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Why is (x,(y,z)) equal to ((x,y),z)?

  1. Sep 20, 2008 #1
    I've started learning about set theory and the book I'm using, in the first couple of paragraphs, isn't quite as detailed and rigorous as I would like it to be, so I'm left with quite some questions. A more advanced student of Mathematics has been able to answer most of them for me, but I'm still stuck with one problem, which is the statement that

    [tex](x, (y, z)) = ((x, y), z) = (x, y, z).[/tex]

    The parentheses denote ordered pairs (or a triplet, in the last case). Would anyone show me why this is true? Or is this too advanced to understand after ten pages of set theory?
  2. jcsd
  3. Sep 20, 2008 #2
    Can you give more context? Just at my first glace (I have not taken a formal set theory class), it looks like that's just a definition. If it's an ordered tuple, then grouping does not matter.
  4. Sep 20, 2008 #3
    I did a problem which was to show that

    [tex](A \times B) \times C = A \times (B \times C) = A \times B \times C[/tex]

    except for the placement of the parentheses in the ordered tuples. I managed to and the tuples came out as in my first post. As a side note, it stated that there is no distinction between the three tuples. I'm wondering why.

    I suppose it could be a definition, but the book says "in practice, no distinction is made for products of sets associated in different ways". It didn't really sound like a definition to me, that's why I thought it reasonable to look for an explanation.

    EDIT: There are supposed to be product crosses between those A's, B's and C's. Not sure what I did wrong.
    Last edited: Sep 20, 2008
  5. Sep 21, 2008 #4


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    You showed

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

    which, in terms of sets, means that the action of forming a Cartesian product satisfies the associated property: that is, since a Cartesian product is a binary operation on sets, initially we are justified in asking if, in a situation like this, it makes a difference to which two sets we initially apply the product. Your work showed that it does not.

    This translates as follows: the notation [tex] ((a,b),c) [/tex] corresponds to the first of the products above, the notation [tex] (a,(b,c)) [/tex] to the second, and [tex] (a,b,c) [/tex] to the third. Because the three products give the same set (this is the result you proved) we do not need to distinguish between the three ways of writing the ordered triples.
  6. Sep 21, 2008 #5
    Ah, but I did not prove that. The problem was to prove that those products are equal "except for the placement of the parentheses", which is what I'm asking about.
  7. Sep 21, 2008 #6


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    Sorry - I simply misread your earlier comments.
    In order to show that two sets are equal you need to show that they contain the same elements (a fact of which you are probably currently aware). Think this way:
    Suppose you have

    ((x,y), z) \in \left(X \times Y\right) \times Z

    From the definition of Cartesian product, you know these two things:

    (x,y) \in X \times Y, \quad z \in Z

    Can you go from this to showing that [tex] (x, (y,z)) \in X \times \left(X \times Y \right) [/tex]

    (A comment: I'm getting a 'database error' now when I try to preview my work, so I won't be able to proof-read my Latex until this is posted. I will apologize now for any "male typing syndrome" that results in a formatting error.)

    Similar steps can be used to show the other set containments you need.
    Please post again if this doesn't help. Good luck
  8. Sep 21, 2008 #7
    They're not really equal in the strictest sense under any definitions I know (I'm no expert here), but there's an obvious bijection. So while ((1,2),3) is not an element of Rx(RxR), it obviously corresponds to (1,(2,3)) in Rx(RxR). That's why your book says no distinction is made in practice.
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