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I am losing my mind over this:

(a) The relation $A=\{(1,1),(2,2),(3,3),(4,4),(3,2),(2,1),(3,1),(4,1)\}$ on the set $S=\{1,2,3,4\}.$

I'm having trouble figuring out if it's reflexive, symmetric, antisymmetric and transitive, because I don't know which ordered pairs to use (I know it has to be antisymmetric, not symmetric, but I want to try to understand all of them).

Like, I want to say it's reflexive, because for every element $S,$ we have an ordered pair that is $a\leq b$ and $b \leq a$, which are $(1,1), (2,2), (3,3), (4,4)$

no idea how to find out if it's transitive.

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You are correct it is reflexive.Not symmetric as$(2,1)\in A$ but $(1,2) \notin A$. transitive is satisfied as there is only one pair in $A$ such that $(a,b)\in A$ and $(b,c) \in A$i.e.$(3,2)$ and $(2,1)$.for antisymmetric it is vacuosly true as there are no elements in $ A$ such that both $(a,b)$ and $(b,a) \in A$

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  • $\begingroup$ Can you please explain symmetry. For example. say there's a set A = {1,2,3} with a subset {(1,2),(2,1), (1,1), (2,2)} We say it is NOT reflexive, as (3,3) is not in the subset? So for every single element of A, it MUST be reflexive in the subset? And for symmetry, it IS symmetrical? because we only care for the values of the subset, aka IF xRy, THEN yRx, but does not have to be true for ALL elements of A? $\endgroup$ – Test Nov 10 '14 at 5:30
  • $\begingroup$ I thought for transitive, it has to be if aRb and bRc, it implies aRc. How do we know that if (3,2) and (2,1) this implies (3,1)? $\endgroup$ – Test Nov 10 '14 at 5:32
  • $\begingroup$ Check the conditions of partially ordered set once more.It says $(a,a)$ should be in $A$ forall $a$.For symmetry if $(a,b)\in A$ then we must have $(b,a)\in A$ $\endgroup$ – Learnmore Nov 10 '14 at 5:35
  • $\begingroup$ Only for 1 pair the hypothesis for transitive is satisfied isn't it? $\endgroup$ – Learnmore Nov 10 '14 at 5:37
  • $\begingroup$ Yeah that's what I was attempting to imply. Was I incorrect for symmetry? Since we have (1,2) and (2,1) but NOT (1,3) or (3,2) or anything else, we are good to go, and it's symmetrical, yes? $\endgroup$ – Test Nov 10 '14 at 5:38

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