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[Symmetry is easy enough for me (assuming I'm even correct)... $$(\forall(a,b)\in\mathcal{R})[(b,a)\in\mathcal{R}]$$

but I don't know how to say it similarly for transitivity and reflexivity.. There's no single $a\in\mathcal{R}$, nor is there $a,b,c\in\mathcal{R}$

Basically, I want to say that $\mathcal{R}=\{(1,1),(2,2),(3,3),(4,4),(1,2),(2,1),(2,3),(3,2),(3,4),(4,3)\}$ is reflexive and symmetric, but not transitive.

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up vote 2 down vote accepted

A relation $\mathcal R\subset A\times A$ is

  • reflexive if $\forall a\in A\colon (a,a)\in\mathcal R$
  • symmetric if $\forall a\in A, b\in A\colon (a,b)\in\mathcal R\to (b,a)\in \mathcal R$ or equivalently $\forall (a,b)\in\mathcal R\colon (b,a)\in\mathcal R$
  • transitive if $\forall a,b,c\in A\colon ((a,b)\in\mathcal R\land (b,c)\in\mathcal R)\to(a,c)\in\mathcal R$ or equivalently $\forall (a,b),(b,c)\in \mathcal R\colon (a,c)\in\mathcal R$
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