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 $1 \neq 2$.