Let's say we have $A=\{1,2,3,4\}$, $S \subseteq A^2$, and $S = \{(1,1),(2,2)\}$ Is it transitive and symmetric? Why is it so?
I have had a doubt about it being symmetric as I understood the symmetry that it would need two pairs like (a,b) (b,a), and I am not sure it applies for pair (a,a).
As for transitivity, I've seen it needs (a,b) (b,c) and (a,c) but in this case we do not have the element for 'b', so (a,c) pair doesn't exist and my question is, if the rule does apply when you do not have appropriate pair for it.
And about transitivity if $B=\{2,3,4,5,6\}$, $R \subseteq B^2$, and $R= \{(2,4), (3,6)\}$ if it is transitive or not? Why?
A relation $\mathcal R$ is transitive over the set $\mathcal A$ if $$\forall x{\in}\mathcal A~\forall y{\in}\mathcal A~\forall z{\in}\mathcal A: \big((x,y){\in}\mathcal R\wedge (y,z){\in}\mathcal R~\to~ (x,z){\in}\mathcal R\big)$$
A relation $\mathcal R$ would not be transitive over the set $\mathcal A$ if a counter example exists in the form: $$\exists x{\in}\mathcal A~\exists y{\in}\mathcal A~\exists z{\in}\mathcal A: \big((x,y){\in}\mathcal R\wedge (y,z){\in}\mathcal R\wedge (x,z){\notin}\mathcal R\big)$$
So, can you find such a counter example in either case?
A relation $\mathcal R$ is symmetric (over the base set $\mathcal A$) if: $$\forall x{\in}\mathcal A~\forall y{\in}\mathcal A:\big((x,y){\in}\mathcal R \to (y,x){\in}\mathcal R\big) $$
A relation $\mathcal R$ would not be symmetric (over the base set $\mathcal A$) if a counter example exists in the form: $$\exists x{\in}\mathcal A~\exists y{\in}\mathcal A:\big((x,y){\in}\mathcal R\wedge (y,x){\notin}\mathcal R\big) $$
So, can you find such a counter example in either case?