Talk:Equivalence Classes are Disjoint

If and only if
CMIIW but to prove disjointness one only needs to prove:


 * $\tuple {x, y} \notin \RR \implies \eqclass x \RR \cap \eqclass y \RR = \O$

The converse is covered here Equivalence Class is not Empty Oliver (talk) 17:55, 16 October 2015 (UTC)


 * Not obviously it isn't. Maybe indirectly, but the page you cited (much easier to follow it as an internal link, btw) does not state and prove $\eqclass x \RR \cap \eqclass y \RR = \O \implies \tuple {x, y} \notin \RR$. --prime mover (talk) 18:01, 16 October 2015 (UTC)


 * I agree with that, but my point is that it isn't even necessary to prove $\eqclass x \RR \cap \eqclass y \RR = \O \implies \tuple {x, y} \notin \RR$ to prove disjointness, as it is defined here anyway Definition:Pairwise Disjoint. Oliver (talk) 18:33, 16 October 2015 (UTC)


 * This page proves both directions. Yes I know, the title doesn't exactly match the contents but I think it would be suboptimal to try and word it accurately. --prime mover (talk) 18:39, 16 October 2015 (UTC)

Question: In Definition of Symmetric Relation in Proof 1, it should be (z,y) instead of (x,y), shouldn't it? I mean if (y,z) are in relation, as stated before, then by symmetry, (z,y) are in relation. I didn't edit because I don't want to accidentally edit something correct. It should work in my way because if (x,z) are in relation and (z,y) are in relation, then (x,y) are in relation by transitivity. --Daffeen (talk) 20:40, 4 December 2020 (CET)


 * Good call. Well noticed. Fixed. --prime mover (talk) 20:49, 4 December 2020 (UTC)