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Real analysis chapter 0

  1. Aug 28, 2008 #1
    1. The problem statement, all variables and given/known data

    where A and B are sets and ' \ ' means set difference

    under what conditions does A \ (A \ B) = B

    2. Relevant equations

    S\A = {X | X in S and X not in A}
    B < A (subset) means if x is in B then x is in A

    3. The attempt at a solution
    I get that if B is a subset of A then A \ (A \ B) = B. But I don't know how to prove it. Here's my attempt thus far

    assume x is in A \ (A \ B)
    to be proved x is in B

    from assumption x is in A and x is not in (A \ B) , then
    x is in A and not { L | L in A and L not in B}
    because B is a subset of A
    x is in A and . . .

    . . .
    It's like I know that A \ B contains nothing from B but because A includes B, subtracting it from A will give back B. I don't know how to phrase it right for making a mathematical argument. Any guidance will be strongly appreciated.
  2. jcsd
  3. Aug 28, 2008 #2
    A\(A\ B) = B

    For A\B, Let [tex] x \in A \wedge x \notin B [/tex]
    Then, A \ (A \ B) implies that

    [tex] x \in A \wedge (x \notin A \vee x \in B) [/tex] by DeMorgan's Law

    Then, by associative law,

    [tex] (x \in A \wedge x \notin A) \vee ( x \in A \wedge x \in B) \implies ( x \in A \vee x \in B) [/tex]

    So, [tex]A \cup B = B[/tex] if [tex] A \subset B[/tex] or if A is the empty set.
    Last edited: Aug 28, 2008
  4. Aug 29, 2008 #3


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    Assuming B is a subset of A:

    If x is in A\(A\B) then x is in A and x is not in A\B. x not in A\B means either x is not in A (which is impossible) or x is in B. Therefore x in A\(A\B) implies x is in B.
    If x is in B, since B is a subest of A, x is in A. Since x is in B, it is NOT in A\B and therefore is in A\(A\B).

    Those two prove that, if B is a subset of A, A\(A\B)= B.

    But you also have to prove the other way: if A\(A\B)= B, then B is a subset of A.

    if x is in B, then x is NOT in A\B but is in A\(A\B) because A\(A\B)= B. Therefore x is in A and so B is a subset of A.
  5. Aug 29, 2008 #4
    Thank you for your help. I actually got it just after posting here, when I read the chapter and found that if B is a subset of A, A\B can be rewritten as B* or B's Complement in A. Then my modeling of the problem became more crystaline, and I went from there, only slightly different than yours HallsOfIvy

    For Konthelion, I have not seen the step you made to get to "x in A and (x not in A or x in B)". Can you explain why you were able to do this?
  6. Aug 31, 2008 #5
    A\B=A[tex]\cap[/tex]B[tex]^{c}[/tex] will help o:)
    Well, Konthelion made a mistake: that should be "so, A[tex]\cap[/tex]B=B , blah blah blah"
    Last edited: Aug 31, 2008
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