0
$\begingroup$

i think i have to use these facts

(a,b)$\in$R* and a$\neq$b

(a,b)$\in$R or a=b

to show they are both subsets of each other .?? but im not sure how.

1 Answers 1

3

If I understand your question correctly, you have a relation $R$ on $A$ that is reflexive, antisymmetric, and transitive, $S=\{\langle a,b\rangle\in R:a\ne b\}$ is the strict order corresponding to $R$, and $R^*$ is the non-strict order extending $S$, and you want to prove that $R^*=R$.

By definition $R^*=S\cup\{\langle a,a\rangle:a\in A\}$, so you’re trying to show that

$\{\langle a,b\rangle\in R:a\ne b\}\cup\{\langle a,a\rangle:a\in A\}=R\;.$

You can indeed do this by showing that each of the sets $\{\langle a,b\rangle\in R:a\ne b\}\cup\{\langle a,a\rangle:a\in A\}$ and $R$ is a subset of the other. I’ll do one direction.

If $\langle x,y\rangle\in\{\langle a,b\rangle\in R:a\ne b\}\cup\{\langle a,a\rangle:a\in A\}$, then either $\langle x,y\rangle\in\{\langle a,b\rangle\in R:a\ne b\}$, or $\langle x,y\rangle\in\{\langle a,a\rangle:a\in A\}$. If $\langle x,y\rangle\in\{\langle a,b\rangle\in R:a\ne b\}$, then certainly $\langle x,y\rangle\in R$. If, on the other hand, $\langle x,y\rangle\in\{\langle a,a\rangle:a\in A\}$, then $\langle x,y\rangle=\langle a,a\rangle$ for some $a\in A$, and then $\langle x,y\rangle\in R$ because $R$ is reflexive.

Now you show that $R\subseteq\{\langle a,b\rangle\in R:a\ne b\}\cup\{\langle a,a\rangle:a\in A\}$: assume that $\langle x,y\rangle\in R$, and show that either $\langle x,y\rangle\in\{\langle a,b\rangle\in R:a\ne b\}$, or $\langle x,y\rangle\in\{\langle a,a\rangle:a\in A\}$.

  • 0
    @Joshua: Yes, that’s exactly right.2012-09-22