My textbook says that "if a weak preference relation is complete and transitive, the strict preference relation MUST be asymmetric and negatively transitive". Now, I think I understand why it must be asymmetric - all strict preference relations are asymmetric, correct? However, I don't understand why it is negatively transitive. If the weak preference is transitive, couldn't the strict preference also be transitive?

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    Commented Feb 9, 2019 at 6:50

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I think your confusion here stems from misinterpreting negative transitivity to mean "not transitive". But that's not what it means. It is perfectly possible for a binary relation to be both transitive and negatively transitive, as is the case here with the strict preference relation.

Definition 1. A binary relation $\mathcal{R}$ is transitive if: $$x \mathcal{R} y \text{ and } y \mathcal{R} z \implies x \mathcal{R} z.$$

Definition 2. A binary relation $\mathcal{R}$ is negatively transitive if: $$x \not\mathcal{R} y \text{ and } y \not\mathcal{R} z \implies x \not\mathcal{R} z.$$

(Those are forward slashes over the $\mathcal{R}$'s meaning "not". The LaTeX "not" here isn't quite working as well as I'd like.)

The strict preference relation $\succ$ is both transitive and negatively transitive.

  • It is transitive because $(x \succ y \text{ and } y \succ z)$ implies $x \succ z.$
  • It is negatively transitive because $(x \not\succ y \text{ and } y \not\succ z)$ is equivalent to $(y \succsim x \text{ and } z \succsim y)$, which, by the transitivity of $\succsim$, implies $z \succsim x$ or $x \not\succ z$.

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