<p>No. And yes. For any set <span class="math-container">$X$</span> we have (by definition) <span class="math-container">$$X^k=\underbrace{X\times\cdots\times X}_{k\text{-times}}=\{(x_1,x_2,\ldots,x_k)\mid x_i\in X\text{ for }i=1,\ldots,k\}.$$</span></p>
<p>Now let, for example, <span class="math-container">$m=2$</span> and <span class="math-container">$n=3$</span>. Then
<span class="math-container">$$(\mathbb{R}^m\big)^n=(\mathbb{R}^m\big)^n$$</span> <span class="math-container">$$=\big(\mathbb{R}^2\big)^3=\Big\{\big((x_1,x_2),(x_3,x_4),(x_5,x_6)\big)\mid (x_1,x_2)\in\mathbb{R}^2,(x_3,x_4)\in\mathbb{R}^2, (x_5,x_6)\in\mathbb{R}^2\Big\}$$</span> <span class="math-container">$$=\Big\{\big((x_1,x_2),(x_3,x_4),(x_5,x_6)\big)\mid x_i\in\mathbb{R}\text{ for }i=1,\ldots,6\Big\}.$$</span>
This is different from <span class="math-container">$$\mathbb{R}^{m\cdot n}=\mathbb{R}^{2\cdot 3}=\mathbb{R}^6$$</span>
<span class="math-container">$$=\Big\{(x_1,x_2,x_3,x_4,x_5,x_6\big)\mid x_i\in\mathbb{R}\text{ for }i=1,\ldots,6\Big\}.$$</span>
So they are not the same, formally <span class="math-container">$(\mathbb{R}^m\big)^n\neq \mathbb{R}^{m\cdot n}$</span> 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.</p>

CC BY-SA 4.0
2021-02-12T11:13:21.657Z
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.
Nomista
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2021-02-12T20:31:22.403Z
@Nomista: Please consider **accepting** the answer by clicking the check mark on the left, if you think it answers your question.
Herr K.
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2021-02-13T09:55:07.683Z
Thanks, for letting me know! I didn't know about the check mark.
Nomista
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2021-02-12T09:04:53.067Z
2021-02-12T09:31:30.047Z
2021-02-12T09:31:30.047Z
Michael Greinecker
<p>Very simple question here: say that there are <em>n</em> individuals and each individual <span class="math-container">$i\leq n$</span> has a consumption bundle <span class="math-container">$x_i\in \mathbb{R}^m$</span> (i.e. there are <span class="math-container">$m$</span> types of goods). Suppose that social welfare is a function <span class="math-container">$f(x_1, x_2, ..., x_n)$</span>. The domain of <span class="math-container">$f$</span> is clearly <span class="math-container">$(\mathbb{R}^m)^n$</span>. My question whether the set <span class="math-container">$(\mathbb{R}^m)^n$</span> is the same as <span class="math-container">$\mathbb{R}^{m\cdot n}$</span>?</p>
<p>In other words, is <span class="math-container">$(\mathbb{R}^m)^n$</span> the real coordinate space of dimension <span class="math-container">$m\cdot n$</span>?</p>

CC BY-SA 4.0
2021-02-11T16:40:56.453Z
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
afreelunch
CC BY-SA 4.0
2021-02-11T19:28:23.810Z
Thanks! I hadn't seen the notation either until I saw the notation in this paper (see second equation on page 7): https://itzhakgilboa.weebly.com/uploads/8/3/6/3/8363317/argenziano_et_al-2018-the_economic_journal.pdf
Nomista
<mathematical-economics><consumption><welfare-economics><social-choice>
Is $(\mathbb{R}^m)^n$ the real coordinate space of dimension $m\cdot n$?
Nomista
# Is $(\mathbb{R}^m)^n$ the real coordinate space of dimension $m\cdot n$?

(Asked by: username_0 on 2021-02-11)

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 $m\cdot n$?

**Tags:** mathematical-economics, consumption, welfare-economics, social-choice
> **Comments:**
> **username_1**: 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  
> **username_0**: Thanks! I hadn't seen the notation either until I saw the notation in this paper (see second equation on page 7): https://itzhakgilboa.weebly.com/uploads/8/3/6/3/8363317/argenziano_et_al-2018-the_economic_journal.pdf  

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## 1 answers

### Answer by username_2 on 2021-02-12. Score: 5 | Accepted answer

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.

> **Comments:**
> **username_0**: 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.  
> **username_3**: @Nomista: Please consider **accepting** the answer by clicking the check mark on the left, if you think it answers your question.  
> **username_0**: Thanks, for letting me know! I didn't know about the check mark.  

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