Is there a difference between $\mathbb{R}^{2n}$ and $\mathbb{R}^n \times \mathbb{R}^n$ or are these interchangeable notations? Is there a difference between $\mathbb{R}^{2n}$ and $\mathbb{R}^n \times \mathbb{R}^n$ or are these interchangeable notations?
Essentially I am confused about whether they are the same when you equip the same metrics or topologies, or if you were to denote a vector from either sets. Do you get the same results in every possible cases?
 A: Edit: Also, see difference between $\mathbb{R}^2$ and $\mathbb{R} \times \mathbb{R}$
They are going to be isomorphic (e.g., topologically, metrically) in almost any way possible via $$((x_1,x_2,\ldots, x_n),(y_1,y_2,\ldots,y_n)) \mapsto (x_1,x_2,\ldots,x_n,y_1,y_2,\ldots,y_n).$$
A: Almost in every circumstance, they are the same.  Certainly they are homeomorphic and diffeomorphic if you are looking at smooth structures.  But really it just comes down to notation.  They are not equal on the nose though.  A point in $\mathbb{R}^n$ has the form $(x_1, \ldots, x_n)$ while a point in $\mathbb{R}^m$ has the form $(y_1, \ldots, y_m)$.  So if you are being extra precise, a point in $\mathbb{R}^n\times \mathbb{R}^m$ will be an ordered pair:  $((x_1, \ldots, x_n), (y_1, \ldots, y_m))$, but we basically always immediately ignore this and call $y_i$ the coordinate $x_{n+i}$, giving you the homeomporhism $\mathbb{R}^n\times \mathbb{R}^m \to \mathbb{R}^{n+m}$, where $((x_1, \ldots, x_n), (y_1, \ldots, y_m))\mapsto(x_1, \ldots, x_{n+m}) $.
A: Perhaps this answer would fit better in the Mathematics Educators stack exchange, but they do have a subtle difference from a teaching perspective.  I might use $\mathbb{R}^n \times \mathbb{R}^n$ as part of an initial category theory proof regarding products with the intent to generalize it to other general categories once the underlying understanding was in place.  Converting it to $\mathbb{R}^{2n}$ and back would involve leveraging the structure of $\mathbb{R}^n$ which I didn't want to be dependent on in the more general case.
