0
$\begingroup$

How do I show that considering the preference relation $\succsim$, then $\succ$ is not complete?

I tried the following (which I don't know if it's right) but I'd also like to know if it's possible to generalize and say that $\succ$ will never be complete.

Suppose $X = \{a,b\}$ with preference relation $\{ a \succsim a, a \succsim b, b \succsim b\}$. Then, we $\succ$ can't be a complete relation on $X$ because neither $b \succsim a$ nor $b \not{\succsim} a$ are in $X$ and therefore it's impossible to conclude that either $a \succ b$ or $b \succ a$ is true.

$\endgroup$
1
  • $\begingroup$ Thanks for the input! Is there a set $X$ such that $\succ$ is complete on it? For instance, would $X = \{a,b\}$ with preferences $\{ a \succsim a, a \succsim b, b \succsim b, b \not{\succsim} a\}$ be a valid set for which $\succ$ is complete? $\endgroup$ – Pedro Cunha May 13 '20 at 1:04
1
$\begingroup$

Your counterexample looks correct.

But an even simpler example would be $X=\{a\}$. We have $a \succsim a$, so $a \not \succ a$.


More generally, we usually define

  1. $\succsim$ to be reflexive (i.e. $a \succsim a$ for all $a$); and
  2. $\succ$ by $a \succ b$ if $a \succsim b$ and $b \not \succsim a$.

By the above definitions, $\succ$ cannot be complete on any non-empty set $X$ because for any $a\in X$, we have $a \not \succ a$.

$\endgroup$
2
  • $\begingroup$ Oh, I think I see the flaw in my reasoning. I was thinking that merely having $a \succ b$ to mean the set would be complete. By the definition of completeness, $\forall x,y \in X$ either $x \succ y, y \succ x$ or both, right? So I should check that the definition of $\succ$ holds for each element of the set and not for each pair in the set. Is that correct? $\endgroup$ – Pedro Cunha May 13 '20 at 1:17
  • 1
    $\begingroup$ You should check $x \sim y$ for both $x=y$ and $x\neq y$. $\endgroup$ – user18 May 13 '20 at 1:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.