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.

increasingsequences of infinitesimals, i.e. converging to the gap. So if we take $1/\varepsilon$, then we are looking adecreasingsequence of infinite numbers. $\endgroup$ – Monroe Eskew Mar 29 at 14:56