# Are there any other rational preference relations without utility function representations, besides Lexicographic?

It seems like lexicographic isn't that "special". Like yes it is special in that supposing it has a utility function gives you a bijection from the rationals to the reals, but I mean unique in some sense. I know that it satisfies a bunch of nice properties, but it still feels like there should be plenty of other (rational) preference relations that don't have utility representations.

## 1 Answer

Yes, there are many. Here are some examples :

Consider the weak preference relation $$\succsim$$ defined over $$\mathbb{R}^2$$ as:

• Example 1

$$(x_1, y_1) \succsim (x_2, y_2)$$

if and only if

either ($$x_1+y_1 > x_2 + y_2$$) or ($$x_1+y_1 = x_2 + y_2$$ and $$x_1 \geq x_2$$)

Claim : $$\succsim$$ cannot be represented by a utility function.

Proof : Suppose by contradiction that there existed a utility function $$u$$ representing these preferences. For each $$a > 0$$, we have $$(a, 0) \succ (0, a)$$, and therefore, $$u(a, 0) > u(0, a)$$. We can therefore assign to $$a$$ a non-degenerate interval of values satisfying the above inequality $$I(a) = [u(0, a), u(a, 0)]$$. For any $$a > b > 0$$, all commodity bundles generating utilities in the interval $$I(a)$$ are strictly preferred to those in the disjoint interval $$I(b)$$ and should therefore be assigned a greater utility level. Then in each of these intervals we can pick a distinct rational number in increasing order to represent preferences. Since $$a \in \mathbb{R_{++}}$$, there are uncountably many such intervals, but set of rational numbers are countable. This results in a contradiction.

• Example 2

$$(x_1, y_1) \succsim (x_2, y_2)$$

if and only if

either ($$\min(x_1,y_1) > \min(x_2,y_2)$$) or ($$\min(x_1,y_1) = \min(x_2,y_2)$$ and $$x_1+y_1 \geq x_2 + y_2$$)

Claim : $$\succsim$$ cannot be represented by a utility function.

Proof : Suppose by contradiction that there existed a utility function $$u$$ representing these preferences. For each $$a > 0$$, we have $$(a+1, a) \succ (a, a)$$, and therefore, $$u(a+1, a) > u(a, a)$$. We can therefore assign to $$a$$ a non-degenerate interval of values satisfying the above inequality $$I(a) = [u(a, a), u(a+1, a)]$$. For any $$a > b > 0$$, all commodity bundles generating utilities in the interval $$I(a)$$ are strictly preferred to those in the disjoint interval $$I(b)$$ and should therefore be assigned a greater utility level. Then in each of these interval we can pick a distinct rational number in increasing order to represent preferences. Since $$a \in \mathbb{R_{++}}$$, there are uncountably many such intervals, but set of rational numbers are countable. This results in a contradiction.

• @Giskard, I've updated example 1 with the proof that there is no utility representation. – Amit Oct 23 '19 at 21:46
• arent these just variations of lexiographic preference relations? – EconJohn Oct 24 '19 at 4:37
• Yes, these can be thought of as variations, but they are not called Lexicographic. – Amit Oct 24 '19 at 4:49