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If A~B, then P(A)~P(B) proof

  1. Sep 1, 2008 #1
    1. The problem statement, all variables and given/known data
    If A~B, then P(A)~P(B), which means the same as,

    If |A|=|B|, then |P(A)|=|P(B)|


    2. Relevant equations



    3. The attempt at a solution
    This problem is difficult for me because I am trying to learn the methods of proof while at the same time taking Intro Analysis. Anyway, here are my thoughts...

    Assume f: A->B is a bijection. From Cantor's Thm. we know that |A|<|P(A)| and |B|<|P(B)|.

    I also know that since |A|=|B|, |B|<|P(A)| and |A|<|P(B)|.

    After this I feel like saying |P(A)|=|P(B)|, but I feel kind of guilty with that.

    Any proof, insight, hint, or a one-liner would be appreciated. Thank you
     
  2. jcsd
  3. Sep 1, 2008 #2

    morphism

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    You should! It's like saying 3=4 because 1<3 and 1<4!

    Let's back-track. We know there is a bijection f:A->B. Can you use this to write down a bijection F:P(A)->P(B)?
     
  4. Sep 1, 2008 #3
    This is a try,

    Let a1,a2,...,ai, be a sequence of elements of P(A).

    Let b1,b2,...,bi, be a sequence of elements of P(B).

    Define F:P(A) --> P(B) as the function that takes an to bn such that n [tex]\in[/tex] I, the index function of i.

    Would this be a bijection?
     
  5. Sep 2, 2008 #4

    HallsofIvy

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    ?? You have not used the fact that A~ B! Also I don't know why you are dealing with sequences. If X is a member of P(A), what can you map it to in P(B)? (X is a set of things in A, and there is a mapping from A to B.)
     
  6. Sep 2, 2008 #5
    This is another try,

    Define F: A --> B, we know that F is bijective.
    Define G:P(A) --> P(B). Let Xi be any set in P(A).
    Then G(Xi) maps to an element of P(B), say Yi.
    Since A~B, then P(A)~P(B)

    Would this be a bijection?
     
  7. Sep 5, 2008 #6

    morphism

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    But it doesn't appear that you've actually defined what G is!
     
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