<p>Inspired by <a href="https://economics.stackexchange.com/questions/32392/are-there-any-other-rational-preference-relations-without-utility-function-repre">this question</a>. The original question was answered by Amit with some nice examples. I would like to know the generalized answer:</p>

<p>Suppose we have a preference ordering <span class="math-container">$\succeq$</span>, which is weakly monotone, thus for any basket of goods <span class="math-container">$x$</span>, <span class="math-container">$y \in \mathbb{R}^n$</span>
<span class="math-container">$$
x &gt;&gt; y \Rightarrow x \succ y,
$$</span>
and which only has singleton indifference curves, thus
<span class="math-container">$$
x \sim y \Rightarrow x = y.
$$</span>
Is there any such preference ordering which can be represented by a utility function?</p>

<hr>

<p>A remark: without monotonicity one can simply take any <a href="https://math.stackexchange.com/questions/243590/bijection-from-mathbb-r-to-mathbb-rn">bijection from <span class="math-container">$\mathbb{R}^n \to \mathbb{R}$</span></a> and claim that as a utility function. This would have singleton indifference curves.</p>

CC BY-SA 4.0
2019-10-24T22:03:03.117Z
Just a clarification: what if neither $x>>y$ nor $y>>x$ hold? I mean, is the preference relation you described a complete order over $\mathbb{R}^n$? I suspect it is not...
GabMac
CC BY-SA 4.0
2019-10-24T22:35:03.110Z
@GabMac It is a complete order, but this is unspecified. E.g. in case of lexicographic preference, there are plenty of cases when neither $x >> y$ nor $y >> x$ hold, yet there is always a preference, and a strict one when $x \neq y$. E.g. $(2,1) \succ_{lexi} (1,2)$.
Giskard
CC BY-SA 4.0
2019-10-25T02:20:37.817Z
I believe it is impossible but do not have a proof yet. If indifference curves are singletons, then the quotient space of $(\mathbb{R}^n,\succ)$ is itself. If it admits a utility representation, it must be Herden Separable. I feel as though there is no countable $Z \subset \mathbb{R}^n$ such that for any $x \succ y \in \mathbb{R}^n - Z$, there is a $z \in Z$ such that $x \succ z \succ y$.
Walrasian Auctioneer
<microeconomics><utility><preferences>
Weakly monotone preferences with singleton indifference curves: do any of them admit a utility representation?
Giskard
# Weakly monotone preferences with singleton indifference curves: do any of them admit a utility representation?

(Asked by: username_0 on 2019-10-24)

Inspired by [this question](https://economics.stackexchange.com/questions/32392/are-there-any-other-rational-preference-relations-without-utility-function-repre). The original question was answered by Amit with some nice examples. I would like to know the generalized answer:

Suppose we have a preference ordering $\succeq$, which is weakly monotone, thus for any basket of goods $x$, $y \in \mathbb{R}^n$
$$
x >> y \Rightarrow x \succ y,
$$
and which only has singleton indifference curves, thus
$$
x \sim y \Rightarrow x = y.
$$
Is there any such preference ordering which can be represented by a utility function?

---

A remark: without monotonicity one can simply take any [bijection from $\mathbb{R}^n \to \mathbb{R}$](https://math.stackexchange.com/questions/243590/bijection-from-mathbb-r-to-mathbb-rn) and claim that as a utility function. This would have singleton indifference curves.

**Tags:** microeconomics, utility, preferences
> **Comments:**
> **username_1**: Just a clarification: what if neither $x>>y$ nor $y>>x$ hold? I mean, is the preference relation you described a complete order over $\mathbb{R}^n$? I suspect it is not...  
> **username_0**: @GabMac It is a complete order, but this is unspecified. E.g. in case of lexicographic preference, there are plenty of cases when neither $x >> y$ nor $y >> x$ hold, yet there is always a preference, and a strict one when $x \neq y$. E.g. $(2,1) \succ_{lexi} (1,2)$.  
> **username_2**: I believe it is impossible but do not have a proof yet. If indifference curves are singletons, then the quotient space of $(\mathbb{R}^n,\succ)$ is itself. If it admits a utility representation, it must be Herden Separable. I feel as though there is no countable $Z \subset \mathbb{R}^n$ such that for any $x \succ y \in \mathbb{R}^n - Z$, there is a $z \in Z$ such that $x \succ z \succ y$.  

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## 0 answers

