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Cartesian product of R^n and R^m

  1. Aug 3, 2010 #1
    This is going to be a weird question, but in textbooks when we're given the two spaces R^n and R^m, and they say something about R^(n+m), then are they referring to ordered pairs of ordered pairs? That is, if x is in R^n and y is in R^m, then R^(n+m) is the set of all ordered pairs (x,y). So for example if n = 1 and m = 2, then all ordered pairs of ordered pairs: (x, (y,z)) where x is in R and (y,z) is in R^2?

    Or do they just mean an (n+m)tuple of real numbers?
     
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  3. Aug 3, 2010 #2

    Gib Z

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    You have to tell by context, if you see [tex]\mathbb{R}^{m+n}[/tex] written without anything else, then you have to assume just a (m+n) tuple of real numbers, but when previously talking about the spaces [tex]\mathbb{R}^n[/tex] and [tex]\mathbb{R}^m[/tex], I'm quite sure they mean the Cartesian product.
     
  4. Aug 3, 2010 #3

    HallsofIvy

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    Note that if a= (x, y, z) and b= (u, v, w) then (a, b)= ((x, y, z), (u, v, w)) is equivalent to (x, y, z, u, v, w). If you have addition, scalar multiplication, etc. for Rm and Rn then the two spaces, RnXRm and Rm+ n, are also isomorphic.
     
  5. Aug 3, 2010 #4
    I was asked to prove that if M is a k-manifold without boundary in [tex] R^m [/tex], and if N is an l-manifold in [tex] R^n [/tex], then M * N is a (k+l)-manifold in [tex] R^{m+n} [/tex].

    I'm guessing then they are talking about an m+n tuple of real numbers?
     
  6. Aug 4, 2010 #5

    HallsofIvy

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    Yes, that's what we are saying.
     
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