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A new point of view on Cantor's diagonalization arguments

  1. Mar 12, 2004 #1
    Hi,

    In this pdf (+ its links)http://www.geocities.com/complementarytheory/NewDiagonalView.pdf
    you can find a new point of view on Cantor's diagonalization arguments.

    I really want to send a BIG THANK YOU to Matt grime and Hurkyl for their hard time with me.

    Yours,

    Organic
     
  2. jcsd
  3. Mar 12, 2004 #2

    matt grime

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    One and exactly one post on this this time:

    The alephs in that article are not the alephs of proper mathematics (conventional mathematics if that is what you prefer); they are not equivalence classes of sets modulo bijective correspondence; two sets that are bijective have different alephs associated to them in that article, at least that is the only way to read the sentence aleph-0 is not aleph-0+1 if they are both to be cardinals; whatever they are they do not obey the definitions that every one is used to; do not think that it is a commentary on the use and proof of Cantor's theorem, it is not, as it does not follow the same conventions; I don't know if he's still claiming this but an example would be organic's claim that the 'cardinality' of the reals was strictly greater than the Naturals yet both were enumerable (countable) despite his agreement there was no bijection between them.

    Any issues that are raised are purely a function of refusing to follow the conventions,apparently under the impression that there is some higher pure definition of these things that we as mathematicians are ignoring by putting our dirty meanings on them.

    My new motto will be don't feed the trolls.
     
    Last edited: Mar 12, 2004
  4. Mar 12, 2004 #3
    Matt,

    Ok, prove by your system that my matrix does not have the complete 01 combinations.

    ...0101 and ...1010 are in the list, for example:

    Let us take again our set:
    Code (Text):

     {...,3,2,1,0}=Z*
         2 2 2 2
         ^ ^ ^ ^
         | | | |
         v v v v
    [b]{[/b]...,1,1,1,1[b]}[/b]<--> 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
     ...
     
    Now let us make a little redundancy diet:
    Code (Text):

     {...,3,2,1,0}=Z*
         2 2 2 2
         ^ ^ ^ ^
         | | | |
         v v v v
    ...  [b]1[/b]-1-1-1 <--> 1
         \  \ \0 <--> 2
          \  0-1 <--> 3
           \  \0 <--> 4
           [b]0[/b]-[b]1[/b]-1 <--> 5
            \ \[b]0[/b] <--> 6
             0-1 <--> 7
              \0 <--> 8
     ... [b]0[/b]-[b]1[/b]-1-1 <--> 9
         \  \ \0 <--> 10
          \  [b]0[/b]-[b]1[/b] <--> 11
           \  \0 <--> 12
           0-1-1 <--> 13
            \ \0 <--> 14
             0-1 <--> 15
              \0 <--> 16
     ...
     
    and we get:
    Code (Text):

     {...,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
     ...
     
    My system is reacher then Cantor's transfinite universes bacause:

    1) By my system aleph0+1 > aleph0 , 2^aleph0 < 3^aleph0

    2) By Cantor's system aleph0+1 = aleph0 , 2^aleph0 = 3^aleph0


    By the way, when we move from the 01 matrix representation to the Binary Tree representation, the meaning of the word magnitude become clearer, because several sequential 1 or 0 notations of each column in the matrix, are compressed to a single notation, which its magnitude equivalent to the quantity of the notations that it represents.
     
    Last edited: Mar 15, 2004
  5. Mar 12, 2004 #4
    Good motto!
     
  6. Mar 12, 2004 #5

    Hurkyl

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    Cardinality, as defined by mathematics, is useful because it tells us things about set functions.

    |A| = |B| iff there is a bijection between A and B.
    |A| <= |B| iff there is a surjection from B onto A.
    |A| <= |B| iff there is an injection from A into B.
    |A| < |B| iff |A| <= |B| and not |A| = |B|.


    Yours does not do this, thus it cannot even serve as a substitute for cardinality.


    What's the next digit?
     
  7. Mar 12, 2004 #6
    Hurkyl,

    You asked what is the next?

    Code (Text):

    {...,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
     
    Answer 1: Both cases and their opposites are already in the complete Binary Tree, therefore no sequence has do be added to the tree.
    Code (Text):

     {..4,3,2,1,0}=Z*
       2 2 2 2 2
       ^ ^ ^ ^ ^
       | | | | |
       v v v v v
              /1  
             1
            / \0  
           1  
           /\ /1  
          /  0
         /    \0  
         [b]1[/b]    
        |\    /1  
        | \  [b]1[/b]
        |  \/ \[b]0[/b]  
       /   [b]0[/b]  
       |    \ /1  
       |     0
       |      \0  
     ..[b]1[/b]      
       |      /1  
       |     1
       |    / \0  
       \   [b]1[/b]  
        |  /\ /[b]1[/b]  
        | /  [b]0[/b]
        |/    \0  
         [b]0[/b]    
         \    /1  
          \  1
           \/ \0  
           0  
            \ /1  
             0
              \0  
     
              /1  
             1
            / \0  
           1  
           /\ /1  
          /  0
         /    \0  
         [b]1[/b]    
        |\    /1  
        | \  [b]1[/b]
        |  \/ \[b]0[/b]  
       /   [b]0[/b]  
       |    \ /1  
       |     0
       |      \0  
     ..[b]0[/b]    
       |      /1  
       |     1
       |    / \0  
       \   [b]1[/b]  
        |  /\ /[b]1[/b]  
        | /  [b]0[/b]
        |/    \0  
         [b]0[/b]    
         \    /1  
          \  1
           \/ \0  
           0  
            \ /1  
             0
              \0  
     ...

     
    Shortly speaking, this tree has the magnitude of 2^aleph0 enumerable unique combinations of infinitely wide (= aleph0 magnitude) 01 sequences.


    --------------------------------------------------------------------------

    Answer 2: Maybe this time you are going to understand the beauty of redundancy and uncertainty as inherent fundamental properties of Math language.

    1) Please this time look and read carefully this pdf:

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


    2) Also please read this pdf about the symmetry proprty:

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

    --------------------------------------------------------------------------

    Because I proved that there exists an enumerable list with 2^aleph0 magnitude, all what you wrote holds only between collections with finitely many objects.

    Another alternative is to accept my dynamic point of view on collections of infinitely many elements saying that aleph0 is a general and flexible quantity, which its particular magnitude determinates by operations that are based on finite and/or infinite values, for example:

    a=aleph0+1 > b=aleph0 means that there is always 1 element in a that cannot be covered by b.

    Also 2^aleph0 < 3^aleph0, 2*aleph0 > aleph0, aleph0^aleph0 > 2^aleph0,
    and so on.

    Shortly speaking the elements are based on unknown or incomplete quantity.
     
    Last edited: Mar 13, 2004
  8. Mar 12, 2004 #7

    Hurkyl

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    Where?

    Last time, you said that ...101010 was row #6. However, we now see that is not row #6.

    You now seem to assert it is row #22. However, if you go out 2 more digits, you'll find that it is not row #22.

    And it is not row #54. Nor is it row #118.


    In fact, for every natural number M, I can tell you a specific digit in which the sequence in row #M does differs from ...10101010.

    In other words, for any natural number M, ...10101010 is not row #M.



    What I wrote are the actual definitions of the symbols involved.
     
  9. Mar 13, 2004 #8
    Hurkyl,

    And this is exactly what happens when we try to find a mapping between infinitely long enumerable collections with different unique structural properties.

    For example: aleph0 < 2^aleph0 < 3^aleph0 ... but each one of them can be represented by its own unique enumerable list.

    Shortly speaking, when we deal with collections with infinitely many elements, their unique structural properties can't be ignored.

    So when we try to compare between two collections with infinitely many elements, first we have to compare between their unique invariant structural properties, and if they are not the same, there cannot be a bijection between these infinitely long collections.

    Cantor did not pay attention to the invariant structural property that exists in any collection of infinitely many elements.

    An example of 2^aleph0 and 3^aleph0:
    Code (Text):

     {...,3,2,1,0}=Z*           {...,3,2,1,0}=Z*
         2 2 2 2                    3 3 3 3
         ^ ^ ^ ^                    ^ ^ ^ ^
         | | | |                    | | | |
         v v v v                    v v v v
    {...,1,1,1,1}<--> 1        {...,2,2,2,2}<--> 1
     ...,1,1,1,0 <--> 2         ...,2,2,2,1 <--> 2
     ...,1,1,0,1 <--> 3         ...,2,2,2,0 <--> 3
     ...,1,1,0,0 <--> 4         ...,2,2,1,2 <--> 4
     ...,1,0,1,1 <--> 5     /   ...,2,2,1,1 <--> 5
     ...,1,0,1,0 <--> 6    /    ...,2,2,1,0 <--> 6
     ...,1,0,0,1 <--> 7    \    ...,2,2,0,2 <--> 7
     ...,1,0,0,0 <--> 8     \   ...,2,2,0,1 <--> 8
     ...,0,1,1,1 <--> 9         ...,2,2,0,0 <--> 9
     ...,0,1,1,0 <--> 10        ...,2,1,2,2 <--> 10
     ...,0,1,0,1 <--> 11        ...,2,1,2,1 <--> 11
     ...,0,1,0,0 <--> 12        ...,2,1,2,0 <--> 12
     ...,0,0,1,1 <--> 13        ...,2,1,1,2 <--> 13
     ...,0,0,1,0 <--> 14        ...,2,1,1,1 <--> 14
     ...,0,0,0,1 <--> 15        ...,2,1,1,0 <--> 15
     ...,0,0,0,0 <--> 16        ...,2,1,0,2 <--> 16
     ...                        ...
     
    Another very important conclusion:

    From this point of view there is no fixed platonic realm waiting for us to discover it.

    For example:

    In base 2 there can be at least to different results to this mapping
    Code (Text):

    ...[b]0[/b] 101010 <--> 6    XOR    ...[b]1[/b] 101010 <--> 6
     
    In this case we have to choose between more than one alternatives, therefore the "right" mapping depends on our decisions as living creatures.
     
    Last edited: Mar 13, 2004
  10. Mar 13, 2004 #9

    Hurkyl

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    Would you agree that the following two statements are true:

    For any binary sequence I choose, there exists a list of binary sequences that contains said sequence.

    For any list of binary sequences I choose, there exists a binary sequence not on that list.
     
  11. Mar 13, 2004 #10
    Hurkyl,
    Can you choose any infinitely long binary sequence wich is not a trivial one like ...01010 or ...111010 and so on?
    What do you mean when you say "I choose"?
     
    Last edited: Mar 13, 2004
  12. Mar 13, 2004 #11

    Hurkyl

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    Such as the sequence [itex]<s_n>[/itex] where:

    [tex]
    s_n := \left\{
    \begin{array}{ll}
    0 \quad & \mbox{n is even} \\
    1 \quad & \mbox{n is odd}
    [/tex]

    Or

    [tex]
    s_n := \left\{
    \begin{array}{ll}
    0 \quad & n = m^2 \mbox{(where m is some integer)} \\
    1 \quad & \mbox{otherwose}
    [/tex]

    Or, given a list L,

    [tex]s_n := 1 - \mbox{(the n-th digit of the n-th row of L)}[/tex]


    Or what about this nifty sequence:

    If [itex]n = p^m[/itex] for some prime p and some integer m, and p is the k-th prime, and you will have chosen at least k lists in your lifetime, then then [itex]s_n[/itex] is one minus the [itex]p^m[/itex]-th digit of the [itex]m[/itex]-th row of the k-th list you have (or will have) chosen. Otherwise, [itex]s_n = 0[/itex].
     
  13. Mar 13, 2004 #12

    Hurkyl

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    I mean that if, by any method, we happen to have a list in our consideration, one for which there is no "choice" to be made in constructing it (so it is really a list, and not just a method for generating lots of lists), then we can find a sequence not on that list.
     
    Last edited: Mar 13, 2004
  14. Mar 13, 2004 #13
    Hurkyl,

    But what you show is the general structure that someone has to "break" and give a specific 01 sequence as a result.

    Your tools cannot do that, because you cannot describe a result witch is not a trivial 01 repetitions.
     
  15. Mar 13, 2004 #14
    Hurkyl,
    By the way I used to construct my 01 list, we can find any 01 unique sequence and its opposite in the list.

    But again you have no mathod to define a non-trivial sequence.
     
    Last edited: Mar 13, 2004
  16. Mar 13, 2004 #15

    Hurkyl

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    If you've constructed the list, then there should be a method to compute the n-th digit of the n-th row of the list. I can then use this method to construct the sequence whose n-th digit is 1 - the n-th digit of the n-th row of your list. *shrug*
     
  17. Mar 13, 2004 #16
    Hurkyl,

    But first you have to define some input, can you do that?
     
  18. Mar 13, 2004 #17

    Hurkyl

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    You said you had a list. I'm using it as "input" to create my sequence.
     
  19. Mar 13, 2004 #18
    Hurkyl,

    Can you use a matrix of aleph0 x 2^aleph0 as an input?

    All you can do is first choose your unique 01 path until some finite place, and then it is easy to find this finite 01 sequence and its opposite in infinitely many places in the above matrix.

    (By the way why did you move my thread to theory development?)
     
  20. Mar 13, 2004 #19

    Hurkyl

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    You said it was a list. (which, by definition, has only aleph0 rows)


    And yes, if you output this list, I don't see why I cannot use it as an input.


    I moved it here because you're not doing mathematics. You may be intent on studying the topics that mathematics likes to study, but you're not doing it in a mathematical fashion. I don't remember the circumstances, but you seemed to prefer theory development to philosophy, so I move your posts here once I think it's clear that you don't want to do things in a mathematical fashion.
     
  21. Mar 13, 2004 #20
    Hurkyl,

    Please look again on this tree and tell me exactly how to you want to use it as an input.

    Code (Text):

     {..4,3,2,1,0}=Z*
       2 2 2 2 2
       ^ ^ ^ ^ ^
       | | | | |
       v v v v v
              /1  
             1
            / \0  
           1  
           /\ /1  
          /  0
         /    \0  
         [b]1[/b]    
        |\    /1  
        | \  [b]1[/b]
        |  \/ \[b]0[/b]  
       /   [b]0[/b]  
       |    \ /1  
       |     0
       |      \0  
     ..[b]1[/b]      
       |      /1  
       |     1
       |    / \0  
       \   [b]1[/b]  
        |  /\ /[b]1[/b]  
        | /  [b]0[/b]
        |/    \0  
         [b]0[/b]    
         \    /1  
          \  1
           \/ \0  
           0  
            \ /1  
             0
              \0  
     
              /1  
             1
            / \0  
           1  
           /\ /1  
          /  0
         /    \0  
         [b]1[/b]    
        |\    /1  
        | \  [b]1[/b]
        |  \/ \[b]0[/b]  
       /   [b]0[/b]  
       |    \ /1  
       |     0
       |      \0  
     ..[b]0[/b]    
       |      /1  
       |     1
       |    / \0  
       \   [b]1[/b]  
        |  /\ /[b]1[/b]  
        | /  [b]0[/b]
        |/    \0  
         [b]0[/b]    
         \    /1  
          \  1
           \/ \0  
           0  
            \ /1  
             0
              \0  
     ...

     
     
  22. Mar 13, 2004 #21

    Hurkyl

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    How exactly are you using it for output?
     
  23. Mar 13, 2004 #22
    Hurkyl,

    What exactly do you want to check about this tree?
     
  24. Mar 13, 2004 #23

    Hurkyl

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    A list of properties that uniquely specify it would be nice.


    Just to be entirely clear, let me ask this question:
    Can I label each leaf with a unique natural number?
     
  25. Mar 13, 2004 #24
    Hurkyl,

    The property of my Binary tree is based on this invariant structure:
    Code (Text):

     
                1 = child
               /
              /
    Father = ?
              \
               \
                0 = child

     
    The number of the Childs depends on any existing Z* member = {0,1,2,3,...} used as the power_value of each level in the tree.

    Because |{0,1,2,3,...}| = aleph0, and these members are used as power_values for each level in the tree, the result can't be but a tree width with aleph0 magnitude and a tree length with 2^aleph0 magnitude.

    Each child is the beginning of infinitely long sequence of 01 unique combinations.

    We can label each child with a unique natural number but this is only an illusion of a bijection that can clearly shown here:

    http://www.geocities.com/complementarytheory/Countable.pdf
     
    Last edited: Mar 13, 2004
  26. Mar 13, 2004 #25
    here it is organic, plain and simple.

    these guys just aren't going to buy into your theories until you can show them how cantor's arguments fail using their language. you have to start with the axioms of set theory. you have to define functions and such. you have to define onto functions. you have to look at powersets. you have to use their language or else they won't believe you. and they're not going to necessarily try to learn your language, which ain't math (no offense intended), so you have to come to their level and do the following: write out cantor's argument as he wrote it and tell them exactly which line or axiom or whatever you think is wrong. and you may not convince them until you give a countexample they can believe. it's just not credible to draw a tree with dots on it and call that a proof. mathematicians eschew such "proofs." they worry a heck of a lot about what hidden assumptions you might be making when you write three little dots.

    my problem with your three little dots is that each "dot" represents an infinite enumerable set.

    i on the other hand have written them something very similar to what you are intending. in my article, i show the following:
    1. there is a set x such that there is a function f that maps x onto the powerset of x.
    2. if there is a function that maps x onto its powerset then its powerset contains at least one "fuzzy set".
    3. if a set's powerset contains no fuzzy sets then there is no function that maps x onto its powerset. (this and 2 are logically equivalent)

    i spell out all my assumptions and all that good stuff. i also claim that what i do fits with set theory rather than being a replacement that no one should bother looking at. (i could, for example, have a set theory where there is only one axiom, the universal set axiom, but that wouldn't be too interesting.) i contend that my tuzfc is an interesting set theory and it has some cool implications.

    my problem is that no one reads it, for whatever reason, and gives me feedback. so for all i know it's complete trash. i've stared at it so many times i don't know heads from tails. it looks fine to me but what do i know?

    so these three results and my article i think are what you want to accomplish: an ammendment to the cantor argument. a revision. i fully agree that cantor needs revision but you're not going to convince anyone the way you're trying to do it. but that shouldn't be the point. you do it because you enjoy discovering mathematics as do i. i don't honestly really care if my theory is right or publishable because it was so fun to create. if it was a waste of time, then c'est la vie. not the first time i've had a set (lol) back.

    i urge all of you, organic and hurkyl especially, to really give my paper a chance and read it. all feedback is welcome.

    hurkyl, i hope i've given organic enough feedback so that i can throw that plug in for my thread; hope this won't be considered trolling his thread.
     
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