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Does an ordered field $F$ satisfying the Cauchy-Cantor axiom but not the Archimedean axiom exist?

Cauchy-Cantor axiom: Every system of nested intervals $I_1 \supset I_2 \supset \cdots \supset I_n \supset \cdots$ have a common point. Here, an interval $I$ is any set of the form $I= \{ x \in F \mid a \leqslant x \leqslant b\}$ for some $a\le b$ in $F$.

Archimedean axiom: For every $x,y \in F$ with $y>0$ there exists a unique $n \in \mathbb{Z}$ such that $ny \leqslant x < (n+1)y$.

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  • $\begingroup$ Start post has been changed. $\endgroup$ – kp9r4d Jul 17 '14 at 16:31
  • $\begingroup$ Thanks for your answer! "does the nested interval property imply that an ordered field is Archimedean", - how can I prove that? $\endgroup$ – kp9r4d Jul 17 '14 at 17:10
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    $\begingroup$ One can construct an example using formal Laurent series. For a discussion of every conceivable related question, please see this article by James Propp. $\endgroup$ – André Nicolas Jul 17 '14 at 17:16
  • $\begingroup$ "The nested interval property statement has the inclusions running the wrong way.", - my fault, I changed. Thanks for article! $\endgroup$ – kp9r4d Jul 17 '14 at 17:21
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    $\begingroup$ The question seems clear enough to me. $\endgroup$ – Michael Hardy Jul 17 '14 at 19:15
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An ordered field that satisfies the Nested Interval Property need not be Archimedean.

A very nice paper that answers all questions of this general character that one might ask, and more, is James Propp's Real Analysis in Reverse.

For an early class of examples, please see the classic Rings of Continuous Functions (Gillman and Jerison).

There are also constructions based on ultrapowers of the reals.

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  • $\begingroup$ In the paper you link, on p.13 it says "Note however that the field of formal Laurent series is not a counterexample; although it satisfies the Shrinking Interval Property, it does not satisfy the Nested Interval Property". So your answer cannot possibly be "the standard example". $\endgroup$ – Asaf Karagila Jul 17 '14 at 21:31
  • $\begingroup$ Thank you, I was thinking of Cantor's version, which is shrinking interval, and is not the one OP uses. $\endgroup$ – André Nicolas Jul 17 '14 at 21:37

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