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Suppose $\kappa$ is an infinite cardinal and $U$ is a countably incomplete uniform ultrafilter over $\kappa$. Then $\mathbb R^\kappa/U$ is nonstandard. What is the cofinality of the set of infinitesimals of this field? What can we say when $U$ is $\kappa$-regular?

Background information: Recall that $U$ is $\kappa$-regular when there exists a sequence $\langle X_\alpha : \alpha < \kappa \rangle \subseteq U$ such that for any $\beta < \kappa$, $\{ \alpha : \beta \in X_\alpha \}$ is finite. If $U$ is $\kappa$-regular, then I can show that the cofinality of $\mathbb R^\kappa/U$ (rather than infinitesimals) is $>\kappa$. Furthermore, if $\mathbb R^\kappa/U$ is $\delta$-saturated, then the cofinality of the infinitesimals is $\geq\delta$. $\omega_1$-saturation is automatic for ultrapowers by countably incomplete ultrafilters. If the ultrafilter satisfies a property stronger than regularity called goodness, then the ultrapower is $\kappa^+$-saturated.

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  • $\begingroup$ I imagine it's the cofinality of $\omega^\kappa/U$, as a linear order. $\endgroup$ – Asaf Karagila Mar 29 at 14:43
  • $\begingroup$ @AsafKaragila This is the answer if we are looking at the set of things below a fixed element $[f]_U$, since for each $\alpha$ can choose a cofinal $\omega$-sequence in $f(\alpha)$. But there is no supremum to the set of infinitesimals, so basically I am asking about possible “gaps.” $\endgroup$ – Monroe Eskew Mar 29 at 14:52
  • $\begingroup$ Well, if you look at $1/\varepsilon$, then you're looking at the cofinality of the linear order $\Bbb R^\kappa/U$, so the fact they are infinitesimals is irrelevant here. $\endgroup$ – Asaf Karagila Mar 29 at 14:54
  • $\begingroup$ The thing is I am looking at increasing sequences of infinitesimals, i.e. converging to the gap. So if we take $1/\varepsilon$, then we are looking a decreasing sequence of infinite numbers. $\endgroup$ – Monroe Eskew Mar 29 at 14:56
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    $\begingroup$ It should be the cofinality of the reverse order on $(\omega^{\kappa}/U) \setminus \omega$, right? Not that that necessarily makes the question easier. $\endgroup$ – James Hanson Mar 29 at 15:02
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As pointed out in a comment by James Hanson, the cofinality of the infinitesimals is the same as the coinitiality (i.e., cofinality or the reverse order) $\mu$ of the nonstandard part of $\omega^\kappa/U$.

Even for $\kappa=\omega$, this coinitiality $\mu$ is not decided by the axioms of set theory. Furthermore, even within a single model of set theory, $\mu$ can depend on the particular ultrafilter $U$.

Specifically, if one starts with a model of CH and adds $\lambda$ Cohen reals, the resulting model has nonprincipal ultrafilters $U$ on $\omega$ for which $\mu$ is any regular uncountable cardinal $\leq\lambda$. (The same holds for the cofinality of the whole ultrapower $\omega^\omega/U$, and in fact this cofinality and $\mu$ can be chosen independently.) Similarly, if one adds $\lambda$ random reals to a model of CH, every regular uncountable cardinal $\leq\lambda$ occurs as $\mu$ for some $U$. (But now the cofinality of $\omega^\omega/U$ is $\aleph_1$ because random forcing is $\omega^\omega$-bounding.)

These results were proved by Mike Canjar in his thesis; the MathSciNet data for the published version are:

MR0924678 (89g:03073) Reviewed

Canjar, Michael

Countable ultraproducts without CH.

Ann. Pure Appl. Logic 37 (1988), no. 1, 1–79.

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    $\begingroup$ Thanks! What about for uncountable $\kappa$? Does it hold in ZFC that $\mu>\kappa$ for regular $U$? $\endgroup$ – Monroe Eskew Mar 29 at 17:34
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    $\begingroup$ @Monroe: Very clearly, if ZFC is consistent, it has a countable model, so all the cardinals there are countable by definition! There is no such thing as uncountable cardinals. $\endgroup$ – Asaf Karagila Mar 29 at 18:43
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    $\begingroup$ @MonroeEskew When I first started to think about your question, my feeling was that regularity of $U$ is just what you'd need to get $\mu>\kappa$, Unfortunately, my attempt to prove it had a gap (euphemism for "it was wrong"), and I still don't see how to fix it. $\endgroup$ – Andreas Blass Apr 4 at 15:36

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