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Thread: Set proof

  1. #1
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    Set proof

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

    Thoughts/Work: I tried proof by contradiction. Suppose that $\displaystyle B \cap C \neq \emptyset $. Then $\displaystyle x \in B $ and $\displaystyle x \in C $. So $\displaystyle x \in (A \cup C) - B \Rightarrow x \in A $ or $\displaystyle x \in C $ and $\displaystyle x \not \in B $. Then $\displaystyle x \in A $ and $\displaystyle x \not \in B $ or $\displaystyle x \in C $. But this contradicts the fact that $\displaystyle B \cap C \neq \emptyset $. So $\displaystyle B \cap C = \emptyset $ for equality between the two sets. Suppose that $\displaystyle (A \cup C) - B \neq (A - B) \cup C $. Then $\displaystyle x \in A $ or $\displaystyle x \in C $ and $\displaystyle x \not \in B \Rightarrow x \not \in A, x \in B $ or $\displaystyle x \not \in C $. So $\displaystyle x \in B $ and $\displaystyle x \not \in C $ or $\displaystyle x \in C $ and $\displaystyle x \not \in B $. This is a contradiction and so $\displaystyle (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; Aug 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 $\displaystyle (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 $\displaystyle \left( {A \cup C} \right) - B = \left( {A - B} \right) \cup C$, what if $\displaystyle \exists z \in \left( {B \cap C} \right)$ then because $\displaystyle z \in B$ we can conclude that $\displaystyle z \notin \left( {A \cup C} \right) - B$. But by the equality that means that $\displaystyle z \notin \left( {A - B} \right) \cup C$ and that contradicts the fact that $\displaystyle z \in C$; thus $\displaystyle \left( {B \cap C} \right) = \emptyset $.

    Second: Suppose that $\displaystyle \left( {B \cap C} \right) = \emptyset$.
    This means that $\displaystyle C \subseteq B'\quad \Rightarrow \quad C \cup B' = B'$.
    $\displaystyle \left( {A - B} \right) \cup C = \left( {A \cap B'} \right) \cup C$
    $\displaystyle = \left( {A \cup C} \right) \cap \left( {B' \cup C} \right)$
    $\displaystyle = \left( {A \cup C} \right) \cap \left( {B'} \right)$
    $\displaystyle = \left( {A \cup C} \right) - B$.
    This means that $\displaystyle \left( {A \cup C} \right) - B = \left( {A - B} \right) \cup C$
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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 $\displaystyle (A \cup C) - B = (A-B) \cup C \Leftrightarrow B \cap C = \emptyset $.

    Thoughts/Work: I tried proof by contradiction. Suppose that $\displaystyle B \cap C \neq \emptyset $. Then $\displaystyle x \in B $ and $\displaystyle 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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  6. #6
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    Well I assumed that $\displaystyle B \cap C \neq \emptyset $ for the '$\displaystyle \Rightarrow $' direction and arrived at a contradiction. So in that direction, I have shown that $\displaystyle B \cap C = \emptyset $. So I don't think (2) is necessary. In the '$\displaystyle \Leftarrow $' direction I arrived at a contradiction by assuming that $\displaystyle (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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