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Every $(\lambda^+,\varkappa^+)$-regular ultrafilter is $(\lambda,\varkappa)$-regular


Author: Paolo Lipparini
Journal: Proc. Amer. Math. Soc. 128 (2000), 605-609
MSC (1991): Primary 03C20, 04A20
DOI: https://doi.org/10.1090/S0002-9939-99-05025-X
Published electronically: July 8, 1999
MathSciNet review: 1623032
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Abstract: We prove the following:

Theorem A. If $D$ is a $(\lambda^+,\varkappa)$-regular ultrafilter, then either

(a)
$D$ is $(\lambda,\varkappa)$-regular, or
(b)
the cofinality of the linear order $\prod _D\langle\lambda,<\rangle$ is $\operatorname{cf}\varkappa$, and $D$ is $(\lambda,\varkappa')$-regular for all $\varkappa'<\varkappa$.

Corollary B. Suppose that $\varkappa$ is singular, $\varkappa>\lambda$ and either $\lambda$ is regular, or $\operatorname{cf}\varkappa<\operatorname{cf}\lambda$. Then every $(\lambda^{+n},\varkappa)$-regular ultrafilter is $(\lambda,\varkappa)$-regular.

We also discuss some consequences and variations.


References [Enhancements On Off] (What's this?)

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Additional Information

Paolo Lipparini
Affiliation: Dipartimento di Matematica, Viale della Ricerca Scientifica, II Università di Roma (Tor Vergata), I-00133 Rome, Italy
Email: lipparin@axp.mat.uniroma2.it, lipparini@unica.it

DOI: https://doi.org/10.1090/S0002-9939-99-05025-X
Keywords: $(\alpha, \varkappa)$-regular ultrafilter, cofinality of ultrapowers, almost $(\lambda, \varkappa)$-regular extensions
Received by editor(s): November 20, 1997
Received by editor(s) in revised form: April 8, 1998
Published electronically: July 8, 1999
Additional Notes: This work was performed under the auspices of G.N.S.A.G.A
Communicated by: Carl G. Jockusch, Jr.
Article copyright: © Copyright 1999 American Mathematical Society

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