Is $(\mathbb{R}^m)^n$ the real coordinate space of dimension $m\cdot n$? Very simple question here: say that there are n individuals and each individual $i\leq n$ has a consumption bundle $x_i\in \mathbb{R}^m$ (i.e. there are $m$ types of goods). Suppose that social welfare is a function $f(x_1, x_2, ..., x_n)$. The domain of $f$ is clearly $(\mathbb{R}^m)^n$. My question whether the set $(\mathbb{R}^m)^n$ is the same as $\mathbb{R}^{m\cdot n}$?
In other words, is $(\mathbb{R}^m)^n$ the real coordinate space of dimension $m\cdot n$?
No. And yes. For any set $X$ we have (by definition) $$X^k=\underbrace{X\times\cdots\times X}_{k\text{-times}}=\{(x_1,x_2,\ldots,x_k)\mid x_i\in X\text{ for }i=1,\ldots,k\}.$$
Now let, for example, $m=2$ and $n=3$. Then
$$(\mathbb{R}^m\big)^n=(\mathbb{R}^m\big)^n$$ $$=\big(\mathbb{R}^2\big)^3=\Big\{\big((x_1,x_2),(x_3,x_4),(x_5,x_6)\big)\mid (x_1,x_2)\in\mathbb{R}^2,(x_3,x_4)\in\mathbb{R}^2, (x_5,x_6)\in\mathbb{R}^2\Big\}$$ $$=\Big\{\big((x_1,x_2),(x_3,x_4),(x_5,x_6)\big)\mid x_i\in\mathbb{R}\text{ for }i=1,\ldots,6\Big\}.$$
This is different from $$\mathbb{R}^{m\cdot n}=\mathbb{R}^{2\cdot 3}=\mathbb{R}^6$$
$$=\Big\{(x_1,x_2,x_3,x_4,x_5,x_6\big)\mid x_i\in\mathbb{R}\text{ for }i=1,\ldots,6\Big\}.$$
So they are not the same, formally $(\mathbb{R}^m\big)^n\neq \mathbb{R}^{m\cdot n}$ However, there is clearly a natural way to associate points in one space with points in the other space and this way to connect them preserves a lot of structure. They are equivalent as vector spaces, topological spaces and whatnot. For this reason, one might often identify these two spaces and unless one does very strange things, little can go wrong.