Timeline for Does antidifferentiability of continuous functions imply Dedekind completeness?
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
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| Feb 8, 2012 at 15:10 | comment | added | Gerald Edgar | Subsequent paper ... full text, open access ... projecteuclid.org/… | |
| Feb 7, 2012 at 23:00 | comment | added | James Propp | It seemed simplest to start a new MathOverflow thread on the subject of Pelling's argument and mine, so I did: mathoverflow.net/questions/87848/… . But I am troubled by the fact that only people with access to JSTOR (or to a copy of the February 1981 issue of the Monthly) will be able to read Pelling's argument. I don't know of a good solution to this problem. What constitutes fair use? Are there precedents for this? | |
| Feb 7, 2012 at 22:24 | comment | added | James Propp | Thanks, Gerald! But I am not sure that Pelling's argument can be correct; doesn't Rolle's Theorem imply completeness? I can't explain the implication in just 600 characters (that's the length-limit for comments), but I'll try to say more about this below. If there's no mistake in my argument, then there must be something wrong with Pelling's. I'd like to know which! | |
| Feb 7, 2012 at 5:11 | comment | added | Gerald Edgar | This reminds me vaguely of an old Monthly problem ... 5861, See the solution by M. Pelling, where he constructs an unusual nonarchimedean ordered field. jstor.org/pss/2321145 ... In that example, Rolle's Theorem holds but not every positive element has a square-root. | |
| Feb 6, 2012 at 22:29 | history | edited | James Propp | CC BY-SA 3.0 |
I fixed a link and corrected a reference
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| Feb 6, 2012 at 8:32 | history | asked | James Propp | CC BY-SA 3.0 |