If X is finite, define this function $u : X \rightarrow \mathbb{R}$ by $u(x) = |\{z\in X:z \prec x \}|$. Prove that $u$ is a utility function for $\succsim$.

Is it sufficient to prove that the relation is transitive and complete?
By Lemma: If $\succsim$ has a utility function, then it is transitive and complete.

  • 1
    $\begingroup$ You're given a very specific utility function and are asked to prove that this particular function represents the preference over a finite choice set. In other words, you're asked to prove $u(x)\ge u(y)\;\Leftrightarrow\; x\succsim y$ for all $x,y\in X$, where $u$ is as defined in the question. $\endgroup$
    – Herr K.
    Commented Oct 8, 2018 at 17:51
  • $\begingroup$ ok thank you. I still get confused on how to prove things like this. $\endgroup$
    – plastico
    Commented Oct 8, 2018 at 21:01

1 Answer 1


You're asked to prove that $u(x)\ge u(y)\;\Leftrightarrow\;x\succsim y$ for any $x,y\in X$, where $u(x)=|\{z\in X:z\prec x\}|$, i.e. the utility of $x$ is measured by the number of other alternatives that rank strictly below it. Since $X$ is finite, let's suppose without loss of generality that $X=\{1,2,\dots,N\}$ where $N$ is some finite number.

I'll prove the case in which there is no indifference among the alternatives, say, $1\succ2\succ\cdots\succ N$. I'll let you finish the proof by establishing the case where there are indifferences among subsets of alternatives.

Step 1. Establishing $u(x)>u(y)\;\Rightarrow\;x\succ y$.

Suppose $u(x)>u(y)$. By the definition of $u$, the number of alternatives strictly worse than $x$ is larger than the number of alternatives strictly worse than $y$. If $y\succsim x$, this would simply contradict the previous statement. Hence, we must have $x\succ y$.

Step 2. Establishing $x\succ y\;\Rightarrow\;u(x)>u(y)$.

Suppose $x\succ y$. Since we assume no indifference among alternatives, the set of strictly worse alternatives to $x$, $\{z\in X:x\succ z\}$, must contain more elements than the set of strictly worse alternatives to $y$, $\{z\in X:y\succ z\}$. In other words, $|\{z\in X:x\succ z\}|>|\{z\in X:y\succ z\}|$. Therefore, we obtain $u(x)>u(y)$ as a result.

Taken together, steps 1 and 2 demonstrate that $x\succ y\;\Leftrightarrow\;u(x)>u(y)$ for any arbitrary $x,y\in X$.


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