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R is a partial order relation on some set A. Which of the following statements are correct?

a) $R\cup R^{-1}$ is an equivalence relation

b) $R^{2}$ is a partial order relation

c) $R\cap R^{-1}$ is an equivalence relation

I've tried solving this. For (a), I think this is true, but the answer I got say false. My logic was:

for every x, $xRx$ since R is reflexive. R is also ant-symmetric, so for every pair $xRy$, we will have $yRx$ because of the union, and so $R\cup R^{-1}$ is symmetric. Now I tried thinking on transitive, I took an example:

$R={(1,1),(2,2),(3,3),(1,2),(2,3),)(1,3)}$

If you do $R\cup R^{-1}$ you get a transitive relation. I can't find an example to break it.

What am I doing wrong ?

For (b) I tried a similar example and go $R^{2}=R$ which shows it's true, but it must have been a private case. How can you explain the fact that (b) is true ? (I am not looking for a formal proof).

For (c), if I use intersection, I get only "reflexive" couples, so this is true. Am I correct on this one at least ?

Thank you.

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    $\begingroup$ Counter example to the transitivity claim in part (a): (1, 2), (3, 2). $\endgroup$ Jun 7, 2017 at 19:53
  • $\begingroup$ What, exactly, is your definition of "partial order relation"? I have never seen a requirement that it be reflexive. $\endgroup$
    – user247327
    Jun 7, 2017 at 19:53
  • $\begingroup$ @user247327, I'd assume they are using the standard reflexive, antisymmetric, and transitive definition (en.wikipedia.org/wiki/Partially_ordered_set). Something like $\leq$. $\endgroup$ Jun 7, 2017 at 19:55
  • $\begingroup$ @user3275222, you may also want to be more specific about what you mean by $R^2$. Is it partial order on $A^2$? Is it an extension of $R$? $\endgroup$ Jun 7, 2017 at 20:03
  • $\begingroup$ Sorry, I'll be more specific. 1. Partial order is indeed reflexive, ant-symmetric and transitive. As for R^2, I mean the composite relation R on R. Thanks. $\endgroup$ Jun 7, 2017 at 20:07

1 Answer 1

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b) If aRRb, bRRa, then some x,y with aRx, xRb, bRy, yRa. Thus...

If aRRb, bRRc, then some x,y with aRx, xRb, bRy, yRc. Thus...

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  • $\begingroup$ I'm not sure whether you are saying (b) is correct or incorrect. Of course it may be helpful to unroll the relation $R^2$ as you outline, but you omit to draw a conclusion one way or the other. $\endgroup$
    – hardmath
    Jun 8, 2017 at 2:25
  • $\begingroup$ William, I don't get what you mean by that. Your first line implies that RR can't be symmetric. But I need it to be anti-symmetric, it's not the same thing. What am I missing ? $\endgroup$ Jun 8, 2017 at 4:11
  • $\begingroup$ @hardmath User didnot want a proof so I gave him a hint. $\endgroup$ Jun 9, 2017 at 3:38
  • $\begingroup$ @user3275222. To continue, aRb and bRa. Hence a = b. How are you defining anti-symmetric? $\endgroup$ Jun 9, 2017 at 3:44

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