Proving ranking using completness and transivity 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,
The exact quote from J&amp;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 $A$.