User talk:Dfeuer/Compact Subspace of Linearly Ordered Space/Converse Proof 2

Can this all be recast in terms of how the elements of the cover compare to the convex components of $Y$ in $X$? Is the so-far-untranslated part of the proof the best we can do? I suspect (based on a vague hint by Kelley) that we can make it one-sided by looking at the supremum of the set of elements whose initial segments have a finite subcover. --Dfeuer (talk) 02:07, 18 February 2013 (UTC)

:Okay, so thinking a drop more about it, I'm thinking the answer is yes: rather than the somewhat awkward partition of the cover in the present proof, use this structure: 1. Prove that $Y$ has only finitely many convex components. 2. Prove that each convex component of $Y$ is covered by a finite subset of the given cover. --04:21, 18 February 2013 (UTC)

Yes, the problem is that $Y$ needn't have finitely many convex components... Cantor space is compact. --Lord_Farin (talk) 08:38, 18 February 2013 (UTC)

Here's another idea that looks better: it's apparently common knowledge that a Generalized Ordered Space/Definition 1 can be embedded (topologically and order-wise) as a dense subspace of a linearly ordered space. If I can actually prove that (which I suspect I can), then it follows immediately that a compact GO-space must be a linearly ordered space (a compact space can never embed densely in a strictly larger Hausdorff space) . To go in the other direction, I can (hopefully) show that if a subspace of a linearly ordered space satisfies the criteria on suprema and infima then its induced order topology is its subspace topology. Then the theorem will follow immediately from the fact that the criteria imply completeness. --Dfeuer (talk) 22:28, 18 February 2013 (UTC)