Definition talk:Pairwise Disjoint

Does it give us an advantage to change the definition from a set of sets to a family of sets? Before you can define a family you need to have defined the concept of a mapping, which is further down the road of complexity. What would be good to strive for would be a definition from as basic a level as possible. --prime mover 02:16, 27 April 2012 (EDT)
 * I can see your point, and I agree. So it would become 'For $S, T \in \Bbb S$ such that $S \ne T$, $S \cap T = \varnothing$.' is it not? --Lord_Farin 04:02, 27 April 2012 (EDT)
 * Effectively. It means reverting back to the last version of Dec 2011. --prime mover 08:35, 27 April 2012 (EDT)
 * Isn't the current version more general? Maybe we could include both cases? --abcxyz 10:33, 27 April 2012 (EDT)

I wouldn't say that a set can be generalised by indexing it; I recall a theorem saying that any set may be indexed, but I can't find it atm. Both cases would offer a solution. Something along the lines of 'when $\Bbb S$ is an indexed set, ... $S_i \cap S_j = \varnothing$ if $i \ne j$' --Lord_Farin 10:57, 27 April 2012 (EDT)
 * My concern was that, for example, using the "family" version, the ordered pair $\left({S, S}\right)$ where $S$ is non-empty would not be considered as pairwise disjoint, whereas if one uses the "set of sets" version, the set under consideration would be just the singleton $\left\{{S}\right\}$, which would be considered as pairwise disjoint.
 * I've changed the definition page. Is it going in the right direction? --abcxyz 11:17, 27 April 2012 (EDT)