Talk:Union of Initial Segments is Initial Segment or All of Woset

Does $b$ need to be distinct from $y$? I'm not sure it does. The argument works well enough even when $b = y$. It makes the argument simpler because you only have to worry about one ordering, $\preccurlyeq$, and not have to worry about its strict counterpart. --prime mover (talk) 11:48, 8 June 2018 (EDT)


 * I thought it needed to be, but reading the proof again I think it does not. --GFauxPas (talk) 11:50, 8 June 2018 (EDT)


 * ... and I'm still trying to work out why the non-existence of a $y$ such that $b \preccurlyeq y$ implies that $J \subseteq S_{x_0}$. I expect it's obvious, but I can't see it. --prime mover (talk) 12:42, 8 June 2018 (EDT)


 * take $x_0 = y$, I changed the variable name to make it clearer. --GFauxPas (talk) 12:52, 8 June 2018 (EDT)


 * But you've just disproved the existence of such a $y$. How does it work that $J$ is a subset of an initial segment of a non-existent object? Sorry, but I must be missing something here, I can't get my head round it. --prime mover (talk) 04:54, 9 June 2018 (EDT)


 * You're correct that this is quite confusing, and I don't know why I thought it was obvious. I was copying Folland,and he's guilty of handwaving (at least by standards). I see what he's doing and I'll fix it. Please continue keeping me on my toes and pushing me, hopefully one day I will be a true Bourbakist. --GFauxPas (talk) 21:34, 9 June 2018 (EDT)

Still pushing ...


 * "By virtue of $J \ne X$, there must be some $a \in A$ with $J \subseteq S_a$."

I still don't get why "there must be". It should be possible to link to some result that defines exactly why this is. It is assumed that the train of thought above it should be needed in its deduction, but I still can't see why, just because there is no $y \in J \setminus X$ such that $b \preccurlyeq y$, then it implies that $J$ is the subset of some initial segment of $X$.

I wonder whether it may be possible to go to a PBC starting "Assume $J$ is not the subset of some initial segment of $X$" or something. I may take a look at Devlin sometime (I gave up a few years ago when my interpretation of his approach was challenged and I could not muster the enthusiasm to continue). --prime mover (talk) 01:16, 10 June 2018 (EDT)