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Functions proof

  1. Sep 18, 2004 #1
    Hi

    let f be a function from set A into X, and Y,Z c X. Prove the following

    f^-1(YandZ)=f^-1(Y)andf^-1(Z);

    any tips would be great
     
  2. jcsd
  3. Sep 19, 2004 #2
    When you need to prove that two sets are equal, [LATEX]A = B[/LATEX], as in your problem the simplest trick you can use is to show that any [LATEX]a \in A[/LATEX] is also an element of B and viceversa.
     
  4. Sep 19, 2004 #3
    I'm really confused with how to do it with function inverses, thanks for the help i'll try to figure it out
     
  5. Sep 19, 2004 #4
    I think you mean that f is a one-to one function from A to X. Is that what you mean by "into"? Also, I think you mean that Y and Z are both subsets of X. Stop me if I'm wrong.

    Let x∈f^-1(Y ^ Z). We wish to show that x∈f^-1(Y)&f^-1(Z). Then once we do that, we wish to start by letting x be in f^-1(Y)&f^-1(Z) and show that that implies x∈f^-1(Y ^ Z).

    This is what the last poster was writing about.

    Now the thing to remember is that x∈f^-1(U) if and only if f(x)∈U.
     
  6. Sep 19, 2004 #5

    mathwonk

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    I think he does not mean the function is one to one, but "and" seems to mean intersection. and f^(-1) just means preimage.


    then this is a corolalary of the usual tautological fact that pullback or inverse image of sets is a boolean homomorphism, i.e. preserves both intersections and unions.
     
  7. Sep 19, 2004 #6
    Hey thanks for youe help I got it now. It's actually really easy thankyou very much :)
     
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