Setting: We have two choices of goods $(x_1,y_1)$ and $(x_2,y_2)$ from the set of choices $[-1,1]^2$. Moreover, we have the following preference relation $$(x_1,y_1)\mathcal{R}(x_2,y_2)\iff |x_1|\geq|x_2|\>\>\text{or}\>\> |y_1|\geq|y_2|$$

Question: We have to check if there exists a utility function reprensation of this preference relation.

My attempt: So from what I have learned, we know that a preference relation admits a utility function representation if it is rational (reflexive, complete, transitive) and continuous. I have found that this preference relation is not transitive, but this does not mean that there does not exist a utility function representation, because the aforementioned statement is not an if and only if statement.

Moreover, I thought we could try to derive a contradiction from the fact that if there exists a utility function $u$ representation of the preference relation, then we have $$(x_1,y_1)\mathcal{R}(x_2,y_2)\iff u(x_1,y_1)\geq u(x_2,y_2)$$ I tried to use the fact that the relation is not transitive to derive a contradiction by using the statement above, but was unsuccessful.

Sadly, these are the two main theorems/propositions that I've learned to solve these problems.

Any help is appreciated!


2 Answers 2


You just need to use the violation of transitivity and proceed by contradiction.

Suppose you have that $(x_1,y_1)R(x_2,y_2)$ and $(x_2,y_2)R(x_3,y_3)$ and a utility function, $u:[-1,1]^2\rightarrow \mathbb{R}$, exists, then $u(x_1,y_1)\geq u(x_2,y_2)$ (these are two reals) and $u(x_2,y_2)\geq u(x_3,y_3)$. (another two reals). Since the reals are transitive, we conclude $u(x_1,y_1)\geq u(x_3,y_3)$ which in turn implies that $(x_1,y_1)R(x_3,y_3)$. However this is a contradiction (if you choose carefully your three bunddles).


Transitivity and completeness are actually necessary for the existence of a utility representation. Whenever you prove that preferences fail to be complete or transitive you can conclude that they do not admit a utility function. For finite choice sets $X$, transitivity and completeness are necessary and sufficient, see Theorem 5 here.

Contrary to what you suggest, continuity is not itself necessary, but it is almost necessary. What you need is a condition called separability. Say that $\succcurlyeq$ is separable if there exists a countable set $Z \subseteq X$ such that for every $x,y\in X$ there exists some $z\in Z$ such that $x\succcurlyeq z \succcurlyeq y$.

Theorem A preference relation $\succcurlyeq$ on $X$ admits a utility representation if and only if it is complete, transitive, and separable.

This is actually a very old result by Cantor that precedes the well-known result by Debreu—which assumes continuity. Cantor's result was brought into Economics by Kreps. You can find a proof here, Theorem 9.

  • 2
    $\begingroup$ "Whenever you prove that preferences fail to be continuous or transitive..." Did you mean "complete or transitive"? $\endgroup$
    – Herr K.
    Aug 6, 2019 at 17:12
  • 1
    $\begingroup$ Let me suggest the edit above. $\endgroup$
    – Bayesian
    Aug 6, 2019 at 18:20

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