1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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.
  2. jcsd
  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

    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


    User Avatar
    Science Advisor
    Homework Helper

    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

    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


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook