<p>This seems like an interesting question but I am not sure I understand all the details here, so this is more a comment than an answer.
Because the preference relation is incomplete, we cannot have a real-valued utility function representing it (recall that indifference is different from indecisiveness because the latter is not transitive).</p>

<p>Moreover, I do not think that you can have an arbitrary set $X$ together with an arbitrary $\succeq$ being represented by a finite-dimensional $u$. To give a simple example, suppose that $X=[1,p_n]$ with $p_n$ being the $n-th$ prime and your preferences are $x\succeq y$ if $x\geq y$ and $x,y\in(p_k,p_{k+1}]$ for some $k&lt;n$. There are $n$ subsets of alternatives, each with a well-defined preference order but alternatives are no comparable across subsets so that you need at least $f(n)$ (not sure about the exact $f$) dimensions in your vector-valued utility function. When $X=[1,\infty)$, the number of dimensions explodes.</p>

<p>An important example is the case of social choice. Suppose that you have $k$ individuals with preferences over $X$ and you want to aggregate them. The Pareto order then gives you $u(x)=(u_i(x))$ for all $i$ the (ordinal) utility of individual $i$ from alternative $x$. Indeed, if $u_i(x)&gt;u_i(y)$ for all $i$, then $u(x)&gt;u(y)$. </p>

<p>Finally, if $X$ was finite, then simply order the elements in some way $x^1,x^2,..,x^k$ and construct $u\in \mathbb R^k$ as follows:
$u_l(x^l)=1$, $u_l(x^k)=1$ iff $x^l\succeq x^k$ and $u_l(x^k)=0$ otherwise. </p>

<p>To see that this represents the preferences, notice that if $x^l\succeq x^{l'}$, $u_l(x^l)=1&gt;u_l(x^{l'})$, $u_{l'}(x^l)=u_{l'}(x^{l'})$; if $x^m\succeq x^l$ it must be that $x^m \succeq x^{l'}$ so that $u_m(x^l)=u_m(x^{l'})$ and viceversa but if $x^l\succeq x^m\succeq x^{l'}$, then $u_m(x^l)&gt;u_m(x^{l'})$; and if $x^m$ is not comparable with $x^l$ it cannot be worse than $x^{l'}$ so that $u_m(x^l)=u_m(x^{l'})=0$, while if it is not comparable with $x^{l'}$ it cannot be better than $x^l$.</p>

CC BY-SA 3.0
2017-09-04T12:50:51.490Z
2017-09-04T12:50:51.490Z
Fato
<blockquote>
  <p>Let $\succeq$ be a preorder, i.e., it is reflexive and transitive on $X$. Prove that there exists an index set $N$ and a multi-utility function $u:X \rightarrow \mathbb{R}^N$ such that $x \succeq y$ if and only if $[u(x)](n) &gt; [u(y)](n)$ for all $n \in N$.</p>
</blockquote>

<p>I know, for example, the relation $x \succeq y$ on $\mathbb{R}^2$ if $x_1 \ge y_1$ and $x_2 \ge y_2$ can be represented with $N= \{1,2\}$ and the multi-utility function defined by $u(x) = (x_1, x_2)$. But, I am unsure of how to do the general proof.</p>

CC BY-SA 3.0
2017-09-04T04:21:41.087Z
The notation confuses me. Why is utility in brackets and the index in parentheses? Why does utility give a vector as its output?
Kitsune Cavalry
<microeconomics><utility>
Preorder and multi-utility function
TeTs
# Preorder and multi-utility function

(Asked by: username_0 on 2017-09-04)

> Let $\succeq$ be a preorder, i.e., it is reflexive and transitive on $X$. Prove that there exists an index set $N$ and a multi-utility function $u:X \rightarrow \mathbb{R}^N$ such that $x \succeq y$ if and only if $[u(x)](n) > [u(y)](n)$ for all $n \in N$.

I know, for example, the relation $x \succeq y$ on $\mathbb{R}^2$ if $x\_1 \ge y\_1$ and $x\_2 \ge y\_2$ can be represented with $N= \{1,2\}$ and the multi-utility function defined by $u(x) = (x\_1, x\_2)$. But, I am unsure of how to do the general proof.

**Tags:** microeconomics, utility
> **Comments:**
> **username_1**: The notation confuses me. Why is utility in brackets and the index in parentheses? Why does utility give a vector as its output?  

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## 1 answers

### Answer by username_2 on 2017-09-04. Score: 1 | Accepted answer

This seems like an interesting question but I am not sure I understand all the details here, so this is more a comment than an answer.
Because the preference relation is incomplete, we cannot have a real-valued utility function representing it (recall that indifference is different from indecisiveness because the latter is not transitive).

Moreover, I do not think that you can have an arbitrary set $X$ together with an arbitrary $\succeq$ being represented by a finite-dimensional $u$. To give a simple example, suppose that $X=[1,p\_n]$ with $p\_n$ being the $n-th$ prime and your preferences are $x\succeq y$ if $x\geq y$ and $x,y\in(p\_k,p\_{k+1}]$ for some $k

An important example is the case of social choice. Suppose that you have $k$ individuals with preferences over $X$ and you want to aggregate them. The Pareto order then gives you $u(x)=(u\_i(x))$ for all $i$ the (ordinal) utility of individual $i$ from alternative $x$. Indeed, if $u\_i(x)>u\_i(y)$ for all $i$, then $u(x)>u(y)$.

Finally, if $X$ was finite, then simply order the elements in some way $x^1,x^2,..,x^k$ and construct $u\in \mathbb R^k$ as follows:
$u\_l(x^l)=1$, $u\_l(x^k)=1$ iff $x^l\succeq x^k$ and $u\_l(x^k)=0$ otherwise.

To see that this represents the preferences, notice that if $x^l\succeq x^{l'}$, $u\_l(x^l)=1>u\_l(x^{l'})$, $u\_{l'}(x^l)=u\_{l'}(x^{l'})$; if $x^m\succeq x^l$ it must be that $x^m \succeq x^{l'}$ so that $u\_m(x^l)=u\_m(x^{l'})$ and viceversa but if $x^l\succeq x^m\succeq x^{l'}$, then $u\_m(x^l)>u\_m(x^{l'})$; and if $x^m$ is not comparable with $x^l$ it cannot be worse than $x^{l'}$ so that $u\_m(x^l)=u\_m(x^{l'})=0$, while if it is not comparable with $x^{l'}$ it cannot be better than $x^l$.


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