Timeline for Is the dual of the product of infinite cyclic groups a free abelian group ?
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
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| Aug 31, 2012 at 5:49 | vote | accept | Ralph | ||
| Aug 30, 2012 at 17:39 | comment | added | Ralph | @Andreas: Very nice, thank you very much (the moment when you posted your edit, I was just reading Cor. 3.7 (ii) that also states the freeness of the dual). BTW: The book of Eklof-Mekler seems to be quite interesting. Thanks for drawing my attention to it. | |
| Aug 30, 2012 at 17:30 | comment | added | Asaf Karagila♦ | Shelah 904 seems somewhat relevant too, the abstract states: "Answering problem (DG) of [EM90], [EM02], we show that there is a reflexive group of cardinality >= first measurable."(shelah.logic.at/904_abs.html) | |
| Aug 30, 2012 at 17:27 | comment | added | Andreas Blass | Ralph, the non-measurable case is Corollary 3.3 of Eklof-Mekler, specialized as in the "Edit" that I added to my answer. The original sources are listed on page 80 of E-M; in particular, they attribute the (specialized) non-measurable case to a 1962 paper of Balcerzyk. | |
| Aug 30, 2012 at 17:19 | history | edited | Andreas Blass | CC BY-SA 3.0 |
added 577 characters in body
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| Aug 29, 2012 at 1:25 | comment | added | Ralph | Andreas, thank you very much for your answer! Do you have a refernce for the "no measurable cardinals $\le \kappa$" case ? I have access to the Eklof-Mekler book and will have a look. But since I don't know much about ultrafilters I would appreciate very much if you will also have a look when you're back. Thanks again. | |
| Aug 28, 2012 at 23:51 | history | answered | Andreas Blass | CC BY-SA 3.0 |