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Math Help - Cartesian product

  1. #1
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    Smile Cartesian product

    My teatcher wasn't able to explain this for me in a clear way. See if you guys can help.
    Let A and B be two random sets. Prove that A \times" alt="\times" /> B (B \cap" alt="\cap" /> C) = (A \times" alt="\times" /> B) \cap" alt="\cap" /> (A \cap" alt="\cap" /> C) by putting in the element (x,y)
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  2. #2
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    Sort of difficult to see what you want to prove like this...

    Did you mean:  A \times (B \times C) = (A \times B) \times (A \times C) ?
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  3. #3
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    Sorry about that. Didn't know how to handle the LaTex-codes right. But I think know what's going on now.

    But yes, that was what I meant. The book wants me to prove the equivalence by putting in the elemet (x,y) into that.
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  4. #4
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    LaTex Help:
    [tex]A \times \left( {B \times C} \right) \ne \left( {A \times B} \right) \times \left( {A \times C} \right)[/tex] gives A \times \left( {B \times C} \right) \ne \left( {A \times B} \right) \times \left( {A \times C} \right).

    Why? A \times \left( {B \times C} \right) is a set of triples.
    Whereas,  \left( {A \times B} \right) \times \left( {A \times C} \right) is a set of pairs made up of pairs.
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  5. #5
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    Ok, this is how the question looks like and what the key says.

    A \times (B \cap C) = (A \times B) \cap (A \times C)

    Let (x,y) be a random element in A \times (B \cap C)
    Then we have that X \epsilon A and y \epsilon B \capC which give y \epsilon B and y \epsilonC.

    (x,y) \epsilon A x B and (x,y) \epsilon A x C so (x,y) \epsilon (A x B) \cap (A x C)
    and now we can prove that (A x B) \cap (A x C) \subseteq A x (B \cap C).

    But I don't get a grip over this. For example why do you have to put the x in A and the y in (B \capC) and not the other way around?

    I would like to see this in a graphic 3D model so I could see what I was doing.
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  6. #6
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    Quote Originally Posted by The Lama View Post
    Ok, this is how the question looks like and what the key says.

    A \times (B \cap C) = (A \times B) \cap (A \times C)

    Let (x,y) be a random element in A \times (B \cap C)
    Then we have that X \epsilon A and y \epsilon B \capC which give y \epsilon B and y \epsilonC.

    (x,y) \epsilon A x B and (x,y) \epsilon A x C so (x,y) \epsilon (A x B) \cap (A x C)
    and now we can prove that (A x B) \cap (A x C) \subseteq A x (B \cap C).

    But I don't get a grip over this. For example why do you have to put the x in A and the y in (B \capC) and not the other way around?

    I would like to see this in a graphic 3D model so I could see what I was doing.
    You need to show x \in A and y \in {B \cap C} because that's simply the order of the cartesian product -- for example:

    X \times Y = \left\{ (x,y) : x \in X ; y \in Y\right\} While
    Y \times X = \left\{ (y,x) : y \in Y; x \in X\right\}

    And those are ordered pairs, so:
    (x,y) \neq (y,x) \Rightarrow X \times Y \neq Y \times X

    Also, your wording is a bit off... You should say "Let (x,y) \in {A \times B}", or "Let (x,y) be some element in A \times B"; random is not quite the right word here!
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  7. #7
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    Quote Originally Posted by The Lama View Post
    (x,y) \epsilon A x B and (x,y) \epsilon A x C so (x,y) \epsilon (A x B) \cap (A x C)
    and now we can prove that (A x B) \cap (A x C) \subseteq A x (B \cap C).
    But I don't get a grip over this. For example why do you have to put the x in A and the y in (B \capC) and not the other way around?
    More LaTex help.
    [tex](x,y)\in A\times (B\cap C)[/tex] gives (x,y)\in A\times (B\cap C).

    The is simply the way a cross product is defined.

    (x,y)\in W\times Z means x\in W\text{ and }y\in Z
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  8. #8
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    Quote Originally Posted by Defunkt View Post
    Also, your wording is a bit off... You should say "Let (x,y) \in {A \times B}", or "Let (x,y) be some element in A \times B"; random is not quite the right word here!
    Alright. Thanks. I just tried to translate it as best I could from a swedish compendium.

    yea. x and \times. Got a bit lazy there I guess.
    Last edited by The Lama; August 27th 2009 at 08:55 AM. Reason: syntax error
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