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Question about the powerset.

  1. Mar 7, 2012 #1
    Im trying to understand this proof by Cantor.
    For every set [itex] X, |X|<|P(x)| [/itex]
    Proof. Let f be a function from X into P(x)
    the set [itex] Y=(x \in X: x \notin f(x) ) [/itex]
    is not in the range of f:
    if [itex] z \in X [/itex] where such that f(z)=Y, then [itex] z \in Y [/itex]
    if and only if [itex] z \notin Y [/itex], a contradiction. Thus f is not
    a function of X onto P(x).
    Hence |P(x)|≠|X|, the function
    f(x)={x} is a one-to-one function of X into P(x) and so
    |X|≤|P(x)|. it follows that
    |X|<|P(x)|.
    I dont understand why z cant be in Y and f(z).
    I guess thats because they defined it that way.
    Is it because we want to find a one-to-one function from
    the set to the power set, and because we want it to be one-to-one
    we want to map every x to a unique element in the power set we don't want x to get mapped to itself. Is that the reason. And do we need z to be in Y and f(z) for it to be onto.
     
  2. jcsd
  3. Mar 7, 2012 #2

    tiny-tim

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    yes
    but x can't get mapped to itself … x and f(x) are in different spaces
     
  4. Mar 7, 2012 #3
    so if x got mapped to itself, it wouldn't be onto.
     
  5. Mar 7, 2012 #4

    micromass

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    We CAN'T map x to itself. Since x is not an element of P(X).
     
  6. Mar 7, 2012 #5
    The element x is not in P(X).
    P(X) contains the set containing x (that is, it contains {x}), but it does not contain x itself. Remember that the power is the set of subsets of X.
     
  7. Mar 7, 2012 #6
    this might be a dumb question, but maybe we didn't do a very good job of defining our funtion from the set to the powerset. why cant x be in P(x)
     
  8. Mar 7, 2012 #7
    The elements of P(X) are subsets, not elements, of X. P(X) contains {x}, but not x itself.
     
  9. Mar 7, 2012 #8

    micromass

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    A simple example will help you. Take X={1,2,3}, then P(X)=\{∅,{1},{2},{3},{1,2},{2,3},{1,3},{1,2,3}}. As you see, 1 is not an element of P(X). But {1} is an element of P(X).
     
  10. Mar 7, 2012 #9
    ok thanks for the responses, it makes more sense now.
     
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