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Let $R$ be a complete binary relation on $U$. $R$ is

  • Transitive if for all $x,y,z \in U$, $xRy \land yRz \implies xRz$,
  • Quasi-transitive if for all $x,y,z \in U$, $xPy \land yPz \implies xPz$

Prove that if $R$ is transitive, it is also quasi-transitive.

I haven't come up with a full solution yet but here's my thinking process. If a relation is transitive, then it deals with $xRy$, which means that $x$ is at least as good as $y$. That means that there two scenarios within this: 1) $x P y$, which means that $x$ is strictly better than $y$, 2) $xIy$ which means that $x$ and $y$ are equal. Therefore, if $R$ is transitive and deals with scenario 1, then it also means that $R$ is quasi-transitive. is this understanding correct?

Again, this is not a full-blown solution or proof. I just wanted to explain my thoughts and try to see if I'm on the right track.

Asaf Karagila
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1 Answers1

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From your discussion and the context I take it that $R$ is a preorder (also called a quasiorder), i.e., a transitive, reflexive relation, and $x\mathrel{P}y$ is equivalent to $x\mathrel{R}y$ and $y\not\mathrel{R}x$. Your $x\mathrel{I}y$ is then $x\mathrel{R}y$ and $y\mathrel{R}x$, the indifference relation on $U$.

Suppose that $R$ is transitive and that $x,y,z\in U$ are such that $x\mathrel{P}y$ and $y\mathrel{P}z$. Then $x\mathrel{R}y$ and $y\mathrel{R}z$, so $x\mathrel{R}z$. To show that $x\mathrel{P}z$, we must show that $z\not\mathrel{R}x$. Suppose, on the contrary, that $z\mathrel{R}x$; then by transitivity $z\mathrel{R}y$. But $y\mathrel{P}z$, so in fact $z\not\mathrel{R}y$. Thus, $z\not\mathrel{R}x$, so $x\mathrel{P}z$, and $R$ is quasi-transitive.

Brian M. Scott
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