Talk:Ordering can be Expanded to compare Additional Pair

Take $S = \{a,b,c\}$, $\preceq$ by putting as only nontrivial comparisons $c \preceq a$ and $b \preceq c$. It is not the case that $\preceq'^+$ is an ordering, since $c \preceq'^+ a \preceq'^+ b \preceq'^+ c$ but $a \ne b \ne c$. I have yet to find where this clashes with your proof(s). It might be the case that you can always add one of $a \preceq b$ and $b \preceq a$, but in current form I think this provides a counterexample to the result. — Lord_Farin (talk) 09:08, 4 March 2013 (UTC)

I'm not understanding you. Your $\preceq$ is already a total ordering. This theorem is about extending a partial ordering. That is, if you had only $a \preceq b$ you could add either $a \preceq c$ or $c \preceq a$. --Dfeuer (talk) 09:13, 4 March 2013 (UTC)

I blame a temporary short-circuiting in my brain. Apologies. — Lord_Farin (talk) 09:15, 4 March 2013 (UTC)

The page should probably be renamed to say Partial Ordering. --Dfeuer (talk) 09:17, 4 March 2013 (UTC)

If you like, it's your page. Makes sense to. --prime mover (talk) 10:55, 4 March 2013 (UTC)