# $<$ on a preorder is a strict partial order

Definition: Suppose $X$ is a preorder. Define $x < y$ as $x \le y$ and $y \not\le x$ for each $x, y \in X$.

Question: Show that this gives a strict partial order on $X$.

-
This is related to an earlier question. I decided to post a new question in order to aggregate everything I have learned in one place without detracting from the answerers earlier contributions. –  Code-Guru Jul 23 '12 at 18:54
Should that be $x<y$ if $x\leq y$ and $y\nleq x$ ? –  Eric Stucky Jul 23 '12 at 19:07
@EricStucky Good catch. I fixed it. –  Code-Guru Jul 23 '12 at 19:10
You should probably slow down on them retags. I'm also not sure that we need two tags for [preorders] and [partial-orders], in fact I'm not 100% certain that either is particularly useful either. Please bring this up to a meta thread before continuing the retagging journey. –  Asaf Karagila Jul 25 '12 at 23:01
@AsafKaragila I finished the 5 or 6 retags already. I apologize for not discussing it in meta first. I was not aware that was part of the protocol here. Thanks for pointing me in the right direction. –  Code-Guru Jul 25 '12 at 23:37

We must show that the relation $<$ is irreflexive and transitive.
Irreflexive: Suppose $x \in X$. Then $x \le x$, so $x \not< x$.
Transitive: Suppose $x, y, z \in X$ such that $x < y$ and $y < z$. Then $x \le y$, $y \not\le x$, $y \le z$, and $z \not\le y$. By the transitivity of $\le$, we immediately have $x \le z$. Now assume that $z \le x$. Then $z \le y$. This is a contradition.