Inspired by this question. 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 $\mathbb{R}^n \to \mathbb{R}$ and claim that as a utility function. This would have singleton indifference curves.

  • $\begingroup$ 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... $\endgroup$
    – GabMac
    Oct 24, 2019 at 22:03
  • $\begingroup$ @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. $(2,1) \succ_{lexi} (1,2)$. $\endgroup$
    – Giskard
    Oct 24, 2019 at 22:35
  • $\begingroup$ 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 $x \succ z \succ y$. $\endgroup$ Oct 25, 2019 at 2:20

1 Answer 1


What you're asking for is equivalent to finding an injective function $f:\mathbb R^n\to\mathbb R$ that is monotone in the sense that if $x$ is coordinate-wise at most as big as $y$, then $f(x)\le f(y)$. As a first step, notice that this is equivalent to finding such a function from $(0,1)^n\to\mathbb (0,1)$. And this is easy by interleaving the digits of the coordinates of $x$, i.e., if $x=(x^1,x^2,\ldots,x^n)=(0.x^1_1x^1_2\ldots,0.x^2_1x^2_2\ldots,\ldots,0.x^n_1x^n_2\ldots)$, then let $f(x)=0.x^1_1x^2_1\ldots x^n_1x^1_2x^2_2\ldots x^n_2x^1_3x^2_3\ldots x^n_3\ldots$. This is clearly injective and we prove monotonicity as follows. If $f(x)>f(y)$, then they differ first at some $x_i^h>y_i^h$ for which $x_j^h=y_j^h$ for all $j<i$. But then $x^h>y^h$, so $x$ cannot be coordinate-wise at most as big as $y$.

ps. Note that $f$ is almost surjective and almost continuous; the only issue is with numbers that have a finite decimal expansion. To see that an injective $f$ cannot be continuous, consider $f^{-1}(\mathbb R_-)$, $f^{-1}(0)$ and $f^{-1}(\mathbb R_+)$ where wlog. $0$ is an inner point of the image of $f$. By continuity, these are two open sets and a point, so they cannot form a partition of $\mathbb R^n$ for $n\ge 2$. (Here we didn't even use monotonicity.)

It is, however, possible to make $f$ into a monotone bijection. This can be achieved as follows. First we show the statement for $f:[0,1)^n\to [0,1)$. Write every number in its finite decimal form, if it has one, i.e., it shouldn't end with $999\ldots$. The number $f(x)$ will start with the first few decimal digits of $x^1$. If $x^1_1\ne 9$, then $f(x)$ will start with $x^1_1$, and then we take the first few digits of $x^2$. If $x^1_1=9$ but $x^1_2\ne 9$, then $f(x)$ will start with $x^1_1x^1_2$, and then we go to $x^2$. And so on, we always go until the first non-$9$ digit of $x^1$ before we start taking digits from $x^2$. Then we repeat this for $x^2$, then for $x^3$, etc. (in a circular order). This finishes the construction for $f:[0,1)^n\to [0,1)$. To make it into a function $\mathbb R^n\to\mathbb R$, just apply a monotone bijection $g:\mathbb Z^n\to\mathbb Z$ to the integer part of the numbers. (Such a $g$ is easy to construct by induction and by taking larger and larger cubes around the origin, $0^n$.) Such a $g$ gives a partition of $\mathbb R^n$ into cubes that are isomorphic to $[0,1)^n$. We can combine $f$ and $g$ to obtain a final function that will map $x$ to $g(\lfloor x\rfloor)+f(x-\lfloor x\rfloor)$.


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