Talk:Union of Connected Sets with Non-Empty Intersections is Connected

I think the indexing sets actually help in this case (although they often do make things harder to follow). I would vote to revert to the last version. --Cynic (talk) 19:14, 7 March 2009 (UTC)

I'd be interested to know why - it seems to me to only add an extra layer of complexity to the statement. Or do you mean you find the proof harder to follow in the new version? I'm sure it is possible to improve it. But I think the statement becomes a lot easier (and the proof doesn't become conceptually harder) if we just talk about a set of connected subsets, and taking their union.

It might also potentially make applications less awkward, since we might otherwise have to make up an "index set" where there isn't one naturally, or otherwise trust to the reader to see the equivalence between this statement and the other one.

But I don't want to be dogmatic about this. lasserempe 21:01, 7 March 2009 (UTC)

It's easy enough to follow, but feels a bit handwavey to me. It'll do, I suppose.

Mind, what I try to do, if doing major surgery on an existing proof, is comment out the existing version of things being changed within so it's more straightforward to compare the different versions. --Matt Westwood 21:29, 7 March 2009 (UTC)

After more thought, I don't think it really matters. On a completely unrelated issue, would $B \cup \bigcup \mathcal{A}$ be more clear with some kind of grouping symbols? I know what it means, but it looks awkward. --Cynic (talk) 21:51, 7 March 2009 (UTC)

Definition of connectedness
This proof uses the first definition of connectedness, which I think makes the proof more awkward than it needs to be. The definition in terms of partitions should do nicely, I think. --Dfeuer (talk) 04:22, 8 December 2012 (UTC)


 * I've set up a page for you to do it. --prime mover (talk) 06:19, 8 December 2012 (UTC)