<p>Note first that for each nontrivial (more than one point) compact interval $I$, there exists a strictly increasing function from $\mathbb{R}$ to $I$.</p>

<p>Let $\langle q_1,q_2, q_3,\ldots\rangle$ be an enumeration of $\mathbb{Q}$. We will define $u:\mathbb{Q}\times\mathbb{R}\to\mathbb{R}$ inductively, one $q_n$ after another.</p>

<p>Let $I_1$ be a nontrivial compact interval and $u_1:\mathbb{R}\to I_1$ be strictly increasing.  </p>

<p>Now assume there are disjoint nontrivial compact intervals $I_1,\ldots,I_n$ that ordered on the line like $q_1,\ldots,q_n$, and strictly increasing functions $u_m:\mathbb{R}\to I_m$ for $m\leq n$. Since these intervals are closed and disjoint, there is space between any two consecutive intervals. So one can put a nontrivial compact interval $I_{n+1}$ in a place corresponding to $q_{n+1}$ among $q_1,\ldots, q_n$. Choose also some strictly increasing function $u_{n+1}:\mathbb{R}\to I_{n+1}$. </p>

<p>This will give you a disjoint sequence $\langle I_1, I_2, \ldots\rangle$ of nontrivial intervals and a sequence of functions $\langle u_1, u_2, \ldots\rangle$ with $u_n:\mathbb{R}\to I$ be strictly increasing. Now define $u:\mathbb{Q}\times\mathbb{R}\to \mathbb{R}$ by $u(q_n,r)=u_n(r)$. </p>

<p>You can verify that $u$ represents the lexicographic ordering on $\mathbb{Q}\times\mathbb{R}$.</p>

<p>Warning: The existence of a countably infinite number of disjoint nontrivial intervals might appear very strange, but is perfectly fine. </p>

CC BY-SA 4.0
2018-09-09T17:17:38.737Z
Thanks very much for this answer. But I am not sure I completely got the idea of the proof.
Rororo
CC BY-SA 4.0
2018-09-09T17:23:25.490Z
The, maybe somewhat surprising, idea is that one can order nontrivial compact intervals in exactly the same way the rationals are ordered. This gives you the ordering of the first coordinate, the numbers within the intervals give you the ordering of the second coordinate.
Michael Greinecker
CC BY-SA 4.0
2018-09-09T17:32:31.893Z
When you assume that "there are disjoint nontrivial compact intervals I1,…,In ordered on the line like q1,…,qn" does this mean that we consider nontrivial compact intervals with q1 and all the reals that may be "coupled" with q1, then I2 being the nontrivial compact interval with all the reals attached to q2 ?

And finally while defining u(q_n,r)=u_n(r) are we done ? Shouldn't we define a function like u=2^(-i) where i would be the number of elements in the compact interval for ex. ?

infinitely many thanks again!
Rororo
CC BY-SA 4.0
2018-09-09T17:39:50.133Z
The ordering assumption means that if $q_l < q_m$ (as rational numbers, not in the order of the inidces), then for all $r_l\in I_l$ and $r_m\in I_m$, we have $r_l<r_m$. The reason that the function $u$ is constructed inductively, and not in the familiar way of "counting smaller elements (in a potentially weighted way)," is that I don't know how one can specify the lenghts and positions of the intervals $I_n$ a priori.
Michael Greinecker
CC BY-SA 4.0
2018-09-09T00:37:07.943Z
2018-09-09T00:37:07.943Z
Michael Greinecker
<p>I would like to ask for your help. I recently learned that the Lexicographic Preference relation can be represented by a utility function <span class="math-container">$u:X\to\mathbb{R}$</span> on <span class="math-container">$\mathbb{Q}\times\mathbb{R}$</span> (but not <span class="math-container">$\mathbb{R}\times\mathbb{Q}$</span>).</p>
<p>To recall, a lexicographic preference relation says that on <span class="math-container">$\mathbb{R^2}$</span>, <span class="math-container">$x\succeq y$</span> if and only if <span class="math-container">$x_1&gt;y_1$</span>  or <span class="math-container">$x_1=y_1$</span> and <span class="math-container">$x_2\geq y_2$</span> where <span class="math-container">$x=(x_1,x_2)$</span> and <span class="math-container">$y=(y_1,y_2)$</span>.</p>
<p>Thus, I would have liked to see a proof of this statement and how is it possible to build such a utility representation. I guess that we must assume first that there exist a utility representation for the Lexicographic preference on <span class="math-container">$\mathbb{Q}\times\mathbb{Q}$</span> since the latter is countable, but I am lost after.</p>
<p>Infinitely many thanks for your help!</p>

<preferences><decision-theory>
Lexicographic Preference Relation on the QxR
Rororo
# Lexicographic Preference Relation on the QxR

(Asked by: username_0 on 2018-09-08)

I would like to ask for your help. I recently learned that the Lexicographic Preference relation can be represented by a utility function $u:X\to\mathbb{R}$ on $\mathbb{Q}\times\mathbb{R}$ (but not $\mathbb{R}\times\mathbb{Q}$).

To recall, a lexicographic preference relation says that on $\mathbb{R^2}$, $x\succeq y$ if and only if $x\_1>y\_1$ or $x\_1=y\_1$ and $x\_2\geq y\_2$ where $x=(x\_1,x\_2)$ and $y=(y\_1,y\_2)$.

Thus, I would have liked to see a proof of this statement and how is it possible to build such a utility representation. I guess that we must assume first that there exist a utility representation for the Lexicographic preference on $\mathbb{Q}\times\mathbb{Q}$ since the latter is countable, but I am lost after.

Infinitely many thanks for your help!

**Tags:** preferences, decision-theory

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## 1 answers

### Answer by username_1 on 2018-09-09. Score: 5

Note first that for each nontrivial (more than one point) compact interval $I$, there exists a strictly increasing function from $\mathbb{R}$ to $I$.

Let $\langle q\_1,q\_2, q\_3,\ldots\rangle$ be an enumeration of $\mathbb{Q}$. We will define $u:\mathbb{Q}\times\mathbb{R}\to\mathbb{R}$ inductively, one $q\_n$ after another.

Let $I\_1$ be a nontrivial compact interval and $u\_1:\mathbb{R}\to I\_1$ be strictly increasing.

Now assume there are disjoint nontrivial compact intervals $I\_1,\ldots,I\_n$ that ordered on the line like $q\_1,\ldots,q\_n$, and strictly increasing functions $u\_m:\mathbb{R}\to I\_m$ for $m\leq n$. Since these intervals are closed and disjoint, there is space between any two consecutive intervals. So one can put a nontrivial compact interval $I\_{n+1}$ in a place corresponding to $q\_{n+1}$ among $q\_1,\ldots, q\_n$. Choose also some strictly increasing function $u\_{n+1}:\mathbb{R}\to I\_{n+1}$.

This will give you a disjoint sequence $\langle I\_1, I\_2, \ldots\rangle$ of nontrivial intervals and a sequence of functions $\langle u\_1, u\_2, \ldots\rangle$ with $u\_n:\mathbb{R}\to I$ be strictly increasing. Now define $u:\mathbb{Q}\times\mathbb{R}\to \mathbb{R}$ by $u(q\_n,r)=u\_n(r)$.

You can verify that $u$ represents the lexicographic ordering on $\mathbb{Q}\times\mathbb{R}$.

Warning: The existence of a countably infinite number of disjoint nontrivial intervals might appear very strange, but is perfectly fine.

> **Comments:**
> **username_0**: Thanks very much for this answer. But I am not sure I completely got the idea of the proof.  
> **username_1**: The, maybe somewhat surprising, idea is that one can order nontrivial compact intervals in exactly the same way the rationals are ordered. This gives you the ordering of the first coordinate, the numbers within the intervals give you the ordering of the second coordinate.  
> **username_0**: When you assume that "there are disjoint nontrivial compact intervals I1,…,In ordered on the line like q1,…,qn" does this mean that we consider nontrivial compact intervals with q1 and all the reals that may be "coupled" with q1, then I2 being the nontrivial compact interval with all the reals attached to q2 ?  And finally while defining u(q_n,r)=u_n(r) are we done ? Shouldn't we define a function like u=2^(-i) where i would be the number of elements in the compact interval for ex. ?  infinitely many thanks again!  
> **username_1**: The ordering assumption means that if $q_l < q_m$ (as rational numbers, not in the order of the inidces), then for all $r_l\in I_l$ and $r_m\in I_m$, we have $r_l<r_m$. The reason that the function $u$ is constructed inductively, and not in the familiar way of "counting smaller elements (in a potentially weighted way)," is that I don't know how one can specify the lenghts and positions of the intervals $I_n$ a priori.  

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