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The construction I've seen of the field of hyperreal numbers considers a non-principal ultrafilter $\mathcal{U}$ on $\mathbb{N}$, then takes the quotient of $\mathbb{R}^{\mathbb{N}}$ by equivalence respect to $\mathcal{U}$, that is, $(a_{n})=(b_{n})$ if the set $S$ of indices $n$ for which $a_{n}=b_{n}$ is inside $\mathcal{U}$.

Is it clear (or even true) that one obtains an isomorphic field with a different choice of ultrafilter?

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  • $\begingroup$ No, it's not clear, and I think that this is either open or blatantly false. I'm not sure, though. $\endgroup$
    – Asaf Karagila
    Mar 20, 2014 at 0:54
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    $\begingroup$ One can get a wide variety of non-isomorphic ultrapowers even if the index set is fixed. There is a large literature. Useful, because we can get additional "nice" properties by appropriate choice of ultrafilter. $\endgroup$ Mar 20, 2014 at 1:00
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    $\begingroup$ related: mathoverflow.net/questions/88292/… $\endgroup$
    – user13618
    Mar 20, 2014 at 1:30
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    $\begingroup$ If the continuum hypothesis holds, then yes. You can read more about this in the following link: mathoverflow.net/questions/136720/… $\endgroup$
    – user136666
    Mar 20, 2014 at 1:37

3 Answers 3

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No, one has to invoke rather strong hypotheses such as the Continuum Hypothesis (CH) to obtain uniqueness. The classic paper on this is the following, where it is proved that it is consistent with ZFC that there are $\,2^{\large \aleph_0}$ non-isomorphic hyperreal fields.

Judy Roitman. Non-Isomorphic Hyper-Real Fields from Non-Isomorphic Ultrapowers.
Math. Z. 181, 93-96 (1982)

See also N. Aldenhoven's 2010 Bachelor's thesis Uniqueness of the Hyperreal Field where you can find an elementary exposition of an earlier result of Erdos, Gillman and Hendrikson (1955) that all real-closed fields of the same cardinality as $\,\Bbb R\,$ with a $\eta_1\!$-ordering are isomorphic.

More recent results can be located by searching for citations of Roitman's paper.

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    $\begingroup$ I wouldn't call CH a "strong hypothesis". (The term "strong" has a consistency strength ring to it, whereas the consistency of CH and its failure are both equivalent to the consistency of ZFC to begin with.) $\endgroup$
    – Asaf Karagila
    Mar 20, 2014 at 1:29
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    $\begingroup$ @Asaf, I think Bill is using "strong" in the sense of "controversial" rather than in the foundational sense. $\endgroup$ Apr 29, 2014 at 8:45
  • $\begingroup$ The second link is broken now. $\endgroup$
    – KCd
    Apr 9, 2017 at 21:55
  • $\begingroup$ @KCd Updated, thanks. $\endgroup$ Apr 10, 2017 at 0:27
  • $\begingroup$ The second link is broken again. $\endgroup$
    – KCd
    Jun 28, 2019 at 0:38
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I would like to add to Bill's answer:

Judy Roitman found a particular model of ZFC$+ \neg$CH where there are $\mathfrak c$ many different hyperreal fields. Very recently, it was shown that in every model of ZFC$+ \neg$CH there actually exist $2^\mathfrak c$ different hyperreal fields... as many as you could hope for since there are exactly $2^\mathfrak c$ (free) ultrafilters on $\omega$!

A dichotomy for the number of ultrapowers

Ilijas Farah & Saharon Shelah

Journal of Mathematical Logic 10 (01n02):45-81 (2010)

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    $\begingroup$ Very recently? [citation needed]. $\endgroup$
    – Asaf Karagila
    Jun 17, 2014 at 1:45
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To obtain the uniqueness of the hyperreal field one needs to add the axiom of saturation which is arguably similar to completeness (in the case of the real field). Furthermore, like the real (complete Archimedean) field, the hyperreals have a definable model. For a discussion see Keisler's text on foundations here.

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