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.

  • $\begingroup$ 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? $\endgroup$ May 13 '20 at 1:04

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 $a \not \succ a$.

  • $\begingroup$ 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? $\endgroup$ 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

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