Timeline for Homomorphisms from powers of Z to Z
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
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| Apr 2, 2019 at 17:38 | comment | added | YCor | I gave a proof at this answer mathoverflow.net/a/326379/14094 (proof of the lemma'), where "reasonable" means "non-Ulam-measurable", and granted the countable case. More precisely it's a proof of the fact that every homomorphism vanishing on the direct sum vanishes on the product. But then the conclusion (that every homomorphism vanishes on all but finitely many factors) is immediate from the countable case. | |
| May 28, 2013 at 19:39 | vote | accept | Michael Barr | ||
| May 28, 2013 at 13:01 | comment | added | Andreas Blass | Some of the older literature uses "measurable" to mean "supporting a non-principal countably complete ultrafilter", which nowadays would be expressed as "greater than or equal to the first measurable cardinal". The same concept is, I believe, sometimes called "Ulam measurable". | |
| May 28, 2013 at 12:59 | comment | added | Andreas Blass | Fernando is right; the intended conclusion is that any homomorphism factors through the standard projection to some finite power, i.e, simply restricting functions $I\to Z$ to a finite subset $F$ of $I$. | |
| May 28, 2013 at 9:12 | comment | added | Fernando Muro | As you write it, the answer is obvious, since the target is $Z^1$. | |
| May 28, 2013 at 8:16 | comment | added | YCor | This is not exactly the statement. Indeed, if $\alpha$ is the smallest measurable cardinal and if $\beta$ is the next cardinal, then $\beta$ is not measurable. The correct statement is: "if $I$ admits no nonprincipal ultrafilter stable by countable intersections", or equivalently "if the cardinal $I$ is smaller than every measurable cardinal", or equivalently "every subset of $I$ is non-measurable". | |
| May 28, 2013 at 7:09 | answer | added | Ralph | timeline score: 12 | |
| May 28, 2013 at 3:45 | comment | added | Allen Knutson | en.wikipedia.org/wiki/Measurable_cardinal | |
| May 28, 2013 at 3:12 | comment | added | Greg Martin | "non-measurable cardinality"? | |
| May 28, 2013 at 2:18 | comment | added | Gerhard Paseman | I remember a similar exercise in "Algebras, Lattices, and Varieties" by McKenzie, McNulty, and Taylor. (Ch 4.4, exercise 20.) I would guess it is a result of Specker, but that it just a guess. I can't tell from a glance at the bibliography where it came from. Maybe someone else can. Gerhard "Someone Take The Baton Now" Paseman, 2013.05.27 | |
| May 28, 2013 at 1:18 | history | asked | Michael Barr | CC BY-SA 3.0 |