Determine whether $R$ is a partial order on set the $S$ and justify your answer. $S=\{1,2,3\}$ and $R=\{(1,1),(2,3),(1,3)\}$.

So the job is to consider if $R$ is reflexive, antisymmetric and transitive. I don't know how to check for any of these in set notation. Could someone please explain? The answer key says it's not reflexive. Is that because it's doesn't have the ordered pairs $(2,2)$ and $(3,3)$?

EDIT: How is it transitive and antisymmetric?

  • 2
    $\begingroup$ I think the link you're missing is: $x~R~y$ is defined as $(x, y) \in R$. So your explanation of reflexive is correct. $\endgroup$ – Henry Swanson Nov 18 '13 at 6:30
  • $\begingroup$ @Celeritas Do you still need help with this? $\endgroup$ – Git Gud Mar 22 '14 at 1:23

This boils down to the definition of binary relation on a set. If $X$ is a set and $R$ is a binary relation on $X$, when people write $xRy$, what they mean is $(x,y)\in R$.

More precisely given a set $X$ and $R$ a subset of $X^2$, by definition $xRy\iff (x,y)\in R$, for all $x,y\in X$.

In your question, $X=\{1,2,3\}$ and $R=\{(1,1). (2,3), (1,3)\}$.

The relation $R$ is reflexive, if, and only if, $\forall x\in X\biggl(\underbrace{(x,x)\in R}_{\large xRx}\biggr)$.

  • $\begingroup$ Comment, downvoter? $\endgroup$ – Git Gud Nov 17 '13 at 10:36

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