# Weakly monotone preferences with singleton indifference curves: do any of them admit a utility representation?

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.

• 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 Oct 24 '19 at 22:03
• @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 Oct 24 '19 at 22:35
• 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 Oct 25 '19 at 2:20

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 $$j. But then $$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)$$.