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,

  • $\begingroup$ Are you sure there are no other conditions? I'm pretty sure this is only true if the set of consumption bundles is finite. $\endgroup$ – Walrasian Auctioneer Apr 24 '20 at 22:12
  • $\begingroup$ This is alll it Said $\endgroup$ – Xenusi Apr 24 '20 at 22:16

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 $A$.

  • $\begingroup$ Thank you very much for the enlightenment! Regarding "Then go to $a_4$and repeat.", shouldn't one prove this by induction? $\endgroup$ – Xenusi Apr 25 '20 at 13:43
  • $\begingroup$ You most definitely could, but since everything is finite I did not go that route. $\endgroup$ – Walrasian Auctioneer Apr 25 '20 at 18:58

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