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I was wondering if partially ordered sets could have loops in their diagrams. For example isn't the $S=\{1,2,3\}$ and relation $R=\{(1,1),(2,2),(3,3),(1,2),(2,3),(3,1)\}$ a partially ordered set that has a cycle? $R$ is reflexive, antisymmetric and transitive.

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up vote 1 down vote accepted

Transitivy fails for your relation $R$.

As for loops on the diagram, that is not possible due to transitivy, that should become apparent from the algorithm to draw diagrams I describe in this answer.

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Oh right because (1,2) and (2,3) is in R but (1,3) is not. Is that why transitivity fails? – Celeritas Nov 24 '13 at 11:54
@Celeritas Exactly. It also fails because $(2,3), (3,1)\in R$, but $(2,1)\not \in R$. – Git Gud Nov 24 '13 at 11:55
I don't get the algorithm because 1)what's Euclidian plane 2)I don't know what it means for x to cover y? – Celeritas Nov 24 '13 at 12:31
@Celeritas Just think of the euclidean place as $\Bbb R^2$. Intuitively, $x$ covers $y$ if there aren't any elements between $x$ and $y$. For instance here, $\{x,y\}$ and $\{x,z\}$ both cover $\{x\}$, but $\{x,y,z\}$ and $\{y,z\}$ don't (for different reasons). – Git Gud Nov 24 '13 at 12:34
@Celeritas It's what denoted here as '$<\cdot$', $a\prec b$ means '$b$ covers $a$'. – Git Gud Nov 24 '13 at 14:37

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