I am currently studying for my Game Theory exam and came across a question that seems pretty basic but somehow can't wrap my head around. So if you could share some insight with me, that would be greatly appreciated.
The question is, "Every transitive relation on three alternatives can be represented by a utility function."
The answer is "False".
However, I thought transitivity means "For all x, y, z in A, if (x >= y) and (y >= z) then (x >= z)" So then, wouldn't a utility function also preserve this relation? That is, wouldn't it also be "if u(x) >= u(y) and u(y) >= u(z), then u(x) >= u(z)?" How come every complete relation on two alternatives can be represented by a utility function (this was another question on the quiz), but a transitive relation cannot be?
I'm confused... Any help would be appreciated! Thanks in advance.
Good question. I suppose they might be considering an example like the following:
$$ x \succeq z, \hspace{1cm} y \succeq z, \hspace{1cm} x \text{ and } y \text{ are incomparable.}$$
Obviously this example is transitive (or doesn't violate transitivity, at least). But it is not complete as $x$ and $y$ are incomparable. If you slap any utility function on this, then either $u(x)>u(y)$, in which case $x \succ y$, $u(y)>u(x)$, in which case $y \succ x$, or $u(x)=u(y)$, in which case $x \sim y$. But each of these cases imply $x$ and $y$ are comparable, when we know they shouldn't be.
As for why a complete relation on two alternatives can definitely be represented, the fact that it is complete means that either $x \succeq y$, $y \succeq x$ or both. If just the first is true, let $u(x)>u(y)$. If just the second is true, let $u(y)>u(x)$. If the third is true, let $u(x)=u(y)$. In each case, we can represent the relation with a utility function.