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Is f-1(A ∪ B) the same as f-1(A) ∪ f-1(B)?

  1. Sep 7, 2011 #1
    1. The problem statement, all variables and given/known data


    2. Relevant equations



    3. The attempt at a solution
     
    Last edited: Sep 7, 2011
  2. jcsd
  3. Sep 7, 2011 #2

    Dick

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    Yes, you are. But I'd feel more comfortable if you tried to prove it rather than taking as an assumption.
     
  4. Sep 7, 2011 #3

    SammyS

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    Have you tried to prove it's true?
     
  5. Sep 7, 2011 #4
    I can prove it using words, not quite there using mathematical symbols, but that part is out of the scope of my class. Thanks guys!
     
  6. Sep 8, 2011 #5

    HallsofIvy

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    You prove two sets are equal by proving that each is a subset of the other. You prove "A" is a subset of "B" by saying "let [itex]x\in A[/itex]", then show "[itex]x\in B[/itex]".

    Here, to show that [itex]f^{-1}(A\cup B)\subset f^{-1}(A)\cup f^{-1}(B)[/itex], start by saying "let [itex]x\in f^{-1}(A\cup B)[/itex]". Then [itex]y= f(x)\in A\cup B[/itex]. And that, in turn, means that either [itex]y\in A[/itex] or [itex]y \in B[/itex]. Consider each of those.

    Note, by the way, that we are considering the inverse image of sets. None of this implies or requires that f actually have an "inverse".
     
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