
Partial Order Relations
Let R be a relation on $\displaystyle \bold{N}^2 $ defined by $\displaystyle (x,y)R(u,v) $ if and only if $\displaystyle x \leq u $ or $\displaystyle y \leq v $. Is R a partial order? Prove your answer.
This is what I came up with so far.
So, just to be courteous to the people who read this, a relation on a set is called a partial ordering if the relation is reflexive, antisymmetric, and transitive.
Therefore, I tried to prove that R is reflexive, antisymmetric, and transitive.
Here's what I did.
Claim: R is a partial order.
Proof: For R to be a partial order, it must be reflexive, antisymmetric, and transitive.
In order to show R is reflexive, let $\displaystyle (x,y) \in \bold{N}^2 $. Notice that $\displaystyle x \leq x $. Therefore, $\displaystyle (x,y)R(x,y) $, and hence, R is reflexive.
In order to show R is antisymmetric, let $\displaystyle (x,y) \in \bold{N}^2 $ and $\displaystyle (u,v) \in \bold{N}^2$. Assume $\displaystyle (x,y)R(u,v) $ and $\displaystyle (u,v)R(x,y) $. Therefore, $\displaystyle x \leq u $ or $\displaystyle y \leq v $, and $\displaystyle u \leq x $ or $\displaystyle v \leq y $.
After this point, I am not sure how to continue. I am thinking that I would have to break this into cases, but I'm not sure how to break it into cases because I have two different statements that both involve "or". It's looking like the transitive proof will also have cases, but I'm not sure how to attempt those either. If I can clear up the antisymmetric one, the transitive one can probably follow pretty easily. I will post a continuation after I get some hints on how to go about continuing. Thanks.

Are both of these true?
$\displaystyle \left( {1,2} \right)R\left( {2,1} \right)\quad \& \quad \left( {2,1} \right)R\left( {1,2} \right)
$

Yes they are, because...
In the first statement, the first elements in the ordered pairs work. $\displaystyle 1 \leq 2 $. In the second case, the second elements work. $\displaystyle 1 \leq 2 $. I guess I'm still not seeing how that's going to help me prove this :(. (we just started relations, so i'm having trouble getting used to the idea right now).


Oh, let's see.
I'll show you what i'm thinking. Critique it and/or confirm it.
Here you go. (i'll put comments in blue.)
So, your example was. Consider the ordered pairs $\displaystyle (x,y) = (1,2)$ and $\displaystyle (u,v) = (2,1) $ in $\displaystyle \bold{N}^2 $.So for R to be antisymmetric, if x ~ y and y ~ x, then x = y. So, for a counterexample, we find x and y such that the hypothesis is true and the conclusion is false. Notice that x = 1 < 2 = u; therefore, (1,2)R(2,1). Also, notice that $\displaystyle v = 1 \leq 2 = y $; therefore, (2,1)R(1,2). Since 1 $\displaystyle \not= 2 $ and $\displaystyle 2 \not= 1 $, $\displaystyle (1,2) \not= (2, 1) $. Thus, R is not antisymmetric, and hence, R is not a partial order.


Ha, thanks a lot.
Disproofs are so much easier than proofs most of the time.
Thanks a lot man. I'll click the thanks button.