What are the differences between $\mathbb{R}^{k+m}$ and $\mathbb{R}^{k}×\mathbb{R}^{m}$ For $k,m \in \mathbb{N}$ are the two sets exactly the same? Or they are same only for $k = m = 1$?
 A: Formally, the two sets are different: the left one consists of elements of the form $\;(x_1,...,x_k,x_{k+1},...x_{k+m})\;$, whether the right one consists of ordered pairs of the form $\;\left((x_1,...,x_k),\,(x_1,...,x_m)\right)\;$
A: They are not exactly the same, since by definition $\mathbb{R}^{k+m}$ is the set of ordered $(k+m)$-tuples of real numbers, while $\mathbb{R}^k\times \mathbb{R}^m$ is the set of ordered pairs $(a,b)$ with $a\in \mathbb{R}^k$ and $b\in\mathbb{R}^m$. Anyway, you may identify the two sets with a bijection, sending an element $(a_1,\ldots, a_k, b_{k+1},\ldots, b_{k+m})$ to the pair $\left((a_1, \ldots, a_k), (b_{k+1},\ldots, b_{k+m})\right)$. In general, you may identify $\mathbb{R}^{k+m}$ with $\mathbb{R}^i\times\mathbb{R}^j$, with $i+j=k+m$. This is the generalized associative law for cartesian product of sets. If $A, B$ and $C$ are sets, you may verify that there is an obvious bijection between $(A\times B)\times C$ and $A\times (B\times C$), which leads to a generalized associative property, that is, given sets $A_1, A_2, \ldots, A_n$, then repeated application of the operation $\times$ produces the same result regardless of how  parentheses are inserted in the expression. Therefore, all products you can form starting form $A_1,\ldots, A_n$, may be identified with the product you get by the following inductive rule:
$$A_1\times\ldots A_n=(A_1\times\ldots A_{n-1})\times A_n$$
which in your case $A_1=\ldots, A_{k+m}=\mathbb{R}$, turns out to be $\mathbb{R}^{k+m}$
