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Thread: HELP with this Proof PLEASE

  1. #1
    Mar 2008

    Question HELP with this Proof PLEASE


    Can anyone help me with this question please ?

    Let X and Y be sets, and let f : X → Y be a function. If A c ("Contained in") X, write f[A] for the set {f(x) | x ∈ A}. Prove that f is a surjection if and only if Y − f[A] c ("Contained in") f[X − A] for all A c ("Contained in") X.

    Note: c means "contained in", I am sorry because I don't know how to type this symbol in and "c" is the closest representation to it.
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  2. #2
    MHF Contributor

    Aug 2006
    You need this lemma. \left( {E \in P(X)} \right) \Rightarrow f\left[ X \right]\backslash f\left[ E \right] \subseteq f\left[ {X\backslash E} \right]

    If f is surjective then by definition f[X]=Y.
    So by the lemma we have at once X\backslash f\left[ A \right] = Y\backslash f\left[ A \right] \subseteq f\left[ {X\backslash A} \right].

    Now if \left( {\forall A \in P(X)} \right)\left[ {Y\backslash f\left[ A \right] \subseteq f\left[ {X\backslash A} \right]} \right] we need to show onto.
    Suppose that f is not onto then \left( {\exists q \in Y} \right)\left[ {q \notin f[X]} \right].
    But that means \left( {\forall A \in P(X)} \right)\left[ {Y\backslash f\left[ A \right] \not \subset f\left[ {X\backslash A} \right]} \right] because \left[ {q \notin f[X\backslash A]} \right].
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