<p>No. Basically, you can encode a form of <a href="https://en.wikipedia.org/wiki/Lexicographic_preferences" rel="nofollow noreferrer">lexicographic preferences</a>, probably the most familiar example of non-representable preferences, as single-peaked preferences on <span class="math-container">$\mathbb{R}$</span>.</p>
<p>Define <span class="math-container">$\succeq$</span> so that <span class="math-container">$x\succeq y$</span> exactly if either <span class="math-container">$|x|&lt;|y|$</span> or <span class="math-container">$|x|=|y|$</span> and <span class="math-container">$x\leq y$</span>. Basically, the closer to the peak of <span class="math-container">$0$</span> a number is, the better, and in case of a tie, the number to the left of <span class="math-container">$0$</span> is better.</p>
<p>Suppose for the sake of contradiction that there is a utility representation <span class="math-container">$v:\mathbb{R}\to \mathbb{R}$</span> of <span class="math-container">$\succeq$</span>. For each <span class="math-container">$r\in\mathbb{R}_{++}$</span> (the strictly positive numbers), let <span class="math-container">$q_r$</span> be a rational number in the interval <span class="math-container">$\big(v(r),v(-r)\big)$</span>. Since for <span class="math-container">$r\neq r'$</span>, <span class="math-container">$\big(v(r),v(-r)\big)\cap\big(v(r'),v(-r')\big)=\emptyset$</span>, we have an injection <span class="math-container">$r\mapsto q_r$</span> from <span class="math-container">$\mathbb{R}_{++}$</span> to <span class="math-container">$\mathbb{Q}$</span>, which is impossible since <span class="math-container">$\mathbb{R}_{++}$</span> is uncountable and <span class="math-container">$\mathbb{Q}$</span> is countable.</p>

CC BY-SA 4.0
2020-09-28T04:36:14.367Z
Did you mean $r\mapsto q_r$ for the injection from $\mathbb R_{++}$ to $\mathbb Q$?
Herr K.
CC BY-SA 4.0
2020-09-28T07:15:53.887Z
@HerrK. Yes; thank you for pointing it out.
Michael Greinecker
CC BY-SA 4.0
2020-09-27T17:15:36.327Z
2020-09-28T07:15:20.400Z
2020-09-28T07:15:20.400Z
Michael Greinecker
<p>Is it true that a single-peaked preference (with the peak at some finite point) over the set of real numbers, always has a utility representation ??</p>
<p>If yes, can you please hint towards the proof or references.</p>
<p>Is this result generalizable to multi-dimensional single-peakedness?</p>

CC BY-SA 4.0
2020-09-27T11:19:31.223Z
What are multi-dimensional single-peaked preferences?
Michael Greinecker
CC BY-SA 4.0
2020-09-30T11:16:16.370Z
By multi-dimensional single peakedness, I was referring to Bossert and Peters' paper titled "Single-peaked choice."
Polime
<microeconomics><choice-theory><social-choice>
Utility representation of single peaked preferences
Polime
# Utility representation of single peaked preferences

(Asked by: username_0 on 2020-09-26)

Is it true that a single-peaked preference (with the peak at some finite point) over the set of real numbers, always has a utility representation ??

If yes, can you please hint towards the proof or references.

Is this result generalizable to multi-dimensional single-peakedness?

**Tags:** microeconomics, choice-theory, social-choice

> **Comments:**
> **username_1**: What are multi-dimensional single-peaked preferences?  
> **username_0**: By multi-dimensional single peakedness, I was referring to Bossert and Peters' paper titled "Single-peaked choice."  

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## 1 answers

### Answer by username_1 on 2020-09-27. Score: 6 | Accepted answer

No. Basically, you can encode a form of [lexicographic preferences](https://en.wikipedia.org/wiki/Lexicographic_preferences), probably the most familiar example of non-representable preferences, as single-peaked preferences on $\mathbb{R}$.

Define $\succeq$ so that $x\succeq y$ exactly if either $|x|<|y|$ or $|x|=|y|$ and $x\leq y$. Basically, the closer to the peak of $0$ a number is, the better, and in case of a tie, the number to the left of $0$ is better.

Suppose for the sake of contradiction that there is a utility representation $v:\mathbb{R}\to \mathbb{R}$ of $\succeq$. For each $r\in\mathbb{R}\_{++}$ (the strictly positive numbers), let $q\_r$ be a rational number in the interval $\big(v(r),v(-r)\big)$. Since for $r\neq r'$, $\big(v(r),v(-r)\big)\cap\big(v(r'),v(-r')\big)=\emptyset$, we have an injection $r\mapsto q\_r$ from $\mathbb{R}\_{++}$ to $\mathbb{Q}$, which is impossible since $\mathbb{R}\_{++}$ is uncountable and $\mathbb{Q}$ is countable.

> **Comments:**
> **username_2**: Did you mean $r\mapsto q_r$ for the injection from $\mathbb R_{++}$ to $\mathbb Q$?  
> **username_1**: @HerrK. Yes; thank you for pointing it out.  

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