Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $X$.

share|improve this question
    
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

1 Answer 1

up vote 0 down vote accepted

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.