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Posers and Puzzles

Posers and Puzzles

  1. 20 Mar '08 14:19
    Assuming the Axiom of Choice:
    Let N_0 denote the size of the Natural numbers ("countably infinite), and let N_1 denote the size of the Real numbers ("uncountably infinite".

    By the continuum hypothesis, N_1=2^N_0.

    Is n^N_0=N_1 for all n>1 (n a Natural number)?
  2. Standard member PBE6
    Bananarama
    20 Mar '08 14:59
    Originally posted by Swlabr
    Assuming the Axiom of Choice:
    Let N_0 denote the size of the Natural numbers ("countably infinite), and let N_1 denote the size of the Real numbers ("uncountably infinite".

    By the continuum hypothesis, N_1=2^N_0.

    Is n^N_0=N_1 for all n>1 (n a Natural number)?
    I think the answer is yes, based on this:

    http://en.wikipedia.org/wiki/Continuum_hypothesis#Implications_of_GCH_for_cardinal_exponentiation

    Of course, I don't have any formal instruction in set theory so I don't actually know what it means, but it fits the first pattern!
  3. Standard member preachingforjesus
    Iron Pillar
    20 Mar '08 15:54
    Originally posted by Swlabr
    Assuming the Axiom of Choice:
    ...
    By the continuum hypothesis, N_1=2^N_0.
    ...
    I am not aware that anyone has proven that N_2 is the size of the real numbers.

    n^N_1=N_2 : n= real | n>1
    number of reals N_R>N_1

    these I'm familliar with; if anyone has proven N_2=N_R please give reference.
  4. Standard member TheMaster37
    Kupikupopo!
    20 Mar '08 20:30
    Originally posted by preachingforjesus
    I am not aware that anyone has proven that N_2 is the size of the real numbers.

    n^N_1=N_2 : n= real | n>1
    number of reals N_R>N_1

    these I'm familliar with; if anyone has proven N_2=N_R please give reference.
    With N_1 you mean Aleph One?

    Then, yes, n^N_0 = N_1

    I used to be able to prove it too (had good fun with Axiomatic Set Theory) , but the knowledge has been collecting dust for some time now :p

    The Continuum Hypothosis claims that N_1 would be equivalent to the reals, so if someone proved that N_2 is equivalent to the reals that person would have a proof against CH.

    I'm not sure, but I seem to recall a proof, proving that CH can't be proven or disproven in Set Theory...
  5. Standard member adam warlock
    Baby Gauss
    20 Mar '08 21:50 / 2 edits
    Originally posted by TheMaster37
    I'm not sure, but I seem to recall a proof, proving that CH can't be proven or disproven in Set Theory...
    I think this is a consequence from Godel's incompleteness theorems... But I have to do some research.

    Edit: This is what I got after a quick search :"He also showed that the continuum hypothesis cannot be disproved from the accepted axioms of set theory, if those axioms are consistent"

    Edit2: Later work by Paul Cohen established that the continuum hypothesis is neither provable nor disprovable from the axioms of Zermelo-Fraenkel set theory with the axiom of choice.
  6. 20 Mar '08 23:21 / 2 edits
    Originally posted by adam warlock
    I think this is a consequence from Godel's incompleteness theorems... But I have to do some research.

    Edit: This is what I got after a quick search :"He also showed that the continuum hypothesis cannot be disproved from the accepted axioms of set theory, if those axioms are consistent"

    Edit2: Later work by Paul Cohen established that the continuum ...[text shortened]... able nor disprovable from the axioms of Zermelo-Fraenkel set theory with the axiom of choice.
    Drat-forgot about that.

    How about asking,

    We know N_0 < 2^N_0 (as CH asks if there exists and S s.t. N_0 < |S| < 2^N_0). Is n^N_0 > 2^N_0? (For all n>2, natural)

    Or another, more general, question: Does there exist a set of cardinality greater than that of the real numbers.
  7. 22 Mar '08 00:01
    Originally posted by Swlabr
    Or another, more general, question: Does there exist a set of cardinality greater than that of the real numbers.
    The power set of the reals, of course, but that's not a very interesting answer. Try the set of all curves.
  8. Standard member TheMaster37
    Kupikupopo!
    25 Mar '08 10:02
    Originally posted by GregM
    The power set of the reals, of course, but that's not a very interesting answer. Try the set of all curves.
    Yup!

    I do have to mention that my set-theory is based on Zermelo-Fraenkel. It was a difficult subject at first, but after a while I got the knack for it wich made the subject infinitely more fun
  9. 26 Mar '08 06:16 / 1 edit
    swlabr, could that have something to do with the rainbow having a beard? or a picture with a mustache?