<p>The exact quote from J&amp;R is as follows:</p>

<blockquote>
  <p>" [Completeness + Transitivity] together imply that the consumer can completely rank any finite number of elements in the consumption set, <span class="math-container">$X$</span>, from best to worst, possibly with some ties."</p>
</blockquote>

<p>Here's a sketch. Take a finite subset of the consumption set, <span class="math-container">$A \subset X$</span>.</p>

<p>Since <span class="math-container">$A$</span> is finite, we can number the elements <span class="math-container">$A = \{a_1,a_2,\dots,a_n\}$</span>. We want to rank elements in <span class="math-container">$A$</span>.</p>

<p>Pick <span class="math-container">$a_1$</span> and <span class="math-container">$a_2$</span>. By completeness, either <span class="math-container">$a_1 \succeq a_2$</span> or <span class="math-container">$a_2 \succeq a_1$</span>. Without loss of generality, suppose <span class="math-container">$a_1 \succeq a_2$</span>.</p>

<p>Now pick <span class="math-container">$a_3$</span>. By completeness, either <span class="math-container">$a_1 \succeq a_3$</span> or <span class="math-container">$a_3 \succeq a_1$</span> and either <span class="math-container">$a_2 \succeq a_3$</span> or <span class="math-container">$a_3 \succeq a_2$</span>.</p>

<p>(i) If <span class="math-container">$a_3 \succeq a_1$</span>, then by transitivity <span class="math-container">$a_3 \succeq a_2$</span>, so <span class="math-container">$a_3 \succeq a_1 \succeq a_2$</span>.</p>

<p>(i) If <span class="math-container">$a_1 \succeq a_3$</span>, then if <span class="math-container">$a_2 \succeq a_3$</span>, <span class="math-container">$a_1 \succeq a_2 \succeq a_3$</span>. Otherwise <span class="math-container">$a_3 \succeq a_2$</span> and thus <span class="math-container">$a_1 \succeq a_3 \succeq a_2$</span></p>

<p>Then go to <span class="math-container">$a_4$</span> and repeat.</p>

<p>Since <span class="math-container">$A$</span> is finite, the process ends, and you have a ranking of elements in <span class="math-container">$A$</span>.</p>

CC BY-SA 4.0
2020-04-25T13:43:27.670Z
Thank you very much for the enlightenment! Regarding "Then go to $a_4$and repeat.", shouldn't one prove this by induction?
Xenusi
CC BY-SA 4.0
2020-04-25T18:58:46.893Z
You most definitely could, but since everything is finite I did not go that route.
Walrasian Auctioneer
CC BY-SA 4.0
2020-04-24T22:49:00.550Z
2020-04-24T22:49:00.550Z
Walrasian Auctioneer
<p>The book by Jehle and Reny says that one can proof the ranking of consumption bundles using the axiom of completness and transivity.
But the have not proved this in the book and I also have a hard time finding it on the internet.</p>

<p>Could someone be kind enough to at least give me at sketch of the proof?</p>

<p>Kind regards,</p>

CC BY-SA 4.0
2020-04-24T22:12:07.353Z
Are you sure there are no other conditions? I'm pretty sure this is only true if the set of consumption bundles is finite.
Walrasian Auctioneer
CC BY-SA 4.0
2020-04-24T22:16:06.450Z
This is alll it Said
Xenusi
<microeconomics>
Proving ranking using completness and transivity
Xenusi
# Proving ranking using completness and transivity

(Asked by: username_0 on 2020-04-24)

The book by Jehle and Reny says that one can proof the ranking of consumption bundles using the axiom of completness and transivity.
But the have not proved this in the book and I also have a hard time finding it on the internet.

Could someone be kind enough to at least give me at sketch of the proof?

Kind regards,

**Tags:** microeconomics

> **Comments:**
> **username_1**: Are you sure there are no other conditions? I'm pretty sure this is only true if the set of consumption bundles is finite.  
> **username_0**: This is alll it Said  

---

## 1 answers

### Answer by username_1 on 2020-04-24. Score: 3 | Accepted answer

The exact quote from J&R is as follows:

> " [Completeness + Transitivity] together imply that the consumer can completely rank any finite number of elements in the consumption set, $X$, from best to worst, possibly with some ties."

Here's a sketch. Take a finite subset of the consumption set, $A \subset X$.

Since $A$ is finite, we can number the elements $A = \{a\_1,a\_2,\dots,a\_n\}$. We want to rank elements in $A$.

Pick $a\_1$ and $a\_2$. By completeness, either $a\_1 \succeq a\_2$ or $a\_2 \succeq a\_1$. Without loss of generality, suppose $a\_1 \succeq a\_2$.

Now pick $a\_3$. By completeness, either $a\_1 \succeq a\_3$ or $a\_3 \succeq a\_1$ and either $a\_2 \succeq a\_3$ or $a\_3 \succeq a\_2$.

(i) If $a\_3 \succeq a\_1$, then by transitivity $a\_3 \succeq a\_2$, so $a\_3 \succeq a\_1 \succeq a\_2$.

(i) If $a\_1 \succeq a\_3$, then if $a\_2 \succeq a\_3$, $a\_1 \succeq a\_2 \succeq a\_3$. Otherwise $a\_3 \succeq a\_2$ and thus $a\_1 \succeq a\_3 \succeq a\_2$

Then go to $a\_4$ and repeat.

Since $A$ is finite, the process ends, and you have a ranking of elements in $A$.

> **Comments:**
> **username_0**: Thank you very much for the enlightenment! Regarding "Then go to $a_4$and repeat.", shouldn't one prove this by induction?  
> **username_1**: You most definitely could, but since everything is finite I did not go that route.  

---

