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

<p>I'll prove the case in which there is no indifference among the alternatives, say, <span class="math-container">$1\succ2\succ\cdots\succ N$</span>. I'll let you finish the proof by establishing the case where there are indifferences among subsets of alternatives.</p>

<hr>

<p><strong>Step 1. Establishing <span class="math-container">$u(x)&gt;u(y)\;\Rightarrow\;x\succ y$</span>.</strong> </p>

<p>Suppose <span class="math-container">$u(x)&gt;u(y)$</span>. By the definition of <span class="math-container">$u$</span>, the number of alternatives strictly worse than <span class="math-container">$x$</span> is larger than the number of alternatives strictly worse than <span class="math-container">$y$</span>. If <span class="math-container">$y\succsim x$</span>, this would simply contradict the previous statement. Hence, we must have <span class="math-container">$x\succ y$</span>. </p>

<p><strong>Step 2. Establishing <span class="math-container">$x\succ y\;\Rightarrow\;u(x)&gt;u(y)$</span>.</strong></p>

<p>Suppose <span class="math-container">$x\succ y$</span>. Since we assume no indifference among alternatives, the set of strictly worse alternatives to <span class="math-container">$x$</span>, <span class="math-container">$\{z\in X:x\succ z\}$</span>, must contain more elements than the set of strictly worse alternatives to <span class="math-container">$y$</span>, <span class="math-container">$\{z\in X:y\succ z\}$</span>. In other words, <span class="math-container">$|\{z\in X:x\succ z\}|&gt;|\{z\in X:y\succ z\}|$</span>. Therefore, we obtain <span class="math-container">$u(x)&gt;u(y)$</span> as a result.</p>

<p>Taken together, steps 1 and 2 demonstrate that <span class="math-container">$x\succ y\;\Leftrightarrow\;u(x)&gt;u(y)$</span> for any arbitrary <span class="math-container">$x,y\in X$</span>. </p>

CC BY-SA 4.0
2018-10-10T00:52:00.007Z
2018-10-10T00:52:00.007Z
Herr K.
<p>If X is finite, define this function <span class="math-container">$u : X \rightarrow  \mathbb{R}$</span> by <span class="math-container">$u(x) = |\{z\in X:z \prec x \}|$</span>. Prove that <span class="math-container">$u$</span> is a utility function for <span class="math-container">$\succsim$</span>.</p>

<hr>

<p>Is it sufficient to prove that the relation is transitive and complete?<br>
By Lemma:  <em>If <span class="math-container">$\succsim$</span> has a utility function, then it is transitive and complete.</em></p>

CC BY-SA 4.0
2018-10-08T17:51:59.610Z
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.
CC BY-SA 4.0
2018-10-08T21:01:33.573Z
ok thank you. I still get confused on how to prove things like this.
plastico
<utility><preferences><self-study><proof>
Prove that $u$ is a utility function for $\succsim$
plastico
# Prove that $u$ is a utility function for $\succsim$

(Asked by: username_0 on 2018-10-08)

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.*

**Tags:** utility, preferences, self-study, proof

> **Comments:**
> **username_1**: 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.  
> **username_0**: ok thank you. I still get confused on how to prove things like this.  

---

## 1 answers

### Answer by username_1 on 2018-10-10. Score: 1 | Accepted answer

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 $x,y\in X$.


---

