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Infinite union and intersection

  1. May 18, 2009 #1
    1. The problem statement, all variables and given/known data

    Given a set A \in R^m, B_n \in R^m for n \in N, show that

    A \ Union {from n = 1 to inf} B_n = Intersection {from n = 1 to inf} (A \ B_n}

    2. Relevant equations

    Same equation as above

    3. The attempt at a solution

    I think I have a solution in mind, but I wanted to make sure it is correct:

    Say, take x \in (Intersection {from n = 1 to inf} (A \ B_n}), in order for x to be in that set, x must be in A \ B_n for all n \in N.

    That implies A \ Union {from n = 1 to inf} B_n, is that correct for the proof? Can I somehow write it out better? I hope someone can fill the gaps in the proof, and any help will be appreciated.
     
  2. jcsd
  3. May 18, 2009 #2

    HallsofIvy

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    How does it imply that? That's the whole point of the proof. Yes, you are right that the fact that x is in A\ B_n for all n means that x is in A. Now you need to show that x is NOT in "union B_n". How does that follow? (I'm not saying it doesn't! I am saying you need to show that.)

    Of course, you also need to prove that "if x is in A\ union B_n, then x is in \intersection (A\B_n)".
     
  4. May 18, 2009 #3
    That's the main trouble, I can't seem to describe the answer...
    If x is in A \ B_n for all n, then x is in a set containing A "minus" B_n for all n, but n goes from 1 to infinity, that's the whole set B_n (union), so x is in A "minus" the union of B_n?

    As for the other way around, take y \in A \ union B_n, for y to exist there must be at least one B_n for some n \in N such that y is in B_n.
    That means the set A does not contain y, but now I'm not sure where to go to show that y is also in the intersecton of (A \ B_n)
     
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