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Union and Intersections proofs for real analysis

  1. Sep 12, 2009 #1
    Hi,

    I have four similar problems that I am not sure how to do: Given: A1 and A2 are in X, B1 and B2 are in Y f: X->Y, g - inverse of f
    I have to either prove or if false find counterargument
    1. f(A1 U A2) = f(A1) U f(A2)
    2. f(A1 n A2) = f(A1) n f(A2)
    3. g(-1)(B1 U B2) = g(B1) U g(B2)
    4. g(B1 n B2) = g(B1) n f(B2)

    I started doing 2. I was able to show that f(A1 n A2) C=(is contained in) f(A1) n f(A2):
    let x € f(A1) and x € f(A2)
    since (A1 n A2) <=A1, x€f(A1)
    since (A1 n A2) <=A2, x€f(A2)
    => x € f(A1 n A2), x € f(A1) n f(A2), i.e. (A1 n A2) C= f(A1) n f(A2)

    But I am not sure how to show the other way, i.e. that f(A1) n f(A2) C= (A1 n A2), in order to conclude that both expressions are equal. Or are they equal at all?
     
  2. jcsd
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