Timeline for finite codimension implies closed?
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
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| Sep 1, 2011 at 22:37 | comment | added | Bill Johnson | No, but IIRC you need something beyond ZF to prove the existence of a discontinuous linear functional on $\ell_2$. To prove the existence of Hamel bases in vector spaces (from which you get a discontinuous linear functional on $\ell_2$) it is enough to have the product of every family of two point sets non empty, and this axiom is strictly weaker than choice. For just $\ell_2$, less is needed. but I think not just ZF. Where are the logicians when we need them? There are least a zillion of them on MO. :) | |
| Sep 1, 2011 at 19:45 | comment | added | Guy Relande | Are you basically saying that one cannot produce non closed finite codimensional subspaces of complete TVS without using axiom of choice? That would be perfect as far as I am concerned (it would, in practice, be stronger than what I was hoping for). | |
| Sep 1, 2011 at 14:49 | comment | added | Bill Johnson | Something is needed in order to get a discontinuous linear functional on $\ell_2$ (which is equivalent to having a dense subspace of codimension one). | |
| Sep 1, 2011 at 6:57 | vote | accept | Guy Relande | ||
| Sep 1, 2011 at 6:56 | vote | accept | Guy Relande | ||
| Sep 1, 2011 at 6:57 | |||||
| Sep 1, 2011 at 6:34 | comment | added | Guy Relande | Very nice, thanks! One could still wonder whether it is possible to construct such a counterexample without using axiom of choice as in your map from $X_2$ to $Y_0$... | |
| Aug 31, 2011 at 20:29 | history | answered | Bill Johnson | CC BY-SA 3.0 |