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Set theory question

  1. Oct 23, 2006 #1
    If P(X) denotes the power set of X. Is |P(A)| = |P(B)| iff A=B true? If so, I have no idea how to prove the |P(A)| => |P(B)| iff A=B direction, so any hints would be great.
     
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  3. Oct 24, 2006 #2
    does |A| mean the cardinality of A? If so, it is clearly false. For example: the sets {0,1,2} and {3,4,5} have power sets with the same cardinality but they are not equal.
     
  4. Oct 24, 2006 #3
    No.

    If A = B, then P(A) = P(B) so of course we have |P(A)| = |P(B)|.

    But if |P(A)| = |P(B)|, A need not be equal to B.
    Just take say A = {1}, B = {2}, then P(A) = {{},{1}} and P(B) = {{}, {2}}, so |P(A)| = |P(B)| but A != B.
     
  5. Oct 25, 2006 #4

    CRGreathouse

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    Does |P(A)| = |P(B)| imply |A| = |B| for infinite sets without the CH?
     
  6. Oct 30, 2006 #5
    Ok never mind on help trying to prove this, is it even true? I suspect it might be equivalent to the SCH.
     
  7. Oct 31, 2006 #6
    Yes.

    Any set [tex]A[/tex] is equinumerous to the set [tex]A_s = \{ \{x\} \in P(A) | x \in A \}[/tex]

    For any set [tex]A[/tex] Let [tex]f_A[/tex] denote the bijective map from [tex]A[/tex] to [tex]A_s[/tex].

    If [tex]|P(A)| = |P(B)|[/tex] then there is a bijection [tex] g [/tex] from [tex]P(A)[/tex] to [tex]P(B)[/tex] and the restriction of [tex] g [/tex] to the set [tex]A_s[/tex], is a bijection from [tex]A_s[/tex] to [tex]B_s[/tex]. Thus a bijection from [tex]A[/tex] to [tex]B[/tex] is given by the composition of functions

    [tex]f_B^{-1} \circ g|A_s \circ f_A[/tex]

    Where [tex] g|A_s [/tex] is the restriction of g to the set [tex]A_s[/tex] and [tex]f_B^{-1}[/tex] is the inverse of the function mapping [tex]B[/tex] onto [tex]B_s[/tex].
     
    Last edited: Oct 31, 2006
  8. Oct 31, 2006 #7

    Hurkyl

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    Why do you know that? That's false for most bijections [itex]\mathcal{P}(A) \rightarrow \mathcal{P}(B)[/itex].

    For example, suppose that A and B actually are both equal to the set {0, 1}. Then, one example of a bijection [itex]\mathcal{P}(A) \rightarrow \mathcal{P}(B)[/itex] is:

    g({}) = {0}
    g({0}) = {}
    g({1}) = {0, 1}
    g({0, 1}) = {1}

    Here, the image of [itex]g \circ f_A[/itex] never takes values in [itex]B_s[/itex].
     
  9. Oct 31, 2006 #8
    I guess I said that because I wasn't thinking clearly. Thanks for the correction.
     
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