Cartesian product of R^n and R^m

[JG89]

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?

 

[Gib Z]

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

 

[HallsofIvy]

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 R^m and Rⁿ then the two spaces, RⁿXR^m and R^{m+ n}, are also isomorphic.

 

[JG89]

I was asked to prove that if M is a k-manifold without boundary in R^m, and if N is an l-manifold in R^n, then M * N is a (k+l)-manifold in R^{m+n}.

I'm guessing then they are talking about an m+n tuple of real numbers?

 

[HallsofIvy]

Yes, that's what we are saying.