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Math Help - discrete question

  1. #1
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    help, i'd be impressed

    Assume that F: A to B is a function and S, a subset of B, is a subset. Prove that f(f-1(S))= S intersected with f(A).
    [ f(f-1)) is f and inverse of f]
    Last edited by ummmm; November 28th 2005 at 01:28 PM.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by ummmm
    Assume that F: A to B is a function and S, a subset of B, is a subset. Prove that f(f-1(S))= S intersected with f(A).
    [ f(f-1)) is f and inverse of f]
    If f(A)\ \neq \ B, \ f^{-1}(S) may not even be defined! (As we may choose S to be B-f(A))

    Or have I missed something.

    Of course I have missed something:

    choose S to be B-f(A),

    then:

    f^{-1}(S)\ =\ f^{-1}(B-f(A))\ =\ \emptyset -the empty set.

    Back to the problem:

    (we need only consider the case where f^{-1}(S)\ \neq\ \emptyset as the result holds trivally otherwise)

    f^{-1}(S)\ =\ \{x;\ x \epsilon A \mbox{ and } f(x) \epsilon S \},

    so if a\  \epsilon \ f^{-1}(S) then f(a)\  \epsilon \ S and f(a)\  \epsilon \ f(A)

    \Rightarrow\ f(f^{-1}(S))\ \subseteq\ S and f(f^{-1}(S))\ \subseteq\ f(A),

    i.e. \ f(f^{-1}(S))\ \subseteq\ S\cap f(A)


    Now suppose s \epsilon\ (S\cap f(A)) then clearly
    s\ =\ f(f^{-1}(s)) and so \ s \epsilon f(f^{-1}(S))\

    \Rightarrow \ S\cap f(A)\ \subseteq\ f(f^{-1}(S))

    hence:

    S\cap f(A)\ =\ f(f^{-1}(S))

    RonL
    Last edited by CaptainBlack; November 29th 2005 at 12:05 AM.
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  3. #3
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    i think f(a)=b

    i think f(a) has to equal b because there if a function from a to b.
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  4. #4
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    anyone?

    anyone?
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  5. #5
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    B is merely the codomain for f
    f(A) may be a proper subset of B

    All that is necessary is that if x and y are in A and f(x) and f(y) are distinct, then x and y are distinct.
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  6. #6
    Grand Panjandrum
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    The proof on its own:

    (we need only consider the case where f^{-1}(S)\ \neq\ \emptyset as the result holds trivally otherwise)

    f^{-1}(S)\ =\ \{x;\ x \epsilon A \mbox{ and } f(x) \epsilon S \},

    so if a\  \epsilon \ f^{-1}(S) then f(a)\  \epsilon \ S and f(a)\  \epsilon \ f(A)

    \Rightarrow\ f(f^{-1}(S))\ \subseteq\ S and f(f^{-1}(S))\ \subseteq\ f(A),

    i.e. \ f(f^{-1}(S))\ \subseteq\ S\cap f(A)


    Now suppose s \epsilon\ (S\cap f(A)) then clearly
    s\ =\ f(f^{-1}(s)) and so \ s \epsilon f(f^{-1}(S))\

    \Rightarrow \ S\cap f(A)\ \subseteq\ f(f^{-1}(S))

    hence:

    S\cap f(A)\ =\ f(f^{-1}(S))

    RonL
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