# Talk:Compact Subspace of Linearly Ordered Space/Reverse Implication

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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)

- 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:

- Compact implies order-complete.
- Closed sublattice implies compact.
- However you want to describe your condition implies compact.
- 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.
- In the event that yours has a valid converse, we can do that too, of course. --Dfeuer (talk) 23:27, 9 February 2013 (UTC)