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Thread: Cardinality

  1. #1 Cardinality 
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    We have card (Real Numbers) = c, and this is also the cardinality of the Real
    numbers in the interval (0,1). The numbers in this interval is constructed from
    arbitrarily long strings of numbers in arbitrary combinations. So:

    c = card (0,1) + card (0,1) + ....

    with a term for every natural number, therefore:

    c = c*(aleph_0)

    while for k a constant: k^(aleph_0) is constructed from arbritarily long strings
    in only one combination, therefore:

    c*(aleph_0) = c > card (k^(aleph_0))

    Then card (k^(aleph_0)) is either < or = aleph_0 since Cantor proved there is no
    other cardinal between the two. Is there another cardinal < aleph_0 except
    card (0)?

    Is the cardinality of an infinite sum equal to the cardinality of it's largest term if
    this term is infinite and the rest are smaller?

    Is the cardinality of a complex variable (s) tending (somehow) to infinity equal to
    the cardinality of |s| for Re (s) tending to infinity?

    For the statement:

    lim (s -> s_0) F(s) > G (s_0)

    if we only have that F(s) has a simple pole at s_0 while G (s_0) may or may not
    have a pole there and card (LS) = c while card (RS) = card k^(aleph_0) can we
    conclude that the statement is true for all F(s) and G (s) with this property no
    matter where the pole s_0 is located?


    It also matters what isn't there - Tao Te Ching interpreted.
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  3. #2  
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    Things gets more complicated (erros of previous post rectified). In fact I left out the

    natural numbers:

    R = {(0,1)}u{(0,1)}u{1}u{(0,1)}u{2}u ...

    R = ((0,1)u(0,1)u ...)uN

    So:

    c = [card (0,1) + card (0,1) + .... ] + aleph_0

    with a term in brackets for every natural number, therefore:

    c = c*(aleph_0) + aleph_0 (3)

    and we have: c = aleph_0^(aleph_0)

    therefore:

    aleph_0(aleph_0^(aleph_0)) + aleph_0 = aleph_0^(aleph_0) (5)

    We have card (N) + card (finite set) = aleph_0 (6)

    5 says:

    card (N)*card (R) + card (N) = card (R) (7)

    so that card (N) acts like the cardinal of a finite set in (6) if taking

    card(N)*card(R) as card (NxR). Since NxR is definable and card (RxR) = c,

    card (NxR) = (aleph_0)*c (by definition of cardinal multiplication). But:

    aleph_0*aleph_0 = aleph_0, so

    card (NxR) = (aleph_0)*c = aleph_0*aleph_0^aleph_0 = c. Therefore

    card (N)*card (R) can be replaced by either c or c*aleph_0 or c*aleph_0.

    So the cardinals needs a logic of it's own.

    As for card {k^aleph_0 | k e N} < c and k^p > 0, for all large p we have:

    0 < card {k^aleph_0 | k e N} = aleph_0 < c.

    c can be proven the second cardinal:

    c is the second cardinal since the largest member of R is made from arbitrary long

    strings of numbers 0 to 9 in arbitrary long strings and aleph_0*aleph_0 = aleph_0,

    therefore the "strings in strings" can only be reflected by aleph_0^(aleph_0), (we cannot

    include operations other than + and x).


    It also matters what isn't there - Tao Te Ching interpreted.
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  4. #3 Re: Cardinality 
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    Quote Originally Posted by talanum1
    c > card (k^(aleph_0))
    Why do you say that? If , then , so actually for . :?

    since Cantor proved there is no other cardinal between the two
    Actually, Cantor only conjectured that there are no cardinals strictly between and . In fact, the conjecture is neither true nor false – it is simply independent of Zermelo–Fraenkel set-theory axioms, as proved by the combined efforts of Gödel and Cohen.
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  5. #4  
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    No check the second posting. card (2^aleph_0) is undefined or zero. I corrected the mistake in the second posting (since 2^aleph_0 it is not an infinite set)

    Is something wrong with the logic of proving c is the second cardinal?
    It also matters what isn't there - Tao Te Ching interpreted.
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  6. #5  
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    is the cardinality of an infinite set all right. In fact it is the cardinality …

    (i) of the power set of the natural numbers

    (ii) of the set of all functions from to {0,1} (i.e. the set of all sequences of 0’s and 1’s)

    and

    (iii) of the real numbers.

    It can indeed be proved that is fine. There’s nothing wrong with that result.
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