Construct a complete and totally ordered ﬁeld that does not have the least upper bound property.

Asked Mar 22 '17 at 05:02

Active Mar 22 '17 at 08:58

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I have known about the uniqueness of real line. But I am told that it is possible to construct a field which is order and Cauchy complete without the least upper bound property in absence of the Archimidean property.

So may I ask how can we do that? Considering all the fields which can be ordered, even a fraction field, could someone explicitly give a construction with some explaination? Thanks in advance!

edited Mar 22 '17 at 08:58

asked Mar 22 '17 at 05:02

Y.X.

3,995

hyperreal fields come to mind, but they're not metric, so we need uniform completeness I presume? – Henno Brandsma Mar 22 '17 at 05:15
Have you tried the surreal numbers to see if they qualify? What about the hyperreal numbers? – Jacob Wakem Mar 22 '17 at 05:15
This question seems to give an answer. – Mark Schultz-Wu Mar 22 '17 at 05:19
@HennoBrandsma Acutally I have not learnt about uniform completeness, the question is just asking about Cauchy completeness. – Y.X. Mar 22 '17 at 05:21
@Y.X. Cauchy sequences can be defined in non-metric spaces, as well as Cauchy filters. – Henno Brandsma Mar 22 '17 at 05:30
@Mark Thanks a lot, it seems to be what I am seaching for.But could you please explan how can we order the elements and why it is complete and does not have the least upper bound property? The book did not give a proof. – Y.X. Mar 22 '17 at 05:38

Construct a complete and totally ordered ﬁeld that does not have the least upper bound property.

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