4
$\begingroup$

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?

$\endgroup$
2
  • 1
    $\begingroup$ What are multi-dimensional single-peaked preferences? $\endgroup$ Commented Sep 27, 2020 at 11:19
  • $\begingroup$ By multi-dimensional single peakedness, I was referring to Bossert and Peters' paper titled "Single-peaked choice." $\endgroup$
    – Polime
    Commented Sep 30, 2020 at 11:16

1 Answer 1

6
$\begingroup$

No. Basically, you can encode a form of 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.

$\endgroup$
2
  • $\begingroup$ Did you mean $r\mapsto q_r$ for the injection from $\mathbb R_{++}$ to $\mathbb Q$? $\endgroup$
    – Herr K.
    Commented Sep 28, 2020 at 4:36
  • $\begingroup$ @HerrK. Yes; thank you for pointing it out. $\endgroup$ Commented Sep 28, 2020 at 7:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.