# 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.

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
Commented Oct 23, 2019 at 21:46
• arent these just variations of lexiographic preference relations?
– EconJohn
Commented Oct 24, 2019 at 4:37
• Yes, these can be thought of as variations, but they are not called Lexicographic.
– Amit
Commented Oct 24, 2019 at 4:49