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Math Help - Set proof

  1. #1
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    Set proof

    Problem: Prove that  (A \cup C) - B = (A-B) \cup C \Leftrightarrow B \cap C = \emptyset .

    Thoughts/Work: I tried proof by contradiction. Suppose that  B \cap C \neq \emptyset . Then  x \in B and  x \in C . So  x \in (A \cup C) - B \Rightarrow x \in A or  x \in C and  x \not \in B . Then  x \in A and  x \not \in B or  x \in C . But this contradicts the fact that  B \cap C \neq \emptyset . So  B \cap C = \emptyset for equality between the two sets. Suppose that  (A \cup C) - B \neq (A - B) \cup C . Then  x \in A or  x \in C and  x \not \in B \Rightarrow x \not \in A, x \in B or  x \not \in C . So  x \in B and  x \not \in C or  x \in C and  x \not \in B . This is a contradiction and so  (A \cup C) - B = (A - B) \cup C . QED

    I am not sure if I made this too complicated. Does this look ok?

    Thanks
    Last edited by shilz222; August 11th 2007 at 07:37 PM.
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  2. #2
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    Quote Originally Posted by shilz222 View Post
    Problem: Prove that  (A \cup C) - B = (A-B) \cup C \Leftrightarrow B \cap C = \emptyset .
    I am not sure if I made this too complicated. Does this look ok?
    Frankly, I do not follow what you have done. I am not sure that you quite understand the process we some times call “pick a point proofs”. Below I have given two types of proof.

    First: Suppose that \left( {A \cup C} \right) - B = \left( {A - B} \right) \cup C, what if \exists z \in \left( {B \cap C} \right) then because z \in B we can conclude that z \notin \left( {A \cup C} \right) - B. But by the equality that means that z \notin \left( {A - B} \right) \cup C and that contradicts the fact that z \in C; thus \left( {B \cap C} \right) = \emptyset .

    Second: Suppose that \left( {B \cap C} \right) = \emptyset.
    This means that C \subseteq B'\quad  \Rightarrow \quad C \cup B' = B'.
    \left( {A - B} \right) \cup C = \left( {A \cap B'} \right) \cup C
     = \left( {A \cup C} \right) \cap \left( {B' \cup C} \right)
     = \left( {A \cup C} \right) \cap \left( {B'} \right)
     = \left( {A \cup C} \right) - B.
    This means that \left( {A \cup C} \right) - B = \left( {A - B} \right) \cup C
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by shilz222 View Post
    Problem: Prove that  (A \cup C) - B = (A-B) \cup C \Leftrightarrow B \cap C = \emptyset .

    Thoughts/Work: I tried proof by contradiction. Suppose that  B \cap C \neq \emptyset . Then  x \in B and  x \in C .
    If I am not mistaken (and I easily may be) what you are saying in the last sentence here is that x must belong either to B or C and this is not true. Since the union of B and C is not empty then there does exist at least one x belonging to both B and C, but an arbitrary x does not have to fulfill this property.

    -Dan
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    Yes TopsQuark that is what I thought. I don't see why my method is incorrect. I arrive at contradictions in both directions.
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by shilz222 View Post
    Yes TopsQuark that is what I thought. I don't see why my method is incorrect. I arrive at contradictions in both directions.
    My thought is that you will have to do two proofs:
    1) x belongs to the intersection of B and C
    2) x does not belong to the intersection of B and C

    You have already completed the first step, now you will need to do the second.

    Or you can be satisfied with Plato's more elegant proof that doesn't share this difficulty.

    -Dan
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    Well I assumed that  B \cap C \neq \emptyset for the '  \Rightarrow ' direction and arrived at a contradiction. So in that direction, I have shown that  B \cap C = \emptyset . So I don't think (2) is necessary. In the '  \Leftarrow ' direction I arrived at a contradiction by assuming that  (A \cup C)-B \neq (A-B) \cup C and showed that there must be equality. But I agree Plato's method is more clear.
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