Whether Cauchy complete and Cantor complete are equivalent in terms of ordered field

Question

For reference, "Cantor complete" means that every nested sequence of bounded closed intervals has non-empty intersection.

It is easy to show that the conditions "Cauchy complete" and "Cantor complete" are equivalent in Archimedean ordered fields. However in terms of non-Archimedean fields, I think those two completeness shouldn’t be equivalent.

In this paper, https://arxiv.org/pdf/1101.5652 ,in theorem 6.6,Cantor k-complete(or Cantor complete) implies Cauchy complete and it’s not a sufficient and necessary condition.

It should be noticed that all the Cauchy complete examples given from that paper and this link Example of a complete, non-archimedean ordered field, are Cantor complete or Spherical complete, like the Hahn series field, Laurent series field and Robinson’s Asymptotic field.

So, can anyone give some examples that only satisfying Cauchy complete but not Cantor complete, or just prove that those two completeness are equivalent in the non-Archimedean case. I would be most grateful for your kind response.

score 2 · Answer 1 · edited Jul 12 '25 at 06:10

2

Finally, I found the answers. In this paper, https://arxiv.org/abs/math/0601722, and another reference, V. Pestov, On a valuation field invented by A. Robinson and certain structures connected with it, Israel J. Math. 74 (1991), p.. 65-79, the answers are showed clearly. Cantor completeness implies spherical completeness implies sequential completeness; but both converse implications are false.

edited Jul 12 '25 at 06:10

John Bentin

20,500

answered Jul 06 '25 at 12:14

dy Deng

299

By “both converse have examples”, do you mean “both converse implications have counterexamples”? – John Bentin Jul 06 '25 at 15:17
yes，my statement is not so clear.I wrote those words without polishing. – dy Deng Jul 12 '25 at 05:09

Whether Cauchy complete and Cantor complete are equivalent in terms of ordered field

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