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I've got five equivalences to prove, and four of them were easily proven; Given K an ordered field, I had to show that the five statements are equivalent:

1 - K is complete (aka every non empty with an upper bound set has a supremum).

2 - K is archimedean and has the property of the nested intervals.

3 - Every non-decrescent sequence with an upper bound converges in K.

4 - Every limited sequence on K has a converging subsequence.

5 - Every Cauchy sequence in K converges.

I showed the implications in the following order: $1 \implies 3 \implies 2 \implies 4 \implies 5$, however I can't at all prove that $5 \implies 1$. To be honest, I can't really figure out how to show that 5 implies on any other of those.

Whatever I try requires me creating a Cauchy sequence so that by hypothesis it converges, and then have some fancy result to prove one of the other equivalences, but the problem is that I can't show that any sequence I try making is a Cauchy sequence to actually make use of the hypothesis because I lack the archimedean property.

So my question is, given K an ordered field and supposing that every Cauchy sequence in K converges (and only that), how can I show that K is complete? Or even any of the other equivalences would help a lot, it would take an extra step, but still.

And for the archimedean + nested intervals part, simply proving that K is archimedean would easily save my day. Whatever I find on the internet uses the archimedean property with the every Cauchy sequence converges property to do so, but I don't think I should be able to use it to prove without having it as an hypothesis in the first place. Any ideas?

Francisco
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    There is a problem indeed. Compare the answers to https://math.stackexchange.com/q/160249/96384 and https://math.stackexchange.com/q/121544/96384. The first one seems to be an exact duplicate, with an answer, which is why I vote to close -- no offense against your well-posed question! – Torsten Schoeneberg Oct 17 '19 at 19:27
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    Could it be that your textbook defines Cauchy sequence in term of a metric (i.e., with $\epsilon\in \Bbb R$ instead of $\epsilon \in K$)? – Hagen von Eitzen Oct 17 '19 at 19:43
  • I see, thank you Torsten, I'll talk to my teacher. – Francisco Oct 17 '19 at 19:56
  • And I am not really using a textbook, this is a kind of homework my teacher made, and he really specifies a ordered field K. – Francisco Oct 17 '19 at 19:58
  • @Francisco: In general, statement 5 does not imply non of statements 1, 2, 3, 4. The Levi-Civita field is a non-archimedean ordered field satisfying 5 but does not satisfy 1, 2, 3 and 4. However, "every Cauchy sequence in K converges AND K is archimedean" is equivalent to 1,2,3, and 4. – Chilote Oct 19 '19 at 16:08

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