# 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)

- 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)

- No worries.--prime mover 16:30, 25 March 2012 (EDT)