15
         0
          \0 <--> 16
 ...
To this tree there cannot be a one common father, because in this case
the tree has a finite size.

By using ... notations we say that we cannot reach the state of the one common father.

Shortly speaking, any collection of infinitely many elements is an open collection of "never ending" story.

My aleph0 is an open collection that its magnitude is unknown.

The open collection can appear in infinitely many magnitudes of different open collections, for example:

When we write a=aleph0+1 > b=aleph0 we mean that a is always bigger then b by one more element, end this ratio between a and b does not change unless we choose to change it.

Shortly speaking there is a relative ratio between open collocations that determinate by us, where aleph0 stands for an open collection.

Through this point of view any arithmetic operation keeps its unique influence on the results for example:

aleph0*2 < 2^aleph0 < 3^aleph0 > aleph0^3 > (aleph0^3)/2 ...

By this attitude our information is richer then the transfinite point of view of aleph0, and it can be used in more interesting ways then the way it is used by the transfinite approach.
 
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  • #78
you're still not writing in the language of math. until you do, mathematicians will ignore you.

having said that, i fully agree with you. it's quite cool, isn't it? but heed the advice in the opening lines of this post.

i believe that i already have formalized what you mean in my tuzfc. so anyone who's half-way interested in what you are writing, including you, should seek to understand my arguments there.
 
  • #79
phoenixthoth


Your wrote:
three dots are not a complete list of R={0,1,...}. as i said earlier, each dot has too much information in it for that to be more than an *infinitesimal* partial list.

three dots don't work in proofs. (they only help you see. so while we may see what you see, or not, that is *not* a proof.)

proof:
let x equal 1+(-1)+1+(-1)+...
1+(-1)+1+(-1)+...=
(1+(-1))+(1+(-1))+...=
0+0+...=0.
therefore, x=0.

also, x=
1+(-1)+1+(-1)+...=
1+((-1)+1)+((-1)+1)+...=
1+0+0+...=1.
therefore, x=1.

therefore, 0=1.

therfore, if we allow three dots to be a proof then we will have to sacrifice the law of identity. do you see this? do you believe 0=1?
The way you use () in (1+(-1))+(1+(-1))+... case, is different from the way you use () in 1+((-1)+1)+((-1)+1)+... case.

Therefore you get different results, and I don't see any connection between these results and ...
 
  • #80
i was wondering if you were going to spot that. you are truly an exceptional student of mathematics.

try this on for size:

x=1+(-1)+1+(-1)+1+(-1)+1+(-1)+1+(-1)+1+(-1)+1+(-1)+1+(-1)+... can be truncated into EITHER this:
x=1+(-1)+1+(-1)+...
OR but NOT XOR
this:
x=1+(-1)+1+(-1)+1+...

ok?

so assume x=x and get a contradiction. copy my original proof from this point forward and you get 0=1.
 
  • #81
Dear phoenixthoth,
x=1+(-1)+1+(-1)+1+(-1)+1+(-1)+1+(-1)+1+(-1)+1+(-1)+1+(-1)+... can be truncated into EITHER this:
x=1+(-1)+1+(-1)+...
OR but NOT XOR
this:
x=1+(-1)+1+(-1)+1+...
What do you mean by truncated?

Nothing is truncated without SASs interference and this is exactly the meaning of ..., which is: x result can be 0 XOR 1.

Shortly speaking, we have to choose the value of x, which means: x value determinate by SASs.

There is no x result out there without SASs determination, when we deal with infinity.

( By the way, please read this paper: http://www.geocities.com/complementarytheory/CQ.pdf )
 
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  • #82
you cannot use conjecture as the basis of a proof. not yet.
 
  • #83
No phoenixthoth,

This invariant information structure

Code:
                             ^
            0 XOR 1 = child  |
           /                 |
          /                  |
father = ?               redundancy    
          \                  |
           \                 |
            0 XOR 1 = child  |
                             v   
        <--uncertainty-->
is not a conjecture but a "rigorous" proof which some of its legal properties are uncertainty and redundancy.

If you don't clearly show this in your theory then you miss the whole point.

Also please read this paper:
http://www.geocities.com/complementarytheory/CuRe.pdf

Thank you,

Organic
 
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  • #84
in no way, shape, or form, is that a rigorous proof.

it is, however, a "RIGOROUS" proof, PErHAps. :P
 
  • #85
Dear phoenixthoth

If you still keep in your theory the stuffed form of infinity and the stuffed proofs of Euclidian mathematics which are based on objective platonic realm, then we have nothing to talk about Math, because from my point of view Math is meaningless without us as its SASs.

For example please read this:

http://www.geocities.com/complementarytheory/Identity.pdf
 
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  • #86
I can't believe I never picked up on this before:


Code:
          /1
         1 
        / \0
       1   
       /\ /1
      /  0
     /    \0
 ... 1    
     \    /1
      \  1
       \/ \0
       0  
        \ /1
         0
          \0

          /1
         1 
        / \0
       1   
       /\ /1
      /  0
     /    \0
 ... 0    
     \    /1
      \  1
       \/ \0
       0  
        \ /1
         0
          \0

You assert that this is a tree with depth aleph0 and 2^aleph0 leaves, right?

What if we remove all of the leaves? We should be left with a tree with depth aleph0-1 and 2^(aleph0-1) leaves, by your reckoning, right?

Removing the leaves gives

Code:
         1 
        / 
       1   
       /\ 
      /  0
     /    
 ... 1    
     \    
      \  1
       \/ 
       0  
        \ 
         0
          
         1 
        / 
       1   
       /\ 
      /  0
     /    
 ... 0    
     \    
      \  1
       \/ 
       0  
        \ 
         0
...


But guess what? This is the exact same tree we started with! (If you don't see it, fill in the next level)

So we must have aleph0-1 = aleph0!
 
  • #87
organic, i think "yoda" just had an ah-ha moment! :P

guess he was right about it not seeming to have any magic properties.

ready for a little breaking out of the box?

how about when someone maybe doesn't rigourously prove but rigourously tries to prove then that should be considered "circumstantial mathematical evidence" when something is correct.
 
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  • #88
Hurkyl,

Your calculation is not right because:

1) aleph0 in my point of view stands for a general notation for any collection of infinitely many elements, and its value is flexible.

2) for example: By saying that a=(aleph0-aleph0) < b=aleph0 we mean that there are aleph0 elements in b that are not covered by a.

Also by a=(aleph0-2^aleph0) < b=aleph0 we mean that there are 2^aleph0 elements in b that are not covered by a.

There are no absolute magnitudes when we deal with collections of infinitely many elements, and no arithmetical operation (finite or infinite) can change their property of being infinitely many elements.

Shortly speaking, no arithmetical operation (finite or infinite) can change the “–“ or “+” sign in a collocation of infinitely many elements.

3) in my Binary tree the aleph0 width magnitude and the 2^aleph0 length magnitude, depends on each other, therefore their relative proportion (notated as width=aleph0 < length=2^aleph0) was not changed by your operation.


You still ignore the inner structure of infinitely many elements, because after your operation we have this list:
Code:
 {...,3,2,1}=N
     2 2 2
     ^ ^ ^
     | | |
     v v v
[b]{[/b]...,1,1,1[b]}[/b]<--> 1
 ...,1,1,1  
 ...,1,1,0 <--> 2 
 ...,1,1,0   
 ...,1,0,1 <--> 3 
 ...,1,0,1   
 ...,1,0,0 <--> 4 
 ...,1,0,0   
 ...,0,1,1 <--> 5 
 ...,0,1,1  
 ...,0,1,0 <--> 6
 ...,0,1,0  
 ...,0,0,1 <--> 7
 ...,0,0,1  
 ...,0,0,0 <--> 8
 ...,0,0,0  
 ...

So, as you see aleph0-1 < aleph0

By the way, the result of your oparation is ((aleph0)-1) < ((2^aleph0)-aleph0)and the reason that it is jusut -aleph0 and not -2^aleph0, can be clearly shown here:
Code:
 <---Arithmetic magnitude 

 {...,3,2,1,0} = Z*
     2 2 2 2  
     ^ ^ ^ ^   
     | | | |   
     v v v v  
{...,[b]1-1-1-1[/b]} <--> 1  Geometric magnitude(based on the 
 ...,1,1,1,[b]0[/b]  <--> 2          |          thin notations)          
 ...,1,1,[b]0[/b]/                   |
 ...,1,1/0,                   |
 ...,1,[b]0[/b], ,                   |
 ...,1/0, ,                   |
 ...,1|0, ,                   |
 ...,1|0, ,                   |
 ...,[b]0[/b]/ , ,                   |
 ...,0, , ,                   |
 ...,0, , ,                   |
 ...,0, , ,                   |
 ...,0, , ,                   |
 ...,0, , ,                   |
 ...,0, , ,                   |
 ...,0, , ,                   |
 ...                          V

After your operation we have:

( ((aleph0)-1) < ((2^aleph0)-aleph0) ) < ( aleph0 < 2^aleph0 )
 
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  • #89
this tree is very much like the tree of knowledge. but i agree with hurkyl, it is not magical.
 
  • #90
phoenixthoth,

What do you mean by "magical"?
 
  • #91
come on organic, don't play this game. you know exactly what "magical" means. magic means that you can prove something by a picture only without a rigorous proof.

btw, organic, my paper was not about SASs.
 
  • #92
phoenixthoth ,

What is proof for you?
 
  • #93
self-evidence. what's it to you?

there are two of me around here. but who is the master and who is the apprentice?
 
  • #94
Then why a self-evidence thing has to be proved?
 
  • #95
Originally posted by Organic
Then why a self-evidence thing has to be proved?

touche.

why don't you ask yourself the same question, or, rather, why you have to go around proving stuff all the time?
 
  • #96
What I understand don't has to be proved to me unless someone (including myself) show me that there is a deeper way to understand this thing.
 
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  • #97
agreement is the seed of salvation, organic. we are in total agreement.
 
  • #99
it looks to me, although I'm just an amateur mathematician, that you have innovation and spirit. yet you lack the ability to articulate yourself within the confines of mathematical rigor.

i would suggest going to school and learning how to do this. this requires a lot of boring homework. i recommend you do your homework.

and, organic, help has arrived.
 
  • #100
phoenixthoth,

For me my pictures are the best tools to understand my ideas.

I don't want to seat in a boring classroom and learn how to use other person's tools of understanding.

I enjoy every moment in my own independent way to give simple shapes to my ideas, and I don't care if the academic persons don't understand it. and specially pure "stuffed" mathematicians that are doing their best to escape from real life influence on their rigorous methods.

I can find persons that can understand my ideas, and these persons do not afraid to open themselves and their methods to the complexity of the real life, for example:

Code:
-----Original Message-----
From: Dr. A.M.Selvam [mailto:amselvam@eth.net]
Sent: Monday, March 01, 2004 10:04 AM
To: Shadmy Doron
Subject: A new approach for the definition of a NUMBER

1 March 2004

Dear Doron  Shadmi 

 I am indebted to you for your email dated 
16 February giving references of your valuable 
research work.
 
 I find your original research work very valuable 
for developing a simple unified theory with ramifications
in the numerical modeling of nonlinear dynamical 
systems/processes. 

 Your research work would benefit many of the scientists 
particularly those who are working in the area of 
numerical modeling.

                                   with best regards
                                    yours sincerely
                              Dr. (Mrs.) A. Mary Selvam

Papers of Dr. (Mrs.) A. Mary Selvam can be found here:

http://www.geocities.com/CapeCanaveral/Lab/5833/pub11.html
 
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  • #101
The problem, as I see it, is that you seem to be spending a lot of effort to assert that your tools talk about things that they don't.

For instance, what ever concept of "quantity" you have in your system, it is different from what a mathematician calls cardinality. You've done a great deal of asserting that your tools are proving that things mathematicians know about cardinality is wrong, when all you're really doing is discovering the differences between cardinality and your concept.

You seem almost fanatical in this pursuit, which is why many people get turned off by your theorizing. To be frank, because of your approach, I lost all interest in your ideas because you just couldn't seem to get past the "Look, I've proved mathematics wrong!" mentality.

The main reason I keep posting in your threads is because I think you're not a hopelessly lost cause like most so-called crackpots. (and that I'm a glutton for punishment!) You still seem to have the "This is so obvious, why can't they see it the way I do?" mentality about things, but you do sometimes seem to learn and adapt in the face of criticism.


Also, I hope you don't think that rigor is all mathematicians do. It is the ultimate standard, but intuition and heuristics have a lot to do with how we do things. For instance, on a great many problems I will start with "this is how my gut says to do this problem", and then if it looks promising, I begin to fill in the details. Filling in the details often gives me a good proof, and other times it illuminates a flaw I made in my intuitive reasoning. There is a lot of mathematics built up on things that are still conjecture; for instance, there are a lot of theorems of the form "If the Riemann hypothesis is true, then this other statement is true". Mathematically, one does not need to have proved the hypothesis with full rigor to reason about things; you just have to acknowledge that you haven't derived an "absolute" fact, but instead a "relative" fact.
 
  • #102
hmm...
 
  • #103
Oh, well, time to feed my habit again.

It appears by 'complete' you mean a set is complete iff you can write down every single element of it on a bit of paper (possibly a very big bit of paper) without omitting anything. Is that a reasonable interpretation of your definition of a 'complete set'?
 
  • #104
Well, assuming you are using all those terms in the correct manner, the infinite binary tree has no leaves; there is not last level.

So perhas you ought to explain what all those things are. Given you didn't understand why a tree wasn't an uncountable totally disconnected set, it's very hard to even begin to say what you think any of these things are. That, and the fact you refuse to state what they are, of course.

By definition the root and leaf set of a tree is in the tree, when they exist: there is the infinite unrooted binary tree which has neither.
 
  • #105
Anyways, I would like to make my point about removing the leaves again:

One of these two diagrams is your "binary tree" with the lables removed.
The other of these two diagrams is your "binary tree" with the rightmost column chopped off, and the labels removed.
To draw each of them, I drew enough of the tree to represent the first 16 rows:

Code:
          /1
         1 
        / \0
       1   
       /\ /1
      /  0
     /    \0
 ... 1    
     \    /1
      \  1 
       \/ \0
       0  
        \ /1
         0
          \0
          
          /1
         1
        / \0
       1  
       /\ /1
      /  0 
     /    \0
 ... 0    
     \    /1
      \  1
       \/ \0
       0  
        \ /1
         0
          \0
 ...

Code:
          /1
         1 
        / \0
       1   
       /\ /1
      /  0
     /    \0
 ... 1    
     \    /1
      \  1 
       \/ \0
       0  
        \ /1
         0
          \0
          
          /1
         1
        / \0
       1  
       /\ /1
      /  0 
     /    \0
 ... 0    
     \    /1
      \  1
       \/ \0
       0  
        \ /1
         0
          \0
 ...

Can you tell which one is which? I can't.
 

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