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My textbook, Microeconomic Theory by Mas-Colell, Whinston, and Green states that given a preference relation $\succsim$ on $X$,

Strict preference relation $\succ$ is defined by $ x \succ y \iff x \succsim y \text{ but not } y \succsim x $

Indifference relation $\sim$ is defined by $ x \sim y \iff x \succsim y \text{ and } y \succsim x $

Completeness property: For all $x,y \in X$, we have either $x \succsim y$ or $y \succsim x$ or both.

Suppose $\succsim$ on $X$ is complete. Define $A$ to be the truth set of $x \succsim y$, $B$ to be the truth set of $y \succsim x$.

Since $\succsim$ is complete, $A \cup B$ is the universal set.

First, $x \succ y \iff x \succsim y \land \lnot (y \succsim x)$, hence the truth set of $x \succ y$ is $A \setminus B$.

Next, $y \succ x \iff y \succsim x \land \lnot (x \succsim y)$, hence the truth set of $y \succ x$ is $B \setminus A$.

Finally, $x \sim y \iff x \succsim y \land y \succsim x$, hence the truth set of $x \sim y$ is $A \cap B$.

Using the identity $A \cup B = (A \setminus B) \sqcup (A \cap B) \sqcup (B \setminus A)$, is it safe to say that

If $\succsim$ is complete, then for any $x,y \in X$ exactly one of $x \succ y$ or $x \sim y$ or $y \succ x$ holds.

user20961
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  • Yes, that seems perfectly valid reasoning to me – J. Dekker Feb 01 '19 at 07:02
  • This just applies the definitions of completeness and relations together. You don't need set theory to show this. You could just work from the mathematical definitions. Anyway...yeah...it is true. – 123 Feb 01 '19 at 22:26

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