Talk:Union of Relations Compatible with Operation is Compatible

Does this result already exist on PW? I don't even know where to look. --Dfeuer (talk) 16:59, 5 January 2013 (UTC)

Why stop here? The same argument seems to prove that any union of relations compatible with $\circ$ is compatible with $\circ$ as well. The current result could be mentioned as a corollary. The whole result seems not to be covered atm, but I refer to PM for a decisive answer. --Lord_Farin (talk) 17:14, 5 January 2013 (UTC)

Very good point. --Dfeuer (talk) 17:54, 5 January 2013 (UTC)

Does that cover it? --Dfeuer (talk) 18:15, 5 January 2013 (UTC)

Yes, seems pretty accurate. As usual, needs to be cast into house style. --Lord_Farin (talk) 18:19, 5 January 2013 (UTC)

I also proved the analogous intersection law. Then I realized that a key thing about groups is that the complement of a relation compatible with a group is compatible with the group. So if $<$ is compatible, so is $\ge$. It's trivial that the inverse of a compatible relation is compatible, so in the case of a group, knowing that $<$, $>$, $\le$, or $\ge$ is compatible proves that they all are. --Dfeuer (talk) 18:46, 5 January 2013 (UTC)

Marvellous. I complete the exercise by noting that the collection of compatible relations on a group is a complemented, distributive and complete lattice. I.e., a complete lattice and also a Boolean algebra. Excellent :). --Lord_Farin (talk) 19:38, 5 January 2013 (UTC)