Partial Order Relations
Let R be a relation on defined by if and only if or . 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 . Notice that . Therefore, , and hence, R is reflexive.
In order to show R is antisymmetric, let and . Assume and . Therefore, or , and or .
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?
Yes they are, because...
In the first statement, the first elements in the ordered pairs work. . In the second case, the second elements work. . 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 and in .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 ; therefore, (2,1)R(1,2). Since 1 and , . 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.