Talk:Equivalence Classes are Disjoint

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


 * $\left({x, y}\right) \notin \mathcal R \implies \left[\!\left[{x}\right]\!\right]_\mathcal R \cap \left[\!\left[{y}\right]\!\right]_\mathcal R = \varnothing$

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 $\left[\!\left[{x}\right]\!\right]_\mathcal R \cap \left[\!\left[{y}\right]\!\right]_\mathcal R = \varnothing \implies \left({x, y}\right) \notin \mathcal R$. --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 $\left[\!\left[{x}\right]\!\right]_\mathcal R \cap \left[\!\left[{y}\right]\!\right]_\mathcal R = \varnothing \implies \left({x, y}\right) \notin \mathcal R$ 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)