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Math Help - Mapping Proof

  1. #1
    is up to his old tricks again! Jhevon's Avatar
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    Mapping Proof

    For some reason i'm having trouble with this problem...

    Prove that a mapping \alpha : S \to T is one-to-one iff \alpha (A \cap B) = \alpha (A) \cap \alpha (B) for every pair of subsets A and B of S.

    we define something like \alpha (A) by \alpha (A) = \{ \alpha (x) : x \in A \}

    Thanks guys
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  2. #2
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    a mapping \alpha : S \to T is one-to-one if \alpha (A \cap B) = \alpha (A) \cap \alpha (B) for every pair of subsets A and B of S.

    The contrapositive of this is:
    if \alpha (A \cap B) \not = \alpha (A) \cap \alpha (B) for every pair of subsets A and B of S then \alpha is not one to one.

    intuitively, <br />
\alpha (A \cap B) \subseteq \alpha (A) \cap \alpha (B)
    So <br />
\alpha (A \cap B) \not = \alpha (A) \cap \alpha (B) implies that there is some element in \alpha (A) \cap \alpha (B) that is not in \alpha (A \cap B). So there is an element a in A and an element b in B such that \alpha (a) = \alpha (b) but a,b \not \in A \cap B so a \not = b so \alpha is not one-to-one.

    To go the other way, we know that the function is not one-to-one so for some a \not = b,  \alpha (a) = \alpha (b). using A={a} and B = {b}, \alpha (A \cap B) = \alpha (\oslash) = \oslash \not = \alpha (A) \cap \alpha (B).
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