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Math Help - logicsets

  1. #1
    Junior Member
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    logicsets

    can anyone do this question

    The sets A, B and C are given by
    A = {1, 2, 3, 4},
    Last edited by srk619; November 19th 2008 at 11:31 AM.
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  2. #2
    Senior Member
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    A diagram is easy but with a pen

    Let see b)

    f (1) = y, f (2) = x, f (3) = z and f:A\rightarrow B=\{x,y,z\}, so f is surjective. But since A has more elements than B, f can't be injective, so can't be bijective.

    g(x) = a, g(y) = b and g:B\rightarrow C=\{a,b\}, so f is surjective. But since B has more elements than C, g can't be injective, so can't be bijective. (yeah the proof is quite similar)

    h is the identity map on C, it is bijective (i.e. injective and surjective).

    k(a) = 2, k(b) = 2 That means k:C\rightarrow A is not injective. Furthermore, as A has more elements than C, Range(k) \neq A so k is not surjective, in conclusion, k is "none of these".
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