# Preorder and multi-utility function

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.

• The notation confuses me. Why is utility in brackets and the index in parentheses? Why does utility give a vector as its output? Sep 4, 2017 at 4:21

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 $k<n$. 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.
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 $x^l$.