Talk:Compact Subspace of Linearly Ordered Space/Reverse Implication

I'd like to make the premise be that $Y$ is a closed sublattice of $X$, but for some reason I can't imagine, the Internet suggests that term is only used if $X$ is complete. There doesn't seem to be any harm in extending it. Dfeuer (talk) 23:11, 9 February 2013 (UTC)


 * Current premise entails $Y$ closed (I draft-proved that) and a complete lattice. --Lord_Farin (talk) 23:12, 9 February 2013 (UTC)


 * Hmm? I don't understand. The premise of my theorem is that $Y$ is a "closed sublattice" (if that term is extended in the obvious way). I don't see how topological closure enters into it. The original theorem claimed that but I disproved that, no? --Dfeuer (talk) 23:16, 9 February 2013 (UTC)


 * And your own draft mentions topological closure in the premises, but does not, as far as I can tell, use it anywhere. --Dfeuer (talk) 23:18, 9 February 2013 (UTC)


 * In any case, $Y$ is compact, $X$ Hausdorff so $Y$ closed. I see now what you mean by "closed sublattice". Nvm, then. --Lord_Farin (talk) 23:20, 9 February 2013 (UTC)

It does happen to be true that $Y$ is topologically closed, but that is not the least bit interesting and I would suggest we remove it from this altogether. It may be (semi-educated guess) that we could use it in the context of $X$ being a linear continuum, maybe, since such are much more like the reals. In any case, proposed refactor:


 * 1) Compact implies order-complete.
 * 2) Closed sublattice implies compact.
 * 3) However you want to describe your condition implies compact.
 * 4) In the event that compact implies closed sublattice, then we can combine that with 1 to make a twofer, but I haven't even considered it yet.
 * 5) In the event that yours has a valid converse, we can do that too, of course. --Dfeuer (talk) 23:27, 9 February 2013 (UTC)