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Can the ultrapower construction (used for extending the field of real numbers to get the field of hyperreals) be applied to the field $\mathbb{Q} $ of rational numbers?

In my view it should be possible to start with the set $\mathbb{Q} ^{\mathbb {N}} $ of all sequences of rationals and then use equivalence between these sequences based on a free ultrafilter. And my guess is that it should lead to another non-archimedean ordered field $ {} ^*\mathbb{Q} $. More specifically if ${} ^*\mathbb{R} $ is the field of hyperreals then ${} ^*\mathbb {Q} $ should be isomorphic to the set $$\{x\mid x\in{} ^*\mathbb {R}, \text{ standard part of }x\text{ is rational or }x\text{ is an infinite hyperreal number} \} $$

Do the axioms like Extension Principle, Transfer Principle and Standard Part Principle apply in analogous manner to ${} ^*\mathbb{Q} $?

Update: Since there are sequences of rationals which converge to an irrational number, most likely the Standard Part Principle will not be valid here. But I don't see if the other two principles have any issues.

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One can apply the ultrapower construction to any mathematical object $X$ and get the corresponding object ${}^\ast\!X$ which is then helpful in studying the properties of the original object $X$ itself.

In particular one can apply the ultrapower construction to $\mathbb Q$ producing the ordered field of hyperrational numbers ${}^\ast \mathbb Q$ which can be naturally identified with a subfield of ${}^\ast \mathbb R$.

The transfer principle applies to all ultraproducts; this is essentially a theorem of Jerzy Los from 1955 (i.e., six years before Robinson's first paper on the subject).

Here is an interesting observation developed more fully e.g., in

Davis, Martin. Applied nonstandard analysis. Pure and Applied Mathematics. Wiley-Interscience [John Wiley & Sons], New York-London-Sydney, 1977.

Consider the subring of $F\subseteq{}^\ast \mathbb Q$ consisting of all finite hyperrationals, and let $I\subseteq F$ be the subring of all infinitesimals (with the usual definitions of "finite" and "infinitesimal"; I can clarify if necessary). Then $I$ is a maximal ideal of $F$ and the quotient $F/I$ is naturally isomorphic to the field $\mathbb R$.

In other words, this approach gives an alternative construction of the real field from the rationals that does not require either Dedekind cuts or equivalence classes of Cauchy sequences.

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  • $\begingroup$ Your answer has cleared even those doubts which I was not able to formulate clearly. Especially the fact that this construction leads to the reals. Thanks a lot!! $\endgroup$
    – Paramanand Singh
    Jan 29 '18 at 15:34
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    $\begingroup$ And I am going through Keisler's book so I am aware of "finite and infinitesimals" as they are used in this theory. The more I read the book I will ask questions on these topics and hopefully be able to understand the whole picture in a much better way. $\endgroup$
    – Paramanand Singh
    Jan 29 '18 at 15:38
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Yes, of course. In particular you can regard ${}^*\Bbb Q$ as a subfield of ${}^*\Bbb R$. But it is not the same as the set of hyperreals with rational standard part. Neither inclusion holds; there is an infinite $x\notin {}^*\Bbb Q$. Then $1/x\notin {}^*\Bbb Q$ but has standard part zero. Also there are infinite elements in ${}^*\Bbb Q$ which have no standard part.

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  • $\begingroup$ But infinite hyperreals also don't have standard part. $\endgroup$
    – Paramanand Singh
    Jan 29 '18 at 7:21
  • $\begingroup$ I think I can understand your argument about why ${} ^*\mathbb{Q} $ would not be the set I described in my post. But I think the principle of standard part should apply. $\endgroup$
    – Paramanand Singh
    Jan 29 '18 at 7:25
  • $\begingroup$ I have updated the description in my post. Does it make sense now? $\endgroup$
    – Paramanand Singh
    Jan 29 '18 at 7:29

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