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Math Help - Sets

  1. #16
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by shilz222 View Post
    I was asked to prove:  (A-B) \cap C = \emptyset \Leftrightarrow A \cap C \subseteq B

    So there are two directions.

    In one direction (  \Leftarrow ) we are given:  A \cap C \subseteq B and  (A-B) \cap C \neq \emptyset for proof by contradiction.

    In the other direction (  \Rightarrow ) we are given:  (A-B) \cap C = \emptyset and then do what you said in your earlier post.
    I apologize if I am misinterpreting something. All of my comments are directed toward your \implies proof by contradition. I have not addressed any comments to the proof in the other direction.

    -Dan
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  2. #17
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    Quote Originally Posted by shilz222 View Post
    I was asked to prove:  (A-B) \cap C = \emptyset \Leftrightarrow A \cap C \subseteq B So there are two directions.
    YES, YES, YES! That is correct.

    Quote Originally Posted by shilz222 View Post
    In one direction (  \Leftarrow ) we are given:  A \cap C \subseteq B and  (A-B) \cap C \neq \emptyset
    NO, NO, NO This not correct.

    One direction: Given \left( {A \cap C} \right) \subseteq B prove that \left( {A - B} \right) \cap C = \emptyset .

    The other direction: Given \left( {A - B} \right) \cap C = \emptyset prove that \left( {A \cap C} \right) \subseteq B.
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  3. #18
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    Why is it not correct?

    We assume that  (A-B) \cap C \neq \emptyset for proof by contradiction.

    One direction: Given that  (A \cap C) \subseteq B and  (A-B) \cap C \neq \emptyset use proof by contradiction to show that  (A-B) \cap C = \emptyset (my thinking).

    Your other direction looks good.

    So now is this correct?
    Last edited by shilz222; August 13th 2007 at 09:28 AM.
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  4. #19
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by shilz222 View Post
    Why is it not correct?

    We assume that  (A-B) \cap C \neq \emptyset for proof by contradiction.

    One direction: Given that  (A \cap C) \subseteq B and  (A-B) \cap C \neq \emptyset use proof by contradiction to show that  (A-B) \cap C = \emptyset (my thinking).

    Your other direction looks good.
    Okay, so to be clear. You agree that your proof of the \implies direction is incorrect? And now you are working on the \leftarrow proof. And you have decided to try to prove it by contradiction. Is this all correct?

    -Dan
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  5. #20
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    I asked help for the  \Longrightarrow direction.

    Quote Originally Posted by Plato View Post
    This is a very important concept: X \cap Y = \emptyset \text{ if an only if }X \subseteq Y'.
    That is: two sets are disjoint iff each is a subset of the complement of the other.

    \left( {A - B} \right) \cap C = \left( {A \cap B'} \right) \cap C = \left( {A \cap C} \right) \cap B', thus if \left( {A - B} \right) \cap C = \emptyset then \left( {A \cap C} \right) \subseteq B.

    Now you do the other way.

    Trying to teach oneself this material can be very dangerous if one wants to do higher mathematics.
    I agree with this method that Plato has done.

    For the  \Longleftarrow direction I already proved it by contradiction.

    Suppose that  (A-B) \cap C \neq \emptyset . If  x \in A \cap C then  x \in A and  x \in C \Rightarrow x \in B . Then  \exists x \in (A-B) \cap C such that  x \in A and  x \in C but  x \not \in B . But this is a contradiction and so  (A-B) \cap C = \emptyset .
    So is this not correct?
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  6. #21
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    Basically I used 2 different techniques to prove the directions.
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  7. #22
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    Here it is.
    Given \left( {A \cap C} \right) \subseteq B that means \left( {A \cap C} \right) \cap B' = \emptyset .
    But \left( {A \cap C} \right) \cap B' = \left( {A \cap B'} \right) \cap C = \left( {A - B} \right) \cap C = \emptyset
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  8. #23
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    I already know that is correct. But why isnt the proof by contradiction correct?
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  9. #24
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    I guess what I am asking is that is it because my proof is awkward that its not correct (i.e. using direct and proof by contradiction)?
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  10. #25
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by shilz222 View Post
    I guess what I am asking is that is it because my proof is awkward that its not correct (i.e. using direct and proof by contradiction)?
    Simply because the direct proof is shorter. From what I have seen (I may be wrong) proof by contradiction is usually only used if a direct proof is either lacking or obscenely difficult.

    -Dan
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