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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 $X$.

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    This is related to [an earlier question](http://math.stackexchange.com/questions/170517/in-a-preorder). 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.2012-07-23
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    Should that be $x if $x\leq y$ and $y\nleq x$ ?2012-07-23
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    @EricStucky Good catch. I fixed it.2012-07-23
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    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.2012-07-25
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    @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.2012-07-25

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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.