The flaw in Cantor's Diagonalization Argument

Discussion in 'Number Theory' started by Seff, May 8, 2023.

  1. Seff

    Seff

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    Diagonaliztion as a process involves constructing a number that cannot possibly exist in an infinite list of numbers of a set such as the reals, then because that list was assumed to have a bijection with the naturals it concludes that a bijection is impossible. This conclusion however is flawed in that it is never tests if diagonalization will also create a new natural number not in the list of natural numbers that we can then use to continue the bijection.

    Say we have a list of all natural numbers:
    3948593...
    1085483...
    7688312...
    ...
    we can add one to the first digit of the first number, the second of the second number, and so on diagonally in order to construct a natural number not in the list: 4290337...
    Because diagonalization can always create a new number in the naturals we can continue the bijection that Cantor abandoned and show that there is no contradiction.
     
    Seff, May 8, 2023
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    conway and TorusField like this.
  2. Seff

    Seff

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    Can someone please tell me I'm not crazy and this makes sense?
     
    Seff, May 8, 2023
    #2
  3. Seff

    Seff

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    Rephrased for readability:

    Cantor assumes a bijection between the reals and the naturals is possible.
    Cantor shows a surjection from the reals to the naturals is impossible using diagonalization.
    Cantor concludes his assumption leads to a contradiction and must be false.

    I assume a bijection between the reals and the naturals is possible.
    I show a surjection from the naturals to the reals is impossible using diagonalization.
    I conclude Cantor's assumption of a contradiction leads to a contradiction itself because if two sets are not surjective into each other then both must be strictly larger than each other which is impossible.

    I conclude there cannot be a contradiction in Cantor's argument and so there must be a bijection between the reals and the naturals.
     
    Seff, May 8, 2023
    #3
  4. Seff

    TorusField

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    I might be wrong, but wouldn't there be an issue if you get to a natural number in the list which has less digits than you need? For example,

    3948593...
    1085483...
    7688312...
    2...
    ...

    How could we create a new natural number with the fourth digit as one greater than the fourth digit of 2?

    I also think that a number with infinite digits is not a natural number, and therefore one could not create a new natural number by infinitely adding one to each diagonal digit in the list.
     
    TorusField, Jun 19, 2023
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    Tommiy likes this.
  5. Seff

    Tommiy

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    Hi, How to find the sum of the elements of a series (from 1 to 1000 ) for example, Is it possible to find the sum of the elements, 1+ (n^2)/(n^3)-((m+1)^2)/((m+1)^3 )+((n+2)^2)/((n+2)^3)-?. (from 1 - to - 1000.) n=m. ciao. Привет, как найти сумму элементов ряда (от 1 до 1000 ), например, можно ли найти сумму элементов, 1+ (n ^ 2)/(n ^ 3)-((m+ 1)^2)/((т+1)^3)+((п+2)^2)/((п+2)^3)-?. n=m. пока..
     
    Tommiy, Jul 27, 2023
    #5
  6. Seff

    HallsofIvy

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    If m= n why use different letters? Also, n^/n^3= 1/n, (n+ 1)^2/(n+ 1)^3= 1/(n+ 1), and (n+ 2)/(n+ 2)^3= 1/(n+ 2). So you are asking for [math]1+ \sum_{n=1}^{1000} 1/n+ 1/(n+ 1)+ 1/(n+ 2)[/math].
     
    HallsofIvy, Aug 4, 2023
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  7. Seff

    HallsofIvy

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    I see no reason to think you are crazy but, no, that does NOT make sense. First, what does the ".,," at the end of "3948593...", etc mean? If it has the usual meaning, that the number continues without end, then those are NOT natural numbers. Every natural number has a finite number of digits and there is a "natural" order- 1, 2, 3, 4, 5, ...

    No, we CANNOT "always create a new number" in the naturals.
     
    HallsofIvy, Aug 5, 2023
    #7
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