Lexicographic preference in \(\mathbb{R}^2\) is as following:
\((a_1, a_2) \succ (b_1, b_2)\) if and only if \(a_1> b_1\) or \(a_1=b_1\) and \(a_2 > b_2\) .
For example, \((1, 0) \succ (0, 5)\) and \((1, 1) \succ (1, 0)\) .
Although lexicographic preference satisfies
- Complete
- Transitive it does not have numerical representation. In other world, there is not any function \(\mathbb{R}^2 \rightarrow \mathbb{R}\) can represent it.
Here is the proof:
First of all, we restrict the domain to \([0 ,1] \times [0,1]\) only. We are going to proof that the utility is not exist even in this subset.
Suppose there is an utility function \(U: \mathbb{R}^2 \rightarrow \mathbb{R}\) represents lexicographic preference. We construct a function \(r(\cdot): \mathbb{R} \rightarrow \mathbb{Q}\) . For every \(x \in \mathbb{R}\), we choose a rational number \(r(x)\) satisfies \(U((x,0)) < r(x) < U((x,1))\). According to the dense of real number, the rational number must exists.
Moreover, if \(x_1,x_2 \in \mathbb{R}\) and \(0 \leq x_1 < x_2 \leq1\) . Then \(r(x_1) < U((x_1,1)) < U((x_2,0)) < r(x_2)\) . To sum up, \(x_1< x_2\) implies \)r(x_1) < r(x_2)\). It also implies that \(r(\cdot)\) is a one-to -one function. It is a contradiction because ration number and real number do not share the same cardinal number.
Related question:
Utility function of this kind of preference on \(\mathbb{R}\). How about on \(\mathbb{Q}\)?