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Thread: 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; Nov 19th 2008 at 10:31 AM.
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  2. #2
    Senior Member
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    A diagram is easy but with a pen

    Let see b)

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

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

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

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