Timeline for Critical Growth of Dimension for Dense Cover by Linear Subspaces
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
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| Aug 9, 2020 at 11:03 | comment | added | Pietro Majer | How can the first condition be satisfied, if X is infinite dimensional? We can pick each $L_n$ within a fixed closed hyperplane, with any finite dimension $N_n$ we like. | |
| Aug 9, 2020 at 9:23 | history | edited | ABIM | CC BY-SA 4.0 |
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| Aug 9, 2020 at 9:08 | comment | added | ABIM | @TarasBanakh Yes, instead if we ask if there is a sequence of dimensions of these subspaces must grow at then I think its possible. I put down the example in the finite-dimensional case also. | |
| Aug 9, 2020 at 9:06 | history | edited | ABIM | CC BY-SA 4.0 |
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| Aug 9, 2020 at 8:58 | history | edited | ABIM | CC BY-SA 4.0 |
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| Aug 9, 2020 at 8:48 | comment | added | Taras Banakh | I think nothing like you wish cannot hold because of the existence of basic sequences in any infinite-dimensional Banach space. Using a basic sequence you can form many finite-dimensional subspaces with desirable properties. | |
| Aug 9, 2020 at 8:44 | comment | added | ABIM | @TarasBanakh Yes I just noticed the bad formulation. I meant to ask if there is a "critical dimension" above which any collection of subspaces of that dimension must have dense union. | |
| Aug 9, 2020 at 8:43 | history | edited | ABIM | CC BY-SA 4.0 |
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| Aug 9, 2020 at 8:43 | comment | added | Taras Banakh | Thanks. But your question should be more concrete anyway. Because you can just take a collection of 1-dimensional subspaces with dense union, in which case the second condition holds and the first does not. Also $N$ can be taken to be equal to 1 and then the first part is trivial. | |
| Aug 9, 2020 at 8:39 | history | edited | ABIM | CC BY-SA 4.0 |
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| Aug 9, 2020 at 8:32 | history | asked | ABIM | CC BY-SA 4.0 |