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Sep 26, 2013 at 8:38 history edited Ben CC BY-SA 3.0
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Sep 26, 2013 at 7:25 comment added Ben Joel, that's a helpful observation, since it allows you to begin with an arbitrary enumeration and try to use some sort of inductive procedure.
Sep 26, 2013 at 3:00 comment added Joel David Hamkins Incidentally, you can reduce to the countable case, since the connected components of $S$ are each countable, and you can treat each connected component separately, afterwards merging these discrete orders into one giant discrete order. (There is a complication when most of the discrete orders arising have a minimal but no maximal element, or conversely, but this can be handled by allowing those bottom parts to overlap.)
Sep 25, 2013 at 23:51 comment added Joel David Hamkins Your edit makes this an extremely interesting question!
Sep 25, 2013 at 19:58 comment added Joel David Hamkins By the way, these kinds of orders are called discrete orders, a linear order where every non-least point has an immediate predecessor and every non-greatest point has an immediate successor.
Sep 25, 2013 at 19:45 history edited Ben CC BY-SA 3.0
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Sep 25, 2013 at 12:53 answer added Joel David Hamkins timeline score: 1
Sep 25, 2013 at 8:27 history asked Ben CC BY-SA 3.0