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Aleph null !

  1. Apr 30, 2008 #1
    How might one show that (aleph_null)! = aleph_1?
  2. jcsd
  3. Apr 30, 2008 #2


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    One might start by defining the "factorial function" for aleph_null!
  4. May 1, 2008 #3
    Ok, so the person who proposed this problem to me gave me a way to understand (aleph_null)!.

    So, consider two sets, A and B. Then |A|*|B|=|A x B|, where AxB is the cartesian product of A and B.

    Thus, consider N_m={1,2,3,...m}, and |N_m|=m.

    Then (aleph_null)! = |N_1 x N_2 x N_3 x ...... |.

    So how can I find a bijection from N_1 x N_2 x N_3 x ...... to, say, P(N), the power set of the naturals?
  5. May 1, 2008 #4


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    Consider {1} x {1, 2} x {1, 2, 3} x ... as the base 1-2-3-... expansion of a number in [0, 1), then biject [0, 1) with the reals by your favorite method. You have at most a countable number of issues with rational numbers. which you can likewise deal with in your preferred method.

    This shows that the set has cardinality [itex]\beth_1[/itex], not [itex]\aleph_1[/itex] unless you have the CH.
  6. May 1, 2008 #5
    Or show that 1x2x3x4.........is greater than or equal to 2x2x2x2x2.... and less than or equal to aleph null^aleph null. Still, this only indicates the factorial is equal to beth_1 without CH, like CRGreathouse said.
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