<p>Yes, there are many. Here are some examples :</p>

<p>Consider the weak preference relation <span class="math-container">$\succsim$</span> defined over <span class="math-container">$\mathbb{R}^2$</span> as:</p>

<ul>
<li><strong>Example 1</strong>  </li>
</ul>

<p><span class="math-container">$(x_1, y_1) \succsim (x_2, y_2)$</span> </p>

<p><em>if and only if</em> </p>

<p>either (<span class="math-container">$x_1+y_1 &gt; x_2 + y_2$</span>) or (<span class="math-container">$x_1+y_1 = x_2 + y_2$</span> and <span class="math-container">$x_1 \geq x_2$</span>)</p>

<p><strong>Claim</strong> : <span class="math-container">$\succsim$</span> cannot be represented by a utility function.</p>

<p><strong>Proof</strong> : Suppose by contradiction that there existed a utility function <span class="math-container">$u$</span> representing these preferences. For each <span class="math-container">$a &gt; 0$</span>, we have <span class="math-container">$(a, 0) \succ (0, a)$</span>, and therefore, <span class="math-container">$u(a, 0) &gt; u(0, a)$</span>. We can therefore assign to <span class="math-container">$a$</span> a non-degenerate interval of values satisfying the above inequality
<span class="math-container">$I(a) = [u(0, a), u(a, 0)]$</span>. For any <span class="math-container">$a &gt; b &gt; 0$</span>, all commodity bundles generating utilities in the interval <span class="math-container">$I(a)$</span> are strictly preferred to those in the disjoint interval <span class="math-container">$I(b)$</span> and should therefore be assigned a greater utility level. Then in each of these intervals we can pick a distinct rational number in increasing order to represent preferences. Since <span class="math-container">$a \in \mathbb{R_{++}}$</span>, there are uncountably many such intervals, but set of rational numbers are countable. This results in a contradiction.</p>

<ul>
<li><strong>Example 2</strong>  </li>
</ul>

<p><span class="math-container">$(x_1, y_1) \succsim (x_2, y_2)$</span> </p>

<p><em>if and only if</em> </p>

<p>either (<span class="math-container">$\min(x_1,y_1) &gt; \min(x_2,y_2)$</span>) or (<span class="math-container">$\min(x_1,y_1) =  \min(x_2,y_2)$</span> and <span class="math-container">$x_1+y_1 \geq x_2 + y_2$</span>)</p>

<p><strong>Claim</strong> : <span class="math-container">$\succsim$</span> cannot be represented by a utility function.</p>

<p><strong>Proof</strong> : Suppose by contradiction that there existed a utility function <span class="math-container">$u$</span> representing these preferences. For each <span class="math-container">$a &gt; 0$</span>, we have <span class="math-container">$(a+1, a) \succ (a, a)$</span>, and therefore, <span class="math-container">$u(a+1, a) &gt; u(a, a)$</span>. We can therefore assign to <span class="math-container">$a$</span> a non-degenerate interval of values satisfying the above inequality
<span class="math-container">$I(a) = [u(a, a), u(a+1, a)]$</span>. For any <span class="math-container">$a &gt; b &gt; 0$</span>, all commodity bundles generating utilities in the interval <span class="math-container">$I(a)$</span> are strictly preferred to those in the disjoint interval <span class="math-container">$I(b)$</span> and should therefore be assigned a greater utility level. Then in each of these interval we can pick a distinct rational number in increasing order to represent preferences. Since <span class="math-container">$a \in \mathbb{R_{++}}$</span>, there are uncountably many such intervals, but set of rational numbers are countable. This results in a contradiction.</p>

CC BY-SA 4.0
2019-10-23T21:46:02.927Z
@Giskard, I've updated example 1 with the proof that there is no utility representation.
Amit
CC BY-SA 4.0
2019-10-24T04:37:37.140Z
arent these just variations of lexiographic preference relations?
EconJohn
CC BY-SA 4.0
2019-10-24T04:49:18.280Z
Yes, these can be thought of as variations, but they are not called Lexicographic.
Amit
CC BY-SA 4.0
2019-10-23T07:34:55.090Z
2019-10-24T12:31:43.403Z
2019-10-24T12:31:43.403Z
Amit
<p>It seems like lexicographic isn't that "special". Like yes it is special in that supposing it has a utility function gives you a bijection from the rationals to the reals, but I mean unique in some sense. I know that it satisfies a bunch of nice properties, but it still feels like there should be plenty of other (rational) preference relations that don't have utility representations. </p>

<consumer-theory>
Are there any other rational preference relations without utility function representations, besides Lexicographic?
542goweast
# Are there any other rational preference relations without utility function representations, besides Lexicographic?

(Asked by: username_0 on 2019-10-23)

It seems like lexicographic isn't that "special". Like yes it is special in that supposing it has a utility function gives you a bijection from the rationals to the reals, but I mean unique in some sense. I know that it satisfies a bunch of nice properties, but it still feels like there should be plenty of other (rational) preference relations that don't have utility representations.

**Tags:** consumer-theory

---

## 1 answers

### Answer by username_1 on 2019-10-23. Score: 4

Yes, there are many. Here are some examples :

Consider the weak preference relation $\succsim$ defined over $\mathbb{R}^2$ as:

* **Example 1**

$(x\_1, y\_1) \succsim (x\_2, y\_2)$

*if and only if*

either ($x\_1+y\_1 > x\_2 + y\_2$) or ($x\_1+y\_1 = x\_2 + y\_2$ and $x\_1 \geq x\_2$)

**Claim** : $\succsim$ cannot be represented by a utility function.

**Proof** : Suppose by contradiction that there existed a utility function $u$ representing these preferences. For each $a > 0$, we have $(a, 0) \succ (0, a)$, and therefore, $u(a, 0) > u(0, a)$. We can therefore assign to $a$ a non-degenerate interval of values satisfying the above inequality
$I(a) = [u(0, a), u(a, 0)]$. For any $a > b > 0$, all commodity bundles generating utilities in the interval $I(a)$ are strictly preferred to those in the disjoint interval $I(b)$ and should therefore be assigned a greater utility level. Then in each of these intervals we can pick a distinct rational number in increasing order to represent preferences. Since $a \in \mathbb{R\_{++}}$, there are uncountably many such intervals, but set of rational numbers are countable. This results in a contradiction.

* **Example 2**

$(x\_1, y\_1) \succsim (x\_2, y\_2)$

*if and only if*

either ($\min(x\_1,y\_1) > \min(x\_2,y\_2)$) or ($\min(x\_1,y\_1) = \min(x\_2,y\_2)$ and $x\_1+y\_1 \geq x\_2 + y\_2$)

**Claim** : $\succsim$ cannot be represented by a utility function.

**Proof** : Suppose by contradiction that there existed a utility function $u$ representing these preferences. For each $a > 0$, we have $(a+1, a) \succ (a, a)$, and therefore, $u(a+1, a) > u(a, a)$. We can therefore assign to $a$ a non-degenerate interval of values satisfying the above inequality
$I(a) = [u(a, a), u(a+1, a)]$. For any $a > b > 0$, all commodity bundles generating utilities in the interval $I(a)$ are strictly preferred to those in the disjoint interval $I(b)$ and should therefore be assigned a greater utility level. Then in each of these interval we can pick a distinct rational number in increasing order to represent preferences. Since $a \in \mathbb{R\_{++}}$, there are uncountably many such intervals, but set of rational numbers are countable. This results in a contradiction.

> **Comments:**
> **username_1**: @Giskard, I've updated example 1 with the proof that there is no utility representation.  
> **username_2**: arent these just variations of lexiographic preference relations?  
> **username_1**: Yes, these can be thought of as variations, but they are not called Lexicographic.  

---

