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Cartesian product of cartesian products

  1. May 15, 2008 #1


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    So if we have sets A X B where A is (a_n, b_m) and C X D where C is (c_o, d_p), the cartesian product of the sets is (A X B) X (C X D) [(a_n, b_m, c_o, d_p)]. Is this correct? And thus, do parenthesis matter at all in Cartesian products? What about order? Is (a_n, b_m) equivalent to (b_m, a_n)?
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  3. May 17, 2008 #2
    I believe these are in fact ordered pairs. So order does matter.

    If we use the definition that an ordered pair [itex](x,y) = \{\{x\}, \{x,y\}\}[/itex] then clearly [itex](y,x) = \{\{y\}, \{x,y\}\} \neq (x,y)[/itex]

    As for the first thing say [itex]A[/itex] contains elements of the form [itex](x,y)[/itex] then [itex]A\times B = \{ ((x,y), b) | (x,y) \in A, b \in B \}[/itex]
    Last edited: May 17, 2008
  4. May 17, 2008 #3


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    The cartesian product is a symmetric monoidal operation (both on sets, and on elements) -- that means it has an identity, is associative, and is commutative... but only up to a natural isomorphism.

    An identity set is any set with a single element. I'll call such a set '1'.
    For all sets A, B, C, we have:
    An isomorphism 1xA --> A ('left identity')
    An isomorphism A --> Ax1 ('right identity')
    An isomorphism (AxB)xC --> Ax(BxC) (the 'associator')
    An isomorphism AxB --> BxA (the 'braiding')

    For example, the associator is the function:
    [itex]\alpha((a, b), c) = (a, (b, c))[/itex]

    None of these isomorphisms need be equalities. In fact, for the usual set-theoretic model of the ordered pair, none of these will be equalities. (I think the only exception is when the empty set is involved)

    If we have a function A --> B, then we have functions
    AxC --> BxC
    CxA --> CxB

    And each of these natural isomorphisms 'commute' with applying a function. i.e. the two compositions

    AxC --> BxC --> CxB
    AxC --> CxA --> CxB

    both yield the same function AxC --> CxB.
    Last edited: May 17, 2008
  5. May 17, 2008 #4


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    Order is important but the "parentheses" don't- the Cartesian product is "associative" but not "commutative".
  6. May 17, 2008 #5


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    I don't know. Using the standard (a, b) := {{a}, {a, b}}, [itex](a, (b, c))\neq((a, b), c)[/itex]. There is a natural bijection between them, but then again there's a natural bijection between (a, b) and (b, a) too -- so why do you say that they associate but not commute?
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