1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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?
     
  2. jcsd
  3. Aug 3, 2010 #2

    Gib Z

    User Avatar
    Homework Helper

    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

    User Avatar
    Science Advisor

    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

    User Avatar
    Science Advisor

    Yes, that's what we are saying.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook