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    Lol trying to prove that
    |X|<|P(X)|

    So yeah, to show its not a surjection I'm having a problem.

    Define A=\{ x \in X| x \not \in f(x)\}

    But, lol let A=f(a)

    But, I don't see understand the argument that if a \in f(a), then contradiction and then if a \not \in f(a).

    But, lol its confusing. Anyone got an easier proof.

    P.S. Also, what if A was empty?
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    (Original post by Simplicity)

    But, I don't see understand the argument that if a \in f(a), then contradiction and then if a \not \in f(a).
    hello brother ..here it is http://en.wikipedia.org/wiki/Axiom_of_power_set
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    (Original post by simplexity)
    hello brother ..here it is http://en.wikipedia.org/wiki/Axiom_of_power_set
    Wikipedia article on power set was a help.

    P.S. But, what if T or A is empty?
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    Even if A is empty, since f is a surjection there exists x such that f(x) = A. So either x is in A or x is not in A. In the former case we have a contradiction since A is empty it can't contain A, in the later case, we then have found an element x such that x is not in f(x) but A (the set of such elements) is empty. Contradiction!
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    (Original post by SimonM)
    Even if A is empty, since f is a surjection there exists x such that f(x) = A. So either x is in A or x is not in A. In the former case we have a contradiction since A is empty it can't contain A, in the later case, we then have found an element x such that x is not in f(x) but A (the set of such elements) is empty. Contradiction!
    Scratching head.

    Hmm, but if A is empty, how can you validly sat let f(x)=A, as yeah, since that would be mapping an element to the empty set?

    arrghhhhhhhhhhh. That didn't help.

    Got some picture?
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    OK, so the empty set belongs to the power set (since it is by definition a subset). We've assumed f to exist, and that it is surjective, so we know there should exist an element x which maps to it
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    (Original post by SimonM)
    OK, so the empty set belongs to the power set (since it is by definition a subset). We've assumed f to exist, and that it is surjective, so we know there should exist an element x which maps to it
    Yes, but if A was empty you would have found that set that isn't mapped onto, hence not a bijection?

    P.S. Hmm, I'm still annoyed. Stupid proof.
 
 
 

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