Timeline for Counterexample for the Open Mapping Theorem
Current License: CC BY-SA 2.5
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 12, 2013 at 0:12 | comment | added | Michael Renardy | Nevertheless, this "proof" actually proves something if suitably corrected. You need an extra condition, but it is one that would usually be true in applications. | |
| Feb 10, 2012 at 22:02 | comment | added | Pietro Majer | (Kamran, the explanation is written above, cyphered in rot 13 not to spoil the riddle for other people. You can read it using e.g. rot13.com/index.php) | |
| Feb 10, 2012 at 21:20 | comment | added | Kamran Reihani | You cannot "deduce that $\|.\|_3$ is complete"! | |
| Feb 14, 2011 at 12:49 | comment | added | arsmath | Note that the existence of an actual counterexample to the "theorem" requires the Axiom of Choice. Eric Schechter in his analysis book talks about negations of choice where the quoted "theorem" actually becomes true. | |
| Nov 4, 2010 at 8:51 | comment | added | Pietro Majer | Thanks. Abgr gung abez 3 vf Onanpu vss gurer rkvfgf n Unhfqbess gbcbybtl ba K juvpu vf jrnxre guna obgu gur gbcbybtl bs abez1 naq gur gbcbybtl bs abez 2. | |
| Nov 3, 2010 at 23:24 | comment | added | Nate Eldredge | Took me a bit, but I see it now. [rot13] Gurer znl or n frdhrapr gung pbairetrf gb bar irpgbe va gur svefg abez, naq gb nabgure va gur frpbaq abez. Fhpu n frdhrapr vf Pnhpul va gur guveq abez ohg qbrf abg pbairetr. | |
| Nov 3, 2010 at 22:35 | history | answered | Pietro Majer | CC BY-SA 2.5 |