What is the difference between $\Bbb{R^n}\times\Bbb{R^m}$ and $\Bbb{R^{m+n}}$? My book specifies a function: $$\Bbb{R^n}\times\Bbb{R^m}\to\Bbb{R^m}$$ What is the difference between $\Bbb{R^n}\times\Bbb{R^m}$ and $\Bbb{R^{m+n}}$? And if there is none, what are the relative advantages of writing $\Bbb{R^n}\times\Bbb{R^m}$ over $\Bbb{R^{m+n}}$?
Thanks.
 A: One way to look at the isomorphism question is to note that both $\mathbb R^n\times \mathbb R^m$  and $\mathbb R^m\times\mathbb R^n$ are isomorphic with $\mathbb R^{n+m}$ in the above sense. There is another "obvious" isomorphism between $\mathbb R^n\times\mathbb R^m$ and $\mathbb R^m\times\mathbb R^n$. These two do not mix. That is if we compose the isomorphisms:
$$\mathbb R^m\times\mathbb R^n\to\mathbb R^{m+n}\to\mathbb R^{n}\times\mathbb R^{m}$$ 
this composition does not result in the same "swap" isomorphism between $$\mathbb R^{m}\times\mathbb R^n\to\mathbb R^n\times\mathbb R^m$$
So this particular isomorphism is a specific way of mapping the two vector spaces isomorphically. It is not the only way, and it doesn't have any "primacy." It might seem an obvious isomorphism, but it is not actually in any sense the "only right isomorphism." This becomes more and more meaningful as you get deeper into mathematics.
A: Formally speaking, $A\times B$ is the set of ordered pairs $\langle a,b\rangle$ such that $a\in A$ and $b\in B$. And if $A$ is a set, then $A^n$ is the set of $n$-tuples from $A$, which are functions from $\{0,\ldots,n-1\}$ into $A$ (that are much simpler to write as $\langle a_1,\ldots,a_n\rangle$, of course).
So $\Bbb R^n\times\Bbb R^m$ is the set of ordered pairs, $\langle x,y\rangle$ such that $x$ is an $n$-tuple of real numbers, and $y$ is an $m$-tuple of real numbers. Whereas $\Bbb R^{n+m}$ is the set of $(n+m)$-tuples.
However we have a very good way of identifying the two spaces, and so considering them essentially the same. Given a pair $\langle x,y\rangle$ as above, we can "concatenate" the two functions by shifting the domain of $y$ by $n$. So now we have a function $$(x,y)(i)=\begin{cases}x(i) & i<n\\ y(i-n) & n\leq i\end{cases}$$
The domain of this function is $\{0,\ldots,n+m-1\}$, and it's easy to verify that this map, sending $\langle x,y\rangle$ to $(x,y)$ preserves pretty much every structure that we want, including (but not limited to) pointwise addition, scalar multiplication, and so on.

So, why would we separate the two cases, $\Bbb R^n\times\Bbb R^m$ and $\Bbb R^{n+m}$? Well, it's a matter of setting your mind into context. In the first case you want to think about something as a function taking two vectors from (possibly) different spaces, whereas in the latter you want to consider a function taking a single vector from a larger space.
Or perhaps the definition of the function is such that it takes a vector from $\Bbb R^m$, manipulates it using the data from the vector from $\Bbb R^n$ and returns the result. In this case it is much easier the understand this as a function from the product of the spaces. Conceptually.
Other times, you might want to "uniformize" your input, and forget the fact that it came from two different spaces. You'd rather forget that $\Bbb R^n$ represented this type of data, and $\Bbb R^m$ represented that type of data. Now you're only concerned with the values of the said data. So you would rather think about this as a single vector in $\Bbb R^{n+m}$ instead.
A: I imagine the convention being used is: if $X$ is a set (or linear space) then $X^n$ is the set (or space) of $n$-tuples from $X$. That is, an element of $X^n$ looks like
$$(x_1, x_2, \cdots, x_n)$$
where all the $x_i \in X$.
Now if $X$ and $Y$ are sets then $X \times Y = \{ (x,y) : x \in X,\ y \in Y \}$.

Given these two definitions, we have
$$\mathbb{R}^m \times \mathbb{R}^n = \{ ((x_1, x_2, \cdots, x_m), (y_1, y_2, \cdots, y_n)) : x_i, y_i \in \mathbb{R} \}$$
and
$$\mathbb{R}^{m+n} = \{ (x_1, x_2, \cdots, x_m, x_{m+1}, x_{m+2}, \cdots, x_{m+n}) : x_i \in \mathbb{R} \}$$
That is, elements of $\mathbb{R}^m \times \mathbb{R}^n$ are pairs, where the first component is an $m$-tuple and the second is an $n$-tuple. By contrast, the elements of $\mathbb{R}^{m+n}$ are $(m+n)$-tuples. We can 'naturally' translate between these two sets by simply inserting or removing brackets.
More precisely still, there is a natural bijection $\mathbb{R}^m \times \mathbb{R}^n \to \mathbb{R}^{m+n}$, which is also a (natural) linear isomorphism, given by
$$((x_1, x_2, \cdots, x_m), (y_1, y_2, \cdots, y_n)) \mapsto (x_1, x_2, \cdots, x_m, y_1, y_2, \cdots, y_n)$$
as can easily be verified.
