# Completeness of the strict binary relation

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.

• 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? – Pedro Cunha May 13 '20 at 1:04

But an even simpler example would be $$X=\{a\}$$. We have $$a \succsim a$$, so $$a \not \succ a$$.
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 $$a \not \succ a$$.
• 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? – Pedro Cunha May 13 '20 at 1:17
• You should check $x \sim y$ for both $x=y$ and $x\neq y$. – user18 May 13 '20 at 1:24