Talk:Outer Measure of Limit of Increasing Sequence of Sets/Proof 1

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As for the 'explain', the statement there is correct. There should be a page Increasing Sequence of Sets Well-Ordered by Inclusion or st. like that. It follows that in fact the subdivision by $S_{n+1} \setminus S_n$ is a partition (i.e., the sets are pairwise disjoint). --Lord_Farin 06:24, 25 March 2012 (EDT)

Unless I'm mistaken, the proof does not assume the well-ordering of $\left\{{S_n : n \in \N}\right\}$; it only assumes the well-ordering of $\Z_{\ge 0}$. After all, the proof includes choosing "the least $n \in \Z_{\ge 0}$ such that...", not "the least $S_n$...". –Abcxyz (talk | contribs) 10:59, 25 March 2012 (EDT)
Technically correct; still, the sets form a partition and thus the minimal is also unique. Maybe there'd better be a page Increasing Sequence of Sets induces Partition on Limit instead. --Lord_Farin 11:03, 25 March 2012 (EDT)
The fact that the above needs to be explained and discussed in the talk page indicates that this part of the proof needs to be expanded upon. As LF says, a page with an appropriately-named result should be enough. --prime mover 11:54, 25 March 2012 (EDT)
As I said before, Increasing Sequence of Sets induces Partition on Limit is not used in this proof, as far as I know. Any ideas about how to make the proof clearer? (I'm asking because I don't have any ideas.) –Abcxyz (talk | contribs) 13:47, 25 March 2012 (EDT)
Yes, explain why "it follows that $x \in S_{n+1} \setminus S_n$." --prime mover 13:50, 25 March 2012 (EDT)
If $n \ge 0$ is the least possible value such that $x \in S_{n+1}$, then $x \notin S_n$. So $x \in S_{n+1} \setminus S_n$. –Abcxyz (talk | contribs) 15:07, 25 March 2012 (EDT)
Yes, very good. But when I said "explain why ..." I didn't mean "explain to me here", I meant "explain on the main page where the question is raised." That's what an "explain" tag is for. It means: there's a detail here that has been glossed over and needs to be explained in more detail. --prime mover 15:25, 25 March 2012 (EDT)
I just wanted to make sure that my explanation is okay to put on the page. Thanks. –Abcxyz (talk | contribs) 15:28, 25 March 2012 (EDT)
No worries.--prime mover 16:30, 25 March 2012 (EDT)