6

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?

Martin Sleziak
  • 53,687
jamaicanworm
  • 4,494
  • I believe it is not the case that the converse is not true, but rather that it is not strong enough. For example we may need the Archimedean property to prove the completeness axiom. If however you start with the monotone convergence theorem you can show that it implies the completeness axiom –  Mar 18 '12 at 00:50
  • 2
    Its not really clear what you mean by "bounded above" or "sup E." These aren't properties of subsets of metric spaces, generally. – Tim kinsella Mar 18 '12 at 00:51
  • Is your metric space ordered somehow? How does the order relate to the metric? This doesn't really make a lot of sense to me as written... – Arturo Magidin Mar 18 '12 at 00:56
  • @ArturoMagidin Whoops I tacitly assumed the OP was talking about $\mathbb{R}$...maybe that is what the OP means? –  Mar 18 '12 at 01:26
  • @jamaicanworm http://math.stackexchange.com/questions/62087/least-upper-bound-property-iff-convergence-of-cauchy-sequences –  Mar 18 '12 at 01:27
  • 1
    Sorry I meant ordered field! – jamaicanworm Mar 18 '12 at 01:30
  • To have Cauchy sequences, you need to be working in a metric space. What metric space structure can you put on, say, the linearly ordered field $\mathcal{L}$ of real Laurent series with only finitely many fractional terms? – Neal Mar 18 '12 at 01:41
  • This is Exercise 8 (with lots of hints) in this answer of mine – kahen Mar 18 '12 at 01:55
  • @Neal no, you don't need a metric to have Cauchy sequences. You just need to be working in a topological group which is preferably a 1st-countable, such that sequences are enough to describe the topology. Try reading Definition 10 in the answer I linked in my previous comment. – kahen Mar 18 '12 at 01:56
  • In the question, what's the link between the order and the topology? In principle you can have an order and a topology that have no relation to each other. – Martin Argerami Mar 18 '12 at 07:59
  • @kahen That is a WONDERFUL answer. Thank you! – Neal Mar 21 '12 at 05:43

1 Answers1

14

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.

Community
  • 1
Pete L. Clark
  • 97,892