Least Upper Bound Property $\implies$ Complete I want to show that if an ordered field $X$ has the least upper bound property (meaning, every nonempty set $E$ which is bounded above has $\sup E \in X$), then it is complete (meaning, every Cauchy sequence converges in $X$).
I know the converse is not true, but how do I prove this direction?
 A: In an arbitrary ordered field one has the notion of Dedekind completeness -- every nonempty bounded above subset has a least upper bound -- and also the notion of sequential completeness -- every Cauchy sequence converges.
The main theorem here is as follows:
For an ordered field $F$, the following are equivalent:
(i) $F$ is Dedekind complete.
(ii) $F$ is sequentially complete and Archimedean.
Since the Archimedean hypothesis often goes almost without saying in calculus / analysis courses, many otherwise learned people are unaware that there are non-Archimedean sequentially complete fields.  In fact there are rather a lot of them, and they can differ quite a lot in their behavior: e.g. some of them are first countable in the induced (order) topology, and some of them are not.  (In particular there are some ordered fields in which a sequence is Cauchy if and only if it is eventually constant!  This is a case where one should consider Cauchy nets if one is serious about exploring the topology...)
These issues are treated in $\S 1.7$ and $1.8$ of these notes.
