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Power sets and cardinalities (proof)

  1. Jun 11, 2012 #1
    1. The problem statement, all variables and given/known data

    Let A be a set. Show that there is no surjective function phi: A --> P(A), where P(A) is the power set of A. What does this say about the cardinalities of A and P(A)?

    2. Relevant equations
    Assume that phi is a surjection of A onto P(A) and consider the set U= {a in A : a not in phi(a)} of A. Since phi is a surjection there must be a u in A satisfying phi(u)=U.


    3. The attempt at a solution
     
  2. jcsd
  3. Jun 11, 2012 #2

    HallsofIvy

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    Okay, so how does "phi(u)= U" lead to a contradiction?
     
  4. Jun 11, 2012 #3

    micromass

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    That's not an effort, that is the hint that was given.
    Anyway, think about whether u is an element of U.
     
  5. Jun 11, 2012 #4
    From those 2 replies i feel as if im trying to overthink this problem. Is it that u does not need to be an element of U but it has to be an element of A? or something to that extent?
     
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