I'm learning about relations on sets, and I'm trying to figure out what exactly antisymmetric means. The way we represent a relation is like a adjacency matrix. In my textbook I see that symmetric relation is symmetrical with respect to the diagonal (of the adj. matrix) and that is logical to me, but they also mention that an antisymmetric relation is a relation with members on only one side of the diagonal. This does not make sense to me. I agree that the relations with members only above the diagonal are indeed antisymmetric but not all antisymmetric need to be like that.

Since the condition for antisymmetric is: (x,y)eR and (y,x)eR implies x = y. If I understand this correctly than a relation like this is antisymmetric: R = { (0,0), (1,1), (2,2), (2,0), (1,2) } on set X = { 0, 1, 2 }.

Am I misunderstanding something?


closed as too localized by Andrés E. Caicedo, Angelo, Pete L. Clark, Franz Lemmermeyer, Joel David Hamkins Feb 26 '11 at 10:52

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The "A beats B in roshambo" relation is antisymmetric in your sense. There does not exist a permutation of the set {rock, paper, scissors} for which the adjacency matrix is supported above the diagonal. So you are correct and your textbook is wrong.

For future reference, this question and related ones are probably better suited at math.SE, because it is about homework/textbook question rather than about research mathematics. I have left this answer as an answer rather than a comment so that the question can be removed from the front page without all the "close as inappropriate" stuff that we usually go through.


Unlike rows and columns of a matrix, elements of a set have no preferred enumeration. If the adjacency matrix would place some instances of the relation below the main diagonal, you can then swap the individual elements' position in the matrix representation, thus yielding a matrix where the instance is above the diagonal. By repeating this for all elements, it is easy to obtain a matrix which respects your criteria.

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    $\begingroup$ I'm not sure about that. You cannot change aRb to bRa so that all of the pairs are above or below the diagonal because that would change the relation. $\endgroup$ – synepis Feb 26 '11 at 0:20

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