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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 $m\cdot n$?

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    $\begingroup$ Clearly, the domain of $f$ is $\mathbb{R}^{mn}$ (to obtain `social welfare', you plug in $m$ goods for each of the $n$ consumers, so $nm$ numbers in total). What does $({\mathbb{R}^{m}})^n$ mean? (The same?) I have never seen this notation $\endgroup$
    – afreelunch
    Feb 11 at 16:40
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    $\begingroup$ Thanks! I hadn't seen the notation either until I saw the notation in this paper (see second equation on page 7): itzhakgilboa.weebly.com/uploads/8/3/6/3/8363317/… $\endgroup$
    – Nomista
    Feb 11 at 19:28
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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.

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  • $\begingroup$ Thanks! This makes a lot of sense and was in one sense what I was worried about (I need the domain to be a real coordinate space). However, given that they are equivalent as vector spaces and topological spaces, it should all work fine for my purposes. $\endgroup$
    – Nomista
    Feb 12 at 11:13
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    $\begingroup$ @Nomista: Please consider accepting the answer by clicking the check mark on the left, if you think it answers your question. $\endgroup$
    – Herr K.
    Feb 12 at 20:31
  • $\begingroup$ Thanks, for letting me know! I didn't know about the check mark. $\endgroup$
    – Nomista
    Feb 13 at 9:55

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