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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.

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1 Answer 1

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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.

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    $\begingroup$ @Giskard, I've updated example 1 with the proof that there is no utility representation. $\endgroup$
    – Amit
    Commented Oct 23, 2019 at 21:46
  • $\begingroup$ arent these just variations of lexiographic preference relations? $\endgroup$
    – EconJohn
    Commented Oct 24, 2019 at 4:37
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    $\begingroup$ Yes, these can be thought of as variations, but they are not called Lexicographic. $\endgroup$
    – Amit
    Commented Oct 24, 2019 at 4:49

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