6
$\begingroup$

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.

$\endgroup$
6
  • $\begingroup$ I imagine it's the cofinality of $\omega^\kappa/U$, as a linear order. $\endgroup$
    – Asaf Karagila
    Mar 29, 2020 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.” $\endgroup$ Mar 29, 2020 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, 2020 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. $\endgroup$ Mar 29, 2020 at 14:56
  • 1
    $\begingroup$ It should be the cofinality of the reverse order on $(\omega^{\kappa}/U) \setminus \omega$, right? Not that that necessarily makes the question easier. $\endgroup$ Mar 29, 2020 at 15:02

1 Answer 1

8
$\begingroup$

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.

$\endgroup$
3
  • 1
    $\begingroup$ Thanks! What about for uncountable $\kappa$? Does it hold in ZFC that $\mu>\kappa$ for regular $U$? $\endgroup$ Mar 29, 2020 at 17:34
  • 1
    $\begingroup$ @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, 2020 at 18:43
  • 1
    $\begingroup$ @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. $\endgroup$ Apr 4, 2020 at 15:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.