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Set Theory Proof

  1. Apr 4, 2009 #1
    1. The problem statement, all variables and given/known data
    Prove that for all sets A, B, and C, (A-C) [tex]\cap[/tex] (B-C) [tex]\cap[/tex] (A-B) = ∅

    2. Relevant equations

    3. The attempt at a solution
    Proof: Suppose A, B, and C are sets
    Let x [tex]\in[/tex] (A-C) [tex]\cap[/tex] (B-C) [tex]\cap[/tex] (A-B)
    Since x [tex]\in[/tex] (A-C), by definition of difference, x [tex]\in[/tex] A and x [tex]\notin[/tex] C
    Since x [tex]\in[/tex] (B-C), x [tex]\in[/tex] B and x [tex]\notin[/tex] C
    Since x [tex]\in[/tex] (A-B), x [tex]\in[/tex] A and x [tex]\notin[/tex] B
    Then by definition of intersection, if x [tex]\in[/tex] A then x [tex]\notin[/tex] C and x [tex]\notin[/tex] B
    Also, if x [tex]\in[/tex] B then x [tex]\notin[/tex] C
    Therefore there is no intersection of sets A, B, and C
    Therefore, the intersection of (A-C) [tex]\cap[/tex] (B-C) [tex]\cap[/tex] (A-B) = ∅

    Is this proof correct, I feel like I am missing something?
  2. jcsd
  3. Apr 4, 2009 #2


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    You are doing fine up to here. Do you see anything contradictory in those conditions you have on x?
  4. Apr 4, 2009 #3
    It states that x [tex]\in[/tex] B and x [tex]\notin[/tex] B, which isn't possible.
    Do I just say that since this is a contradiction, the intersection is the null set?
  5. Apr 4, 2009 #4


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    Sure. There is no x that can satisfy those two conditions.
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