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One of the more common constructions of the hyperreals is $\mathbb{R}^{*} := \mathbb{R}^{\omega}/\mathcal{U}$ where $\mathcal{U}$ is some ultrafilter containing the filter of sequences in $\mathbb{R}^{\omega}$ with finitely many nonzero entries. The existence of such an ultrafilter is proven nonconstructively, by appealing to Zorn's Lemma or some equivalent. Things like the ordering on $\mathbb{R}^{*}$ are then established with reference to $\mathcal{U}$.

My question is, there's nothing, as far as I know, preventing there being more than one such ultrafilter (or is there?). Aren't there then sets of hyperreal numbers with different properties depending on the ultrafilter one "chooses"? Or does it not matter, and whichever ultrafilter you use, you get the same system, in some sense?

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    $\begingroup$ Assuming $\mathsf{CH}$ every two ultrapowers of $\Bbb R$ over a countable index set are isomorphic as ordered fields, I don't know whether this statement is actually equivalent to $\mathsf{CH}$. With no extra assumptions beyond $\mathsf{ZFC}$ we can say that any two such ultrapowers will be elementarily equivalent (because they are all elementarily equivalent to $\Bbb R$ of course) $\endgroup$ – Alessandro Codenotti Jul 4 '19 at 15:59
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    $\begingroup$ "Furthermore, the field obtained by the ultrapower construction from the space of all real sequences, is unique up to isomorphism if one assumes the continuum hypothesis," according to en.wikipedia.org/wiki/Hyperreal_number. I don't know if this really addresses your question. $\endgroup$ – saulspatz Jul 4 '19 at 16:00
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    $\begingroup$ @saulspatz See also this answer on the CH uniqueness result. $\endgroup$ – Bill Dubuque Jul 4 '19 at 16:03
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Yes, you're right - we really shouldn't talk about "the" hyperreals. However, since any two hyperreal fields are elementarily equivalent (an ultrapower of a structure is elementarily equivalent to the structure itself, per Los), this is generally a harmless abuse of terminology. Alternatively we might assume CH, which as noted above does guarantee that any two nontrivial ultrapowers of $\mathbb{R}$ by an ultrafilter on the naturals are isomorphic.

Incidentally, we don't need to use ultrafilters on the naturals - any ultrafilter which is not countably closed will generate an appropriate field. Once we permit these, or indeed look at arbitrary sufficiently saturated fields with the transfer property (these are generally called "hyperreal fields" for obvious reasons), we really can't avoid non-isomorphism.

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