difference between $\mathbb{R}^2$ and $\mathbb{R} \times \mathbb{R}$ I was going through some of notes in regards to Fourier analysis and I noticed that in some cases when dealing with a 2 dimensional transform the function $f \in \mathbb{R}^2$ while other times $f \in \mathbb{R} \times \mathbb{R}$. Is there any difference between both these spaces, or is it just a notation?
 A: They're just two different ways of denoting the same set. Likewise
$$\mathbb{R}^n = \underbrace{\mathbb{R} \times \mathbb{R} \times \cdots \times \mathbb{R}}_{n\ \text{copies}}$$

Added: Since I'm a logician these days, I can't help but add the following.
The above is true for the purposes of analysis. If you're being a bit more formal about things, you might argue that $\mathbb{R} \times \mathbb{R}$ denotes the set of all ordered pairs of the form $\langle x,y \rangle$, where $x$ and $y$ are real numbers, and $\mathbb{R}^2$ is the set of functions $2 \to \mathbb{R}$, where $2$ denotes the set $\{ 0, 1 \}$.
In any case, there is a natural bijection between the two sets. Namely, in one direction you send the function $f$ to the ordered pair $\langle f(0), f(1) \rangle$, and in the other direction you send the pair $ \langle x,y \rangle$ to the function $f$ defined by $f(0)=x$ and $f(1)=y$.
A: I imagine the only reason they might use the $\mathbb{R}\times\mathbb{R}$ notation is to emphasise one of the $\mathbb{R}$ dimensions, for instance time. In a technical sense $\mathbb{R}\times\mathbb{R}$ has more structure than $\mathbb{R}^2$ as it also naturally comes equipped with two maps on to its factors $\mathbb{R}^{(1)}$ and $\mathbb{R}^{(2)}$ which $\mathbb{R}^2$ doesn't automatically come equipped with (until one shows that it's isomorphic to $\mathbb{R}\times\mathbb{R}$).
It is true that these sets are naturally isomorphic in almost any natural category you can think of, but abstractly it's important to note that they are not equal.
