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    Let A and B be sets such that A \cap B = A \cup B. Prove that A=B. [6 marks]

    I'm not really sure how to go about answering questions of this type, so excuse how potentially stupid my approach may look.

    Proof: Suppose that x \in A and y \in B such that x \neq y. Then x \notin (A \cap B) and y \notin (A \cap B). We have that x \in (A \cup B) and y \in (A \cup B). But since A \cap B = A \cup B, we have that x \in (A \cap B) and y \in (A \cap B). This then forces the condition that x=y. So x \in A \equiv y \in A \equiv x \in B \equiv y \in B. So that A=B.
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    (Original post by Zacken)
    Proof: Suppose that x \in A and y \in B such that x \neq y. Then x \notin (A \cap B) and y \notin (A \cap B).
    I don't see how that (final quoted sentence) follows from your definition of x and y.
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    (Original post by ghostwalker)
    I don't see how that (final quoted sentence) follows from your definition of x and y.
    I was thinking of something along the lines of 1 \in A and 2\in B but 1 \notin (A \cap B) and 2 \notin (A \cap B)
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    (Original post by Zacken)
    Let A and B be sets such that A \cap B = A \cup B. Prove that A=B. [6 marks]

    I'm not really sure how to go about answering questions of this type, so excuse how potentially stupid my approach may look.

    Proof: Suppose that x \in A and y \in B such that x \neq y. Then x \notin (A \cap B) and y \notin (A \cap B).
    This doesn't seem to make sense. What stops x \in A \cap B? This has to be true for some elements of A.

    You need to show that:

    1. x \in A \Rightarrow x \in B
    2. x \in B \Rightarrow x \in A

    using the fact that A \cap B = A \cup B
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    (Original post by Zacken)
    I was thinking of something along the lines of 1 \in A and 2\in B but 1 \notin (A \cap B) and 2 \notin (A \cap B)
    Those are specific elements which wouldn't necessarily be members of general sets. There are no specific elements you can use for a general set.

    atsruser's suggestion would be the way I'd do it.
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    (Original post by atsruser)
    This doesn't seem to make sense. What stops x \in A \cap B? This has to be true for some elements of A.

    You need to show that:

    1. x \in A \Rightarrow x \in B
    2. x \in B \Rightarrow x \in A

    using the fact that A \cap B = A \cup B
    I see, yeah. That was a bit of a stupid assumption on my part.

    Would this be better?

    Let x \in A, then it follow that x \in (A \cup B), this implies that x \in (A \cap B), so x\in A and x \in B.

    Let x \in B, then it follow that x \in (A \cup B), this implies that x \in (A \cap B), so x\in A and x \in B.

    Which shows that x\in A \Rightarrow x\in B and x\in B \Rightarrow x \in A.

    Hence A=B.
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    (Original post by ghostwalker)
    Those are specific elements which wouldn't necessarily be members of general sets. There are no specific elements you can use for a general set.

    atsruser's suggestion would be the way I'd do it.
    Yeah, that was a bit of a brain fart on my part. Thanks!
    Is my revised proof better?
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    (Original post by Zacken)
    I see, yeah. That was a bit of a stupid assumption on my part.

    Would this be better?
    Yes, that's correct.

    BTW you don't need the brackets around the A \cap B and so on. They're distracting.
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    (Original post by atsruser)
    Yes, that's correct.

    BTW you don't need the brackets around the A \cap B and so on. They're distracting.
    Yay! Great, thanks to you and Ghost. +Rep.

    Right, brackets. Will keep in mind next time!
 
 
 
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