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

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)
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)