Talk:Closed Set in Linearly Ordered Space

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Would it be better to state this more like "If $s$ is a supremum of $S$, then $s \in Y$"? --Dfeuer (talk) 07:40, 12 February 2013 (UTC)

Actually, I think this thing is getting ready to graduate to a theorem characterizing closed sets in linearly ordered spaces. Something like

$Y$ is closed in $X$ iff: for each nonempty $S \subseteq Y$, if $b \in X$ is a supremum or infimum of $S$ then $b \in Y$.

It's probably also possible to define the closure of a set as the set of suprema and infima of its non-empty subsets. It's 3am, though, so my thinking might be a bit fuzzy. --Dfeuer (talk) 07:56, 12 February 2013 (UTC)