# Cofinality of infinitesimals

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.

• I imagine it's the cofinality of $\omega^\kappa/U$, as a linear order. – Asaf Karagila Mar 29 at 14:43
• @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.” – Monroe Eskew Mar 29 at 14:52
• 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. – Asaf Karagila Mar 29 at 14:54
• 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. – Monroe Eskew Mar 29 at 14:56
• It should be the cofinality of the reverse order on $(\omega^{\kappa}/U) \setminus \omega$, right? Not that that necessarily makes the question easier. – James Hanson Mar 29 at 15:02

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.

• Thanks! What about for uncountable $\kappa$? Does it hold in ZFC that $\mu>\kappa$ for regular $U$? – Monroe Eskew Mar 29 at 17:34
• @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. – Asaf Karagila Mar 29 at 18:43
• @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. – Andreas Blass Apr 4 at 15:36