Proving ranking using completness and transivity
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 $A$.