Timeline for Infinite products of topological groups
Current License: CC BY-SA 2.5
9 events
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| Feb 14, 2010 at 1:17 | comment | added | Pete L. Clark | @Arminius and TL: right, for many students "almost all" is most familiar in the sense of measure theory, so I can imagine someone looking for a suitable measure on the arbitrary index set I. (Of course "almost all" in this sense is equivalent to the set of exceptions having finite counting measure, but that seems overly convoluted...) | |
| Feb 13, 2010 at 23:59 | comment | added | Tom Leinster | Arminius, I guess the point is that "almost all" to mean "all but finitely many" is a slightly informal usage. At least, that's been my own experience. So if someone didn't know what it meant, they might have trouble looking it up. Indeed, they'd probably find lots of stuff defining "almost all" in the measure theory sense. | |
| Feb 13, 2010 at 23:43 | comment | added | user717 | No, no, it's fine. But just for my own education: Is there a difference between "almost all" and "all but finitely many"? For me it's (at the moment) exactly the same... | |
| Feb 13, 2010 at 23:15 | comment | added | Pete L. Clark | I took the liberty of editing as GE suggested above, an instance of my "easy question deserves easily read answer" philosophy. @Arminius, if for any reason you truly prefer your phrasing, please feel free to change it back; you needn't defend your reasons for doing so. | |
| Feb 13, 2010 at 23:12 | history | edited | Pete L. Clark | CC BY-SA 2.5 |
added 9 characters in body
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| Feb 13, 2010 at 18:32 | history | edited | user717 | CC BY-SA 2.5 |
edited body
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| Feb 13, 2010 at 18:29 | comment | added | Gerald Edgar | ...where, in this context, "almost all" means "all but finitely many". | |
| Feb 13, 2010 at 18:29 | vote | accept | commonname | ||
| Feb 13, 2010 at 18:20 | history | answered | user717 | CC BY-SA 2.5 |