1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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


    User Avatar
    Science Advisor

    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
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook