I am having trouble understanding where A x A is anti symmetric.

Say A = {1,2,3}. We get A x A to be {(1,1)(1,2)...(3,3)}

As I understand it, a different relation {(1,1)(2,2)(3,3)} is both symmetric and anti symmetric. Symmetric because xRy implies yRx, in this case 1R1 implies 1R1. It's anti symmetric because 1R1 and 1R1, then 1=1.

Is the same true for the relation A x A? It's symmetric because all xRy implies yRx, but in the case of (1,1)(2,2)(3,3) it has antisymmetric properties.

I am also asking this because I read in my book that A x A is a total order, which implies that it's a partial order, which means that this relation must also be a anti symmetric?


If $|A| > 1$, then $R = A \times A$ is not a totally ordered relation on $A$ because it is not anti-symmetric. Indeed, for your example, we have that $1R2$ and $2R1$ but $1 \neq 2$.

  • $\begingroup$ Does that mean relations (a,a)(b,b)(c,c)... are the only relations that are both antisymmetric and symmetric (if whatever set they came from has more than 1 element)? $\endgroup$ – pajkatt Oct 31 '16 at 5:35
  • $\begingroup$ Yes, a relation is both antisymmetric and symmetric iff it is a subset of the identity relation. $\endgroup$ – Adriano Oct 31 '16 at 5:38
  • $\begingroup$ Ah, then I have a problem because a practice quiz from my book has the solution that given set A = {1,2,3} and R = A x A, that R is a total function (from A to A). Am I misunderstanding something somewhere or is the book just wrong? $\endgroup$ – pajkatt Oct 31 '16 at 5:46
  • $\begingroup$ $R$ is indeed a total function, but it is not a total order. $\endgroup$ – Adriano Oct 31 '16 at 5:52

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