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C(reals) = C(P(naturals))?

  1. Nov 26, 2003 #1
    C(reals) = C(P(naturals))??

    Could someone help me please.
    I am studying Cantor's set theory at present, but am a little confused as to why he concludes that the cardinality of the set of real numbers is equal to the cardinal number of the power set of naturals (2^aleph0).

  2. jcsd
  3. Nov 26, 2003 #2


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    The elements of 2N can be treated as sequences of 1's and 0's where a sequence corresponding to a subset A of N has a 1 in the n th position if n is in A.

    You should have no trouble seeing that this mapping is bijective.

    Now, if you look at the real numbers on [0,1] base 2, you get numbers like:
    which are also sequences of ones and zeros. So there's a natural mapping.

    Unfortunately there is a problem because
    in the reals.

    But that only occurs a countable number of (N) times. So we can certainly construct a bijection to [0,1] + N.

    So we have |2N| = |[0,1] + N|

    but you should already know that |[0,1] + N|=|[0,1]|=|R|
    (Cantor certainly did)

    So by substitution we get:
    |2N |=|R|
    Last edited: Nov 26, 2003
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