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May 18, 2011 at 20:52 vote accept Asaf Karagila
Sep 15, 2010 at 5:37 comment added Pete L. Clark @AK: Yes, I believe so. (This is not really my area of core expertise.) See for instance emis.de/proceedings/TopoSym2001/20.pdf
Sep 15, 2010 at 4:07 comment added Asaf Karagila @Pete: So there are no real closed fields which are Dedekind-closed except the real numbers?
Sep 15, 2010 at 3:57 comment added Pete L. Clark @AK: I think he means that the order is "bounded complete": every set which is bounded above has a least upper bound.
Sep 15, 2010 at 3:47 comment added Asaf Karagila When you say complete, do you mean metrically? Because that'd be obvious since we define metrics using the reals. Or do you mean in the sense that it is a complete order (i.e. all the Dedekind cuts are realized)?
Sep 15, 2010 at 3:37 history answered Richard Borcherds CC BY-SA 2.5