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We know that every complete ordered field is isomorphic to $\mathbb R$, but are there examples of complete ordered fields different, not isomorphically different of course, from $\mathbb R$?

user5826
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    Every complete Archimedian ordered field is isomorphic to $\mathbb{R}$. – Daniel Fischer Nov 04 '14 at 21:31
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    @DanielFischer: The distinction depends on the exact definition of completeness. At our university, completeness (in the context of this definition) is (most of the times) defined as every bounded (from above) subset having a supremum. In this case, the Archimedes property is a consequence. If you define completeness in terms of Cauchy sequences, you are correct of course. – PhoemueX Nov 05 '14 at 06:21
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    As we know, there are different ways to construct the real number system in set theory. If we knew which one you consiteded to be the real $\mathbb R,$ then I guess any of the others would answer your question? – bof Apr 16 '16 at 05:06

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Any Dedekind-complete ordered field can be defined to be the reals, $R$, although it is sometimes useful to have some other relationships between the members of $R. $ Examples: Assume that we have "the" field $Q$ of rationals, we can define $R$ as the set of equivalence classes of Cauchy sequences in $Q$, or as the union of $Q$ with the set of its proper Dedekind cuts, or by the usual set of decimal representations....

DanielWainfleet
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If $\mathbb R$ denotes a complete ordered field, then the set of pairs $(x,0)$, where $x\in\mathbb R$, can also be regarded as a complete ordered field via the operations

  • $(x,0)+(y,0)=(x+y,0)$
  • $(x,0)\cdot(y,0)=(xy,0)$
  • $(x,0)\le (y,0)$ iff $x\le y$

A less trivial example would be: as additive groups, $\mathbb R$ and $\mathbb C$ are isomorphic. Let $\varphi:\mathbb R\to\mathbb C$ be a group isomorphism. Use this isomorphism to define $\cdot$ and $\le$ on $\mathbb C$ in the following way: $\varphi(a+b)=\varphi(a)+\varphi(b)$, and $\varphi(x)\le\varphi(y)\iff x\le y$. Since $\varphi$ is bijective, these operations are well-defined. Then, $\varphi$ is an isomorphism of ordered fields.

More generally, any set with the same cardinality as $\mathbb R$ can be endowed with operations that turn it into a complete order field, via transport of structure.

Joe
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