# Prove that $u$ is a utility function for $\succsim$

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.

• 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. – Herr K. Oct 8 '18 at 17:51
• ok thank you. I still get confused on how to prove things like this. – Zhang_anlan Oct 8 '18 at 21:01

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 $$x,y\in X$$.