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Cartesian product

  1. May 19, 2005 #1

    quasar987

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    Is it true that

    [tex]\mathbb{R}\times \mathbb{R}^2 = \mathbb{R}^2 \times \mathbb{R} = \mathbb{R}^3[/tex]

    ?
     
  2. jcsd
  3. May 19, 2005 #2

    Hurkyl

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    Yes. (with a but)
     
  4. May 19, 2005 #3
    Wouldn't the elements of [tex]\mathbb{R}\times\mathbb{R}^2[/tex] all be of the form (a, (b, c)) whereas the elements of [tex]\mathbb{R}^2\times\mathbb{R}[/tex] all be of the form ((a, b), c)?
     
  5. May 19, 2005 #4
    (Assuming you meant the 2 to have precedence over the x)
     
  6. May 19, 2005 #5

    Hurkyl

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    That's the essense of the "but".
     
  7. May 19, 2005 #6

    quasar987

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    But how important is this "but"?
     
  8. May 19, 2005 #7

    Hurkyl

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    It's so unimportant, I didn't feel the need to elaborate on it in my response. :biggrin:

    In the strictest sense, they aren't equal, but I don't think I've ever really seen anyone distinguish between them.
     
  9. May 20, 2005 #8

    matt grime

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    Strictly speaking they are isomorphic not equal, however the isomoprphism is fairly 'canonical' and in general there are canonical isomorphisms between Ax(BxC) and (AxB)xC, and we will by commonly accepted abuse of notation refer to it as AxBxC.

    This is one of the "modern" ways of saying it in the laguage of category theory.
     
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