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Thread: Partial Order Relations

  1. #1
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    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.
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  2. #2
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    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)
    $
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  3. #3
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    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).
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  4. #4
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    Is R antisymmetric?
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  5. #5
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    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.
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  6. #6
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    By George, you have.
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  7. #7
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    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.
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